# CBSE Class 10 Math Question Paper 2016 | Q11

###### Areas Related to Circles | 3 Mark Question | Area of semicircle and right triangle

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).

## NCERT Solution to Class 10 Maths

### Explanatory Answer | CBSE 2016 Maths Paper Q11

AB is a diameter to the circle.

Property
Angle in a semicircle is a right angle.

∴ ∠ACB = 90°
∴ Δ ACB is a right triangle.

##### Step 1: Compute length of side BC

By Pythagoras theorem, AB2 = AC2 + BC2
AB = 13 cm, AC = 12 cm
132 = 122 + BC2
Or BC2 = 169 - 144 = 25
∴ BC = 5 cm

##### Step 2: Compute Area of Shaded Region

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

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