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class 10th Mathematics Chapter 9 Test 2 | Tangents and Angles of a Circle | Exercise 9.2 & Review Exercise | Punjab Board 2027
class 10th Math Chapter 9 exercise 9.2
Class 10 Mathematics Test 2 based on Tangents and Angles of a Circle is prepared for students following the Punjab Board 2027 Model Papers. This test covers Exercise No. 9.2 and the Review Exercise and includes important objective and subjective questions for Mathematics exam preparation. Students can use this Class 10 Mathematics Unit 9 Test to revise key concepts, practice questions, and improve their preparation for the Punjab Boards 2027 Mathematics examination.
Exercise 9.2 & Review Exercise
T-2
Unit 9: Tangents and Angles of a circle
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer.
(1 × 5 = 5)
(i)
Length ℓ of an arc which subtends an angle of
θ radius of a circle of radius
r is calculated by the formula
ℓ =
SOLUTION
Step 1: The radian measure of an angle at the centre is θ = ℓ / r (arc length divided by radius).
Step 2: Multiply both sides by r: ℓ = rθ (θ in radians).
Answer: (b) rθ
Step 1: The radian measure of an angle at the centre is θ = ℓ / r (arc length divided by radius).
Step 2: Multiply both sides by r: ℓ = rθ (θ in radians).
Answer: (b) rθ
(ii)
Area A of a sector having angle
θ radius of a circle of radius
r is calculated by the formula
A =
SOLUTION
Step 1: The sector is the fraction θ/2π of the whole circle (area πr²).
Step 2: A = (θ / 2π) × πr² = ½ r²θ (θ in radians).
Answer: (c) ½r²θ
Step 1: The sector is the fraction θ/2π of the whole circle (area πr²).
Step 2: A = (θ / 2π) × πr² = ½ r²θ (θ in radians).
Answer: (c) ½r²θ
(iii)
If r = 4 and
ℓ = 14 then angle subtends by the arc
in radian measure is θ =
SOLUTION
Step 1: θ = ℓ / r.
Step 2: θ = 14 / 4 = 3.5 radians.
Answer: (a) 3.5
Step 1: θ = ℓ / r.
Step 2: θ = 14 / 4 = 3.5 radians.
Answer: (a) 3.5
(iv)
If r = 6 and
θ = 60° then length of the arc is
ℓ =
SOLUTION
Step 1: Convert to radians: 60° = 60 × π/180 = π/3.
Step 2: ℓ = rθ = 6 × π/3 = 2π.
Answer: (b) 2π
Step 1: Convert to radians: 60° = 60 × π/180 = π/3.
Step 2: ℓ = rθ = 6 × π/3 = 2π.
Answer: (b) 2π
(v)
Angle inscribed in a semi-circle is:
SOLUTION
Step 1: A semicircle subtends a central angle of 180° on its diameter.
Step 2: An inscribed angle is half the central angle on the same arc: ½ × 180° = 90°.
Answer: (b) 90°
Step 1: A semicircle subtends a central angle of 180° on its diameter.
Step 2: An inscribed angle is half the central angle on the same arc: ½ × 180° = 90°.
Answer: (b) 90°
Exercise 9.2 & Review Exercise
T-2
Unit 9: Tangents and Angles of a circle
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions:
(6 × 2 = 12)
(i)
A projector casts an image over an arc on a circular wall.
Two observers on the arc segment report one observing a 50° angle.
What is the other angle in the same segment?
SOLUTION
Step 1: Both observers stand on the same arc and look at the two ends of the same chord (the image).
Step 2: Angles in the same segment of a circle are equal.
The other angle = 50°
Step 1: Both observers stand on the same arc and look at the two ends of the same chord (the image).
Step 2: Angles in the same segment of a circle are equal.
The other angle = 50°
(ii)
A cable in a semicircle bridge forms a triangle with the diameter
of the arch. If the cable meets the endpoints of the diameter and
the top of the arch, what is the angle at the top?
SOLUTION
Step 1: The top of the arch lies on the semicircle, and the diameter is the chord joining its end points.
Step 2: The angle in a semicircle is a right angle (central angle 180°, inscribed angle = ½ × 180°).
Angle at the top = 90°
Step 1: The top of the arch lies on the semicircle, and the diameter is the chord joining its end points.
Step 2: The angle in a semicircle is a right angle (central angle 180°, inscribed angle = ½ × 180°).
Angle at the top = 90°
(iii)
Prove that area of a sector is
½r²θ, where
θ is in radians.
PROOF
Step 1: Let a sector of a circle of radius r have central angle θ radians. The whole circle has angle 2π radians and area πr².
Step 2: The area of a sector is proportional to its central angle: (Area of sector) / (Area of circle) = θ / 2π.
Step 3: Area of sector = (θ / 2π) × πr².
Step 4: Cancel π: Area = (θ / 2) × r² = ½ r²θ. Proved.
A = ½ r²θ (also A = ½ rℓ, since ℓ = rθ)
Step 1: Let a sector of a circle of radius r have central angle θ radians. The whole circle has angle 2π radians and area πr².
Step 2: The area of a sector is proportional to its central angle: (Area of sector) / (Area of circle) = θ / 2π.
Step 3: Area of sector = (θ / 2π) × πr².
Step 4: Cancel π: Area = (θ / 2) × r² = ½ r²θ. Proved.
A = ½ r²θ (also A = ½ rℓ, since ℓ = rθ)
(iv)
In a circular arch over a monument entrance, two spotlight fixtures
are placed such that they each shine from two different points on
the arch to the same chord PQ on the base of the arch. If the angle
formed at one light is 55°, what is the angle at the second light
on the same side of chord PQ?
SOLUTION
Step 1: Both lights are on the arch, on the same side of chord PQ, and both look at the ends of PQ.
Step 2: Angles in the same segment (standing on the same chord) are equal.
The angle at the second light = 55°
Step 1: Both lights are on the arch, on the same side of chord PQ, and both look at the ends of PQ.
Step 2: Angles in the same segment (standing on the same chord) are equal.
The angle at the second light = 55°
(v)
Uzma cut a pizza of radius 14cm into 8 equal slices.
What is the area of one slice (sector)?
SOLUTION
Step 1: Central angle of each slice: θ = 2π / 8 = π/4 radians.
Step 2: Area = ½ r²θ = ½ × 14² × π/4 = ½ × 196 × π/4.
Step 3: = 24.5π.
Step 4: With π = 3.1416: 24.5 × 3.1416 = 76.97.
Check: area of the whole pizza ÷ 8 = π(14)² / 8 = 196π/8 = 24.5π ✓
Area of one slice = 24.5π ≈ 76.97 cm² (≈ 77 cm² with π = 22/7)
Step 1: Central angle of each slice: θ = 2π / 8 = π/4 radians.
Step 2: Area = ½ r²θ = ½ × 14² × π/4 = ½ × 196 × π/4.
Step 3: = 24.5π.
Step 4: With π = 3.1416: 24.5 × 3.1416 = 76.97.
Check: area of the whole pizza ÷ 8 = π(14)² / 8 = 196π/8 = 24.5π ✓
Area of one slice = 24.5π ≈ 76.97 cm² (≈ 77 cm² with π = 22/7)
(vi)
A circular garden has a walking path forming a quadrilateral
inscribed in it. If one angle is 87°, what is the opposite angle?
SOLUTION
Step 1: A quadrilateral inscribed in a circle is a cyclic quadrilateral.
Step 2: Its opposite angles are supplementary (add to 180°).
Step 3: Opposite angle = 180° − 87° = 93°.
Opposite angle = 93°
Step 1: A quadrilateral inscribed in a circle is a cyclic quadrilateral.
Step 2: Its opposite angles are supplementary (add to 180°).
Step 3: Opposite angle = 180° − 87° = 93°.
Opposite angle = 93°
Note: Answer the following Questions.
(4 + 4 = 8)
3.
The perimeter and area of a sector are 14cm and 10cm² respectively.
Find the radius of the circle and the central angle of the sector.
SOLUTION
Let the radius be r and the arc length be ℓ.
Step 1: Perimeter of a sector = 2r + ℓ = 14, so ℓ = 14 − 2r. …(1)
Step 2: Area of a sector = ½ r ℓ = 10. …(2)
Step 3: Put (1) in (2): ½ r (14 − 2r) = 10, i.e. 7r − r² = 10.
Step 4: r² − 7r + 10 = 0, so (r − 2)(r − 5) = 0, giving r = 2 or r = 5.
Case 1: r = 5 cm
ℓ = 14 − 10 = 4 cm, and θ = ℓ / r = 4/5 = 0.8 radian (≈ 45.8°).
Case 2: r = 2 cm
ℓ = 14 − 4 = 10 cm, and θ = ℓ / r = 10/2 = 5 radians (≈ 286.5°).
Check: Case 1: ½ × 5 × 4 = 10 ✓ and 2(5) + 4 = 14 ✓. Case 2: ½ × 2 × 10 = 10 ✓ and 2(2) + 10 = 14 ✓
r = 5 cm, θ = 0.8 rad or r = 2 cm, θ = 5 rad
Let the radius be r and the arc length be ℓ.
Step 1: Perimeter of a sector = 2r + ℓ = 14, so ℓ = 14 − 2r. …(1)
Step 2: Area of a sector = ½ r ℓ = 10. …(2)
Step 3: Put (1) in (2): ½ r (14 − 2r) = 10, i.e. 7r − r² = 10.
Step 4: r² − 7r + 10 = 0, so (r − 2)(r − 5) = 0, giving r = 2 or r = 5.
Case 1: r = 5 cm
ℓ = 14 − 10 = 4 cm, and θ = ℓ / r = 4/5 = 0.8 radian (≈ 45.8°).
Case 2: r = 2 cm
ℓ = 14 − 4 = 10 cm, and θ = ℓ / r = 10/2 = 5 radians (≈ 286.5°).
Check: Case 1: ½ × 5 × 4 = 10 ✓ and 2(5) + 4 = 14 ✓. Case 2: ½ × 2 × 10 = 10 ✓ and 2(2) + 10 = 14 ✓
r = 5 cm, θ = 0.8 rad or r = 2 cm, θ = 5 rad
4.
A ferris wheel has a radius of 12m. If a passenger travels a distance
of 18m along the circumference of the ferris wheel, what is the angle
(in radians) swept by the passenger’s position from the starting point?
SOLUTION
Step 1: Given: radius r = 12 m, arc length ℓ = 18 m.
Step 2: Use ℓ = rθ, so θ = ℓ / r.
Step 3: θ = 18 / 12 = 1.5.
Angle swept = 1.5 radians (≈ 85.9°)
Step 1: Given: radius r = 12 m, arc length ℓ = 18 m.
Step 2: Use ℓ = rθ, so θ = ℓ / r.
Step 3: θ = 18 / 12 = 1.5.
Angle swept = 1.5 radians (≈ 85.9°)
