Test: Area of Combination of Figures - Class 10 MCQ

Let side of square be x cms inscribed in a circle.

Radius of circle (r) = 1/2(diagonal of square)

AC=54cm, Radius=AO=54/2=27cm
Area of the bigger circle= πR²= 22/7 * 27 * 27=2291 (approx)
Diameter of the smaller circle=54-10=44
Radius=44/2=22
Area of the smaller circle= πr²=22/7 * 22 * 22=1521 (approx)
Area of the shaded region=Area of the bigger circle-Area of the smaller circle
= 2291-1521=770 cm²

Perimeter is the boundary of the figure
So we have two semicircles of radius 14cm which makes one complete circle. Also there are two semicircles of radius 7 cm which makes one circle.
So perimeter= 2πR+2πr=2(R+r)=2*22/7(14+7)=22*6=132cm

ANSWER :- b

Solution :- Area of rec. park=(120×90)m²=10800m²

Area of rec. park excluding lawn=2950m²

Area of circular lawn=(10800 - 2950)m²

=7850m²

Let the radius be = r

7850/3.14 = 3.14r²/3.14

→2500 = r²

→√2500 = √r²

→50 = r

→r = 50

Radius = 50m

Four circles of radius 5 cm are cut out.
Area cut = 4*(π5²) = 100π
Total area of circular sheet = π(20)² = 400π
Area uncut = 300 π
The ratio of uncut to cut portion is 3 : 1