The given question is a medium difficulty 3-mark question from the chapter Areas Related to Circles. Computing area of a semi circle and area of a right triangle to find the area of a shaded region. It is an important question and appeared in Section C of the 2016 CBSE class 10 board paper.

Question 11: O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14).

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AB is a diameter to the circle.

Property

Angle in a semicircle is a right angle.

∴ ∠ACB = 90°

∴ Δ ACB is a right triangle.

By Pythagoras theorem, AB^{2} = AC^{2} + BC^{2}

AB = 13 cm, AC = 12 cm

13^{2} = 12^{2} + BC^{2}

Or BC^{2} = 169 - 144 = 25

∴ BC = 5 cm

Area of shaded region = Area of semicircle of radius \\frac{13}{2})cm - Area of right triangle ACB

Sides of right triangle ACB are 5, 12, and 13.

∴ Area of shaded region = \\frac{1}{2}) × (3.14) × \(\frac{13}{2}{)}^{2}) - \\frac{1}{2}) × 12 × 5

= \\frac{1}{2}) × 3.14 × \\frac{169}{4}) - 30

= \\frac{3.14 × 169-240}{8}) = 36.3325 sq cm

Class 10 Maths

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CBSE Class 10 Maths - 2022

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