Question

. A circle is inscribed in ABC as shown below. Find the area of ABC.

Answer 1

Explanation:

x+y=8cm ..(1)

y+z=12cm ...(2)

x+z=10cm ...(3)

Adding (1), (2) and (3), we get

2(x+y+z)=30

x+y+z=30/2

x+y+z=15 ..(4)

Equation (4)−(2), we get

x+12=15

x=15−12

x=3

Equation (4)−(3), we get

10+y=15

y=15−10

y=5

(4)−(1), we get

8+z=15

z=15−8

z=7

So,

AD=3 cm

BE=5 cm

CF=7 cm Answer.

Answer 2

Answer:

a2 =b2+ c2 – 2bc cos A

72 = 102 + 82 - (2 x 10 x 8) cos A

49=164- 60 cos A

-116 =160 cos A

Cos A =116/160

Cos A = 0.725

Cos1 0.725 = 43.53115

<BAC = 43.53

a/(sin sin A ) = 2R

7/(sin 43.53) = 2R

R =5.082cm

Sin 43.53 = 3.5/r

r = 3.5/(sin sin 43.53 )= 5.082 cm

Area of △OCB =1/2 ab sin θ

= 1/2 x 5.082 sin 87.06

= 12.896cm2

Area of sector ACB

=θ/360 πr2

=87.06/360 x 22/7 x 5.082 = 19.630

Shaded region

(19.630 — 12.896) = 6.734 cm2