Question

hey friends
Good noon
Plz answer this question
☺️☺️☺️

Answer 1

good afternoon.

area of AOCB = πr²¢/360° = 22/7×7×7×90°/360

=> 22×7/4

=> 11×7/2 = 77/2 cm²

Area of this traingle DOC=1/2×4×7

=> 2×7 = 14cm

and area of sheded region = 77/2 -14

=> 77-28/2 = 49/2 = 24.5 cm² Answer

Answer 2

heya dear

here is your answer

Given,

OABC is a quadrant of a circle
and radius of circle = 7cm

1st lets find the area of sector

area of sector = {(pi) r^2 } / 4

( 22/7 × 7^2 ) / 4

(22 × 49) / (7×4 )

1078/28

38.5cm^2

now area of triangle = 1/2 × b × h

here base = 7cm ( radius of circle act as base)

height = 4cm

=> area = 1/2 × 7 × 4

=> 1/2 × 28

=> 14cm^2

therefore, area of remaining portion = area of sector - area of triangle

= 38.5 - 14

=24.5cm^2

i hope it helped you