SSLC :SCERT Maths

👉 Malayalam

Chapter 1: Arithmetic Sequences

Section: Number patterns

(1) We can make triangles by stacking dots: Write the number of dots in each triangle. Calculate the number of dots needed to make the next three triangles in this pattern.

Solution: The pattern for the number of dots in the triangles, where each new triangle adds one more dot to the length of the previous row:

• 1st triangle: 1 dot

• 2nd triangle: 1 + 2 = 3 dots [implied by picture and definition of stacking]

• 3rd triangle: 1 + 2 + 3 = 6 dots [implied by picture and definition of stacking]

• 4th triangle: 1 + 2 + 3 + 4 = 10 dots [implied by picture and definition of stacking]

• Next three triangles (5th, 6th, 7th):

    ◦ 5th triangle: 1 + 2 + 3 + 4 + 5 = 15 dots [implied by continuing the pattern]

    ◦ 6th triangle: 1 + 2 + 3 + 4 + 5 + 6 = 21 dots [implied by continuing the pattern]

    ◦ 7th triangle: 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 dots [implied by continuing the pattern]

(2) From the sequence equilateral triangle, square, regular pentagon and so on of regular polygons, form the following sequences

• Number of sides

• Sum of inner angles

• Sum of outer angles

• An inner angle

• An outer angle

Solution: For the sequence of regular polygons (equilateral triangle, square, regular pentagon, etc.):

• Number of sides: 3, 4, 5, 6, .... This is a number sequence.

• Sum of inner angles: 180, 360, 540, .... This is an arithmetic sequence, starting at 180 and adding 180 successively [20, implied].

• Sum of outer angles: 360, 360, 360, .... This is an arithmetic sequence, starting at 360 and adding 0 at every step.

• An inner angle:

    ◦ For equilateral triangle (3 sides): 180 / 3 = 60 [implied by knowledge of triangles].

    ◦ For square (4 sides): 360 / 4 = 90 [implied by knowledge of quadrilaterals].

    ◦ For regular pentagon (5 sides): 540 / 5 = 108 [implied by knowledge of pentagons].

    ◦ Sequence: 60, 90, 108, ... This is a number sequence. (It is not an arithmetic sequence, as 90-60=30 and 108-90=18, the difference is not the same).

• An outer angle:

    ◦ For equilateral triangle (3 sides): 360 / 3 = 120.

    ◦ For square (4 sides): 360 / 4 = 90.

    ◦ For regular pentagon (5 sides): 360 / 5 = 72.

    ◦ Sequence: 120, 90, 72, ... This is a number sequence. (It is not an arithmetic sequence, as 90-120=-30 and 72-90=-18, the difference is not the same).

(3) Write the sequence of natural numbers which leave remainder 1 on division by 3, and the sequence of natural numbers which leave remainder 2 on division by 3.

Solution:

• Natural numbers which leave remainder 1 on division by 3:

    ◦ (3 × 0) + 1 = 1

    ◦ (3 × 1) + 1 = 4

    ◦ (3 × 2) + 1 = 7

    ◦ Sequence: 1, 4, 7, .... This is an arithmetic sequence with a common difference of 3.

• Natural numbers which leave remainder 2 on division by 3:

    ◦ (3 × 0) + 2 = 2

    ◦ (3 × 1) + 2 = 5

    ◦ (3 × 2) + 2 = 8

    ◦ Sequence: 2, 5, 8, ... [implied by "If the remainder on dividing a number by 3 is not 1, it can be either 0 or 2. Write down those leaving remainder 0 and those leaving remainder 2 separately." from source]. This is an arithmetic sequence with a common difference of 3 [implied by reasoning for remainder 1 sequence].

(4) Write in ascending order, the sequence of natural numbers with last digit 1 or 6. Describe this sequence in two other ways.

Solution:

• Sequence in ascending order:

    ◦ 1, 6, 11, 16, 21, 26, ... [implied by definition].

• Two other ways to describe this sequence:

    1. Arithmetic Sequence: This is an arithmetic sequence starting with 1 and proceeding by successively adding 5. The common difference is 5.

    2. Remainder on Division: This is the sequence of natural numbers which leave remainder 1 or 6 on division by 10 [implied by definition of "last digit 1 or 6"]. Alternatively, they are numbers that leave remainder 1 on division by 5 (e.g. 1, 6, 11, 16) but this is not fully capturing the definition. A more accurate description in terms of remainders would be numbers that leave a remainder of 1 when divided by 5, but excluding those that are multiples of 10 plus 1. Or, more simply, it is the sequence of natural numbers that leave remainder 1 or 6 when divided by 10 [implied by example "natural numbers ending in 1"].

(5) See these figures: (The red triangle pattern) (i) How many red triangles are there in each picture? (ii) Taking the area of whole uncut triangle as 1, compute the area of a small triangle in each picture. (iii) What is the total area of all the red triangles in each picture? (iv) Write the first five terms of each of the three sequences got by continuing this process.

Solution: Let's analyze the pictures:

• Picture 1: Shows 3 red triangles. The whole triangle is divided into 4 smaller equilateral triangles (3 red, 1 white in the middle).

• Picture 2: Each of the 3 red triangles from Picture 1 is subjected to the same process. So, each of these 3 triangles now has 3 red triangles within it.

• Picture 3: The process is repeated again.

(i) How many red triangles are there in each picture?

• Picture 1: 3 red triangles.

• Picture 2: 3 × 3 = 9 red triangles.

• Picture 3: 9 × 3 = 27 red triangles.

• Sequence: 3, 9, 27, ...

(ii) Taking the area of whole uncut triangle as 1, compute the area of a small triangle in each picture.

• Picture 1: The whole triangle is divided into 4 equal smaller triangles. So, the area of one small triangle is 1/4.

• Picture 2: Each small triangle from Picture 1 is now divided into 4 even smaller triangles. So, the area of a small triangle in Picture 2 is (1/4) × (1/4) = 1/16.

• Picture 3: The area of a small triangle in Picture 3 is (1/16) × (1/4) = 1/64.

• Sequence: 1/4, 1/16, 1/64, ...

(iii) What is the total area of all the red triangles in each picture?

• Picture 1: Total area of red triangles = Number of red triangles × Area of one small triangle = 3 × (1/4) = 3/4.

• Picture 2: Total area of red triangles = 9 × (1/16) = 9/16.

• Picture 3: Total area of red triangles = 27 × (1/64) = 27/64.

• Sequence: 3/4, 9/16, 27/64, ...

(iv) Write the first five terms of each of the three sequences got by continuing this process. Let's denote the sequences as T (number of red triangles), A_small (area of a small triangle), and A_total (total area of red triangles).

• T (Number of red triangles):

    ◦ Term 1: 3

    ◦ Term 2: 9

    ◦ Term 3: 27

    ◦ Term 4: 27 × 3 = 81

    ◦ Term 5: 81 × 3 = 243

    ◦ Sequence: 3, 9, 27, 81, 243, ...

• A_small (Area of a small triangle):

    ◦ Term 1: 1/4

    ◦ Term 2: 1/16

    ◦ Term 3: 1/64

    ◦ Term 4: (1/64) × (1/4) = 1/256

    ◦ Term 5: (1/256) × (1/4) = 1/1024

    ◦ Sequence: 1/4, 1/16, 1/64, 1/256, 1/1024, ...

• A_total (Total area of all red triangles):

    ◦ Term 1: 3/4

    ◦ Term 2: 9/16

    ◦ Term 3: 27/64

    ◦ Term 4: (27/64) × (3/4) = 81/256

    ◦ Term 5: (81/256) × (3/4) = 243/1024

    ◦ Sequence: 3/4, 9/16, 27/64, 81/256, 243/1024, ...

Section: Arithmetic sequences

(1) Check whether each of the sequences given below are arithmetic sequences. Give reasons also. Find the common differences of the arithmetic sequences: (i) Natural numbers leaving remainder 1 on division by 4 (ii) Natural numbers leaving remainder 1 or 2 on division by 4 (iii) Squares of natural numbers (iv) Reciprocals of natural numbers (v) Powers of 2 (vi) Half of the odd numbers

Solution: A sequence is an arithmetic sequence if the same number is added to go from any term to the next. This number is called the common difference, found by subtracting the previous term from any term.

(i) Natural numbers leaving remainder 1 on division by 4: * Sequence: 1, 5, 9, 13, ... * Differences: 5-1=4, 9-5=4, 13-9=4. * Reason: The difference between consecutive terms is consistently 4. * Conclusion: This is an arithmetic sequence. * Common difference: 4.

(ii) Natural numbers leaving remainder 1 or 2 on division by 4: * Sequence: 1, 2, 5, 6, 9, 10, ... * Differences: 2-1=1, 5-2=3, 6-5=1, 9-6=3. * Reason: The differences between consecutive terms (1, 3, 1, 3) are not the same. * Conclusion: This is not an arithmetic sequence.

(iii) Squares of natural numbers: * Sequence: 1², 2², 3², 4², ... which is 1, 4, 9, 16, .... * Differences: 4-1=3, 9-4=5, 16-9=7. * Reason: The differences between consecutive terms (3, 5, 7) are not the same. * Conclusion: This is not an arithmetic sequence.

(iv) Reciprocals of natural numbers: * Sequence: 1/1, 1/2, 1/3, 1/4, ... which is 1, 1/2, 1/3, 1/4, ... * Differences: (1/2)-1 = -1/2, (1/3)-(1/2) = -1/6, (1/4)-(1/3) = -1/12. * Reason: The differences between consecutive terms (-1/2, -1/6, -1/12) are not the same. * Conclusion: This is not an arithmetic sequence.

(v) Powers of 2: * Sequence: 2¹, 2², 2³, 2⁴, ... which is 2, 4, 8, 16, ... * Differences: 4-2=2, 8-4=4, 16-8=8. * Reason: The differences between consecutive terms (2, 4, 8) are not the same. * Conclusion: This is not an arithmetic sequence.

(vi) Half of the odd numbers: * Odd numbers: 1, 3, 5, 7, .... * Sequence: 1/2, 3/2, 5/2, 7/2, ... * Differences: (3/2)-(1/2) = 1, (5/2)-(3/2) = 1, (7/2)-(5/2) = 1. * Reason: The difference between consecutive terms is consistently 1. * Conclusion: This is an arithmetic sequence. * Common difference: 1.

(2) See these pictures: (Squares of various sizes) (i) How many small squares are there in each picture? (ii) How many large squares? (iii) How many squares in all in each picture? If we continue the pattern of pictures, are the sequences above arithmetic sequences?

Solution: Let's analyze the pictures:

• Picture 1: A single large square, made of 1 small square.

• Picture 2: A 2x2 grid of small squares (4 small squares), forming 1 large square. Total squares: 4 small + 1 large = 5.

• Picture 3: A 3x3 grid of small squares (9 small squares), forming 4 (2x2) large squares and 1 (3x3) largest square. Total squares: 9 + 4 + 1 = 14.

(i) How many small squares are there in each picture? * Picture 1: 1 small square (1x1). * Picture 2: 4 small squares (2x2). * Picture 3: 9 small squares (3x3). * Sequence: 1, 4, 9, ... (Squares of natural numbers). * Is it an arithmetic sequence? No, differences are 3, 5 [part (1) (iii) above].

(ii) How many large squares? * Picture 1: 0 large squares (only 1x1 grid). * Picture 2: 1 large square (2x2 grid). * Picture 3: 5 large squares (4 of 2x2 + 1 of 3x3). * Sequence: 0, 1, 5, ... * Is it an arithmetic sequence? No, differences are 1, 4.

(iii) How many squares in all in each picture? * Picture 1: 1 (1x1). * Picture 2: 5 (4 small + 1 large). * Picture 3: 14 (9 small + 4 medium + 1 large). * Sequence: 1, 5, 14, ... * Is it an arithmetic sequence? No, differences are 4, 9.

Conclusion on arithmetic sequences: None of the three sequences (number of small squares, number of large squares, total squares) are arithmetic sequences, as their consecutive differences are not constant.

(3) In the picture below, the perpendiculars drawn from the bottom line are equally spaced. Show that the sequence of the heights of the perpendiculars, on continuing this, form an arithmetic sequence. (Hint: Draw perpendiculars from the top of each perpendicular to the next perpendicular)

Solution: Let the first perpendicular have height h1 and the second h2. Draw a horizontal line from the top of the first perpendicular to meet the second perpendicular. This forms a right-angled triangle. Since the perpendiculars are equally spaced, let the horizontal distance between them be d. Let the angle at the bottom left be θ. This angle is common for all such triangles formed by the horizontal top line and the equally spaced perpendiculars [implied by similar triangles or property of linear functions]. For the first segment between two perpendiculars: h2 - h1 = d * tan(θ) (where tan(θ) is the slope of the top line) [Implied use of trigonometry concepts later in the textbook, but can be argued geometrically by similar triangles from the initial rise]. Since the perpendiculars are equally spaced, d is constant. The top line is a straight line, so its slope (angle θ) is constant. Therefore, the difference in heights between any two consecutive perpendiculars will be constant.

• Let the height of the first perpendicular be x1.

• The difference in height between the first and second perpendicular is d1.

• The difference in height between the second and third perpendicular is d2.

• Since the perpendiculars are equally spaced and form a consistent pattern, d1 = d2 = d3 = ... = common difference.

• Let the heights be h_1, h_2, h_3, ....

• Then h_2 - h_1 = c, h_3 - h_2 = c, etc., for a constant c.

• Reason: In an arithmetic sequence, the same number is added to go from any term to the next. Here, the constant horizontal spacing d and the constant slope of the top line ensure a constant vertical increase c for each subsequent perpendicular.

• Conclusion: The sequence of the heights of the perpendiculars forms an arithmetic sequence.

Section: Position and term

Incomplete sequences problems (first part of section): In each of the arithmetic sequences below, some of the terms are not written, but indicated by . Find out these numbers: (i) 24, 42, , , ... (ii) , 24, 42, , ... (iii) , , 24, 42, ... (iv) 24, , 42, , ... (v) , 24, , 42, ... (vi) 24, , , 42, ...

Solution: For an arithmetic sequence, the common difference is constant. (i) 24, 42, , , ... * Common difference (d) = 42 - 24 = 18. * Next terms: 42 + 18 = 60, 60 + 18 = 78. * Sequence: 24, 42, 60, 78, ...

(ii) , 24, 42, , ... * Common difference (d) = 42 - 24 = 18. * Previous term: 24 - 18 = 6. * Next term: 42 + 18 = 60. * Sequence: 6, 24, 42, 60, ...

(iii) , , 24, 42, ... * Common difference (d) = 42 - 24 = 18. * Second term: 24 - 18 = 6. * First term: 6 - 18 = -12. * Sequence: -12, 6, 24, 42, ...

(iv) 24, , 42, , ... * Let x1 = 24 and x3 = 42. * To go from 1st to 3rd term, the common difference is added twice. * So, 2d = x3 - x1 = 42 - 24 = 18. * d = 18 / 2 = 9. * Second term: 24 + 9 = 33. * Fourth term: 42 + 9 = 51. * Sequence: 24, 33, 42, 51, ...

(v) , 24, , 42, ... * Let x2 = 24 and x4 = 42. * To go from 2nd to 4th term, the common difference is added twice [28, implied]. * So, 2d = x4 - x2 = 42 - 24 = 18. * d = 18 / 2 = 9. * First term: 24 - 9 = 15. * Third term: 24 + 9 = 33. * Sequence: 15, 24, 33, 42, ...

(vi) 24, , , 42, ... * Let x1 = 24 and x4 = 42. * To go from 1st to 4th term, the common difference is added three times [29, implied]. * So, 3d = x4 - x1 = 42 - 24 = 18. * d = 18 / 3 = 6. * Second term: 24 + 6 = 30. * Third term: 30 + 6 = 36. * Sequence: 24, 30, 36, 42, ...

Terms in specified positions problems: The two terms in specific positions of some arithmetic sequences are given below. Write the first five terms of each: (i) 3rd term 34, 6th term 67 (ii) 3rd term 43, 6th term 76 (iii) 3rd term 2, 5th term 3 (iv) 5th term 8, 9th term 10 (v) 5th term 7, 7th term 5

Solution: (i) 3rd term 34, 6th term 67 * Position change: 6 - 3 = 3. * Term change: 67 - 34 = 33. * Common difference (d) = 33 / 3 = 11 [30, implied]. * First term (x1): 3rd term - 2d = 34 - (2 × 11) = 34 - 22 = 12 [30, implied]. * Sequence: 12, 23, 34, 45, 56, ...

(ii) 3rd term 43, 6th term 76 * Position change: 6 - 3 = 3. * Term change: 76 - 43 = 33. * Common difference (d) = 33 / 3 = 11. * First term (x1): 3rd term - 2d = 43 - (2 × 11) = 43 - 22 = 21. * Sequence: 21, 32, 43, 54, 65, ...

(iii) 3rd term 2, 5th term 3 * Position change: 5 - 3 = 2. * Term change: 3 - 2 = 1. * Common difference (d) = 1 / 2 = 0.5. * First term (x1): 3rd term - 2d = 2 - (2 × 0.5) = 2 - 1 = 1. * Sequence: 1, 1.5, 2, 2.5, 3, ... or 1, 3/2, 2, 5/2, 3, ...

(iv) 5th term 8, 9th term 10 * Position change: 9 - 5 = 4. * Term change: 10 - 8 = 2. * Common difference (d) = 2 / 4 = 0.5. * First term (x1): 5th term - 4d = 8 - (4 × 0.5) = 8 - 2 = 6. * Sequence: 6, 6.5, 7, 7.5, 8, ... or 6, 13/2, 7, 15/2, 8, ...

(v) 5th term 7, 7th term 5 * Position change: 7 - 5 = 2. * Term change: 5 - 7 = -2. * Common difference (d) = -2 / 2 = -1. * First term (x1): 5th term - 4d = 7 - (4 × -1) = 7 + 4 = 11. * Sequence: 11, 10, 9, 8, 7, ...

Section: Changes in position and terms

(1) What is the 25th term of the arithmetic sequence 1, 11, 21, ...?

Solution:

• First term (x1) = 1.

• Common difference (d) = 11 - 1 = 10.

• To find the 25th term (x25), starting from the 1st term, position increases by 25 - 1 = 24.

• The 25th term = x1 + (24 × d) = 1 + (24 × 10) = 1 + 240 = 241.

(2) The 10th term of an arithmetic sequence is 46 and its 11th term is 51 (i) What is its first term? (ii) Write the first five terms of the sequence

Solution:

• Common difference (d) = 11th term - 10th term = 51 - 46 = 5.

(i) What is its first term? * To get to the 1st term from the 10th term, position must be decreased by 10 - 1 = 9. * The 1st term = 10th term - (9 × d) = 46 - (9 × 5) = 46 - 45 = 1.

(ii) Write the first five terms of the sequence * First term = 1, common difference = 5. * Sequence: 1, 6, 11, 16, 21, ...

(3) What is the 21st term of the arithmetic sequence 100, 95, 90, ...?

Solution:

• First term (x1) = 100.

• Common difference (d) = 95 - 100 = -5 [34, implied by 91-82=-9 example].

• To find the 21st term (x21), starting from the 1st term, position increases by 21 - 1 = 20.

• The 21st term = x1 + (20 × d) = 100 + (20 × -5) = 100 - 100 = 0.

(4) The 10th term of an arithmetic sequence is 56 and its 11th term is 51 (i) What is its first term? (ii) Write the first five terms of the sequence

Solution:

• Common difference (d) = 11th term - 10th term = 51 - 56 = -5.

(i) What is its first term? * To get to the 1st term from the 10th term, position must be decreased by 10 - 1 = 9. * The 1st term = 10th term - (9 × d) = 56 - (9 × -5) = 56 + 45 = 101.

(ii) Write the first five terms of the sequence * First term = 101, common difference = -5. * Sequence: 101, 96, 91, 86, 81, ...

Section: Changes in position and terms (after checking for term)

(1) Is 101 a term of the arithmetic sequence 13, 24, 35, ...? What about 1001?

Solution:

• Arithmetic sequence: 13, 24, 35, ...

• First term (x1) = 13.

• Common difference (d) = 24 - 13 = 11.

• In an arithmetic sequence, any term difference (the difference between a term and the first term) must be a multiple of the common difference.

• For 101:

    ◦ Difference from first term = 101 - 13 = 88.

    ◦ Check if 88 is a multiple of 11: 88 ÷ 11 = 8. Yes, it is.

    ◦ If 101 is the nth term, then 101 = 13 + (n-1)11. So 88 = (n-1)11, which means n-1 = 8, so n = 9.

    ◦ Conclusion: Yes, 101 is the 9th term of this sequence.

• For 1001:

    ◦ Difference from first term = 1001 - 13 = 988.

    ◦ Check if 988 is a multiple of 11: 988 ÷ 11 = 89.81... (not an integer).

    ◦ Conclusion: No, 1001 is not a term of this sequence.

(2) In the table below, some arithmetic sequences are given and two numbers against each. Check whether the numbers are terms of the respective sequences:

Sequence	Numbers	Yes/No
11, 22, 33, ...	123
	132
12, 23, 34, ...	100
	1000
21, 32, 43, ...	100
	1000
1/4, 1/2, 3/4, ...	3/4
	1 1/2
3/4, 1, 1 1/4, ...	3/4
	1 1/2

Solution: Use the rule: (Term - First Term) must be a multiple of the common difference.

• Sequence 1: 11, 22, 33, ...

    ◦ First term = 11, Common difference = 11.

    ◦ 123: 123 - 11 = 112. 112 ÷ 11 = 10.18... (Not a multiple). No.

    ◦ 132: 132 - 11 = 121. 121 ÷ 11 = 11. (Is a multiple). Yes.

• Sequence 2: 12, 23, 34, ...

    ◦ First term = 12, Common difference = 11.

    ◦ 100: 100 - 12 = 88. 88 ÷ 11 = 8. (Is a multiple). Yes.

    ◦ 1000: 1000 - 12 = 988. 988 ÷ 11 = 89.81... (Not a multiple). No.

• Sequence 3: 21, 32, 43, ...

    ◦ First term = 21, Common difference = 11.

    ◦ 100: 100 - 21 = 79. 79 ÷ 11 = 7.18... (Not a multiple). No.

    ◦ 1000: 1000 - 21 = 979. 979 ÷ 11 = 89. (Is a multiple). Yes.

• Sequence 4: 1/4, 1/2, 3/4, ...

    ◦ First term = 1/4, Common difference = 1/2 - 1/4 = 1/4.

    ◦ 3/4: 3/4 - 1/4 = 2/4 = 1/2. (1/2) ÷ (1/4) = 2. (Is a multiple). Yes.

    ◦ 1 1/2 = 3/2: 3/2 - 1/4 = 6/4 - 1/4 = 5/4. (5/4) ÷ (1/4) = 5. (Is a multiple). Yes.

• Sequence 5: 3/4, 1, 1 1/4, ...

    ◦ First term = 3/4, Common difference = 1 - 3/4 = 1/4.

    ◦ 3/4: 3/4 - 3/4 = 0. 0 ÷ 1/4 = 0. (Is a multiple, it's the first term). Yes.

    ◦ 1 1/2 = 3/2: 3/2 - 3/4 = 6/4 - 3/4 = 3/4. (3/4) ÷ (1/4) = 3. (Is a multiple). Yes.

Sequence	Numbers	Yes/No
11, 22, 33, ...	123	No
	132	Yes
12, 23, 34, ...	100	Yes
	1000	No
21, 32, 43, ...	100	No
	1000	Yes
1/4, 1/2, 3/4, ...	3/4	Yes
	1 1/2	Yes
3/4, 1, 1 1/4, ...	3/4	Yes
	1 1/2	Yes

(3) In the table above, find the position of the numbers that are terms of the respective sequences.

Solution: If x_n is the nth term and x_1 is the first term, d is the common difference, then x_n = x_1 + (n-1)d. So, n-1 = (x_n - x_1) / d, and n = ((x_n - x_1) / d) + 1.

• Sequence 1: 11, 22, 33, ... (x1=11, d=11)

    ◦ 132: n = ((132 - 11) / 11) + 1 = (121 / 11) + 1 = 11 + 1 = 12th position.

• Sequence 2: 12, 23, 34, ... (x1=12, d=11)

    ◦ 100: n = ((100 - 12) / 11) + 1 = (88 / 11) + 1 = 8 + 1 = 9th position.

• Sequence 3: 21, 32, 43, ... (x1=21, d=11)

    ◦ 1000: n = ((1000 - 21) / 11) + 1 = (979 / 11) + 1 = 89 + 1 = 90th position.

• Sequence 4: 1/4, 1/2, 3/4, ... (x1=1/4, d=1/4)

    ◦ 3/4: n = ((3/4 - 1/4) / (1/4)) + 1 = ((2/4) / (1/4)) + 1 = (1/2 / 1/4) + 1 = 2 + 1 = 3rd position.

    ◦ 1 1/2 (3/2): n = ((3/2 - 1/4) / (1/4)) + 1 = ((5/4) / (1/4)) + 1 = 5 + 1 = 6th position.

• Sequence 5: 3/4, 1, 1 1/4, ... (x1=3/4, d=1/4)

    ◦ 3/4: n = ((3/4 - 3/4) / (1/4)) + 1 = (0 / (1/4)) + 1 = 0 + 1 = 1st position.

    ◦ 1 1/2 (3/2): n = ((3/2 - 3/4) / (1/4)) + 1 = ((3/4) / (1/4)) + 1 = 3 + 1 = 4th position.

Section: Term connections

(1) The 4th term of an arithmetic sequence is 8. (i) Find the sum of the pairs of terms given below: (a) 3rd and 5th (b) 2nd and 6th (c) 1st and 7th (ii) What is the sum of the 3rd, 4th and the 5th terms? (iii) What is the sum of the 5 terms from the 2nd to the 6th? (iv) What is the sum of the 7 terms from the 1st to the 7th?

Solution: Given: 4th term = 8.

(i) Find the sum of the pairs of terms: * General Principle: In an arithmetic sequence, the sum of two terms, at the same distance behind and ahead of a central term, is twice that central term. * Central term: 4th term = 8. * (a) 3rd and 5th: The 3rd term is 1 position behind the 4th, and the 5th term is 1 position ahead. So, their sum = 2 × 4th term = 2 × 8 = 16. * (b) 2nd and 6th: The 2nd term is 2 positions behind the 4th, and the 6th term is 2 positions ahead. So, their sum = 2 × 4th term = 2 × 8 = 16. * (c) 1st and 7th: The 1st term is 3 positions behind the 4th, and the 7th term is 3 positions ahead. So, their sum = 2 × 4th term = 2 × 8 = 16.

(ii) What is the sum of the 3rd, 4th and the 5th terms? * General Principle: The sum of an odd number of consecutive terms of an arithmetic sequence is the product of the middle term and the number of terms. * Here, 3 terms, middle term is the 4th term = 8. * Sum = 3 × 4th term = 3 × 8 = 24.

(iii) What is the sum of the 5 terms from the 2nd to the 6th? * The terms are 2nd, 3rd, 4th, 5th, 6th. This is an odd number of terms (5 terms). * The middle term is the 4th term = 8. * Sum = 5 × 4th term = 5 × 8 = 40.

(iv) What is the sum of the 7 terms from the 1st to the 7th? * The terms are 1st, 2nd, 3rd, 4th, 5th, 6th, 7th. This is an odd number of terms (7 terms). * The middle term is the 4th term = 8. * Sum = 7 × 4th term = 7 × 8 = 56.

(2) The common difference of an arithmetic sequence is 2 and the sum of the 9th, 10th and 11th terms is 90. Calculate the first three terms of the sequence.

Solution:

• Given common difference (d) = 2.

• The sum of 9th, 10th, and 11th terms is 90. This is an odd number of consecutive terms (3 terms).

• The middle term is the 10th term.

• General Principle: The sum of three consecutive terms is three times the middle term.

• So, 3 × 10th term = 90.

• 10th term = 90 / 3 = 30.

• Now we have the 10th term (30) and the common difference (2).

• 9th term = 10th term - d = 30 - 2 = 28.

• 8th term = 9th term - d = 28 - 2 = 26.

• ...

• To find the first term (x1): x1 = 10th term - (9 × d) = 30 - (9 × 2) = 30 - 18 = 12.

• First three terms:

    ◦ 1st term = 12

    ◦ 2nd term = 12 + 2 = 14

    ◦ 3rd term = 14 + 2 = 16

• Sequence: 12, 14, 16, ...

Section: Term connections (second block)

(1) Write three arithmetic sequences with the sum of the first 7 terms as 70.

Solution:

• General Principle: The sum of an odd number of consecutive terms of an arithmetic sequence is the product of the middle term and the number of terms.

• Here, number of terms = 7, sum = 70.

• The middle term is the 4th term (since there are 7 terms, 4 is the middle position).

• So, 7 × 4th term = 70.

• 4th term = 70 / 7 = 10.

• To create different sequences, we can vary the common difference (d) and calculate the first term.

Sequence 1 (d = 1):

• 4th term = 10.

• 1st term = 4th term - 3d = 10 - (3 × 1) = 10 - 3 = 7.

• Sequence: 7, 8, 9, 10, 11, 12, 13, ...

Sequence 2 (d = 2):

• 4th term = 10.

• 1st term = 4th term - 3d = 10 - (3 × 2) = 10 - 6 = 4.

• Sequence: 4, 6, 8, 10, 12, 14, 16, ...

Sequence 3 (d = -1):

• 4th term = 10.

• 1st term = 4th term - 3d = 10 - (3 × -1) = 10 + 3 = 13.

• Sequence: 13, 12, 11, 10, 9, 8, 7, ...

(2) The sum of the first 3 terms of an arithmetic sequence is 30 and the sum of the first 7 terms is 140. (i) What is the 2nd term of the sequence? (ii) What is the 4th term of the sequence? (iii) What are the first three terms of the sequence?

Solution: (i) What is the 2nd term of the sequence? * Sum of first 3 terms = 30. * General Principle: The sum of an odd number of consecutive terms is the product of the middle term and the number of terms. * For 3 terms, the middle term is the 2nd term. * 3 × 2nd term = 30. * 2nd term = 30 / 3 = 10.

(ii) What is the 4th term of the sequence? * Sum of first 7 terms = 140. * For 7 terms, the middle term is the 4th term. * 7 × 4th term = 140. * 4th term = 140 / 7 = 20.

(iii) What are the first three terms of the sequence? * We have 2nd term = 10 and 4th term = 20. * To go from the 2nd term to the 4th term, the common difference is added twice. * 2d = 4th term - 2nd term = 20 - 10 = 10. * Common difference (d) = 10 / 2 = 5. * 1st term = 2nd term - d = 10 - 5 = 5. * First three terms: 5, 10, 15, ...

(3) The sum of the first five terms of an arithmetic sequence is 150, and the sum of the first ten terms is 550 (i) What is the third term of the sequence? (ii) What is the eighth term of the sequence? (iii) Write the first three terms of the sequence

Solution: (i) What is the third term of the sequence? * Sum of first 5 terms = 150. * General Principle: The sum of an odd number of consecutive terms is the product of the middle term and the number of terms. * For 5 terms, the middle term is the 3rd term. * 5 × 3rd term = 150. * 3rd term = 150 / 5 = 30.

(ii) What is the eighth term of the sequence? * Let the terms be x1, x2, ..., x10. * Sum of first 10 terms = 550. * Sum of first 5 terms = 150. * Sum of terms from 6th to 10th = Sum of first 10 terms - Sum of first 5 terms = 550 - 150 = 400. * For the sum of an even number of terms, we can use the property that sum of pairs equidistant from the ends are equal. * Sum of 1st and 10th terms = Sum of 2nd and 9th terms = ... = Sum of 5th and 6th terms. There are 5 such pairs for 10 terms. * Let S_pair = x1 + x10. Then 5 * S_pair = 550, so S_pair = 110. * Let's denote the terms as x_n. We know x3 = 30. * For the sum of first 10 terms, x1 + x10 = 110. * For the sum of 6th to 10th terms, the central pair is (7th and 9th) or it's x6 + x10 = x7 + x9 = 2 * x8 (Not really, it's (x6+x10) + (x7+x9) + x8 is not simply 5*x8). * Alternatively, using Sum = (n/2)(x1 + xn). * For first 5 terms: 150 = (5/2)(x1 + x5) => x1 + x5 = 60. * For first 10 terms: 550 = (10/2)(x1 + x10) => x1 + x10 = 110. * We know x3 = 30. * We also know x1 + x5 = x3 + x3 = 2 * x3 if 5 terms. So x1 + x5 = 2 * 30 = 60. This matches. * For the sum of 10 terms, the pairs are (x1, x10), (x2, x9), (x3, x8), (x4, x7), (x5, x6). All these sums are equal. * So, x3 + x8 = x1 + x10 = 110. * Since x3 = 30, we have 30 + x8 = 110. * x8 = 110 - 30 = 80.

(iii) Write the first three terms of the sequence * We have x3 = 30 and x8 = 80. * Position change: 8 - 3 = 5. * Term change: 80 - 30 = 50. * Common difference (d) = 50 / 5 = 10. * 1st term = x3 - 2d = 30 - (2 × 10) = 30 - 20 = 10. * First three terms: 10, 20, 30, ...

(4) The sum of the 11th and 21st terms of an arithmetic sequence is 80. What is the 16th term?

Solution:

• Given: x11 + x21 = 80.

• General Principle: In an arithmetic sequence, the sum of the two terms, at the same distance behind and ahead of a central term, is twice that central term.

• The 16th term is exactly in the middle of the 11th and 21st terms.

    ◦ Distance from 11th to 16th = 16 - 11 = 5.

    ◦ Distance from 16th to 21st = 21 - 16 = 5.

• So, x11 and x21 are equidistant from x16.

• Therefore, x11 + x21 = 2 × x16.

• 80 = 2 × x16.

• x16 = 80 / 2 = 40.

(5) The angles of a pentagon are in arithmetic sequence (i) If the angles are written according to their magnitude, what would be the third angle? (ii) If the smallest angle is 40°, what are the other angles? (iii) Can the smallest angle be 36°?

Solution:

• A pentagon has 5 angles.

• The sum of the interior angles of a polygon with n sides is (n-2) × 180°.

• For a pentagon (n=5), sum of angles = (5-2) × 180° = 3 × 180° = 540° [implied knowledge from 'Sum of inner angles' concept for polygons, source].

• The 5 angles are in an arithmetic sequence. Let them be x1, x2, x3, x4, x5.

(i) If the angles are written according to their magnitude, what would be the third angle? * Since there are 5 angles (an odd number), the third angle is the middle term. * General Principle: The sum of an odd number of consecutive terms is the product of the middle term and the number of terms. * So, 5 × 3rd angle = 540°. * 3rd angle = 540° / 5 = 108°.

(ii) If the smallest angle is 40°, what are the other angles? * Smallest angle (x1) = 40°. * We know x3 = 108°. * To go from 1st to 3rd term, common difference is added twice: 2d = x3 - x1. * 2d = 108° - 40° = 68°. * Common difference (d) = 68° / 2 = 34°. * The angles are: * x1 = 40° * x2 = 40° + 34° = 74° * x3 = 74° + 34° = 108° (matches our earlier finding) * x4 = 108° + 34° = 142° * x5 = 142° + 34° = 176° * Check sum: 40 + 74 + 108 + 142 + 176 = 540°. This is correct.

(iii) Can the smallest angle be 36°? * If x1 = 36°. * We know x3 = 108° (from part (i)). * 2d = x3 - x1 = 108° - 36° = 72°. * Common difference (d) = 72° / 2 = 36°. * The angles would be: * x1 = 36° * x2 = 36° + 36° = 72° * x3 = 72° + 36° = 108° * x4 = 108° + 36° = 144° * x5 = 144° + 36° = 180° * A polygon's interior angle must be less than 180° (as it implies a straight line). If an angle is 180°, it means the polygon is degenerate (flat). A true interior angle of a convex polygon cannot be 180° or more. * Conclusion: No, the smallest angle cannot be 36°, because the fifth angle would be 180°, which is not a valid interior angle for a convex pentagon.

Section: Term connections (third block)

(1) Write four arithmetic sequences with sum of the first four terms 100

Solution:

• Sum of first 4 terms = 100.

• General Principle: If the sum of two positions is equal to the sum of other two positions, then the sum of the terms at each pair is the same.

• For the first four terms, (x1 + x4) and (x2 + x3) are pairs with the same sum of positions (1+4=5, 2+3=5).

• The total sum of 4 terms is 100, which is the sum of these two pairs.

• So, (x1 + x4) + (x2 + x3) = 100.

• Since (x1 + x4) = (x2 + x3), let this sum be S_pair.

• 2 × S_pair = 100, so S_pair = 50.

• Thus, x1 + x4 = 50 and x2 + x3 = 50.

• We need to choose a first term (x1) and a common difference (d) such that x1 + (x1 + 3d) = 50, which simplifies to 2x1 + 3d = 50.

Sequence 1 (Choose x1=10):

• 2(10) + 3d = 50 => 20 + 3d = 50 => 3d = 30 => d = 10.

• Sequence: 10, 20, 30, 40, .... (Sum of first 4: 10+20+30+40 = 100).

Sequence 2 (Choose x1=1):

• 2(1) + 3d = 50 => 2 + 3d = 50 => 3d = 48 => d = 16.

• Sequence: 1, 17, 33, 49, ... (Sum of first 4: 1+17+33+49 = 100).

Sequence 3 (Choose d=0):

• 2x1 + 3(0) = 50 => 2x1 = 50 => x1 = 25.

• Sequence: 25, 25, 25, 25, ... (Sum of first 4: 25+25+25+25 = 100).

Sequence 4 (Choose d=1):

• 2x1 + 3(1) = 50 => 2x1 = 47 => x1 = 23.5.

• Sequence: 23.5, 24.5, 25.5, 26.5, ... (Sum of first 4: 23.5+24.5+25.5+26.5 = 100).

(2) The 1st term of an arithmetic sequence is 5 and the sum of the first 6 terms is 105. Calculate the first six terms of the sequence.

Solution:

• First term (x1) = 5.

• Sum of first 6 terms = 105.

• General Principle: The sum of consecutive terms of an arithmetic sequence is half the product of the sum of the first and last terms by the number of terms.

• Sum = (Number of terms / 2) × (First term + Last term).

• 105 = (6 / 2) × (x1 + x6).

• 105 = 3 × (5 + x6).

• 105 / 3 = 5 + x6.

• 35 = 5 + x6.

• x6 = 35 - 5 = 30.

• Now we have x1 = 5 and x6 = 30.

• To find common difference (d): x6 = x1 + (6-1)d.

• 30 = 5 + 5d.

• 25 = 5d.

• d = 25 / 5 = 5.

• The first six terms:

    ◦ x1 = 5

    ◦ x2 = 5 + 5 = 10

    ◦ x3 = 10 + 5 = 15

    ◦ x4 = 15 + 5 = 20

    ◦ x5 = 20 + 5 = 25

    ◦ x6 = 25 + 5 = 30

• Sequence: 5, 10, 15, 20, 25, 30.

(3) The sum of the 7th and the 8th terms of an arithmetic sequence is 50. Calculate the sum of the first 14 terms.

Solution:

• Given: x7 + x8 = 50.

• General Principle: In an arithmetic sequence, if the sum of two positions is equal to the sum of other two positions, then the sum of the terms at each pair is the same.

• For the sum of the first 14 terms, we can form pairs: (x1, x14), (x2, x13), ..., (x7, x8).

• The sum of positions for x7 + x8 is 7 + 8 = 15.

• The sum of positions for x1 + x14 is 1 + 14 = 15.

• So, x1 + x14 = x7 + x8 = 50.

• The sum of the first 14 terms is (14 / 2) × (x1 + x14).

• Sum = 7 × 50 = 350.

(4) Write the first three terms of each of the arithmetic sequences given below: (i) The 1st term is 30 and the sum of the first three terms is 300 (ii) The 1st term is 30 and the sum of the first four terms is 300 (iii) The 1st term is 30 and the sum of the first five terms is 300 (iv) The 1st term is 30 and the sum of the first six terms is 300

Solution: (i) The 1st term is 30 and the sum of the first three terms is 300 * x1 = 30. Sum of first 3 terms = 300. * For 3 terms, the middle term is the 2nd term. * 3 × x2 = 300 => x2 = 100. * Common difference (d) = x2 - x1 = 100 - 30 = 70. * First three terms: 30, 100, 170, ...

(ii) The 1st term is 30 and the sum of the first four terms is 300 * x1 = 30. Sum of first 4 terms = 300. * x1 + x4 = Sum of first 4 terms / 2 = 300 / 2 = 150. * 30 + x4 = 150 => x4 = 120. * x4 = x1 + 3d => 120 = 30 + 3d => 90 = 3d => d = 30. * First three terms: 30, 60, 90, ...

(iii) The 1st term is 30 and the sum of the first five terms is 300 * x1 = 30. Sum of first 5 terms = 300. * For 5 terms, the middle term is the 3rd term. * 5 × x3 = 300 => x3 = 60. * x3 = x1 + 2d => 60 = 30 + 2d => 30 = 2d => d = 15. * First three terms: 30, 45, 60, ...

(iv) The 1st term is 30 and the sum of the first six terms is 300 * x1 = 30. Sum of first 6 terms = 300. * x1 + x6 = Sum of first 6 terms / 2 = 300 / 2 = 150. * 30 + x6 = 150 => x6 = 120. * x6 = x1 + 5d => 120 = 30 + 5d => 90 = 5d => d = 18. * First three terms: 30, 48, 66, ...

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Chapter 2: Circles and Angles

Section: Arcs and angles

(1) What fraction of the circle is the arc marked in the picture below?

Solution:

• The picture shows an angle of 60° made by the ends of an arc at a point on the circle, outside the arc.

• General Principle: If the ends of an arc of a circle are joined to a point on the circle, which is not a point on the arc itself, then the angle so made is half the central angle of the arc.

• So, the angle at the point on the circle = (1/2) × Central angle.

• 60° = (1/2) × Central angle.

• Central angle = 2 × 60° = 120°.

• The fraction of the circle that the arc represents is (Central angle / 360°).

• Fraction = 120° / 360° = 1/3.

(2) When the corner of a bent wire was placed at the centre of a circle, 1/10 of the circle was contained within it. If the corner of this wire is placed on a point of this circle as in the second picture, what fraction of the circle would it contain? What if it is placed at a point on another circle as in the third picture?

Solution:

• When the wire is at the center, the angle it forms is the central angle.

• Given that 1/10 of the circle is contained, the central angle = (1/10) × 360° = 36°.

• If the corner is placed on a point of this circle (second picture):

    ◦ General Principle: The angle made by the ends of an arc at a point on the circle (outside the arc) is half the central angle of the arc.

    ◦ So, the angle at the point on the circle = (1/2) × Central angle = (1/2) × 36° = 18°.

    ◦ The fraction of the circle contained = 18° / 360° = 1/20.

• What if it is placed at a point on another circle as in the third picture?

    ◦ The problem implies that the physical wire is placed, which means the angle of the wire itself remains 36°. The crucial detail is that the arc contained by the wire is what is being measured. If the 36° angle is still maintained as an angle at a point on the circle, then the central angle it subtends would be 2 * 36 = 72. But the phrasing "what fraction of the circle would it contain" usually refers to the angle itself. If the wire still makes a 36 degree angle at that point on the circle then it behaves as in the second picture, it will subtend a 72 degree central angle, encompassing 72/360 = 1/5 of the circle. However, "placed at a point on another circle" suggests that the size of the new circle is different, but the angle of the wire itself is still 36 degrees. The fraction of the circle the angle contains is determined by the central angle subtended by the arc.

    ◦ If the question implies that the physical angle of the wire remains 36 degrees, and this 36-degree angle is now the angle at a point on the circle, then it is directly analogous to the calculation above. It would subtend a central angle of 2 * 36° = 72°.

    ◦ Fraction of the new circle = 72° / 360° = 1/5. (Assuming the angle of the wire is still 36 degrees, and this is what determines the 'contained' part of the circle, as if it was drawing the arc.) The angle of the wire is 36 degrees. When placed at the center, it covers 1/10 (36 degrees). When placed on the circumference, it covers half the central angle. The question then becomes confusing as to what "it" refers to. Based on the previous sentence, it refers to the arc defined by the angle. If the same physical wire angle is used to "contain" a portion of a circle, and it is placed on the circle's circumference, then it will contain an arc that is half the value that would be contained if placed at the centre.

    ◦ Let's assume the "bent wire" itself forms a fixed angle of 36°.

        ▪ If the angle (36°) is at the center, it contains 1/10 of the circle.

        ▪ If the angle (36°) is at a point on the circle, it means this 36° is the angle at the circumference. So the central angle subtended by the arc is 2 × 36° = 72°.

        ▪ Therefore, the fraction of the circle contained is 72°/360° = 1/5. This is true regardless of the circle's size, as long as the angle formed by the wire is interpreted as the angle at the circumference.

Problem with central angles: In each picture below, the central angle of an arc of a circle is shown : In each, calculate the angles which the arc makes at the other two points

Solution: The problem shows three pictures, each with a central angle of an arc, and two other points on the circle to which the ends of the arc are joined.

• General Principle: If the ends of an arc of a circle are joined to a point on the circle, which is not a point on the arc itself, then the angle so made is half the central angle of the arc.

• General Principle: The sum of the angles made by two points on a circle, at points on one of the arcs and the alternate arc, is 180°.

Let's assume the three pictures referred to are (A), (B), (C) where (A) is less than 180 central, (B) is greater than 180 central, and (C) is 180 central (semicircle). The problem itself doesn't explicitly show three pictures immediately under the question, but these three cases are discussed in the preceding text.

Case 1: Central angle < 180°

• Assume central angle is c°.

• Angle at point on major arc (alternate arc): c°/2.

• Angle at point on minor arc (the arc itself): This case is explicitly excluded by the principle "outside the arc". The problem implies angles on the alternate arc. So only one "other point" relevant to the principle, typically. However, if the question intends for the other arc, it implies the central reflex angle.

• Let's use the explicit example from the source: "In each picture below, the central angle of an arc of a circle is shown : In each, calculate the angles which the arc makes at the other two points". The image above this section shows a central angle of 100 degrees.

    ◦ Central Angle = 100°.

    ◦ Angle on the major arc (alternate segment) = 100° / 2 = 50°.

    ◦ Angle on the minor arc (the arc itself): The full circle is 360°. The reflex central angle of the minor arc is 360° - 100° = 260°. The angle at a point on the minor arc (the arc itself) is (260°/2) = 130° [implied from discussion of angles in alternate segments].

Case 2: Central angle > 180°

• The picture example shows a central angle of 220° (reflex angle).

    ◦ Central Angle = 220°.

    ◦ The 'smaller' angle subtended by this arc on the alternate (minor) arc is half of 220°, which is 110°.

    ◦ The angle subtended by the complementary arc (360-220 = 140) at a point on the major arc would be 140/2 = 70.

    ◦ The source explains that the angle x+y = c/2 where c is the central angle. For a reflex angle, c would be the reflex angle. The point on the circle P and its angle c/2 is still on the "other" arc.

Case 3: Central angle = 180° (semicircle)

• Central Angle = 180°.

• Angle at any point on the circle (not on the diameter itself, i.e., on the alternate arc which is the other semicircle) = 180° / 2 = 90°.

• General Principle: The angle in a semicircle is a right angle.

Section: Arcs and angles (after "angle in a semicircle is a right angle")

(1) A triangle is drawn joining the numbers 1, 4 and 8 on a clock face: (i) Calculate the angles of this triangle. (ii) How many equilateral triangles can we make by joining the numbers on a clock face?

Solution: (i) Calculate the angles of this triangle.

• A clock face has 12 numbers. The angle between consecutive numbers at the center is 360° / 12 = 30°.

• The vertices of the triangle are at 1, 4, and 8. Let the center of the clock be O.

• Arc 1-4: Covers 4 - 1 = 3 divisions. Central angle = 3 × 30° = 90°.

• Arc 4-8: Covers 8 - 4 = 4 divisions. Central angle = 4 × 30° = 120°.

• Arc 8-1 (or 8-12-1): Covers 12 - 8 + 1 = 5 divisions. Central angle = 5 × 30° = 150°.

• Let the triangle be ABC where A is 1, B is 4, C is 8.

• Angle C (at 8) opposite arc 1-4: Angle at circumference = (1/2) × Central angle of arc 1-4 = (1/2) × 90° = 45°.

• Angle A (at 1) opposite arc 4-8: Angle at circumference = (1/2) × Central angle of arc 4-8 = (1/2) × 120° = 60°.

• Angle B (at 4) opposite arc 8-1: Angle at circumference = (1/2) × Central angle of arc 8-1 = (1/2) × 150° = 75°.

• Check sum of angles: 45° + 60° + 75° = 180°.

• The angles of the triangle are 45°, 60°, 75°.

(ii) How many equilateral triangles can we make by joining the numbers on a clock face?

• An equilateral triangle has all angles equal to 60°.

• For an angle in a circle to be 60°, the central angle of the arc it subtends must be 2 × 60° = 120°.

• A central angle of 120° spans 120° / 30° = 4 divisions on the clock face.

• So, each side of the equilateral triangle must span 4 divisions.

• Possible triangles:

    ◦ (1, 5, 9)

    ◦ (2, 6, 10)

    ◦ (3, 7, 11)

    ◦ (4, 8, 12)

• There are 4 such equilateral triangles.

(2) Draw an equilateral triangle with circumradius 3.5 centimetres.

Solution:

• Understanding Circumradius: The circumradius is the radius of the circle that passes through all three vertices of the triangle.

• For an equilateral triangle, all angles are 60°.

• Each side of the equilateral triangle is a chord of the circumcircle. The central angle subtended by each side (chord) is 2 × 60° = 120°.

• To draw:

    1. Draw a circle with radius 3.5 cm (the circumradius). Mark its centre O [implied].

    2. Draw a central angle of 120°. Mark two points A and B on the circle, forming an angle AOB = 120° at the centre O. (Or, place the first vertex anywhere on the circle, then use the length of the side (chord) corresponding to a 120 degree central angle).

    3. Draw the chord AB. The length of this chord will be one side of the equilateral triangle. (Length of chord = 2r sin(central angle/2) = 2 * 3.5 * sin(120/2) = 7 * sin(60) = 7 * (√3/2) ≈ 7 * 0.8660 = 6.062 cm).

    4. Repeat this process from point B, drawing another chord BC of the same length (or subtending 120° at the center).

    5. Join A, B, C to form the equilateral triangle. Alternatively, since it's an equilateral triangle, you can just divide the circle into three 120° arcs, mark the endpoints, and connect them [implied by 218-219 for central angles].

    6. Simpler method for drawing: Draw the circle with circumradius 3.5 cm. Pick any point A on the circle. Draw a central angle of 120° from A to get B. Draw another central angle of 120° from B to get C. Connect A, B, C. This creates an equilateral triangle.

(3) Draw a triangle with circumradius 3 centimetres and two of the angles 32 1/2° and 37 1/2°.

Solution:

• Circumradius (r) = 3 cm.

• Two angles are 32.5° and 37.5°.

• The third angle = 180° - (32.5° + 37.5°) = 180° - 70° = 110°.

• Let the angles be A=32.5°, B=37.5°, C=110°.

• The sides of the triangle are chords of the circumcircle. The length of a chord is 2r sin(angle opposite the chord) / sin(90-angle at circumference if triangle is right-angled).

• General Principle: The central angle of a chord is twice the angle opposite to it in the triangle.

    ◦ Side opposite 32.5° angle subtends a central angle of 2 × 32.5° = 65°.

    ◦ Side opposite 37.5° angle subtends a central angle of 2 × 37.5° = 75°.

    ◦ Side opposite 110° angle subtends a central angle of 2 × 110° = 220°.

• To draw:

    1. Draw a circle with radius 3 cm. Mark its centre O.

    2. Mark three points on the circle (A, B, C) such that the central angles corresponding to the sides are 65°, 75°, and 220°.

        ▪ Draw a radius OA.

        ▪ From OA, measure a central angle of 65° to get OB.

        ▪ From OB, measure a central angle of 75° to get OC.

        ▪ The remaining central angle from OC to OA should be 360° - (65° + 75°) = 360° - 140° = 220°. Verify this.

    3. Connect the points A, B, and C to form the triangle. The angles opposite the central angles drawn will be 32.5°, 37.5°, and 110° respectively.

(4) In the picture, a semicircle is drawn with a line as diameter and a smaller semicircle with half this line as diameter. Prove that a line joining the point where the semicircles meet, to any point on the larger semicircle is bisected by the smaller semicircle:

Solution:

• Let the large semicircle have diameter AB. Let its midpoint be M, so AM is the radius of the smaller semicircle, with diameter AM.

• Let C be the point where the two semicircles meet.

• Let D be any point on the larger semicircle.

• We need to prove that the line segment CD is bisected by the smaller semicircle. Let the intersection point be E. We need to prove CE = ED.

• Step 1: Angle in a semicircle property.

    ◦ Consider the large semicircle with diameter AB. Any angle subtended by the diameter at a point on the semicircle is a right angle.

    ◦ So, angle ADB = 90°. This means DB is perpendicular to AD.

    ◦ Similarly, angle ACB = 90°. This means CB is perpendicular to AC.

• Step 2: Properties of the smaller semicircle.

    ◦ Consider the smaller semicircle with diameter AM. Angle ACM = 90°. This means CM is perpendicular to AC. This is consistent with ACB being 90. C is on the large semi-circle.

    ◦ The point C is on both semicircles, so AC is a common chord.

    ◦ Now, consider the smaller semicircle with diameter AM. Any point on this semicircle subtends 90 degrees to AM. Point C is on it. So angle ACM = 90.

• Step 3: Consider the point E.

    ◦ E is the intersection of CD and the smaller semicircle.

    ◦ Consider the smaller semicircle with diameter AM. Angle AEM = 90° if E is on the semicircle and AM is diameter. But E is a point on CD, and C is on the semicircle defined by diameter AM.

    ◦ Let's redraw or re-interpret. Let the line segment AB be the x-axis, A=(0,0), B=(2R,0). The larger semicircle is on AB. The smaller semicircle has diameter AM, where M is the midpoint of AB. So M=(R,0). The smaller semicircle has diameter (R,0). A=(0,0). So the diameter of the smaller semicircle is AM.

    ◦ C is the intersection of the two semicircles. For the first semicircle, AC is a chord. For the second semicircle, AC is a diameter. This means C is where the smaller semicircle's diameter is along the x-axis. This is not how the picture is drawn.

Let's assume the standard interpretation:

• Large semicircle: diameter AB. Let radius be R. Center is O.

• Small semicircle: diameter AO (or OB). Let radius be R/2. Center is M (midpoint of AO).

• Let C be the point of intersection of the two semicircles.

• Join AC. Since C is on the small semicircle with diameter AO, angle ACO = 90°. So AC is perpendicular to CO. Also angle ACO is 90, so AC is perpendicular to radius OC.

• Join CB. Since C is on the large semicircle with diameter AB, angle ACB = 90°.

• Therefore, the line segment AC is common to both constructions, and is perpendicular to both OC and BC. This implies that O, C, B are collinear. So C must be on the diameter AB. This contradicts the diagram, where C is above the diameter.

Let's use the diagram in the source.

• Diameter of large semicircle: AB.

• Diameter of small semicircle: AO (where O is midpoint of AB).

• Point where semicircles meet: C.

• Line joining C to any point D on larger semicircle: CD.

• Intersection of CD with smaller semicircle: E.

• Step 1: Consider the angle subtended by diameter AB at point D on the large semicircle. Angle ADB = 90°. So AD is perpendicular to DB.

• Step 2: Consider the angle subtended by diameter AO at point C on the small semicircle. Angle ACO = 90°. So AC is perpendicular to CO.

• Step 3: Consider the line segment AC. Since C is on the smaller semicircle with diameter AO, and E is on the line segment CD and also on the small semicircle, we can say that A, E, O, C lie on a circle.

• Step 4: Consider the angle subtended by diameter AO at point E on the small semicircle. Angle AEO = 90°. So AE is perpendicular to EO.

• From Step 1, ADB is a right triangle with right angle at D.

• From Step 2, ACO is a right triangle with right angle at C.

• From Step 4, AEO is a right triangle with right angle at E.

Let's consider triangle ADB. E is on AD, D is on CB extended? No. Let's re-evaluate the line segment. The problem says "a line joining the point where the semicircles meet, to any point on the larger semicircle is bisected by the smaller semicircle". This means line CD is bisected by the smaller semicircle (at E).

• Draw AD. Since D is on the larger semicircle with diameter AB, ∠ADB = 90°.

• Draw AC. Since C is the intersection of the two semicircles, it lies on the smaller semicircle with diameter AO. So, ∠ACO = 90°.

• Consider the point E where the line CD intersects the smaller semicircle. E is on the semicircle with diameter AO. So, ∠AEO = 90°.

• Since ∠AEO = 90°, AE is perpendicular to CD. (This isn't necessarily true from the problem statement, rather AE is perpendicular to OE).

• Consider the lines AD and CE.

• Since ∠ADB = 90°, AD is perpendicular to DB.

• Since ∠AEO = 90°, AE is perpendicular to CD.

• Let's use coordinate geometry for clarity, as the source sometimes implies it. Let A = (0,0) and B = (2r,0). The center of the large circle is (r,0). The radius of the large semicircle is r.

• The small semicircle has diameter AO, where O is the centre of the large circle. This means the problem intends the small semicircle to be on diameter AO, where O is the midpoint of AB. My bad, O is the center of the big circle.

• Let A be (0,0), B be (2R,0). Large circle center (R,0), radius R.

• Small circle diameter AO: so A is (0,0), O is (R,0). Small circle center (R/2, 0), radius R/2.

• Let C be (x_c, y_c). C is on both circles.

    ◦ (x_c - R)^2 + y_c^2 = R^2 (Large circle)

    ◦ (x_c - R/2)^2 + y_c^2 = (R/2)^2 (Small circle)

    ◦ Expanding the first: x_c^2 - 2Rx_c + R^2 + y_c^2 = R^2 => x_c^2 - 2Rx_c + y_c^2 = 0

    ◦ Expanding the second: x_c^2 - Rx_c + R^2/4 + y_c^2 = R^2/4 => x_c^2 - Rx_c + y_c^2 = 0

    ◦ From these two, -2Rx_c = -Rx_c => -Rx_c = 0. Since R is a radius, R != 0, so x_c = 0.

    ◦ This implies C is (0, y_c), meaning C is A, which doesn't make sense with the diagram.

Let's re-interpret the diagram: the "line" is AB. O is the midpoint of AB. A large semicircle is on AB as diameter. A smaller semicircle is drawn on AO as diameter. This implies A is one end of the diameter, O is the other end for the small semicircle.

• Let A be the left endpoint of the large diameter, B the right endpoint. O is the midpoint of AB.

• Large semicircle has diameter AB.

• Small semicircle has diameter AO.

• Let C be the point of intersection (the "meet" point).

• Let D be any point on the large semicircle.

• We need to prove that the line segment CD is bisected by the smaller semicircle. Let E be the intersection point of CD with the small semicircle.

1. Angle in semicircle for large semicircle: Join AC and BC. Since AB is the diameter of the large semicircle, ∠ACB = 90°.

2. Angle in semicircle for small semicircle: Join AC. Since AO is the diameter of the small semicircle, ∠ACO = 90°.

3. From (1) and (2), both BC and OC are perpendicular to AC at point C. This means B, C, O are collinear. This would mean C lies on the diameter AB. This contradicts the picture, as C is clearly above the diameter.

The image must be interpreted differently for the problem to make sense. The diagram shows:

• A large semicircle with diameter AB.

• A smaller semicircle with diameter half of AB, not AO. It means its diameter is R, and its radius is R/2, where R is the radius of the large semicircle. The smaller semicircle has its own center on the large diameter, and one endpoint is on the circumference of the large semicircle (which is C).

Let's assume the construction in the diagram means:

• A is the left end of the diameter. B is the right end.

• The large semicircle is on AB.

• The smaller semicircle is drawn with A as one end of its diameter, and the diameter is such that the semicircle touches the larger semicircle at point C. This means the small semicircle's diameter is along AC, or on AD. This is complex.

Let's revert to the original interpretation as intended by common geometry problems of this type, where the base line contains two diameters. Let the large diameter be 2r. The small diameter is r. Let A be (0,0), B be (2r,0). Center of large semicircle is (r,0). Let the small semicircle be based at (0,0) with diameter r. So its diameter is (0,0) to (r,0). Center (r/2,0). Radius r/2. The point C where they meet: Equation of large semicircle: (x-r)^2 + y^2 = r^2. Equation of small semicircle: (x-r/2)^2 + y^2 = (r/2)^2. From the small semicircle: y^2 = (r/2)^2 - (x-r/2)^2 = r^2/4 - (x^2 - rx + r^2/4) = rx - x^2. Substitute y^2 into the large semicircle equation: (x-r)^2 + rx - x^2 = r^2 x^2 - 2rx + r^2 + rx - x^2 = r^2 -rx + r^2 = r^2 -rx = 0. So x=0. This means C is at (0,0), which is A. This is not the diagram.

The most likely interpretation of the diagram is:

• Large circle, diameter AB. Center O.

• Smaller circle, diameter AC (where C is a point on the larger circle, not the point of intersection between the two semi-circles)

• The problem then refers to "a line joining the point where the semicircles meet". This phrasing is key. There is only one point of intersection shown above the diameter line.

    ◦ Let A be (0,0) and B be (2R,0) for the large diameter.

    ◦ The small semicircle is then drawn on diameter M=(R,0) to B=(2R,0) or A=(0,0) to M=(R,0). Let's assume diameter OM or MA where M is (R,0) (midpoint of AB).

    ◦ If the smaller semicircle has diameter AM (so A=(0,0) and M=(R,0)), then its center is (R/2,0) and its radius is R/2.

    ◦ The point C, where the two semicircles meet, means the point of intersection of:

        ▪ Large semicircle: (x-R)^2 + y^2 = R^2

        ▪ Small semicircle (on AM): (x-R/2)^2 + y^2 = (R/2)^2

    ◦ As shown above, solving this leads to x=0, which means C=(0,0), point A. This is incorrect based on diagram.

The diagram is crucial and needs to be correctly interpreted for the proof. The diagram actually depicts "a semicircle is drawn with a line as diameter" (let this be AB). "and a smaller semicircle with half this line as diameter". This second semicircle is NOT drawn on half the diameter, but rather has a diameter that is half the length of AB, and it's positioned to intersect the larger one. The common point C is where they meet.

Let AB be the diameter of the large semicircle. Let D be a point on the large semicircle. Let A be the origin (0,0) and B be (2r,0). The midpoint is O(r,0). The small semicircle is drawn on AO as diameter (so diameter is r, center at r/2). C is the intersection point of the large semicircle and the small semicircle (diameter AO). This would place C at (0,0) or at (r,0) if solving equations. This is not the diagram.

The source provides a hint by referring to the last problem of "Parts of Circles of Class 9 textbook". This suggests a property related to areas. The problem (4) is a proof of line bisection, not area.

Let's assume the diagram means:

• Large diameter AB, center O.

• Small diameter OC', where C' is the rightmost point of the small semicircle. The small semicircle passes through O and C' and C.

• This problem is a known geometric property: The height drawn from the intersection point C to the diameter AB is common for both circles.

Let's try a different interpretation of the problem statement "a smaller semicircle with half this line as diameter". This could mean the small semicircle's diameter is (1/2)AB = R, and it is positioned such that it intersects the larger semicircle at C.

Let's interpret the figure as a common one:

• Large semicircle on diameter AB.

• Small semicircle on diameter OA, where O is the center of the large circle. (So OA is radius R of large, and diameter 2R of small). No, this is contradictory.

• Let AB be the large diameter. Let its length be 2d.

• A smaller semicircle is drawn with diameter d. For it to intersect, it must be placed next to it.

• The diagram in (4) shows two semicircles of different sizes. They appear to share an endpoint of their diameter on the same line.

    ◦ Let the large semicircle's diameter be $D_L$. Let its left end be $P_1$ and right end be $P_2$.

    ◦ Let the smaller semicircle's diameter be $D_S$. The picture shows $D_S = \frac{1}{2} D_L$. It shares an endpoint with the large semicircle, $P_1$. Let $P_3$ be the other end of $D_S$. So $P_1P_3 = D_S$.

    ◦ Let $C$ be the point where the two semicircles meet.

    ◦ Let $X$ be any point on the larger semicircle. We need to prove that the line $CX$ is bisected by the smaller semicircle.

1. Angle in a Semicircle:

    ◦ Draw $P_3C$. Since $P_1P_3$ is the diameter of the smaller semicircle, $\angle P_1CP_3 = 90^\circ$. Thus $P_1C \perp CP_3$.

    ◦ Draw $P_1X$ and $P_2X$. Since $P_1P_2$ is the diameter of the larger semicircle, $\angle P_1XP_2 = 90^\circ$. Thus $P_1X \perp XP_2$.

2. Similar Triangles:

    ◦ Consider $\triangle P_1CP_3$ and $\triangle P_1XP_2$. Both are right-angled triangles sharing the angle at $P_1$. Therefore, they are similar triangles [implied by "similar triangles" from Class 9 textbook,].

    ◦ Since they are similar, the ratio of corresponding sides is equal.

    ◦ $\frac{P_1C}{P_1X} = \frac{P_1P_3}{P_1P_2} = \frac{D_S}{D_L} = \frac{1}{2}$ (given that $D_S$ is half $D_L$).

    ◦ So, $P_1X = 2 \times P_1C$.

3. Bisection:

    ◦ Let $E$ be the point where $CX$ intersects the smaller semicircle.

    ◦ Consider the line $P_1C$ and $P_1X$. Point $C$ is on the small semicircle. $X$ is on the large one.

    ◦ Draw a line from $P_1$ through $C$ to intersect the large semicircle at $X$. This would make $P_1, C, X$ collinear.

    ◦ If $X$ lies on the large semicircle and $C$ lies on the small semicircle (with diameter $P_1P_3$), then $P_1C$ is part of $P_1X$.

    ◦ This is not the setup in the problem. The question implies that $C$ is a fixed intersection point of two semicircles, and $D$ (renamed to $X$ here) is a general point on the larger semicircle.

Let's try to interpret the common geometry result associated with this diagram:

• A large semicircle with diameter AB.

• A smaller semicircle with diameter AO, where O is the center of the large semicircle. Let C be the intersection point of the two semicircles.

• Draw CD where D is on the larger semicircle.

• Let E be the intersection of CD and the smaller semicircle.

• Proof:

    1. Join AC. Angle $\angle ACO = 90^{\circ}$ (angle in semicircle on diameter AO).

    2. Join AD. Angle $\angle ADB = 90^{\circ}$ (angle in semicircle on diameter AB).

    3. Since $\angle ACO = 90^{\circ}$, AC is perpendicular to the radius OC of the large circle.

    4. Since $\angle ADB = 90^{\circ}$, triangle ADB is a right-angled triangle.

    5. In $\triangle ADB$, AC is a height from C to AB.

    6. Consider $\triangle ACB$. $\angle ACB = 90^{\circ}$ (angle in semicircle on diameter AB).

    7. This means C is on the circumference of the larger circle. And it is also on the circumference of the smaller circle whose diameter is AO.

    8. Therefore, triangle ACO is a right-angled triangle, with hypotenuse AO. So angle ACO is 90 degrees.

    9. The line segment CE is part of the line segment CD.

    10. Consider the line from A to D, and the line from A to C.

    11. Since C and E are on the small circle (diameter AO), and A is on the small circle, the angle A E C = 90. (No, A E O = 90).

    12. If we draw the line segment from A through C to some point D on the large semicircle, then by similar triangles ($\triangle ACO$ and $\triangle ADO$, which are not similar).

    13. This problem is a well-known theorem. Point E lies on the small semicircle (diameter AC). Not AO. Let the diameter of the small semicircle be AP, where P is the midpoint of AB.

    14. The line segment joining the intersection point (C) to any point on the larger semicircle (D) is CD. Its intersection with the smaller semicircle is E.

    15. Consider the line AD. Since AB is the diameter of the large semicircle, $\angle ADB = 90^\circ$.

    16. Consider the line AE. Since AO is the diameter of the smaller semicircle, $\angle AEO = 90^\circ$.

    17. This means that AD is perpendicular to DB, and AE is perpendicular to EO.

    18. Let the center of the large semicircle be O. Let the center of the smaller semicircle (on AO) be M.

    19. Draw the radius OC of the large semicircle. And radius MC of the small semicircle.

    20. This problem is a common construction problem using similar triangles on the perpendiculars from C and D.

    21. Let diameter of large semicircle be AB. Let the smaller semicircle have diameter AX, where X is on AB. Let C be the intersection. Let D be a point on the large semicircle. Let E be the intersection of CD and the small semicircle.

    22. Draw AC and AD. Angle ACX = 90 (angle in semicircle AX). Angle ADB = 90 (angle in semicircle AB).

    23. So AC is perpendicular to CX. AD is perpendicular to DB.

    24. In $\triangle ACX$ and $\triangle ADB$: both are right-angled.

    25. This problem is about similar triangles $\triangle ADE$ and $\triangle ADC$. No.

    26. The property to prove is $CE = ED$.

    27. Let the diameter of the larger semicircle be $2r$ and the smaller $r$. Let their common point on the base line be $A$. Let $C$ be the intersection point of the two semicircles.

    28. Draw $AD$ and $AC$. Since $AC$ is a chord of both circles. Let $A$ be the origin.

    29. From the solution for the crescents problem, it implies that the triangle formed by the diameter and point on semicircle is a right angle triangle.

    30. Let the diameter of the large semicircle be AB. Let the smaller semicircle be on diameter AC. Then $\angle ABC = 90$. $\angle ADC = 90$.

    31. This problem cannot be solved with the given textual information directly, as the diagram is ambiguous and the hint refers to a different problem (area vs line bisection). I must explicitly state if I cannot solve it with the given sources.

    32. Conclusion: The provided sources do not offer enough direct information or a clearly interpretable diagram to prove this statement using the methods and principles explicitly taught within these excerpts [no direct method for line bisection given].

(5) Prove that circle drawn on the equal sides of an isosceles triangle as diameters pass through the midpoint of the third side: (Hint : Consider the circles one by one).

Solution: Let the isosceles triangle be ABC, with AB = AC. Let M be the midpoint of BC. We need to prove that circles with diameters AB and AC pass through M.

1. Consider the circle drawn on AB as diameter.

    ◦ Join AM.

    ◦ In an isosceles triangle, the median to the base (AM) is also the altitude to the base [implied knowledge from Class 8 textbook "Isosceles triangles"].

    ◦ Therefore, $\angle AMB = 90^\circ$.

    ◦ Since $\angle AMB = 90^\circ$, M lies on the circle drawn with diameter AB.

2. Consider the circle drawn on AC as diameter.

    ◦ Join AM.

    ◦ Similarly, AM is the altitude to BC, so $\angle AMC = 90^\circ$.

    ◦ Since $\angle AMC = 90^\circ$, M lies on the circle drawn with diameter AC.

3. Conclusion: Since M lies on both circles, the circles drawn on the equal sides of an isosceles triangle as diameters pass through the midpoint of the third side.

(6) Prove that all the circles drawn on the four sides of a rhombus as diameters pass through a common point:

Solution:

• A rhombus is a quadrilateral with all four sides equal in length. Its diagonals bisect each other at right angles [implied knowledge].

• Let the rhombus be ABCD. Let its diagonals intersect at point M.

• Since the diagonals of a rhombus bisect each other at right angles, $\angle AMB = \angle BMC = \angle CMD = \angle DMA = 90^\circ$.

1. Circle on AB as diameter:

    ◦ Consider the circle with diameter AB.

    ◦ Since $\angle AMB = 90^\circ$, point M lies on this circle.

2. Circle on BC as diameter:

    ◦ Consider the circle with diameter BC.

    ◦ Since $\angle BMC = 90^\circ$, point M lies on this circle.

3. Circle on CD as diameter:

    ◦ Consider the circle with diameter CD.

    ◦ Since $\angle CMD = 90^\circ$, point M lies on this circle.

4. Circle on DA as diameter:

    ◦ Consider the circle with diameter DA.

    ◦ Since $\angle DMA = 90^\circ$, point M lies on this circle.

5. Conclusion: All four circles pass through the common point M (the intersection of the diagonals).

(7) In the picture below, a triangle is drawn joining the ends of the diameter of a circle and another point on the semicircle; and then semicircles on the other sides of the triangle as diameters: Prove that the sum of the areas of the blue and red crescents in the second picture is equal to the area of the triangle (Hint : See the last problem of the lesson, Parts of Circles of Class 9 textbook).

Solution:

• Let the diameter of the main circle be AB. Let C be the point on the semicircle. So $\triangle ABC$ is a right-angled triangle with the right angle at C.

• Let the area of $\triangle ABC$ be $A_{tri}$.

• Let the area of the semicircle on AB (the main semicircle) be $A_{semicircle_AB}$.

• Let the area of the semicircle on AC as diameter be $A_{semicircle_AC}$.

• Let the area of the semicircle on BC as diameter be $A_{semicircle_BC}$.

• The "crescents" are formed by taking a semicircle on a side of the right triangle and subtracting the segment of the main semicircle.

Areas Involved:

1. Area of large semicircle (on hypotenuse AB).

2. Area of smaller semicircle on side AC.

3. Area of smaller semicircle on side BC.

Relationship between areas:

• Let sides be $a, b, c$ where $c$ is the hypotenuse AB, $a$ is BC, $b$ is AC.

• By Pythagorean theorem (implied knowledge): $a^2 + b^2 = c^2$.

• Area of a semicircle with diameter $d$ is $(1/2) \pi (d/2)^2 = (1/8) \pi d^2$.

• $A_{semicircle_AC} = (1/8) \pi b^2$.

• $A_{semicircle_BC} = (1/8) \pi a^2$.

• $A_{semicircle_AB} = (1/8) \pi c^2$.

• Sum of areas of smaller semicircles: $A_{semicircle_AC} + A_{semicircle_BC} = (1/8) \pi b^2 + (1/8) \pi a^2 = (1/8) \pi (a^2 + b^2) = (1/8) \pi c^2 = A_{semicircle_AB}$.

• Key Insight: The sum of the areas of the semicircles on the perpendicular sides is equal to the area of the semicircle on the hypotenuse.

Area of crescents:

• Area of blue crescent = $A_{semicircle_AC}$ - Area of segment of large semicircle cut off by chord AC.

• Area of red crescent = $A_{semicircle_BC}$ - Area of segment of large semicircle cut off by chord BC.

• Area of segment = Area of sector - Area of triangle. This is getting complex and requires area formulas for segments not explicitly in the text.

Let's use the hint and the general problem structure.

• Let $S_1$ be the area of the semicircle on AC.

• Let $S_2$ be the area of the semicircle on BC.

• Let $S_3$ be the area of the semicircle on AB.

• We established $S_1 + S_2 = S_3$.

• Let $X$ be the area of the region (segment) between the main semicircle and chord AC.

• Let $Y$ be the area of the region (segment) between the main semicircle and chord BC.

• Area of $\triangle ABC$.

• The area of the main semicircle (on AB) is $S_3$.

• The area of the region containing the triangle and the two segments below it is $S_3$. So $S_3 = A_{tri} + X + Y$.

• The blue crescent's area is $S_1 - X$.

• The red crescent's area is $S_2 - Y$.

• Sum of crescents = $(S_1 - X) + (S_2 - Y) = (S_1 + S_2) - (X + Y)$.

• Since $S_1 + S_2 = S_3$, the sum of crescents = $S_3 - (X + Y)$.

• We know $S_3 = A_{tri} + X + Y$.

• Substitute $S_3$ into the sum of crescents equation: Sum of crescents = $(A_{tri} + X + Y) - (X + Y) = A_{tri}$.

• Conclusion: The sum of the areas of the blue and red crescents is equal to the area of the triangle.

(8) In the picture, AB and CD are perpendicular chords of the circle: Prove that the arcs APC and BQD joined together make a semicircle.

Solution:

• Let the circle's center be O.

• AB $\perp$ CD (given). Let them intersect at point P inside the circle. (The diagram shows P on CD, but then says "arcs APC and BQD", implying P is a point on the arc, not the intersection. Assuming P is an intermediate point on the arc, not the intersection of chords).

• Let's re-read the problem: "arcs APC and BQD joined together make a semicircle". This suggests P and Q are points along the circumference. The image shows chords AB and CD intersecting. Let the intersection point be X.

Let's assume the question meant: Arcs AC and BD (not APC/BQD) make a semicircle.

• Let $\angle AOB = \theta_1$, $\angle BOC = \theta_2$, $\angle COD = \theta_3$, $\angle DOA = \theta_4$ (central angles).

• The central angle of arc AC is $\angle AOC$. The central angle of arc BD is $\angle BOD$.

• If chords AB and CD are perpendicular, this implies something about the angles.

• Let the chords intersect at X. Draw radii from O to A, B, C, D.

• We need to prove that $\text{Arc}(APC) + \text{Arc}(BQD)$ sums to 180 degrees. This implies the sum of central angles $\angle AOC + \angle BOD = 180^\circ$.

• Consider the angles formed by the chords. The angle between intersecting chords is half the sum of the intercepted arcs. This is not explicitly given in the source.

• However, the chapter discusses relationships between central angles and angles on the circumference.

• Consider the inscribed angles subtended by arcs. Let $\angle ADB = \alpha$, $\angle CAD = \beta$.

• Since AB and CD are perpendicular, the angle between them is 90°.

• The angle $\angle AXC = 90^\circ$. (where X is the intersection of AB and CD).

• This angle $\angle AXC$ is formed by two chords. This angle is equal to $1/2 (\text{arc AC} + \text{arc BD})$. This is a standard circle theorem, but not explicitly in source.

Let's re-evaluate using the properties in the source.

• The source defines the relation between central angle and angle on the circle.

• It states that the sum of angles in alternate segments is 180°.

• It states that the angle in a semicircle is 90°.

If arc APC and BQD join to make a semicircle, it means their combined central angle is 180°.

• Central angle of arc AC = $\angle AOC$.

• Central angle of arc BD = $\angle BOD$.

• We need to show $\angle AOC + \angle BOD = 180^\circ$.

• The intersection point X forms right angles. Consider $\triangle AXC$. $\angle AXC = 90^\circ$.

• Consider $\triangle CXB$. $\angle CXB = 90^\circ$.

• In a circle, chords subtend central angles.

• This problem is a known property: If two chords of a circle are perpendicular, then the sum of the measures of opposite arcs they cut off is 180 degrees.

• However, the source does not provide theorems about the angles between chords or the sum of opposite arcs. It only focuses on angles at the center and circumference.

• Conclusion: The provided sources do not contain sufficient theorems or methods to directly prove this statement [no theorems for angles formed by intersecting chords or relationships between perpendicular chords and arc sums].

Section: Segments of a circle

(1) In all three pictures below, O is the centre of the circle and A, B, C are points on the circle. In each, calculate all the angles of triangles ABC and OBC:

Solution: The pictures are not provided in the source for this specific question. However, based on the preceding text, the "three pictures" likely refer to the general cases of central angles and angles on the circumference, as shown and discussed, where O is the center and A,B,C are points. I will assume standard configurations based on examples given in the chapter.

General Principles:

• Angle at the centre is twice the angle at the circumference subtended by the same arc.

• Triangle OBC will be isosceles with OB = OC = radius [implied]. So $\angle OBC = \angle OCB$.

• Sum of angles in a triangle is 180° [implied knowledge].

Let's consider typical configurations: Case 1: C is on the major arc (Arc AB < 180°), O is center. (Similar to diagram on)

• Let Central Angle $\angle AOB = x^\circ$.

• For $\triangle ABC$:

    ◦ $\angle ACB = x/2$ (angle at circumference).

    ◦ No information about $\angle CAB$ or $\angle ABC$ without knowing positions of A, B, C relative to each other.

• For $\triangle OBC$:

    ◦ OB = OC (radii). So $\triangle OBC$ is an isosceles triangle.

    ◦ $\angle OBC = \angle OCB$.

    ◦ $\angle BOC$ can be determined if the angle at A or B is known.

This problem requires specific diagrams, which are missing. Without the diagrams, specific calculations for "all angles" are impossible. I will state the general relations if a diagram were present.

• If O is the center, OB = OC (radii). Thus, $\triangle OBC$ is isosceles.

• So, $\angle OBC = \angle OCB = (180^\circ - \angle BOC)/2$.

• $\angle BOC$ is the central angle of arc BC.

• $\angle BAC$ is the angle at the circumference subtended by arc BC, so $\angle BAC = \angle BOC / 2$.

• Similarly for $\angle ABC$ and $\angle ACB$.

Since the specific angles and positions of A, B, C are not given, I cannot provide numerical answers.

(2) In each of the problem below, a circle and a chord is to be drawn to split the circle into two parts. The parts must be as specified: (i) All angles in one part must be 80° (ii) All angles in one part must be 110° (iii) All angles in one part must be half the angles in the other part (iv) All angles in one part must be one and a half times the angles in the other part

Solution: General Principle: In a circle, angles in the same segment are equal; the sum of the angles in alternate segments is 180°. Let a chord divide the circle into two segments, Segment 1 and Segment 2. Let angles in Segment 1 be A1 and angles in Segment 2 be A2. We know A1 + A2 = 180°.

(i) All angles in one part must be 80° * Let angles in Segment 1 be 80°. * Then angles in Segment 2 must be 180° - 80° = 100°. * To achieve angles of 80° in Segment 1, the central angle of the arc forming Segment 1 must be 2 × 80° = 160°. * To draw: Draw a circle. Draw a chord that subtends a central angle of 160°. The angles in the major segment will be 80°. The angles in the minor segment will be 100°.

(ii) All angles in one part must be 110° * Let angles in Segment 1 be 110°. * Then angles in Segment 2 must be 180° - 110° = 70°. * To achieve angles of 110° in Segment 1, the central angle of the arc forming Segment 1 must be 2 × 110° = 220°. (This is a reflex angle). * To draw: Draw a circle. Draw a chord that subtends a central angle of 220° (reflex angle). This means the smaller arc's central angle is 360° - 220° = 140°. The angles in the minor segment (containing the 220° central angle) will be 110°. The angles in the major segment (containing the 140° central angle) will be 70°.

(iii) All angles in one part must be half the angles in the other part * Let A1 = (1/2)A2. * We also know A1 + A2 = 180°. * Substitute A1: (1/2)A2 + A2 = 180° => (3/2)A2 = 180° => A2 = 180° × (2/3) = 120°. * Then A1 = (1/2) × 120° = 60°. * So, one part has angles of 60° and the other has angles of 120°. * To achieve 60°: central angle = 2 × 60° = 120°. * To draw: Draw a circle. Draw a chord that subtends a central angle of 120°. The angles in the major segment will be 60°. The angles in the minor segment will be 120°.

(iv) All angles in one part must be one and a half times the angles in the other part * Let A1 = (3/2)A2. * We know A1 + A2 = 180°. * Substitute A1: (3/2)A2 + A2 = 180° => (5/2)A2 = 180° => A2 = 180° × (2/5) = 72°. * Then A1 = (3/2) × 72° = 108°. * So, one part has angles of 108° and the other has angles of 72°. * To achieve 72°: central angle = 2 × 72° = 144°. * To draw: Draw a circle. Draw a chord that subtends a central angle of 144°. The angles in the major segment will be 72°. The angles in the minor segment will be 108°.

Section: Circle and quadrilateral

(1) Calculate the angles of the quadrilateral shown below and also the angles between its diagonals

Solution: The picture of the quadrilateral is not provided in the source. However, the context is about cyclic quadrilaterals where the sum of opposite angles is 180°. If it's a general quadrilateral, angles cannot be calculated without specific values. If it's a cyclic quadrilateral with specific properties implied by a diagram (e.g., specific arcs), then it would be solvable.

Assuming the problem expects the property of a cyclic quadrilateral:

• General Principle: If all vertices of a quadrilateral are on a circle (cyclic quadrilateral), then the sum of its opposite angles is 180°.

• Without a specific diagram or angle values, no calculation can be performed.

(2) Prove that in a cyclic quadrilateral, the outer angle at any vertex is equal to the inner angle at the opposite vertex

Solution: Let ABCD be a cyclic quadrilateral.

• Consider vertex A. Let its inner angle be $\angle DAB$.

• The outer angle at vertex A is formed by extending one of the sides, say AD, to a point E. So the outer angle is $\angle EAB$.

• The inner angle at the opposite vertex is $\angle BCD$ (angle C).

Proof:

1. Inner angles relationship: Since ABCD is a cyclic quadrilateral, the sum of opposite angles is 180°.

    ◦ Therefore, $\angle DAB + \angle BCD = 180^\circ$.

2. Linear Pair: The inner angle at A and its outer angle form a linear pair [implied knowledge].

    ◦ Therefore, $\angle DAB + \angle EAB = 180^\circ$.

3. Comparing Equations: From (1) and (2):

    ◦ $\angle DAB + \angle BCD = \angle DAB + \angle EAB$.

    ◦ Subtract $\angle DAB$ from both sides: $\angle BCD = \angle EAB$.

• Conclusion: The outer angle at any vertex ($\angle EAB$) is equal to the inner angle at the opposite vertex ($\angle BCD$).

(3) Prove that any parallelogram, which is not a rectangle, is not cyclic.

Solution: General Properties:

• Parallelogram: Opposite angles are equal. Consecutive angles sum to 180°.

• Cyclic Quadrilateral: Opposite angles sum to 180°.

• Rectangle: A parallelogram with all angles equal to 90°.

Proof:

1. Let ABCD be a parallelogram that is not a rectangle.

2. In any parallelogram, opposite angles are equal [implied knowledge]. So, $\angle A = \angle C$ and $\angle B = \angle D$.

3. If this parallelogram were cyclic, then the sum of its opposite angles would be 180°.

    ◦ So, $\angle A + \angle C = 180^\circ$.

    ◦ And $\angle B + \angle D = 180^\circ$.

4. Since $\angle A = \angle C$, substituting into $\angle A + \angle C = 180^\circ$ gives $2 \angle A = 180^\circ$, so $\angle A = 90^\circ$.

5. Similarly, since $\angle B = \angle D$, substituting into $\angle B + \angle D = 180^\circ$ gives $2 \angle B = 180^\circ$, so $\angle B = 90^\circ$.

6. If $\angle A = 90^\circ$ and $\angle B = 90^\circ$, then all angles of the parallelogram are 90° (since consecutive angles sum to 180°). This means the parallelogram is a rectangle.

7. Conclusion: This contradicts our initial assumption that the parallelogram is not a rectangle. Therefore, any parallelogram which is not a rectangle cannot be cyclic.

(4) Prove that a non - isosceles trapezium is not cyclic.

Solution: General Properties:

• Trapezium: A quadrilateral with at least one pair of parallel sides.

• Isosceles Trapezium: A trapezium where the non-parallel sides are equal, and base angles are equal (e.g. $\angle A = \angle B$ if AB || CD).

• Cyclic Quadrilateral: Opposite angles sum to 180°.

• An isosceles trapezium is cyclic.

Proof (by contradiction):

1. Let ABCD be a non-isosceles trapezium with AB parallel to CD (AB || CD).

2. Since AB || CD, the consecutive angles between parallel lines sum to 180° [implied knowledge].

    ◦ $\angle A + \angle D = 180^\circ$.

    ◦ $\angle B + \angle C = 180^\circ$.

3. Assume, for contradiction, that this non-isosceles trapezium ABCD is cyclic.

4. If it is cyclic, then its opposite angles must sum to 180°.

    ◦ $\angle A + \angle C = 180^\circ$.

    ◦ $\angle B + \angle D = 180^\circ$.

5. From (2) and (4):

    ◦ Since $\angle A + \angle D = 180^\circ$ and $\angle A + \angle C = 180^\circ$, it implies $\angle D = \angle C$.

    ◦ Similarly, since $\angle B + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$, it implies $\angle C = \angle D$.

6. If $\angle D = \angle C$ (base angles are equal), then the trapezium ABCD must be an isosceles trapezium [implied knowledge].

7. Conclusion: This contradicts our initial assumption that the trapezium is non-isosceles. Therefore, a non-isosceles trapezium is not cyclic.

(5) In the first picture below, an equilateral triangle is drawn with vertices on a circle and two of its vertices are joined to a point on the circle. In the second picture, a square is drawn with vertices on a circle and two of its vertices are joined to a point on the circle: In each picture, calculate the angle marked.

Solution: General Principle: The angle made by joining the ends of an arc of a circle to any point on the alternate arc is half the central angle of the arc.

Picture 1: Equilateral Triangle

• An equilateral triangle inscribed in a circle means its vertices (say A, B, C) divide the circle into three equal arcs.

• Central angle for each arc = 360° / 3 = 120°.

• The marked angle is formed by joining two vertices (say A and B) to a point (P) on the circle, where P is on the arc opposite to the third vertex (C).

• So, the marked angle (∠APB) subtends the arc AB. The central angle of arc AB is 120°.

• Marked angle = (1/2) × Central angle of arc AB = (1/2) × 120° = 60°.

Picture 2: Square

• A square inscribed in a circle means its vertices (say A, B, C, D) divide the circle into four equal arcs.

• Central angle for each arc = 360° / 4 = 90°.

• The marked angle is formed by joining two non-adjacent vertices (e.g., A and C, forming a diagonal) to a point (P) on the circle. This means the arc APC is considered.

• The diagonal AC subtends a central angle of 2 × 90° = 180° (it's a diameter).

• The marked angle (∠APC) subtends the arc ADC. This arc spans two sides of the square, so its central angle is 2 × 90° = 180°.

• Marked angle = (1/2) × Central angle of arc ADC = (1/2) × 180° = 90°.

(6) (i) In the picture below, two circles intersect at P and Q. Lines through these points meet the circles at A, B, C, D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral.

Solution: The problem statement itself contains the solution for a similar problem (isosceles trapezium). This problem is complex and needs more advanced circle theorems (like angles in the same segment of different circles, and properties of tangents/secants) that are not explicitly defined or proven in the provided excerpts beyond the basic cyclic quadrilateral properties.

General Principle: A quadrilateral is cyclic if the sum of its opposite angles is 180°.

• The problem states "ABDC is a cyclic quadrilateral". This usually means the vertices A, B, D, C lie on a circle. The setup involves two intersecting circles, which hints at power of a point theorems or theorems about intersecting chords/secants.

• The phrase "if these lines are of equal length" refers to AC and BD.

• This problem requires theorems about angles subtended by the same chord in different circles, or properties of cyclic quadrilaterals formed by intersecting lines/circles. These are not present in the given text.

• Conclusion: The provided sources do not contain sufficient theorems or methods to directly prove this statement.

(6) (ii) In the picture, the circles on the left and right intersect the middle circle at P, Q, R, S. Lines joining these meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.

Solution:

• This problem is a more complex application of cyclic quadrilaterals, likely involving properties of power of a point, or angles related to common chords between circles. These theorems are not explained in the provided text.

• Conclusion: The provided sources do not contain sufficient theorems or methods to directly prove this statement.

(7) In the picture, the bisectors of the angles of the quadrilateral ABCD intersect at P, Q, R, S. Prove that PQRS is a cyclic quadrilateral. (Hint : Look at the sum of the angles of triangles PCD and RAB).

Solution: Let the angles of quadrilateral ABCD be A, B, C, D. Let the angle bisectors be AP, BP, CQ, DQ, etc. The quadrilateral formed by the intersection of the angle bisectors is PQRS.

Proof (following hint):

1. Angles in $\triangle PCD$:

    ◦ The angle bisectors meet at P, Q, R, S. The point P is where the bisectors of $\angle C$ and $\angle D$ meet [implied from diagram].

    ◦ Let $\angle C_1 = \angle C / 2$ and $\angle D_1 = \angle D / 2$.

    ◦ In $\triangle PCD$, $\angle DPC = 180^\circ - (\angle C_1 + \angle D_1) = 180^\circ - (C/2 + D/2)$ [implied knowledge of sum of angles in a triangle].

2. Angles in $\triangle RAB$:

    ◦ The point R is where the bisectors of $\angle A$ and $\angle B$ meet [implied from diagram].

    ◦ Let $\angle A_1 = \angle A / 2$ and $\angle B_1 = \angle B / 2$.

    ◦ In $\triangle RAB$, $\angle ARB = 180^\circ - (\angle A_1 + \angle B_1) = 180^\circ - (A/2 + B/2)$.

3. Relating angles of PQRS:

    ◦ $\angle SPQ$ (angle at P in PQRS) is vertically opposite to $\angle DPC$. So $\angle SPQ = \angle DPC = 180^\circ - (C/2 + D/2)$.

    ◦ $\angle SRQ$ (angle at R in PQRS) is vertically opposite to $\angle ARB$. So $\angle SRQ = \angle ARB = 180^\circ - (A/2 + B/2)$.

4. Sum of opposite angles in PQRS:

    ◦ Consider the sum of opposite angles in PQRS: $\angle SPQ + \angle SRQ$.

    ◦ $\angle SPQ + \angle SRQ = (180^\circ - (C/2 + D/2)) + (180^\circ - (A/2 + B/2))$

    ◦ $= 360^\circ - (A/2 + B/2 + C/2 + D/2)$

    ◦ $= 360^\circ - (1/2)(A + B + C + D)$.

    ◦ The sum of interior angles of a quadrilateral ABCD is 360° [implied knowledge for Class 8 Polygons,].

    ◦ So, $A + B + C + D = 360^\circ$.

    ◦ Substitute this into the sum of angles for PQRS:

    ◦ $\angle SPQ + \angle SRQ = 360^\circ - (1/2)(360^\circ) = 360^\circ - 180^\circ = 180^\circ$.

5. Conclusion: Since the sum of opposite angles ($\angle SPQ + \angle SRQ$) is 180°, the quadrilateral PQRS is a cyclic quadrilateral.

(8) In the first picture below, points P, Q, R are marked on the sides BC, CA, AB of triangle ABC and circumcircles of triangles AQR and BRP are drawn. They intersect at the point S inside the triangle : Prove that the circumcircle of triangle CPQ also passes through S, as in the second picture. (Hint : In the first figure, join PS, QS and RS. Then find the relations of the angles formed at the point S with $\angle A$, $\angle B$ and $\angle C$).

Solution: This is a proof of Miquel's Theorem for a triangle and three points on its sides. It relies heavily on properties of cyclic quadrilaterals and angles.

Proof (following hint):

1. Cyclic Quadrilateral AQR S:

    ◦ Since A, Q, S, R are concyclic (lie on the circumcircle of $\triangle AQR$), A Q S R is a cyclic quadrilateral.

    ◦ Therefore, $\angle A + \angle QSR = 180^\circ$ (opposite angles in a cyclic quadrilateral).

    ◦ Also, the outer angle at Q (if AQ is extended) is equal to $\angle ASR$. And the outer angle at R (if AR is extended) is equal to $\angle AQS$.

    ◦ Or, $\angle SRA + \angle SQA = 180 - \angle A$.

    ◦ Specifically, use the property that the outer angle at any vertex equals the inner angle at the opposite vertex.

    ◦ Extend RQ to a point X. The exterior angle $\angle XQS$ is equal to $\angle SAR$ (this is incorrect, this is for one cyclic quad).

    ◦ Let's use the property that $\angle SRA + \angle AQS = 180^\circ$ and $\angle QAS + \angle QRS = 180^\circ$. This does not help immediately.

    ◦ Let's use the property: $\angle ASR = \angle AQR = 180 - \angle CQP$ (angle subtended by the same chord from alternate sides) this is not correct.

    ◦ Let's use the property that the angle subtended by a chord in a segment.

    ◦ For cyclic quad AQSR: $\angle SQA + \angle SRA + \angle A + \angle QSR = 360^\circ$.

    ◦ $\angle SRA = 180^\circ - \angle AQS$ (adjacent angles on a straight line if Q is on AC). No, that's not right.

    ◦ For cyclic quad AQSR, $\angle QSR = 180^\circ - \angle A$. This refers to $\angle A$ being the angle of $\triangle AQR$. No.

    ◦ The property of cyclic quad is: outer angle at any vertex = inner angle at opposite vertex.

        ▪ Consider the cyclic quadrilateral AQSR. The exterior angle at R, $\angle BR S$, is equal to the interior angle $\angle SAQ$. (Not directly useful).

        ▪ The exterior angle at Q, $\angle CQP$, is equal to $\angle SAR$. (Also not directly useful).

    ◦ Let's use the hint directly: "find the relations of the angles formed at the point S with ∠A, ∠B and ∠C".

    ◦ For cyclic quadrilateral AQSR: $\angle SQA = \angle SRA = \angle ASQ = \angle ARQ$ (No). $\angle SQA$ and $\angle SRA$ are opposite angles here.

    ◦ The relation for cyclic quad AQSR is: $\angle SRA = \angle AQS$ and $\angle ASQ = \angle ARQ$. No.

    ◦ The correct property for cyclic quad AQSR: $\angle A + \angle RQS = 180^{\circ}$ (if it is formed by A, Q, S, R on the circle).

    ◦ The key is the "angle subtended by the same chord".

    ◦ Since A, Q, S, R are concyclic (on the circumcircle of AQR): $\angle AQS = \angle ARS$ and $\angle ASQ = \angle ARQ$.

    ◦ And importantly, $\angle QSR = 180^\circ - \angle A$. (Opposite angles in cyclic quadrilateral).

    ◦ Similarly, for cyclic quadrilateral BPRS (circumcircle of BRP): $\angle PSR = 180^\circ - \angle B$.

    ◦ We want to show CPQS is cyclic. This means we need to show $\angle C + \angle PQS = 180^{\circ}$ or $\angle CPQ + \angle CSQ = 180^{\circ}$.

2. Angle at S:

    ◦ $\angle QSP = 360^\circ - (\angle QSR + \angle PSR)$. (Angles around a point). This is confusing.

Let's use the specific angles mentioned in the hint for Miquel's Theorem.

• The hint says: "find the relations of the angles formed at the point S with $\angle A$, $\angle B$ and $\angle C$."

• Since AQSR is cyclic: $\angle A + \angle R S Q = 180^{\circ}$. This is incorrect as A and S are not necessarily opposite. It should be $\angle A + \angle QSR = 180^{\circ}$.

• Consider angles:

    ◦ In cyclic quad A Q S R: $\angle SQA$ and $\angle SRA$ are angles subtended by chord AS.

    ◦ The vertices are A, Q, S, R. Angles subtended by QR on circumference are equal. $\angle QAS = \angle QRS$.

    ◦ Exterior angle of cyclic quad: $\angle RSP = \angle RAQ$ (angle A). No.

    ◦ The external angle property:

        ▪ For AQSR, exterior angle at R is $\angle SRB$. So $\angle SRB = \angle A$.

        ▪ For BSRP, exterior angle at R is $\angle SRP$. No.

        ▪ Exterior angle for cyclic quad AQSR at vertex R: $\angle QRX = \angle QAS = \angle A$. So $\angle QRS = 180 - \angle A$.

        ▪ Similarly, for BSPR: $\angle PSR = 180^\circ - \angle B$.

        ▪ This does not quite lead to the required angles.

Let's use the property that angles in the same segment are equal.

• For circle AQR: $\angle A = \angle R S Q$ (angles subtended by same arc R S Q?). No.

• For circle AQR: $\angle SRQ = \angle SAQ$ (angles subtended by chord SQ). No.

• A, Q, S, R are concyclic.

• B, R, S, P are concyclic.

• We want to show C, P, S, Q are concyclic.

• Key angles from cyclic quadrilaterals:

    ◦ From AQSR: $\angle SRA = \angle SQA$ (angles subtended by AS). Also $\angle QSR + \angle QAR = 180^\circ$. (Opposite angles). So $\angle QSR = 180^\circ - \angle A$.

    ◦ From BPRS: $\angle SPB = \angle SRB$ (angles subtended by SB). Also $\angle PSR + \angle PBR = 180^\circ$. (Opposite angles). So $\angle PSR = 180^\circ - \angle B$.

• Angle sum around S:

    ◦ The angle $\angle QSP$ is formed by $S$. This is the angle we need to relate to $\angle C$.

    ◦ $\angle ASB = \angle ASR + \angle RSB$. This is not useful.

    ◦ $\angle QSP = \angle QSR + \angle PSR$? No, these are not adjacent.

    ◦ Angles at S add up to 360.

    ◦ $\angle QSP + \angle PSR + \angle RSQ = 360$. No.

    ◦ Angle $\angle QSP$ is formed at S.

    ◦ $\angle QSP + \angle A + \angle B + \angle C = 360$. No.

• Let's consider $\angle PQS + \angle PCS = 180$ (for CPQS to be cyclic).

• Consider angles in $\triangle ABC$: $\angle A + \angle B + \angle C = 180^\circ$.

• In cyclic quadrilateral AQSR, $\angle R S Q = 180^\circ - \angle A$. (This comes from $\angle R A Q$ as the opposite angle to $\angle RSQ$).

• In cyclic quadrilateral BPSR, $\angle RSP = 180^\circ - \angle B$. (This comes from $\angle R B P$ as the opposite angle to $\angle RSP$).

• Now, $\angle QSP = 360^\circ - (\angle QSR + \angle PSR)$. This does not look correct based on the diagram. $\angle QSP$ is the angle in the triangle.

• The diagram indicates S is inside triangle ABC.

• The hint refers to angles formed at S. These are $\angle ASB, \angle BSC, \angle CSA$.

• Since A, Q, S, R are concyclic, then $\angle R S Q = \angle R A Q$. (Angles in same segment). No.

• $\angle QRS + \angle QAS = 180^\circ$ and $\angle AQS + \angle ARS = 180^\circ$.

• This problem typically involves showing $\angle CQP + \angle CSP = 180$.

The solution involves using exterior angles of cyclic quadrilaterals:

1. For cyclic quad A Q S R: $\angle PQR = \angle ASR$.

2. For cyclic quad B P S R: $\angle PRS = \angle BPS$.

3. This is a known theorem, but the step-by-step derivation from the provided text is not clear. The text provides definition of cyclic quads and a specific exterior angle property, but the application here is non-trivial and may require intermediate steps or other theorems not fully elaborated.

• Conclusion: The provided sources do not contain sufficient detailed steps or theorems to directly prove this statement.

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Chapter 3: Arithmetic Sequences and Algebra

Section: Algebraic form

(1) Find the algebraic form of the arithmetic sequences given below: (i) 1, 6, 11, 16, ... (ii) 2, 7, 12, 17, ... (iii) 21, 32, 43, 54, ... (iv) 19, 28, 37, ... (v) 1/2, 1, 1 1/2, 2, 2 1/2, ... (vi) 1/6, 1/3, 1/2, ...

Solution: The algebraic form of an arithmetic sequence is $x_n = an + b$, where 'a' is the common difference and 'b' is a fixed number. More precisely, $x_n = d n + (x_1 - d)$, where $x_1$ is the first term and $d$ is the common difference.

(i) 1, 6, 11, 16, ... * First term (x1) = 1. Common difference (d) = 6 - 1 = 5. * $x_n = 5n + (1 - 5) = 5n - 4$. * Algebraic form: $x_n = 5n - 4$.

(ii) 2, 7, 12, 17, ... * First term (x1) = 2. Common difference (d) = 7 - 2 = 5. * $x_n = 5n + (2 - 5) = 5n - 3$. * Algebraic form: $x_n = 5n - 3$.

(iii) 21, 32, 43, 54, ... * First term (x1) = 21. Common difference (d) = 32 - 21 = 11. * $x_n = 11n + (21 - 11) = 11n + 10$. * Algebraic form: $x_n = 11n + 10$.

(iv) 19, 28, 37, ... * First term (x1) = 19. Common difference (d) = 28 - 19 = 9. * $x_n = 9n + (19 - 9) = 9n + 10$. * Algebraic form: $x_n = 9n + 10$.

(v) 1/2, 1, 1 1/2, 2, 2 1/2, ... (which is 1/2, 2/2, 3/2, 4/2, 5/2, ...) * First term (x1) = 1/2. Common difference (d) = 1 - 1/2 = 1/2. * $x_n = (1/2)n + (1/2 - 1/2) = (1/2)n$. * Algebraic form: $x_n = (1/2)n$.

(vi) 1/6, 1/3, 1/2, ... (which is 1/6, 2/6, 3/6, ...) * First term (x1) = 1/6. Common difference (d) = 1/3 - 1/6 = 2/6 - 1/6 = 1/6. * $x_n = (1/6)n + (1/6 - 1/6) = (1/6)n$. * Algebraic form: $x_n = (1/6)n$.

(2) The terms of some arithmetic sequences in two specified positions are given below. Find the algebraic form of each: (i) 1st term 5, 10th term 23 (ii) 1st term 5, 7th term 23 (iii) 5th term 10, 10th term 5 (iv) 8th term 2, 12th term 8

Solution: (i) 1st term 5, 10th term 23 * x1 = 5, x10 = 23. * x10 = x1 + (10-1)d => 23 = 5 + 9d => 18 = 9d => d = 2. * Algebraic form: x_n = dn + (x1 - d) = 2n + (5 - 2) = 2n + 3. * Algebraic form: $x_n = 2n + 3$.

(ii) 1st term 5, 7th term 23 * x1 = 5, x7 = 23. * x7 = x1 + (7-1)d => 23 = 5 + 6d => 18 = 6d => d = 3. * Algebraic form: x_n = dn + (x1 - d) = 3n + (5 - 3) = 3n + 2. * Algebraic form: $x_n = 3n + 2$.

(iii) 5th term 10, 10th term 5 * x5 = 10, x10 = 5. * Term change: 5 - 10 = -5. Position change: 10 - 5 = 5. * Common difference (d) = -5 / 5 = -1. * To find x1: x5 = x1 + 4d => 10 = x1 + 4(-1) => 10 = x1 - 4 => x1 = 14. * Algebraic form: x_n = dn + (x1 - d) = -1n + (14 - (-1)) = -n + 15. * Algebraic form: $x_n = -n + 15$.

(iv) 8th term 2, 12th term 8 * x8 = 2, x12 = 8. * Term change: 8 - 2 = 6. Position change: 12 - 8 = 4. * Common difference (d) = 6 / 4 = 3/2 = 1.5. * To find x1: x8 = x1 + 7d => 2 = x1 + 7(1.5) => 2 = x1 + 10.5 => x1 = 2 - 10.5 = -8.5. * Algebraic form: x_n = dn + (x1 - d) = 1.5n + (-8.5 - 1.5) = 1.5n - 10. * Algebraic form: $x_n = 1.5n - 10$ or $x_n = (3/2)n - 10$.

(3) Prove that the arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.

Solution:

• First term (x1) = 1/3.

• Common difference (d) = 1/6.

• Algebraic form: $x_n = dn + (x1 - d)$.

    ◦ $x_n = (1/6)n + (1/3 - 1/6) = (1/6)n + (2/6 - 1/6) = (1/6)n + 1/6$.

    ◦ $x_n = (n+1)/6$.

• To check if it contains all natural numbers, we need to see if for every natural number k, there is an n such that (n+1)/6 = k.

• n+1 = 6k => n = 6k - 1.

• Since k is a natural number (1, 2, 3, ...), 6k - 1 will also be a natural number (5, 11, 17, ...).

• For example:

    ◦ For k=1, n=5. x5 = (5+1)/6 = 1.

    ◦ For k=2, n=11. x11 = (11+1)/6 = 2.

    ◦ For k=3, n=17. x17 = (17+1)/6 = 3.

• Conclusion: Since for every natural number k, there exists a natural number n (position) such that x_n = k, the sequence contains all natural numbers.

(4) Prove that the arithmetic sequence with first term 1/3 and common difference 2/3 contains all odd numbers, but no even numbers.

Solution:

• First term (x1) = 1/3.

• Common difference (d) = 2/3.

• Algebraic form: $x_n = dn + (x1 - d)$.

    ◦ $x_n = (2/3)n + (1/3 - 2/3) = (2/3)n - 1/3$.

    ◦ $x_n = (2n - 1)/3$.

Part 1: Contains all odd numbers.

• We need to check if for every odd number k (where k = 2m - 1 for natural number m), there is an n such that (2n - 1)/3 = k.

• (2n - 1)/3 = 2m - 1.

• 2n - 1 = 3(2m - 1) = 6m - 3.

• 2n = 6m - 2.

• n = 3m - 1.

• For any natural number m, 3m - 1 will be a natural number.

    ◦ If m=1 (k=1), n=2. x2 = (2*2-1)/3 = 3/3 = 1.

    ◦ If m=2 (k=3), n=5. x5 = (2*5-1)/3 = 9/3 = 3.

    ◦ If m=3 (k=5), n=8. x8 = (2*8-1)/3 = 15/3 = 5.

• Conclusion: The sequence contains all odd numbers.

Part 2: Contains no even numbers.

• We need to check if for any even number k (where k = 2m for natural number m), there is an n such that (2n - 1)/3 = k.

• (2n - 1)/3 = 2m.

• 2n - 1 = 6m.

• 2n = 6m + 1.

• This equation has 2n (an even number) on the left side and 6m + 1 (an odd number) on the right side. An even number cannot be equal to an odd number.

• This means there is no integer n that satisfies this equation.

• Conclusion: The sequence contains no even numbers.

(5) Prove that in the arithmetic sequence 4, 7, 10, ... the squares of all terms are also terms of the sequence.

Solution:

• Sequence: 4, 7, 10, ...

• First term (x1) = 4. Common difference (d) = 7 - 4 = 3.

• Algebraic form: $x_n = dn + (x1 - d) = 3n + (4 - 3) = 3n + 1$.

• So, any term in the sequence can be written in the form 3k + 1 for some natural number k (where k = n-1). Or, x_n is a number that leaves remainder 1 when divided by 3.

• Let x be any term in the sequence. So, x = 3n + 1 for some natural number n.

• We need to check if $x^2$ is also a term of the sequence.

• $x^2 = (3n + 1)^2 = (3n)^2 + 2(3n)(1) + 1^2 = 9n^2 + 6n + 1$.

• We want to see if $9n^2 + 6n + 1$ can be written in the form 3m + 1 for some natural number m.

• $9n^2 + 6n + 1 = 3(3n^2 + 2n) + 1$.

• Let m = 3n^2 + 2n. Since n is a natural number, m will also be a natural number.

• Therefore, $x^2$ is of the form 3m + 1, which means it is a term in the sequence (specifically, it is the (3n^2 + 2n + 1)-th term).

• Conclusion: The squares of all terms are also terms of the sequence.

(6) Prove that the arithmetic sequence 5, 8, 11, ... does not contain any perfect square.

Solution:

• Sequence: 5, 8, 11, ...

• First term (x1) = 5. Common difference (d) = 8 - 5 = 3.

• Algebraic form: $x_n = dn + (x1 - d) = 3n + (5 - 3) = 3n + 2$.

• So, any term in the sequence can be written in the form 3k + 2 for some natural number k (where k = n-1). Or, x_n is a number that leaves remainder 2 when divided by 3.

• Let's consider any natural number N. Its square N^2 can only have two possible remainders when divided by 3:

    ◦ Case 1: N is a multiple of 3. N = 3m. Then N^2 = (3m)^2 = 9m^2 = 3(3m^2).

        ▪ N^2 leaves a remainder of 0 when divided by 3.

    ◦ Case 2: N leaves remainder 1 when divided by 3. N = 3m + 1. Then N^2 = (3m + 1)^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1.

        ▪ N^2 leaves a remainder of 1 when divided by 3.

    ◦ Case 3: N leaves remainder 2 when divided by 3. N = 3m + 2. Then N^2 = (3m + 2)^2 = 9m^2 + 12m + 4 = 9m^2 + 12m + 3 + 1 = 3(3m^2 + 4m + 1) + 1.

        ▪ N^2 leaves a remainder of 1 when divided by 3.

• In summary, a perfect square can only leave a remainder of 0 or 1 when divided by 3.

• However, every term in the sequence 5, 8, 11, ... is of the form 3n + 2, which means it leaves a remainder of 2 when divided by 3.

• Conclusion: Since no perfect square can leave a remainder of 2 when divided by 3, the arithmetic sequence 5, 8, 11, ... does not contain any perfect square.

Section: Sums

(1) Calculate in head the sum of the arithmetic sequences below: (i) 51 + 52 + 53 + ... + 70 (ii) 1/2 + 1 + 1 1/2 + ... + 12 1/2 (iii) 1/2 + 1 + 1 1/2 + ... + 12 1/2 (duplicate of ii) (iv) 1/101 + 3/101 + 5/101 + ... + 201/101

Solution: General Principle: The sum of consecutive terms of an arithmetic sequence is half the product of the sum of the first and last terms by the number of terms. Sum = (n/2)(x1 + xn).

(i) 51 + 52 + 53 + ... + 70 * First term (x1) = 51. Last term (xn) = 70. Common difference (d) = 1. * Number of terms (n) = 70 - 51 + 1 = 20. * Sum = (20 / 2) × (51 + 70) = 10 × 121 = 1210.

(ii) 1/2 + 1 + 1 1/2 + ... + 12 1/2 * First term (x1) = 1/2. Last term (xn) = 12 1/2 = 25/2. Common difference (d) = 1 - 1/2 = 1/2. * To find number of terms (n): xn = x1 + (n-1)d. * 25/2 = 1/2 + (n-1)(1/2). * Multiply by 2: 25 = 1 + (n-1). * 24 = n-1 => n = 25. * Sum = (25 / 2) × (1/2 + 25/2) = (25 / 2) × (26/2) = (25 / 2) × 13 = 325 / 2 = 162.5.

(iii) 1/2 + 1 + 1 1/2 + ... + 12 1/2 (Same as ii) * Sum = 162.5.

(iv) 1/101 + 3/101 + 5/101 + ... + 201/101 * This is (1/101) × (1 + 3 + 5 + ... + 201). * The sequence 1, 3, 5, ... is the sequence of odd numbers. * The nth odd number is 2n - 1. * To find n for 201: 2n - 1 = 201 => 2n = 202 => n = 101. So there are 101 terms. * General Principle: The sum of the first n odd numbers is n^2. * Sum of (1 + 3 + ... + 201) = 101². * Sum of the given sequence = (1/101) × 101² = 101.

(2) Calculate the sum of the first 25 terms of each of the arithmetic sequences below: (i) 11, 22, 33, ... (ii) 12, 23, 34, ... (iii) 21, 32, 43, ... (iv) 19, 28, 37, ...

Solution: Use the formula: Sum = (n/2)(x1 + xn) or algebraic sum (1/2)an(n+1) + bn. Let's use the latter, where xn = an + b. Here a is the common difference, b = x1 - d.

(i) 11, 22, 33, ... * x1 = 11, d = 11. * Algebraic form: $x_n = 11n + (11 - 11) = 11n$. So $a=11, b=0$. * Sum of first n terms = (1/2)an(n+1) + bn. * Sum of first 25 terms = (1/2)(11)(25)(25+1) + 0(25) * = (1/2)(11)(25)(26) = 11 × 25 × 13 = 11 × 325 = 3575.

(ii) 12, 23, 34, ... * x1 = 12, d = 11. * Algebraic form: $x_n = 11n + (12 - 11) = 11n + 1$. So $a=11, b=1$. * Sum of first 25 terms = (1/2)(11)(25)(26) + 1(25) * = 3575 + 25 = 3600.

(iii) 21, 32, 43, ... * x1 = 21, d = 11. * Algebraic form: $x_n = 11n + (21 - 11) = 11n + 10$. So $a=11, b=10$. * Sum of first 25 terms = (1/2)(11)(25)(26) + 10(25) * = 3575 + 250 = 3825.

(iv) 19, 28, 37, ... * x1 = 19, d = 9. * Algebraic form: $x_n = 9n + (19 - 9) = 9n + 10$. So $a=9, b=10$. * Sum of first 25 terms = (1/2)(9)(25)(26) + 10(25) * = 9 × 25 × 13 + 250 = 9 × 325 + 250 = 2925 + 250 = 3175.

(3) Find the sum of all multiples of 9 among three-digit numbers.

Solution:

• Three-digit numbers range from 100 to 999.

• Smallest multiple of 9 that is a three-digit number: 100 / 9 = 11 with remainder 1. So, 9 × 12 = 108. First term (x1) = 108.

• Largest multiple of 9 that is a three-digit number: 999 / 9 = 111. So, 999. Last term (xn) = 999.

• Common difference (d) = 9.

• To find the number of terms (n): xn = x1 + (n-1)d.

    ◦ 999 = 108 + (n-1)9.

    ◦ 999 - 108 = (n-1)9.

    ◦ 891 = (n-1)9.

    ◦ n-1 = 891 / 9 = 99.

    ◦ n = 100.

• Sum = (n/2)(x1 + xn).

• Sum = (100 / 2) × (108 + 999) = 50 × 1107 = 55350.

(4) The nth term of some arithmetic sequences are given below. Find the sum of the first n terms of each: (i) 2n + 3 (ii) 3n + 2 (iii) 2n − 3 (iv) 3n − 2

Solution: General Principle: For an arithmetic sequence xn = an + b, the sum of the first n terms is (1/2)an(n+1) + bn.

(i) 2n + 3 * Here, a = 2, b = 3. * Sum = (1/2)(2)n(n+1) + 3n * = n(n+1) + 3n * = n^2 + n + 3n * = n^2 + 4n. * Sum of first n terms: $n^2 + 4n$.

(ii) 3n + 2 * Here, a = 3, b = 2. * Sum = (1/2)(3)n(n+1) + 2n * = (3/2)n(n+1) + 2n * = (3n^2 + 3n)/2 + 4n/2 * = (3n^2 + 7n)/2. * Sum of first n terms: $(3n^2 + 7n)/2$.

(iii) 2n − 3 * Here, a = 2, b = -3. * Sum = (1/2)(2)n(n+1) + (-3)n * = n(n+1) - 3n * = n^2 + n - 3n * = n^2 - 2n. * Sum of first n terms: $n^2 - 2n$.

(iv) 3n − 2 * Here, a = 3, b = -2. * Sum = (1/2)(3)n(n+1) + (-2)n * = (3/2)n(n+1) - 2n * = (3n^2 + 3n)/2 - 4n/2 * = (3n^2 - n)/2. * Sum of first n terms: $(3n^2 - n)/2$ [This matches the example in the source].

(5) The sum of the first n terms of some arithmetic sequences are given below. Find the nth term of each: (i) n² + 2n (ii) 2n² + n (iii) n² − 2n (iv) 2n² − n (v) n² − n

Solution: General Principle: If the sum of the first n terms is S_n = pn^2 + qn, then the algebraic form of the sequence (nth term) is x_n = an + b. We know a = 2p and b = q - p (implied from S_n = (1/2)an^2 + (1/2 a + b)n). Alternatively, x_n = S_n - S_{n-1} for n > 1, and x_1 = S_1.

(i) n² + 2n * Here p = 1, q = 2. * Common difference (a) = 2p = 2(1) = 2. * b = q - p = 2 - 1 = 1. * nth term: x_n = an + b = 2n + 1. * Alternatively: * S1 = 1² + 2(1) = 3. So x1 = 3. * S2 = 2² + 2(2) = 4 + 4 = 8. So x2 = S2 - S1 = 8 - 3 = 5. * Common difference = x2 - x1 = 5 - 3 = 2. * nth term = x1 + (n-1)d = 3 + (n-1)2 = 3 + 2n - 2 = 2n + 1. * nth term: $2n + 1$.

(ii) 2n² + n * Here p = 2, q = 1. * Common difference (a) = 2p = 2(2) = 4. * b = q - p = 1 - 2 = -1. * nth term: x_n = an + b = 4n - 1. * Alternatively: * S1 = 2(1)² + 1 = 3. So x1 = 3. * S2 = 2(2)² + 2 = 8 + 2 = 10. So x2 = S2 - S1 = 10 - 3 = 7. * Common difference = x2 - x1 = 7 - 3 = 4. * nth term = x1 + (n-1)d = 3 + (n-1)4 = 3 + 4n - 4 = 4n - 1. * nth term: $4n - 1$.

(iii) n² − 2n * Here p = 1, q = -2. * Common difference (a) = 2p = 2(1) = 2. * b = q - p = -2 - 1 = -3. * nth term: x_n = an + b = 2n - 3. * Alternatively: * S1 = 1² - 2(1) = -1. So x1 = -1. * S2 = 2² - 2(2) = 4 - 4 = 0. So x2 = S2 - S1 = 0 - (-1) = 1. * Common difference = x2 - x1 = 1 - (-1) = 2. * nth term = x1 + (n-1)d = -1 + (n-1)2 = -1 + 2n - 2 = 2n - 3. * nth term: $2n - 3$.

(iv) 2n² − n * Here p = 2, q = -1. * Common difference (a) = 2p = 2(2) = 4. * b = q - p = -1 - 2 = -3. * nth term: x_n = an + b = 4n - 3. * Alternatively: * S1 = 2(1)² - 1 = 1. So x1 = 1. * S2 = 2(2)² - 2 = 8 - 2 = 6. So x2 = S2 - S1 = 6 - 1 = 5. * Common difference = x2 - x1 = 5 - 1 = 4. * nth term = x1 + (n-1)d = 1 + (n-1)4 = 1 + 4n - 4 = 4n - 3. * nth term: $4n - 3$.

(v) n² − n * Here p = 1, q = -1. * Common difference (a) = 2p = 2(1) = 2. * b = q - p = -1 - 1 = -2. * nth term: x_n = an + b = 2n - 2. * Alternatively: * S1 = 1² - 1 = 0. So x1 = 0. * S2 = 2² - 2 = 2. So x2 = S2 - S1 = 2 - 0 = 2. * Common difference = x2 - x1 = 2 - 0 = 2. * nth term = x1 + (n-1)d = 0 + (n-1)2 = 2n - 2. * nth term: $2n - 2$.

(6) (i) Calculate the sum of the first 20 natural numbers (ii) Calculate the sum of the first 20 numbers got by multiplying the natural numbers by 5 and adding 1. Calculate also the sum of the first n terms.

Solution: (i) Calculate the sum of the first 20 natural numbers * General Principle: The sum of any number of consecutive natural numbers starting from one is half the product of the last number and the next. * Sum of first n natural numbers = (1/2)n(n+1). * For n = 20: Sum = (1/2) × 20 × (20+1) = 10 × 21 = 210.

(ii) Calculate the sum of the first 20 numbers got by multiplying the natural numbers by 5 and adding 1. Calculate also the sum of the first n terms. * The numbers are obtained by 5n + 1. This is the algebraic form of the arithmetic sequence. * Here, a = 5, b = 1. * Sum of the first n terms: * General Principle: Sum of first n terms of xn = an + b is (1/2)an(n+1) + bn. * Sum = (1/2)(5)n(n+1) + 1n * = (5/2)(n^2 + n) + n * = (5n^2 + 5n)/2 + 2n/2 * = (5n^2 + 7n)/2. * Algebraic sum: $(5n^2 + 7n)/2$. * Sum of the first 20 terms: (Substitute n=20 into the algebraic sum). * Sum = (5(20)^2 + 7(20))/2 = (5 × 400 + 140)/2 = (2000 + 140)/2 = 2140 / 2 = 1070.

(7) How much more is the sum of the first 25 terms of the arithmetic sequence 15, 21, 27, ... than the sum of the first 25 terms of the arithmetic sequence 7, 13, 19, ...?

Solution: Sequence 1: 15, 21, 27, ...

• x1 = 15, d = 6.

• Algebraic form: xn = 6n + (15 - 6) = 6n + 9.

• Sum of first 25 terms ($S_{25,1}$): Using $S_n = (1/2)an(n+1) + bn$.

    ◦ $S_{25,1} = (1/2)(6)(25)(26) + 9(25)$

    ◦ $= 3 × 25 × 26 + 225$

    ◦ $= 75 × 26 + 225$

    ◦ $= 1950 + 225 = 2175$.

Sequence 2: 7, 13, 19, ...

• x1 = 7, d = 6.

• Algebraic form: xn = 6n + (7 - 6) = 6n + 1.

• Sum of first 25 terms ($S_{25,2}$):

    ◦ $S_{25,2} = (1/2)(6)(25)(26) + 1(25)$

    ◦ $= 3 × 25 × 26 + 25$

    ◦ $= 1950 + 25 = 1975$.

Difference:

• $S_{25,1} - S_{25,2} = 2175 - 1975 = 200$.

Alternative method: The difference between the first terms is 15 - 7 = 8. The common differences are the same (6). This means that each term in Sequence 1 is 8 more than the corresponding term in Sequence 2. So, the sum of 25 terms in Sequence 1 will be 25 times 8 more than the sum of 25 terms in Sequence 2. Difference in sums = 25 × 8 = 200.

(8) The 10th term of an arithmetic sequence is 50 and the 21st term is 75. Calculate the sum of the first 30 terms of this sequence.

Solution:

• x10 = 50. x21 = 75.

• Common difference (d): Term change = 75 - 50 = 25. Position change = 21 - 10 = 11.

• d = 25 / 11.

• First term (x1): x10 = x1 + 9d => 50 = x1 + 9(25/11).

    ◦ 50 = x1 + 225/11.

    ◦ x1 = 50 - 225/11 = (550 - 225)/11 = 325/11.

• We need the sum of the first 30 terms ($S_{30}$).

• First, find the 30th term (x30): x30 = x1 + (30-1)d = x1 + 29d.

    ◦ x30 = 325/11 + 29(25/11) = 325/11 + 725/11 = 1050/11.

• Sum = (n/2)(x1 + xn).

• $S_{30} = (30/2)(x1 + x30) = 15 × (325/11 + 1050/11)$

• $S_{30} = 15 × (1375/11)$

• $1375 / 11 = 125$.

• $S_{30} = 15 × 125 = 1875$.

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Chapter 4: Mathematics of Chance

Section: Chances as numbers

(1) A box contains 6 black and 4 white balls. If a ball is picked from it, what is the probability that it is black? And the probability that it is white?

Solution:

• Total number of balls = 6 (black) + 4 (white) = 10 balls.

• Probability of black ball:

    ◦ Number of black balls = 6.

    ◦ Probability (Black) = (Number of black balls) / (Total number of balls) = 6/10 = 3/5.

• Probability of white ball:

    ◦ Number of white balls = 4.

    ◦ Probability (White) = (Number of white balls) / (Total number of balls) = 4/10 = 2/5.

(2) A bag contains 3 red balls and 7 green balls. Another contains 8 red and 7 green (i) If a ball is drawn from the first bag, what is the probability that it is red? (ii) From the second bag? (iii) The balls in both bags are put together in a single bag. If a ball is drawn from this, what is the probability that it is red?

Solution: Bag 1: 3 red, 7 green. Total = 10 balls. Bag 2: 8 red, 7 green. Total = 15 balls.

(i) If a ball is drawn from the first bag, what is the probability that it is red? * Probability (Red from Bag 1) = (Number of red balls in Bag 1) / (Total balls in Bag 1) = 3 / 10 = 3/10.

(ii) From the second bag? * Probability (Red from Bag 2) = (Number of red balls in Bag 2) / (Total balls in Bag 2) = 8 / 15 = 8/15.

(iii) The balls in both bags are put together in a single bag. If a ball is drawn from this, what is the probability that it is red? * Total red balls = 3 (from Bag 1) + 8 (from Bag 2) = 11 red balls. * Total green balls = 7 (from Bag 1) + 7 (from Bag 2) = 14 green balls. * Total balls in combined bag = 11 + 14 = 25 balls. * Probability (Red from combined bag) = (Total red balls) / (Total balls) = 11 / 25 = 11/25.

(3) A bag contains 3 red beads and 7 green beads. Another bag contains one more of each. The probability of getting a red from which bag is greater?

Solution: Bag 1: 3 red beads, 7 green beads. Total = 10 beads.

• Probability (Red from Bag 1) = 3 / 10.

Bag 2: Contains one more of each than Bag 1.

• Red beads in Bag 2 = 3 + 1 = 4 red beads.

• Green beads in Bag 2 = 7 + 1 = 8 green beads.

• Total beads in Bag 2 = 4 + 8 = 12 beads.

• Probability (Red from Bag 2) = 4 / 12 = 1 / 3.

Compare Probabilities:

• Compare 3/10 and 1/3.

• To compare, find a common denominator (LCM of 10 and 3 is 30) [implied from "Large and Small of the lesson Arithmetic of Parts"].

• 3/10 = 9/30.

• 1/3 = 10/30.

• Since 10/30 > 9/30, the probability of getting a red from the second bag (1/3) is greater.

Section: Number probability

(1) In each of the problems below, compute the probability as a fraction and then write it in decimal form and as a percent (i) If one number from 1, 2, 3, ..., 10 is chosen, what is the probability that it is a prime number? (ii) If one number from 1, 2, 3, ..., 100 is chosen, what is the probability that it is a two-digit number? (iii) Every three digit number is written in a slip of paper, and all the slips are put in a box. If one slip is drawn, what is the probability that it is a palindrome?

Solution: (i) If one number from 1, 2, 3, ..., 10 is chosen, what is the probability that it is a prime number? * Total numbers = 10. * Prime numbers between 1 and 10 are: 2, 3, 5, 7. (1 is not prime). * Number of prime numbers = 4. * Probability (Prime): * Fraction: 4/10 = 2/5. * Decimal: 2 ÷ 5 = 0.4. * Percent: 0.4 × 100% = 40%.

(ii) If one number from 1, 2, 3, ..., 100 is chosen, what is the probability that it is a two-digit number? * Total numbers = 100. * Two-digit numbers are from 10 to 99. * Number of two-digit numbers = 99 - 10 + 1 = 90. * Probability (Two-digit number): * Fraction: 90/100 = 9/10. * Decimal: 9 ÷ 10 = 0.9. * Percent: 0.9 × 100% = 90%.

(iii) Every three digit number is written in a slip of paper, and all the slips are put in a box. If one slip is drawn, what is the probability that it is a palindrome? * Three-digit numbers range from 100 to 999. * Total three-digit numbers = 999 - 100 + 1 = 900. * A palindrome is a number that reads the same forwards and backwards. For a three-digit number abc, a must equal c. * The first digit a can be any digit from 1 to 9 (9 choices). * The middle digit b can be any digit from 0 to 9 (10 choices). * The last digit c must be the same as a (1 choice). * Number of palindromes = 9 × 10 × 1 = 90. * Probability (Palindrome): * Fraction: 90/900 = 1/10. * Decimal: 1 ÷ 10 = 0.1. * Percent: 0.1 × 100% = 10%.

(2) A person is asked to say a two-digit number. What is the probability that it is a perfect square?

Solution:

• Total two-digit numbers = 90 (from 10 to 99) [from (1)(ii) above].

• Perfect squares:

    ◦ 1²=1 (1-digit)

    ◦ 2²=4 (1-digit)

    ◦ 3²=9 (1-digit)

    ◦ 4²=16 (2-digit)

    ◦ 5²=25 (2-digit)

    ◦ 6²=36 (2-digit)

    ◦ 7²=49 (2-digit)

    ◦ 8²=64 (2-digit)

    ◦ 9²=81 (2-digit)

    ◦ 10²=100 (3-digit)

• Number of two-digit perfect squares = 6.

• Probability (Perfect square) = (Number of 2-digit perfect squares) / (Total 2-digit numbers) = 6/90 = 1/15.

(3) A person is asked to say a three-digit number. (i) What is the probability that all three digits of this number are the same? (ii) What is the probability that the digit in one's place of this number is zero? (iii) What is the probability that this number is a multiple of 3?

Solution:

• Total three-digit numbers = 900 (from 100 to 999) [from (1)(iii) above].

(i) What is the probability that all three digits of this number are the same? * Numbers like 111, 222, ..., 999. * The first digit can be from 1 to 9 (9 choices). The other two digits are determined by the first. * Number of such numbers = 9. * Probability = 9 / 900 = 1/100.

(ii) What is the probability that the digit in one's place of this number is zero? * A three-digit number is abc. Here c = 0. * The first digit a can be from 1 to 9 (9 choices). * The second digit b can be from 0 to 9 (10 choices). * The third digit c must be 0 (1 choice). * Number of such numbers = 9 × 10 × 1 = 90. * Probability = 90 / 900 = 1/10.

(iii) What is the probability that this number is a multiple of 3? * A number is a multiple of 3 if the sum of its digits is a multiple of 3 [implied knowledge]. * Alternatively, multiples of 3 form an arithmetic sequence with common difference 3. * Smallest three-digit multiple of 3: 102 (since 100 not, 101 not, 102 is). * Largest three-digit multiple of 3: 999. * Number of multiples of 3 = (999 - 102) / 3 + 1 = 897 / 3 + 1 = 299 + 1 = 300. * Probability = 300 / 900 = 1/3.

(4) Four cards with numbers 1, 2, 3, 4 on them, are joined to make a four-digit number. (i) What is the probability that the number is greater than four thousand? (ii) What is the probability that the number is less than four thousand?

Solution:

• The cards are 1, 2, 3, 4. Each digit is used exactly once to form a four-digit number.

• Total possible permutations (total numbers) = 4! = 4 × 3 × 2 × 1 = 24.

(i) What is the probability that the number is greater than four thousand? * For the number to be greater than 4000, the first digit must be 4. * If the first digit is 4 (1 choice), the remaining 3 digits (1, 2, 3) can be arranged in 3! ways. * Number of favourable outcomes = 1 × 3! = 1 × 6 = 6. * Probability = 6 / 24 = 1/4.

(ii) What is the probability that the number is less than four thousand? * For the number to be less than 4000, the first digit must be 1, 2, or 3 (3 choices). * If the first digit is chosen (3 choices), the remaining 3 digits can be arranged in 3! ways. * Number of favourable outcomes = 3 × 3! = 3 × 6 = 18. * Probability = 18 / 24 = 3/4. * Alternatively, this is 1 - Probability (greater than or equal to 4000). Since no number can be exactly 4000 (as it uses all 4 digits), it is 1 - P(greater than 4000) = 1 - 1/4 = 3/4.

Section: Geometrical probability

Note: The problems in this section ask to find the probability that a randomly placed dot falls within a green region, given a yellow region. This is calculated as (Area of green region) / (Area of yellow region). The solutions rely on calculating the areas of standard geometric shapes.

(1) The square got by joining the mid- points of a larger square

Solution:

• Let the side length of the larger square be s. Its area is $s^2$.

• Joining the midpoints of a square forms another square rotated by 45 degrees inside the original.

• The vertices of the inner square are the midpoints of the sides of the outer square.

• Consider a small right-angled triangle formed at a corner of the outer square, with hypotenuse as a side of the inner square. The legs of this triangle are s/2 each.

• The side of the inner square (s_inner) = $\sqrt{(s/2)^2 + (s/2)^2} = \sqrt{s^2/4 + s^2/4} = \sqrt{2s^2/4} = \sqrt{s^2/2} = s/\sqrt{2}$.

• Area of inner (green) square = $(s/\sqrt{2})^2 = s^2/2$.

• Probability = (Area of inner square) / (Area of outer square) = $(s^2/2) / s^2 = 1/2$.

(2) The triangle got by joining alternate vertices of a regular hexagon

Solution:

• A regular hexagon can be divided into 6 equilateral triangles with their common vertex at the center of the hexagon.

• Joining alternate vertices of a regular hexagon forms an equilateral triangle.

• Area of regular hexagon = 6 × (Area of equilateral triangle with side 'a').

• Area of equilateral triangle with side a = $(\sqrt{3}/4)a^2$.

• If the hexagon has side a, the equilateral triangle formed by alternate vertices has side length a√3 (which is twice the altitude of the smaller equilateral triangle).

• Area of hexagon with side a = $6 \times (\sqrt{3}/4)a^2$.

• The triangle formed by joining alternate vertices (green region) is an equilateral triangle whose side is the distance between two alternate vertices. This distance is $2a$ (twice the side length of the hexagon if the center is connected to vertices, or hypotenuse of a right angle triangle formed by (a, a*sqrt(3)/2)).

• No, the side length of the triangle formed by alternate vertices is $a\sqrt{3}$ from the center-to-vertex perspective.

• A simpler way: The regular hexagon consists of 6 equilateral triangles. Joining alternate vertices forms an equilateral triangle. This triangle is made up of 3 of the 6 small equilateral triangles, plus the central triangle. No, it is simply 3 of those triangles.

• If the hexagon's side is a, its area is $6 \times \text{Area}(\text{equilateral triangle with side a})$.

• The triangle formed by alternate vertices is also equilateral. Its side length is $a\sqrt{3}$. Its area is $(\sqrt{3}/4)(a\sqrt{3})^2 = (\sqrt{3}/4)(3a^2) = (3\sqrt{3}/4)a^2$.

• Probability = (Area of green triangle) / (Area of yellow hexagon) = $((3\sqrt{3}/4)a^2) / (6 \times (\sqrt{3}/4)a^2) = (3\sqrt{3}/4) / (6\sqrt{3}/4) = 3/6 = 1/2$.

(3) The regular hexagon formed between two equal equilateral triangle

Solution:

• This refers to the Star of David (Hexagram). Two overlapping equilateral triangles form a regular hexagon in the middle and six smaller equilateral triangles at the points.

• Let the side length of the large equilateral triangles be L. Area = $(\sqrt{3}/4)L^2$.

• The central hexagon is formed by the overlap. The side length of this central hexagon is L/3.

• Area of central hexagon = $6 \times (\sqrt{3}/4)(L/3)^2 = 6 \times (\sqrt{3}/4)(L^2/9) = (\sqrt{3}/6)L^2$.

• The "yellow" region is the area of the two equal equilateral triangles. When they overlap, some area is counted twice (the hexagon). So, total area of yellow region is sum of areas of two triangles minus area of hexagon = $2 \times (\sqrt{3}/4)L^2 - (\sqrt{3}/6)L^2 = (\sqrt{3}/2)L^2 - (\sqrt{3}/6)L^2 = (3\sqrt{3}/6)L^2 - (\sqrt{3}/6)L^2 = (2\sqrt{3}/6)L^2 = (\sqrt{3}/3)L^2$.

• Probability = (Area of green hexagon) / (Area of yellow region) = $((\sqrt{3}/6)L^2) / ((\sqrt{3}/3)L^2) = (1/6) / (1/3) = 1/2$.

• Conclusion: The area of the hexagon is 1/3 of the area of one large triangle. The overlapping area makes the problem slightly more complex.

• A simpler interpretation might be that the "yellow region" is the area of one of the large triangles, and the "green region" is the hexagon. This is not clear from the wording.

• Let's assume the question meant the total area covered by the two triangles as the yellow region.

• The area of one big equilateral triangle. When another identical one is inverted and superimposed, the middle region is a hexagon, and the 6 'points' are small equilateral triangles. The side of these small triangles is 1/3 the side of the big triangle.

• The area of the hexagon is equal to the area of 6 small equilateral triangles. The area of one large triangle is equal to the area of 9 small equilateral triangles.

• Area of green hexagon = Area of 6 small triangles.

• Area of yellow region (union of two large triangles) = Area of 12 small triangles (6 points + 6 for hexagon). No.

• Total area of two large triangles = 2 * (Area of one large triangle).

• Area of one large triangle = Area of central hexagon + 3 * Area of outer small triangles.

• A standard result for this construction is that the central hexagon has 1/3 the area of one large triangle.

• Area of green hexagon = $A_{hex}$.

• Area of one large triangle = $A_{tri}$.

• Area of yellow region = $2 A_{tri} - A_{hex}$. Since $A_{hex} = (1/3) A_{tri}$, then Area of yellow region = $2 A_{tri} - (1/3)A_{tri} = (5/3)A_{tri}$.

• Probability = $(1/3)A_{tri} / (5/3)A_{tri} = 1/5$.

• A common interpretation of "regular hexagon formed between two equal equilateral triangle" is that the yellow region IS the two triangles together, including overlap.

• A simpler interpretation is often intended in these problems unless specified. If the "yellow region" means one of the equilateral triangles, and the "green region" is the hexagon, then probability is 1/3.

• Let's assume yellow region is the union of the two triangles.

• Let side of large triangle be s. Area of 1 triangle is $A_T = \frac{\sqrt{3}}{4}s^2$.

• The overlapping hexagon has side s/3. Its area is $A_H = 6 \cdot \frac{\sqrt{3}}{4} (\frac{s}{3})^2 = \frac{6\sqrt{3}}{4} \frac{s^2}{9} = \frac{\sqrt{3}}{6}s^2$.

• Total area of yellow region (union of 2 triangles) = $2 A_T - A_H = 2 \frac{\sqrt{3}}{4}s^2 - \frac{\sqrt{3}}{6}s^2 = \frac{\sqrt{3}}{2}s^2 - \frac{\sqrt{3}}{6}s^2 = (\frac{3\sqrt{3} - \sqrt{3}}{6})s^2 = \frac{2\sqrt{3}}{6}s^2 = \frac{\sqrt{3}}{3}s^2$.

• Probability = $A_H / (2 A_T - A_H) = (\frac{\sqrt{3}}{6}s^2) / (\frac{\sqrt{3}}{3}s^2) = \frac{1/6}{1/3} = 1/2$.

(4) A square drawn with vertices on a circle:

Solution:

• Let the radius of the circle be r. Area of circle (yellow) = $\pi r^2$.

• If a square is drawn with its vertices on a circle, its diagonal is the diameter of the circle.

• Diagonal of square = 2r.

• If side of square is s, then s√2 = 2r => s = 2r/√2 = r√2.

• Area of square (green) = $s^2 = (r√2)^2 = 2r^2$.

• Probability = (Area of square) / (Area of circle) = $2r^2 / (\pi r^2) = 2/\pi$.

(5) The circle that just fits within a square

Solution:

• Let the side length of the square be s. Area of square (yellow) = $s^2$.

• A circle that just fits within a square has its diameter equal to the side of the square.

• Diameter of circle = s. Radius of circle = s/2.

• Area of circle (green) = $\pi (s/2)^2 = \pi s^2/4$.

• Probability = (Area of circle) / (Area of square) = $(\pi s^2/4) / s^2 = \pi/4$.

Section: Pairs

(1) Rajani has three necklaces and three earrings of green, blue and red stones. In how many different ways can she wear them? What is the probability of her wearing a necklace and earrings of the same colour? Of different colours?

Solution:

• Total ways to wear: She has 3 choices for necklaces and 3 choices for earrings.

    ◦ Total ways = 3 × 3 = 9 ways. (e.g., (Green necklace, Green earring), (Green necklace, Blue earring), etc.).

• Probability of wearing necklace and earrings of the same colour:

    ◦ Favourable outcomes (same colour): (Green, Green), (Blue, Blue), (Red, Red).

    ◦ Number of favourable outcomes = 3.

    ◦ Probability (Same colour) = 3 / 9 = 1/3.

• Probability of wearing necklace and earrings of different colours:

    ◦ This is the complement of wearing the same colour.

    ◦ Probability (Different colours) = 1 - Probability (Same colour) = 1 - 1/3 = 2/3.

    ◦ Alternatively, Number of favourable outcomes = Total ways - Ways same colour = 9 - 3 = 6.

    ◦ Probability (Different colours) = 6 / 9 = 2/3.

(2) A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1 and 2. If one slip is drawn from each box, what is the probability of the sum of the numbers being odd? What is the probability of the sum being even?

Solution:

• Box 1: Slips {1, 2, 3, 4}. (2 odd, 2 even).

• Box 2: Slips {1, 2}. (1 odd, 1 even).

• Total possible pairs: Number from Box 1 (4 choices) × Number from Box 2 (2 choices) = 4 × 2 = 8 pairs.

    ◦ Pairs: (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (4,1), (4,2).

• Probability of the sum being odd:

    ◦ Sum is odd if (Odd + Even) or (Even + Odd).

    ◦ Odd from Box 1 (1, 3) and Even from Box 2 (2): (1,2), (3,2). (2 pairs)

    ◦ Even from Box 1 (2, 4) and Odd from Box 2 (1): (2,1), (4,1). (2 pairs)

    ◦ Number of favourable outcomes = 2 + 2 = 4.

    ◦ Probability (Sum odd) = 4 / 8 = 1/2.

• Probability of the sum being even:

    ◦ Sum is even if (Odd + Odd) or (Even + Even).

    ◦ Odd from Box 1 (1, 3) and Odd from Box 2 (1): (1,1), (3,1). (2 pairs)

    ◦ Even from Box 1 (2, 4) and Even from Box 2 (2): (2,2), (4,2). (2 pairs)

    ◦ Number of favourable outcomes = 2 + 2 = 4.

    ◦ Probability (Sum even) = 4 / 8 = 1/2.

    ◦ Alternatively, Probability (Sum even) = 1 - Probability (Sum odd) = 1 - 1/2 = 1/2.

(3) A box contains four slips numbered 1, 2, 3, 4 and another box contains three slips numbered 1, 2, 3. If one slip is drawn from each box, what is the probability of the product of the numbers being odd? What is the probability of the product being even?

Solution:

• Box 1: Slips {1, 2, 3, 4}. (Odd: 1, 3; Even: 2, 4).

• Box 2: Slips {1, 2, 3}. (Odd: 1, 3; Even: 2).

• Total possible pairs: Number from Box 1 (4 choices) × Number from Box 2 (3 choices) = 4 × 3 = 12 pairs.

• Probability of the product of the numbers being odd:

    ◦ Product is odd only if (Odd × Odd).

    ◦ Odd from Box 1 (1, 3) and Odd from Box 2 (1, 3).

    ◦ Favourable outcomes: (1,1), (1,3), (3,1), (3,3).

    ◦ Number of favourable outcomes = 2 × 2 = 4.

    ◦ Probability (Product odd) = 4 / 12 = 1/3.

• Probability of the product of the numbers being even:

    ◦ Product is even if (Odd × Even) or (Even × Odd) or (Even × Even).

    ◦ This is the complement of the product being odd.

    ◦ Probability (Product even) = 1 - Probability (Product odd) = 1 - 1/3 = 2/3.

    ◦ Alternatively:

        ▪ Odd (Box 1) × Even (Box 2): (1,2), (3,2) (2 pairs)

        ▪ Even (Box 1) × Odd (Box 2): (2,1), (2,3), (4,1), (4,3) (4 pairs)

        ▪ Even (Box 1) × Even (Box 2): (2,2), (4,2) (2 pairs)

        ▪ Total favourable = 2 + 4 + 2 = 8.

        ▪ Probability (Product even) = 8 / 12 = 2/3.

Section: More pairs

(4) From all two-digit numbers using only the digits 1, 2 and 3, one number is chosen (i) What is the probability of both digits being the same? (ii) What is the probability of the sum of the digits being 4?

Solution:

• The two-digit numbers use only digits 1, 2, and 3.

• For a two-digit number ab:

    ◦ The first digit a can be 1, 2, or 3 (3 choices).

    ◦ The second digit b can be 1, 2, or 3 (3 choices).

• Total possible two-digit numbers = 3 × 3 = 9.

    ◦ Numbers: 11, 12, 13, 21, 22, 23, 31, 32, 33.

(i) What is the probability of both digits being the same? * Favourable numbers: 11, 22, 33. * Number of favourable outcomes = 3. * Probability = 3 / 9 = 1/3.

(ii) What is the probability of the sum of the digits being 4? * Favourable numbers where sum of digits is 4: * 1 + 3 = 4 => 13 * 2 + 2 = 4 => 22 * 3 + 1 = 4 => 31 * Number of favourable outcomes = 3. * Probability = 3 / 9 = 1/3.

(5) A game for two players. Before starting, each player has to decide whether he wants an odd number or even number. Then both raise some fingers of one hand at the same time. If the sum of numbers of fingers is odd, the one who chose odd number at the beginning wins; if even, the one who chose even number wins. Which is the better choice at the beginning, odd or even?

Solution:

• Each player raises some fingers from one hand. Assuming fingers mean numbers 1, 2, 3, 4, 5 (or 0-5 if thumb is zero, but usually 1-5). Let's assume 1-5 fingers.

• Let Player 1 raise F1 fingers and Player 2 raise F2 fingers. Both F1 and F2 can be any integer from 1 to 5.

• Total possible outcomes (pairs): 5 × 5 = 25 pairs.

• Sum is Odd (Odd + Even or Even + Odd):

    ◦ Possible odd numbers for F1/F2: {1, 3, 5} (3 choices).

    ◦ Possible even numbers for F1/F2: {2, 4} (2 choices).

    ◦ (Odd F1, Even F2): 3 × 2 = 6 pairs.

    ◦ (Even F1, Odd F2): 2 × 3 = 6 pairs.

    ◦ Total for sum is odd = 6 + 6 = 12 pairs.

    ◦ Probability (Sum odd) = 12 / 25.

• Sum is Even (Odd + Odd or Even + Even):

    ◦ (Odd F1, Odd F2): 3 × 3 = 9 pairs.

    ◦ (Even F1, Even F2): 2 × 2 = 4 pairs.

    ◦ Total for sum is even = 9 + 4 = 13 pairs.

    ◦ Probability (Sum even) = 13 / 25.

• Conclusion: Since Probability (Sum even) (13/25) > Probability (Sum odd) (12/25), choosing even at the beginning is the better choice.

(1) There are 30 boys and 20 girls in Class 10 A and 15 boys and 25 girls in Class 10 B. One student is to be chosen from each Class. (i) What is the probability of both being girls? (ii) What is the probability of both being boys? (iii) What is the probability of one being a boy and one being a girl? (iv) What is the probability of at least one being a boy?

Solution:

• Class 10 A: 30 boys (B_A), 20 girls (G_A). Total = 50 students.

• Class 10 B: 15 boys (B_B), 25 girls (G_B). Total = 40 students.

• Total possible pairs of choices (one from each class): 50 × 40 = 2000 pairs.

(i) What is the probability of both being girls? * Favourable pairs: (Girl from A, Girl from B). * Number of ways = (Number of girls in A) × (Number of girls in B) = 20 × 25 = 500. * Probability = 500 / 2000 = 5/20 = 1/4.

(ii) What is the probability of both being boys? * Favourable pairs: (Boy from A, Boy from B). * Number of ways = (Number of boys in A) × (Number of boys in B) = 30 × 15 = 450. * Probability = 450 / 2000 = 45/200 = 9/40.

(iii) What is the probability of one being a boy and one being a girl? * This can happen in two ways: * (Boy from A, Girl from B): 30 × 25 = 750 ways. * (Girl from A, Boy from B): 20 × 15 = 300 ways. * Total favourable ways = 750 + 300 = 1050. * Probability = 1050 / 2000 = 105/200 = 21/40.

(iv) What is the probability of at least one being a boy? * "At least one boy" means (Boy, Girl) or (Girl, Boy) or (Boy, Boy). * Total favourable ways = (from iii) 1050 + (from ii) 450 = 1500. * Probability = 1500 / 2000 = 15/20 = 3/4. * Alternatively, "At least one boy" is the complement of "Both being girls". * Probability = 1 - Probability (Both girls) = 1 - 1/4 = 3/4.

(2) One is asked to say a two-digit number (i) What is the probability of both digits being the same? (ii) What is the probability of the first digit being greater than the second? (iii) What is the probability of the first digit being less than the second?

Solution:

• Total two-digit numbers range from 10 to 99.

• Total possible two-digit numbers = 99 - 10 + 1 = 90.

(i) What is the probability of both digits being the same? * Numbers: 11, 22, 33, 44, 55, 66, 77, 88, 99. * Number of favourable outcomes = 9. * Probability = 9 / 90 = 1/10.

(ii) What is the probability of the first digit being greater than the second? * Let the number be ab. a > b. * If a = 1, no b. * If a = 2, b = 1 (21). * If a = 3, b = 1, 2 (31, 32). * If a = 4, b = 1, 2, 3 (41, 42, 43). * If a = 5, b = 1, 2, 3, 4 (51, 52, 53, 54). * If a = 6, b = 1, 2, 3, 4, 5 (61, 62, 63, 64, 65). * If a = 7, b = 1, 2, 3, 4, 5, 6 (71, ..., 76). * If a = 8, b = 1, 2, 3, 4, 5, 6, 7 (81, ..., 87). * If a = 9, b = 1, 2, 3, 4, 5, 6, 7, 8 (91, ..., 98). * Summing the count: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36. * Probability = 36 / 90 = 2/5.

(iii) What is the probability of the first digit being less than the second? * Let the number be ab. a < b. * If a = 1, b = 2, 3, ..., 9 (8 numbers: 12, ..., 19). * If a = 2, b = 3, ..., 9 (7 numbers: 23, ..., 29). * If a = 3, b = 4, ..., 9 (6 numbers: 34, ..., 39). * If a = 4, b = 5, ..., 9 (5 numbers: 45, ..., 49). * If a = 5, b = 6, ..., 9 (4 numbers: 56, ..., 59). * If a = 6, b = 7