In given figure ABPC is a quadrant of a circle of radius 14 cm
In given figure ABPC 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.
Answer
We know, AC = r
In Δ ACB,
BC2 = AC2 + AB2
BC = AC√2 (Since, AB = AC)
BC = r√2
Required area = ar(Δ ACB) + ar(semicircle on BC as diameter) – ar(quadrant ABPC)
= (½ × r × r) + (½ × π × (r√2/2)2) – (¼ × π × r2)
= r2/2 + πr2/4 – πr2/4
= r2/2
= 196/2 cm2
= 98 cm2