Question

In Fig. 5.33 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.​

Answer 1

Answer:

Heya..... kindly attach the figure mate.

Answer 2

$\huge\overbrace{\underbrace\pink{Answer}}$

Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC2 = AB2 +AC2

⇒ BC2 = 142 +142

⇒ BC = 14√2 cm

Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC =( ½)×14×14 = 98 cm2

Area of quadrant = (¼)×(22/7)×(14×14) = 154 cm2

Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm2

Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant

= 154 +98-154 cm2 = 98cm2