In Fig. 12.33, ABC is a quadrant of a circle of radius `14` cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Explanation

To find the area of semi-circle, we need to find the radius or diameter `(BC)` of the semicircle.
`triangleABC` is a right-angled triangle, right-angled at A (ABC being a quadrant).
`AB = AC = 14 cm` [Radius of the circle]

Using Pythagoras theorem, we can find the hypotenuse `(BC)` of `triangleABC.`

`BC^2 = AB^2 + AC^2`

= `(14 cm)^2 + (14 cm)^2`
`BC` = `(sqrt2) × (14 cm)^2`
= `14sqrt2 cm`
Radius of semicircle `BDC`, `r = (BC)/2 = (14sqrt2)/(2) cm = 7sqrt2 cm`
Area of the shaded region = Area of semicircle - (Area of quadrant ABC - Area `triangleABC`)
= `(pir^2)/2 - [(90^@)/(360^@) × pi(14)^2 - 1/2 xx AC xx AB]`

= `(pi(7sqrt2))^2/2 - [(pi(14))^2/4 - 1/2 xx 14 xx 14]`
= `[(22 xx 7 xx 7 xx 2)/(7 xx 2)] - [(22 xx 14 xx 14)/(7 xx 4) - 7 xx 14]`

= `154 - (154 - 98)`
= `98 cm^2`