Q19 of 52 Page 10

in fig. 4, O is the center of a circle such that diameter AB = 13 am and AC = 12 cm. BC is joined. Find the area of the shaded region. (Taken π = 3.14) (CBSE 2016)


∠ACB = 90°

As angle in a semicircle is a right angle.

So △ABC is a right angled triangle

By Phythagoras theorem, which states, In a right angled triangle

(hypotenuse)2 = (base)2 + (perpendicular)2

In △ABC

AB2 = BC2 + CA2

132 = BC2 + 122

169 = BC2 + 144

BC2 = 25

BC = 5 cm

And we know,

= 30 cm2

= 66.3325 cm2

Area of shaded region = area of semicircle - Area of △ABC

= 66.3325 - 30

= 36.3325 cm2

More from this chapter

All 52 →
17

In Fig. 10.69, AB is a diameter of a circle with centre O and AT is a tangent. If , find .(CBSE 2015)

 18

In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with center O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°. (CBSE 2016)

 20

In Fig. 2, a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E, and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF. (CBSE 2016)

 21

Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. (CBSE 2017)