Chapter 5 - I’m Up and Down, and Round and Round (Circles)

Exercise Set 5.1 & 5.2

Q. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle. Is the centre inside or outside the triangle?

Solution: The third angle is ∠C = 180° - (70° + 60°) = 50°.
Since all angles are less than 90°, it is an acute-angled triangle. Therefore, the circumcentre lies inside the triangle.

Q. What is the least possible radius of a circle passing through two points A and B?

Solution: The least possible radius occurs when the segment AB acts as the diameter of the circle. Thus, the minimum radius is AB / 2.

Q. Show that the triangle formed by a chord and the centre of the circle is isosceles.

Solution: Let the chord be AB and the centre be O. The sides OA and OB connect the centre to the circumference, so they are radii of the same circle. Hence, OA = OB. Since two sides are equal, ΔOAB is an isosceles triangle.

End-of-Chapter Exercises

1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?

Solution: A perpendicular drawn from the centre to the chord bisects it. Let half the chord be x.
Using the Baudhäyana-Pythagoras theorem:

Radius² = (Perpendicular Distance)² + (Half Chord)²
13² = 5² + x²
169 = 25 + x² ⇒ x² = 144 ⇒ x = 12 cm.

The total length of the chord is 2 × 12 = 24 cm.

7. ABCD is a cyclic quadrilateral. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?

Solution: The sum of opposite angles in a cyclic quadrilateral is always 180°.
∠C = 180° - ∠A = 180° - 75° = 105°.
∠D = 180° - ∠B = 180° - 110° = 70°.

8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.

Solution: Since PQRS is cyclic, opposite angles sum to 180°.

∠P + ∠R = 180°
(2x + 10) + (3x - 20) = 180
5x - 10 = 180 ⇒ 5x = 190 ⇒ x = 38

∠P = 2(38) + 10 = 76 + 10 = 86°.
∠R = 3(38) - 20 = 114 - 20 = 94°.

Chapter 6 - Measuring Space: Perimeter and Area

Exercise Set 6.1 & 6.3

Q. The perimeter of a circle is 44 cm. What is its radius? (Use π ≈ 22/7)

Solution: The formula for the perimeter (circumference) is C = 2πr.

44 = 2 × (22/7) × r
44 = (44/7) × r
r = (44 × 7) / 44 = 7 cm.

The radius is 7 cm.

Q. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Solution: Formula: Area = πr² × (θ/360°).

Area = (22/7) × 7² × (60/360)
= (22/7) × 49 × (1/6)
= 154 / 6 ≈ 25.67 cm²

The area of the sector is 25.67 cm².

End-of-Chapter Exercises

4. The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.

Solution: Area = 1/2 × base × height.
54 = 1/2 × 12 × h ⇒ 54 = 6h ⇒ h = 9 cm.
The two legs are 12 cm and 9 cm. Using the Pythagoras theorem for the hypotenuse:
Hypotenuse = √(12² + 9²) = √(144 + 81) = √225 = 15 cm.
Perimeter = 12 + 9 + 15 = 36 cm.

5. The sides of a triangle are in the ratio 2:3:4, and its perimeter is 45 cm. Find its area.

Solution: Let sides be 2x, 3x, 4x.
Perimeter = 2x + 3x + 4x = 9x = 45 ⇒ x = 5.
The sides are 10 cm, 15 cm, 20 cm.
Semi-perimeter s = 45 / 2 = 22.5 cm.
Using Heron's Formula: Area = √[s(s-a)(s-b)(s-c)]

Area = √[22.5 × (22.5 - 10) × (22.5 - 15) × (22.5 - 20)]
Area = √[22.5 × 12.5 × 7.5 × 2.5]
Area = √5273.4375 ≈ 72.62 sq. cm.

The area is approximately 72.62 cm².

Chapter 7 - The Mathematics of Maybe: Introduction to Probability

Exercise Set 7.2 & 7.3

Q. What is the probability of getting an even number when rolling a fair 6-sided die?

Solution: The sample space is {1, 2, 3, 4, 5, 6}.
The favourable outcomes (even numbers) are {2, 4, 6}.
Total possible outcomes = 6; Total favourable outcomes = 3.
P(Even Number) = 3 / 6 = 1/2 or 50%.

End-of-Chapter Exercises

1. Fill in the blanks:
(i) The probability of an impossible event is _______.
(ii) The set of all possible outcomes of a random experiment is called the _________.
(iii) The probability of an event that is certain to happen is _______.
(iv) Tossing a fair coin has a probability of ______ for getting heads.

Solution:
(i) 0
(ii) Sample Space
(iii) 1
(iv) 1/2 (or 50%),

4. Write the sample space and calculate the probability:
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
(ii) A die is rolled once. What is the probability of getting a number greater than 4?

Solution:
(i) Sample space S = {HH, HT, TH, TT}. Outcomes with at least one head = {HH, HT, TH} (3 outcomes). Probability = 3/4.
(ii) Sample space S = {1, 2, 3, 4, 5, 6}. Numbers greater than 4 = {5, 6} (2 outcomes). Probability = 2 / 6 = 1/3.

Chapter 8 - Predicting What Comes Next: Exploring Sequences

Exercise Set 8.1 & 8.2 (Arithmetic Progressions)

1. Find the first five terms of the sequence in which the nth term is given by t_n = 3n − 4.

Solution: Substitute n = 1, 2, 3, 4, 5:

• t₁ = 3(1) - 4 = -1
• t₂ = 3(2) - 4 = 2
• t₃ = 3(3) - 4 = 5
• t₄ = 3(4) - 4 = 8
• t₅ = 3(5) - 4 = 11

The terms are -1, 2, 5, 8, 11.

Q. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …

Solution: The first term a = 3, and the common difference d = 8 - 3 = 5.
Formula: t_n = a + (n - 1)d

t₁₀ = 3 + (10 - 1)5 = 3 + 9(5) = 3 + 45 = 48
t₂₆ = 3 + (26 - 1)5 = 3 + 25(5) = 3 + 125 = 128

The 10th term is 48 and the 26th term is 128.

Q. Which term of the AP: 21, 18, 15, … is −81? Also, is 0 a term of this AP?

Solution: Here a = 21, d = 18 - 21 = -3.
To find the term for -81: t_n = a + (n - 1)d

-81 = 21 + (n - 1)(-3)
-102 = -3(n - 1) ⇒ n - 1 = 34 ⇒ n = 35

-81 is the 35th term.

To check if 0 is a term:

0 = 21 + (n - 1)(-3)
-21 = -3(n - 1) ⇒ n - 1 = 7 ⇒ n = 8

Since 8 is an integer, Yes, 0 is the 8th term.

Exercise Set 8.3 (Geometric Progressions)

1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

Solution: We know t_n = a × r^n-1 and r = 2.
We are given t₈ = a × r⁷ = 192.
To find the 12th term, we can multiply the 8th term by the common ratio 4 times:

t₁₂ = t₈ × r⁴
t₁₂ = 192 × 2⁴ = 192 × 16 = 3072

The 12th term is 3072.

2. Find the 10th and nth terms of the GP: 5, 25, 125, …

Solution: Here, the first term a = 5, and the common ratio r = 25 / 5 = 5.
The nth term formula is t_n = a × r^n-1.

t_n = 5 × 5^n-1 = 5ⁿ
t₁₀ = 5 × 5^10-1 = 5¹⁰

The 10th term is 5¹⁰ and the nth term is 5ⁿ.