Free Areas Related to Circles 01 Practice Test - 10th Grade 

Area of a square = $sid{e}^2$ 
Side of the square = 50 cm 
Area of square Handkerchief = ${50}^2$ = 2500 $c{m}^2$
Area of circles = $\pi \times r^2$
Radius of the circle = 7 cm

Total area of circles = 7 x $\frac{22}{7}$ x $7^2$ = 1078$c{m}^2$

Remaining area = 2500 - 1078 = 1422 $c{m}^2$ 

The circumference i.e , perimeter of a sector of angle $P^{\circ }$ of a circle with radius R is given by

$\frac{P^{\circ }}{360}\times 2\pi r+2R$

$=\frac{P^{\circ }}{360}\times 2\pi R+2R$

$=\frac{{60}^{\circ }}{360}\times 2\pi \left(20\right)+2\left(20\right)$

= 20.93 + 40

= 60.93 cm 

Given that radius r = 6 cm
$\theta$ = ${70}^{\circ }$
Area of sector of an angle $\theta$ with radius r 
=$\frac{\theta }{360}$ $\times$  $\pi$ $\times$ $(r{)}^2$
= $\frac{70}{360}$ ×  $\frac{22}{7}$  ×  $6^2$ 
= 22 $c{m}^2$

Area of a sector with angle $\theta$ is given by =$\frac{\theta }{360°}$ $\times$ $\pi$ $\times$ $r^2$ ( where $\theta$ is the angle made by the sector )
Required area = Area of sector OABO – Area of sector OCDO

The angle formed by the sector OABO and sector OCDO = 90°  
$=\frac{90}{360}\times \frac{22}{7}\times (21{)}^2–\frac{90}{360}\times \frac{22}{7}\times (14{)}^2$
$=\frac{1}{4}\times \frac{22}{7}\times \left\{\right(21{)}^2–\left(14{)}^2\right\}$
$=\frac{1}{4}\times \frac{22}{7}\times \left\{\right(35\left)\right(7\left)\right\}$  $[a^2-b^2=(a+b\left)\right(a-b\left)\right]$
=192.50 $m^2$

A = $\pi r^2$
$\text{Given r =}\frac{7}{\sqrt{\pi }}cm$

$A=\pi \times \frac{7}{\sqrt{\pi }}\times \frac{7}{\sqrt{\pi }}$

   $=49c{m}^2$

Here we are asked to find the area of the shaded area. The area of the shaded area should be the difference between the area of the square and the area of the circle.

Here we are given the side of the square, x = 14 cm
The   area of the square is = $sid{e}^2$ = $x^2$ = ${14}^2$ 

= 196$c{m}^2$ 

Therefore the radius of the circle will be, r =$\frac{d}{2}$

r = $\frac{14}{2}$ = 7 cm

Area of the circle,

A=$\pi \times r^2$                      (use π = 22/7)

= $\left(\frac{22}{7}\right)\times 7^2$   = 154 $c{m}^2$        

Area of the shaded portion = Area of the square - Area of the circle

= 196 - 154 =42 $c{m}^2$ 

Therefore the area of the shaded portion will be 42 $c{m}^2$ 

Area of a circle is directly proportional to the square of the radius of circle.

Radius = 42 cm

8 ribs implies angle subtend between consecutive ribs = $\frac{360}{8}$ = ${45}^{\circ }$ .

Area between consecutive ribs = $\frac{\theta }{360}\times \pi \times (radius{)}^2$
$\frac{45}{360}$ × $\frac{22}{7}$ ×  ${42}^2$ = 693 $c{m}^2$

For the quadrant $\theta =90°$
Area of the quadrant  = $\frac{\theta }{360°}\times \pi (r{)}^2$
Area of quadrant OACB =  $\frac{90}{360}$ $\times$  $\frac{22}{7}$  $(7{)}^2$ =  $\frac{77}{2}$

Area of shaded region = Area of quadrant OACB - area of $\Delta$OAD

                            = $\frac{77}{2}$ - $\frac{1}{2}$ × 7 × 4

                             = 24.5 $c{m}^2$