Table of Contents
Exercise 9.3 Solutions
Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
In a circle of radius 10 cm, a chord subtends a right angle at the centre. Find the area of the corresponding: (use π = 3.14)
i. Minor segment
ii. Major segment
Solution:
Given: Radius (r) = 10 cm, Central angle (θ) = 90°
i. Minor segment area:
Area of sector = \(\frac{θ}{360} × πr^2 = \frac{90}{360} × 3.14 × 10^2 = 78.5 \, \text{cm}^2\)
Area of triangle = \(\frac{1}{2} × r^2 × \sinθ = \frac{1}{2} × 100 × 1 = 50 \, \text{cm}^2\)
Minor segment area = Sector area – Triangle area = 78.5 – 50 = 28.5 cm²
ii. Major segment area:
Total circle area = πr² = 3.14 × 100 = 314 cm²
Major segment area = Total area – Minor segment = 314 – 28.5 = 285.5 cm²
Problem 2
In a circle of radius 12 cm, a chord subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle (use π = 3.14 and √3 = 1.732)
Solution:
Given: Radius (r) = 12 cm, Central angle (θ) = 120°
Area of sector = \(\frac{120}{360} × 3.14 × 12^2 = 150.72 \, \text{cm}^2\)
Area of triangle = \(\frac{1}{2} × 12^2 × \sin120° = 72 × \frac{\sqrt{3}}{2} = 62.352 \, \text{cm}^2\)
Minor segment area = Sector area – Triangle area = 150.72 – 62.352 = 88.368 cm²
≈ 88.37 cm² (rounded to two decimal places)
Problem 3
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (use π = 22/7)
Solution:
For one wiper:
Area swept = \(\frac{115}{360} × \frac{22}{7} × 25^2 ≈ 627.48 \, \text{cm}^2\)
For two wipers (non-overlapping):
Total area = 2 × 627.48 = 1254.96 cm² ≈ 1255 cm²
Problem 4
Find the area of the shaded region in the adjacent figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter (use π = 3.14)
Solution:
Area of square = 10 × 10 = 100 cm²
Area of four semicircles = Area of two full circles (radius = 5 cm)
= 2 × π × 5² = 157 cm²
Shaded area = Square area – Semicircles area = 100 – 157 = -57 cm² (This suggests the shaded area is actually the intersection areas)
Correct interpretation: Shaded area is the area of square not covered by semicircles
Actual shaded area = Square area – (4 semicircles – overlapping lens areas)
More precise calculation needed based on exact figure description
Problem 5
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm, and APD and BPC are semicircles. (use π = 22/7)
Solution:
Area of square = 7 × 7 = 49 cm²
Area of two semicircles (radius = 3.5 cm) = Area of one full circle
= π × (3.5)² = 38.5 cm²
Shaded area = Square area – Semicircles area = 49 – 38.5 = 10.5 cm²
Problem 6
In the figure, OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD=2cm, find the area of the shaded region. (use π = 22/7)
Solution:
Area of quadrant = \(\frac{1}{4} × \frac{22}{7} × (3.5)^2 = 9.625 \, \text{cm}^2\)
Area of ΔOCB = \(\frac{1}{2} × 3.5 × 3.5 = 6.125 \, \text{cm}^2\)
Area between chord CB and radii = 9.625 – 6.125 = 3.5 cm²
Area of ΔODB = \(\frac{1}{2} × 2 × 3.5 = 3.5 \, \text{cm}^2\)
Shaded area = 3.5 – 3.5 = 0 cm² (This suggests the shaded area description needs clarification)
Alternative interpretation: Shaded area might be the difference between sector area and triangle area
Problem 7
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm with centre O. If ∠AOB = 30°, find the area of the shaded region. (use π = 22/7)
Solution:
Area of sector OAB = \(\frac{30}{360} × \frac{22}{7} × 21^2 = 115.5 \, \text{cm}^2\)
Area of sector OCD = \(\frac{30}{360} × \frac{22}{7} × 7^2 = 12.833 \, \text{cm}^2\)
Shaded area = 115.5 – 12.833 ≈ 102.67 cm²
Problem 8
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each. (use π = 3.14)
Solution:
Area of one quadrant = \(\frac{1}{4} × 3.14 × 10^2 = 78.5 \, \text{cm}^2\)
Area of square formed by radii = 10 × 10 = 100 cm²
Area of common region = 2 × quadrant area – square area
= 2 × 78.5 – 100 = 57 cm²
