SSLC: Maths Notes
๐ Malayalam
Chapter 1: Arithmetic Sequences
Key Concepts:
• A number sequence is a collection of numbers ordered as first, second, third, and so on, according to some rule.
• An arithmetic sequence starts with a number and proceeds by adding the same number again and again.
• The numbers in a sequence are called terms.
• In an arithmetic sequence, the same number is added to go from any term to the next. Conversely, the same number is subtracted to go from any term to the previous term.
• This number, found by subtracting the previous term from any term, is called the common difference.
• In any arithmetic sequence, the change in terms is proportional to the change in position.
• Term Connections:
◦ The sum of the terms just before and just after a term in an arithmetic sequence is twice this term.
◦ The sum of three consecutive terms is three times the middle term.
◦ The sum of an odd number of consecutive terms is the product of the middle term and the number of terms.
◦ If one position is increased and another position is decreased by the same amount in an arithmetic sequence, the sum of the terms at these positions does not change.
◦ If the sum of two positions is equal to the sum of two other positions, then the sum of the terms at each pair is the same.
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Solved Problem: Finding Terms and Common Difference
Query: Calculate the first five terms of the arithmetic sequence with the 3rd term as 37 and the 7th term as 73.
Solution:
1. Find the common difference:
◦ To go from the 3rd term to the 7th term, the position increases by 7 – 3 = 4 times.
◦ The numerical difference between the 7th and 3rd terms is 73 – 37 = 36.
◦ Therefore, 4 times the common difference is 36.
◦ The common difference is 36 ÷ 4 = 9.
2. Find the first term:
◦ To go from the 3rd term to the 1st term, we must subtract the common difference 3 – 1 = 2 times.
◦ The 1st term is 37 – (2 × 9) = 37 – 18 = 19.
3. Write the sequence:
◦ Starting with the 1st term (19) and adding the common difference (9) repeatedly, the sequence is: 19, 28, 37, 46, 55, ....
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Practice Questions:
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
◦ (iii) Squares of natural numbers
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
3. The 3rd term of an arithmetic sequence is 15 and the 8th term is 35.
◦ (i) What is its 13th term?
◦ (ii) What is its 23rd term?
4. Is 101 a term of the arithmetic sequence 13, 24, 35, ...? What about 1001?
5. 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?
6. The sum of the 7th and the 8th terms of an arithmetic sequence is 50. Calculate the sum of the first 14 terms.
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Chapter 2: Circles and Angles
Key Concepts:
• The angle made by the ends of an arc at a point on the circle (outside the arc itself) is half the central angle of the arc.
• The angle in a semicircle is a right angle (90°).
• Any two points on a circle split the circle into two arcs, where each is the alternate arc of the other.
• A chord divides the circular region into two segments.
• In a circle, angles in the same segment are equal.
• The sum of the angles in alternate segments is 180°.
• A cyclic quadrilateral is a quadrilateral where all four vertices lie on a circle.
• If all vertices of a quadrilateral are on a circle, then the sum of its opposite angles is 180°.
• Conversely, if the sum of opposite angles of a quadrilateral is 180°, then a circle can be drawn passing through all four of its vertices.
• Any rectangle is a cyclic quadrilateral.
• Any isosceles trapezium is cyclic.
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Solved Problem: Relationship between Central Angle and Angle on the Circle
Query: What angle should we mark at a point on the circle to mark off 1/5 of the circle?
Solution:
1. Find the central angle: If an arc is 1/5 of the circle, its central angle is (1/5) × 360° = 72°.
2. Apply the principle: The angle made by the ends of an arc at a point on the circle (outside the arc) is half the central angle.
3. Calculate the angle: So, the angle at a point on the circle should be (1/2) × 72° = 36°.
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Practice Questions:
1. What fraction of the circle is the arc marked in the picture below, if the angle at the point on the circle is 45°?
2. 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?
3. In each of the problems 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 half the angles in the other part
4. Prove that in a cyclic quadrilateral, the outer angle at any vertex is equal to the inner angle at the opposite vertex.
5. In the 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.
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Chapter 3: Arithmetic Sequences and Algebra
Key Concepts:
• The algebraic form of a sequence uses algebra to concisely state the rule for finding any term based on its position.
• For a sequence,
x_n denotes the term in the nth position.• The algebraic form of any arithmetic sequence is x_n = an + b, where
a and b are specific numbers.• In this form, a is the common difference of the arithmetic sequence.
• The terms of any arithmetic sequence are obtained by multiplying natural numbers (1, 2, 3, ...) by a fixed number (
a) and adding a fixed number (b).• The sum of any number of consecutive natural numbers starting from one is half the product of the last number and the next:
1 + 2 + 3 + ... + n = (1/2)n(n + 1).• The sum of the first n even numbers is
n(n + 1).• The sum of the first n odd numbers (starting with 1) is
n^2.• The sum of the first n terms of an arithmetic sequence x_1, x_2, ..., x_n can be calculated as:
(1/2)n(x_1 + x_n). This means the sum is half the product of the sum of the first and last terms by the number of terms.• The algebraic form of the sum of the first n terms of an arithmetic sequence is pn^2 + qn, where
p and q are specific numbers.
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Solved Problem: Finding the Algebraic Form of a Sequence
Query: Find the algebraic form of the arithmetic sequence 12, 23, 34, ....
Solution:
1. Identify the first term and common difference:
◦ The 1st term (
f) is 12. ◦ The common difference (
d) is 23 - 12 = 11.2. Use the general formula f + (n - 1)d for the nth term:
◦
x_n = 12 + ((n - 1) × 11)3. Simplify the expression:
◦
x_n = 12 + 11n - 11 ◦
x_n = 11n + 14. The algebraic form of this sequence is x_n = 11n + 1.
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Practice Questions:
1. Find the algebraic form of the arithmetic sequences given below:
◦ (i) 1, 6, 11, 16, ...
◦ (iii) 21, 32, 43, 54, ...
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
◦ (iii) 5th term 10, 10th term 5
3. Prove that the arithmetic sequence with first term 1/3 and common difference 2/3 contains all odd numbers, but no even numbers.
4. Calculate the sum of the first 25 terms of each of the arithmetic sequences below:
◦ (i) 11, 22, 33, ...
◦ (ii) 12, 23, 34, ...
5. The sum of the first
n terms of some arithmetic sequences are given below. Find the nth term of each: ◦ (i)
n^2 + 2n ◦ (ii)
2n^2 + n6. The sum of the first three terms of an arithmetic sequence is 30 and the sum of the first seven terms is 140.
◦ (i) What is the 2nd term of the sequence?
◦ (ii) What is the 4th term of the sequence?
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Chapter 4: Mathematics of Chance
Key Concepts:
• Probability is expressed as a fraction, decimal, or percentage, representing the likelihood of an event occurring.
• When calculating probability, consider the total number of possible outcomes and the number of favourable outcomes.
• Geometrical probability relates probability to ratios of areas. If a point is marked randomly within a larger region, the probability that it falls within a specific sub-region is the ratio of the sub-region's area to the larger region's area.
• When events involve choices from multiple sets (e.g., choosing from two boxes), the total number of possible pairs is the product of the number of items in each set.
• The probability of "at least one" event can be found by subtracting the probability of "none" from 1 (complement rule).
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Solved Problem: Probability with Pairs of Items
Query: A basket contains 50 mangoes and 20 of them are not ripe. Another contains 40 mangoes and 15 of them are not ripe. If one mango is picked from each basket, what is the probability of both being ripe?
Solution:
1. Calculate the total number of possible pairs:
◦ There are 50 mangoes in the first basket and 40 in the second.
◦ The total number of ways to choose one mango from each is 50 × 40 = 2000 pairs.
2. Calculate the number of ripe mangoes in each basket:
◦ First basket: 50 total - 20 unripe = 30 ripe mangoes.
◦ Second basket: 40 total - 15 unripe = 25 ripe mangoes.
3. Calculate the number of pairs where both mangoes are ripe:
◦ Multiply the number of ripe mangoes in each basket: 30 × 25 = 750 pairs.
4. Calculate the probability:
◦ Probability (both ripe) = (Number of pairs with both ripe) / (Total number of possible pairs)
◦ Probability (both ripe) = 750 / 2000 = 3/8.
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Practice Questions:
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?
2. If one number from 1, 2, 3, ..., 100 is chosen, what is the probability that it is a two-digit number?
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 this number is a multiple of 3?
4. Cut out a rectangular piece of cardboard and draw a triangle joining the midpoint of one side to the endpoints of the opposite side. If we mark a point within the rectangle, with eyes closed, what is the probability that it would be within the triangle?
5. 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?
6. 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 at least one being a boy?
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Chapter 5: Second Degree Equations
Key Concepts:
• A second-degree equation involves a term with a variable raised to the power of 2 (e.g.,
x^2).• Many geometry problems involving areas or products can be translated into second-degree equations.
• The technique of "completing the square" (using identities like
(x + a)^2 = x^2 + 2ax + a^2 or (x - a)^2 = x^2 - 2ax + a^2) is crucial for solving these equations.• When solving for a variable (e.g.,
x) by taking the square root of both sides of an equation (e.g., y^2 = k), remember to consider both the positive and negative square roots (y = √k or y = -√k).• In practical problems (e.g., finding lengths), some algebraic solutions (e.g., negative values) may not be suitable in the context of the problem. Always check if the solutions make sense in the real-world scenario.
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Solved Problem: Solving a Second Degree Equation by Completing the Square
Query: One side of a rectangle is 20 metres longer than the other and its area is 224 square metres. What are the lengths of the sides?
Solution:
1. Formulate the algebraic equation:
◦ Let the length of the shorter side be
x metres. ◦ Then the length of the longer side is
x + 20 metres. ◦ The area is
x(x + 20) = x^2 + 20x square metres. ◦ Given the area is 224, the equation is:
x^2 + 20x = 224.2. Complete the square:
◦ To make
x^2 + 20x a perfect square in the form (x + a)^2, we compare 20x with 2ax. This means 2a = 20, so a = 10. ◦ We need to add
a^2 = 10^2 = 100 to both sides of the equation. ◦
x^2 + 20x + 100 = 224 + 100 ◦
(x + 10)^2 = 3243. Solve for x:
◦ Take the square root of both sides:
x + 10 = ±√324. ◦
x + 10 = ±18. ◦ This gives two possibilities:
▪
x + 10 = 18 => x = 18 - 10 = 8 ▪
x + 10 = -18 => x = -18 - 10 = -284. Interpret the solutions in context:
◦ Since
x represents a length, it must be a positive number. Therefore, x = -28 is not a suitable answer. ◦ The shorter side
x is 8 metres. ◦ The longer side
x + 20 is 8 + 20 = 28 metres.
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Practice Questions:
1. When each side of a square was reduced by 2 metres to make a smaller square, its area became 49 square metres. What was the length of a side of the original square?
2. The first few terms of the arithmetic sequence 9, 11, 13, ... were added and then 16 added to the sum, to get 256. How many terms were added?
3. One of the perpendicular sides of a right triangle is 5 centimetres more than the other. The area of the triangle is 12 square centimetres. What are the lengths of these sides?
4. The product of two consecutive odd numbers is 195. What are the numbers?
5. Consider the arithmetic sequence 99, 97, 95, ...
◦ How many terms of this, starting from the first, must be added to get 900 as the sum?
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Chapter 6: Trigonometry
Key Concepts:
• In a right triangle with angles 45°, 45°, 90°, the sides are in the ratio 1 : 1 : √2 (shorter side : shorter side : hypotenuse).
• In a right triangle with angles 30°, 60°, 90°, the sides are in the ratio 1 : √3 : 2 (shortest side : medium side : longest side).
• For an angle
a° in a right triangle: ◦ Sine (sin a°) = (Length of the side opposite angle
a°) / (Length of the hypotenuse). ◦ Cosine (cos a°) = (Length of the side adjacent to angle
a°) / (Length of the hypotenuse). ◦ Tangent (tan a°) = (Length of the side opposite angle
a°) / (Length of the side adjacent to angle a°).• These ratios (sin, cos, tan) are fixed for a specific angle size and do not change with the size of the triangle.
• The area of a triangle can be calculated using trigonometry if two sides and the included angle are known (e.g.,
(1/2)ab sin C).• The length of any chord of a circle is twice the product of the radius and the sine of half the central angle:
Chord Length = 2r sin (Central Angle / 2).• In triangles with the same angles, the sides are in the same ratio. This ratio can be expressed using the sine of the opposite angles:
side a : side b : side c = sin A : sin B : sin C.• Angle of elevation is the angle formed by lifting the line of vision from the horizontal to see something above.
• Angle of depression is the angle formed by lowering the line of vision from the horizontal to see something below.
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Solved Problem: Calculating Area of a Triangle using Sine
Query: What is the area of the triangle below? (A triangle with sides 4cm and 6cm, and the included angle is 50°)
Solution:
1. Draw a perpendicular to find the height:
◦ Draw a perpendicular from the top vertex to the base (6 cm side). Let the height be
h.2. Use sine to find the height:
◦ This perpendicular forms a right triangle with the 4 cm side as the hypotenuse and the 50° angle.
◦
sin 50° = h / 4. ◦
h = 4 × sin 50°. ◦ From the trigonometric table,
sin 50° ≈ 0.7660. ◦ So,
h ≈ 4 × 0.7660 = 3.064 cm.3. Calculate the area of the triangle:
◦ Area =
(1/2) × base × height. ◦ Area =
(1/2) × 6 cm × 3.064 cm. ◦ Area
≈ 9.192 square centimetres.
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Practice Questions:
1. Calculate the area of the parallelograms shown below (one with sides 5cm, 8cm and angle 40°).
2. The lengths of two sides of a triangle are 8 centimetres and 10 centimetres and the angle between them 40°.
◦ (i) What is the area of this triangle?
◦ (ii) What is the area of the triangle with lengths of two sides the same, but the angle between them 140°?
3. A circle is to be drawn, passing through the ends of a line 5 centimetres long; and the angle in one of the segments made by the line should be 80°. What should be the radius of the circle?
4. A ladder leans against a wall with its foot 2 metres away from the wall. The angle between the ladder and the ground is 40°. How high is the top of the ladder from the ground?
5. A person, 1.7 metres tall, standing 10 metres away from the foot of a tree sees the top of the tree at an angle of elevation of 40°. What is the height of the tree?
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Chapter 7: Coordinates
Key Concepts:
• Points in a plane can be represented by coordinates, a pair of numbers
(x, y).• The x-axis is the horizontal line, and the y-axis is the vertical line. Their intersection is the origin (0,0).
• The
x-coordinate (or abscissa) represents the horizontal distance from the y-axis, and the y-coordinate (or ordinate) represents the vertical distance from the x-axis.• Properties of Rectangles with Sides Parallel to Axes:
◦ If two points
(x_1, y_1) and (x_2, y_2) are opposite vertices of such a rectangle, then the other two vertices are (x_1, y_2) and (x_2, y_1). ◦ The
x-coordinates of points on a line parallel to the y-axis are the same. ◦ The
y-coordinates of points on a line parallel to the x-axis are the same. ◦ For such a rectangle to exist, the x-coordinates (
x_1, x_2) must be different, and the y-coordinates (y_1, y_2) must be different.• Lengths and Distances:
◦ The length of the horizontal sides of a rectangle (parallel to x-axis) between
x_1 and x_2 is |x_1 - x_2|. ◦ The length of the vertical sides (parallel to y-axis) between
y_1 and y_2 is |y_1 - y_2|. ◦ The distance between two points (x_1, y_1) and (x_2, y_2) is given by the distance formula (derived from Pythagoras' theorem, as the diagonal of a rectangle):
√((x_1 - x_2)^2 + (y_1 - y_2)^2). ◦ The distance between a point (x, y) and the origin (0,0) is
√(x^2 + y^2). ◦ Three points are collinear (on the same line) if the largest distance between any two of them is equal to the sum of the other two distances.
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Solved Problem: Finding Coordinates of a Rectangle and Distance between Points
Query: Given the opposite vertices of a rectangle as (2, 2) and (7, 3) with sides parallel to the axes, find the coordinates of the other two vertices and the length of its diagonal.
Solution:
1. Find the coordinates of the other two vertices:
◦ Let the given opposite vertices be A = (2, 2) and C = (7, 3).
◦ For a rectangle with sides parallel to the axes, the x-coordinate of one vertex is combined with the y-coordinate of the opposite vertex to find the adjacent vertices.
◦ The other two vertices will be B = (7, 2) and D = (2, 3).
◦ (Reasoning: To move from (2,2) to (7,2), the y-coordinate stays the same (parallel to x-axis), and the x-coordinate changes to 7. To move from (2,2) to (2,3), the x-coordinate stays the same (parallel to y-axis), and the y-coordinate changes to 3.)
2. Calculate the lengths of the sides of the rectangle:
◦ Length of horizontal sides (e.g., AB):
|7 - 2| = 5 units. ◦ Length of vertical sides (e.g., AD):
|3 - 2| = 1 unit.3. Calculate the length of the diagonal:
◦ The diagonal is the distance between the opposite vertices, A(2,2) and C(7,3).
◦ Using the distance formula:
√((x_1 - x_2)^2 + (y_1 - y_2)^2). ◦ Distance =
√((2 - 7)^2 + (2 - 3)^2). ◦ Distance =
√((-5)^2 + (-1)^2). ◦ Distance =
√(25 + 1). ◦ Distance =
√26 units.
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Practice Questions:
1. Find the y-coordinates of points on the x-axis.
2. Find the coordinates of the other three vertices of the rectangle in the picture (bottom-left vertex is (2,1), top-right is (7,3)).
3. The sides of the rectangle in the picture are parallel to the axes and the origin is the midpoint (point of intersection of the diagonals) of the rectangle. If one vertex is (3,2), what are the coordinates of the other three vertices of the rectangle?
4. Without drawing the axes, mark these points with the left and right, and top and bottom positions correct. Find the coordinates of the other two vertices of the rectangle with each pair as the coordinates of two opposite vertices and sides parallel to the axes:
◦ (i) (3,5), (7,8)
◦ (iii) (−3, 5), (−7,1)
5. Calculate the lengths of the sides and the diagonals of the quadrilateral in the picture below (vertices are (-2,-2), (2,-2), (3,1), (-1,1)).
6. A circle is drawn with centre at the origin and with radius 10.
◦ (i) Check whether each of the points with coordinates (6, 9), (5, 9), (6, 8) is within the circle, outside the circle or on the circle.
◦ (ii) Write the coordinates of 8 points on this circle.
