Question

Taking vertices of the square as centres four equal circles are drawn in such a way that each circle touches two of the others. If the area of shaded portion enclosed between the circles is 85.7 ${cm}^2$ then, find the radius of the circles. (Take $\pi =\frac{22}{7}$)

• A. 7 cm
• B. 8 cm
• C. 12 cm
• D. 10 cm

Answer 1

The correct option is D 10 cm

Let A, B, C and D be the four corners of the square and P,Q,R,S be the points where circles are touching in pairs.

Let us take the radius of each circle as ‘$r$’ cm

Area of sector = $\frac{x^{\circ }}{{360}^{\circ }}\pi r^2$

$\because$ The angle subtended by each sector (APS, PBQ, QCR and RDS) at their centres = ${90}^{\circ }$

$\therefore$ Area of sector APS = $\frac{{90}^{\circ }}{{360}^{\circ }}$$\times$$\frac{22}{7}$$\times$ $r^2$

$\because$ The radius and angle subtended at the centre by sectors APS, PBQ, QCR and RDS are equal so their areas will be same.

$\therefore$ Area of sector APS = Area of sector PBQ = Area of sector QCR= Area of sector RDS = $\frac{{90}^{\circ }}{{360}^{\circ }}$$\times$$\frac{22}{7}$$\times$ $r^2$

= $\frac{1}{4}$$\times$$\frac{22}{7}$$\times$ $r^2$

Area enclosed between the circles
= [Area of square] – [Sum of Areas of sectors APS, PBQ, QCR and RDS]

Area enclosed between the circles
= ${\left(2r\right)}^2$ – 4$\times$$\frac{1}{4}$$\times$$\frac{22}{7}$$\times$ $r^2$

=0.857$r^2$ ${cm}^2$ ------------(1)

Given area enclosed between the circles = 85.7 ${cm}^2$ ----------(2)

On equating (1) and (2), we get
85.7 = 0.857$r^2$

$\Rightarrow$ $r^2$ = $\frac{85.7}{0.857}$

$\Rightarrow$ $r^2$ = 100

$\Rightarrow$ $r$ = 10 cm