Question

In the given figure, circles of radius 3 cm each are drawn by taking A,B,C and D as centres. Find the area of the shaded region. ​

Answer 1

PRE-REQUISITE KNOWLEDGE??

•Sum of internal angles of a quadrilateral=360°

•Area of Circle =πr² ;r->radius of circle

•Area of Sector of a circle=

$( \frac{x}{360} )\pi {r}^{2}$

x->angle subtended inside the sector

r->radius of sector (/circle)

°spark_plug??

area of circle with radius r =

$( \frac{360}{360} )\pi {r}^{2}$

SOLUTION

Let a,b,c,d be the interior angles of the Quadrilateral ABCD ; at the vertices A,B,C,D. resp.

Hence; a+b+c+d=360° (refer pre-requisite knowledge)

Also; Given,

the radius of each sector is 3cm.

Area of shaded region= Area of Sector at A +Area of Sector at B + Area of Sector at C + Area of Sector at D

=$( \frac{a}{360} )\pi {3}^{2} + ( \frac{b}{360} )\pi {3}^{2} \: \\ + ( \frac{c}{360} )\pi {3}^{2} \: + ( \frac{d}{360} )\pi {3}^{2}$

=

$\pi {3}^{2} ( \frac{a}{360} + \frac{b}{360} +\frac{c}{360} +\frac{d}{360})$

=

$\pi {3}^{2} ( \frac{ a+ b+ c+ d }{360} )$

=

$9\pi( \frac{360}{360} )$

=Area of circle with radius 3cm ...°

=

$9\pi {cm}^{2}$

=9(3.14) cm²

=28.26 cm²

Hope you Understood The geometry behind the question...

Regards;

Leukonov/OLegion.