From figure, area equal to one full circle exists in the semi-circle.

Area of larger semi-circle = $\frac{3.14\times 2^2}{2}$ = 6.28 $c{m}^2$

Area of smaller semi-circles = 2 x $\frac{3.14\times 1^2}{2}$ = 3.14 $c{m}^2$

Area of shaded portion = 3.14 $c{m}^2$

Area of quadrant = $\frac{90}{360}\times \frac{22}{7}\times 14\times 14$  = 154 $c{m}^2$

Area of $△ABC$ = 0.5 x ${14}^2$ = 98 $c{m}^2$

The diameter = 14 × $\sqrt{2}$ = 19.8 cm.  

Area of semi-circle = 0.5 $3.14\times {9.9}^2$ = 153.95 $c{m}^2$

The area of segment = 154 - 98 = 56  $c{m}^2$

The required area = 153.95 - 56 = 97.95$c{m}^2$.

From the figure, two circles of radius 4 cm exist in the rectangle.

Their total area = 2$\pi r^2$ = 2 $\times$ 3.14  x 16 = 100.48 $c{m}^2$

From figure
Length of rectangle is 4r = 16 cm
Breadth of rectangle is 2r = 8 cm
Area of rectangle = 16 x 8 = 128 $c{m}^2$

Required area = [Area of rectangle -  Total area of circles]
= 128 - 100.48 = 27.52 $c{m}^2$.
$\therefore$ The area of the shaded region
 = 27.52 $c{m}^2$.

Diameter = 14 cm

Perimeter of semi circle = $\pi$ $\times$ r = $\frac{22}{7}$ x 7 = 21.99 cm

Total perimeter of protractor is = 21.99 + 14 = 35.99 cm ~ 36 cm.

In the figure shown, the segment CFDC which is bigger is called the major segment and the segment CEDC which is smaller is called the minor segment.

Perimeter of square and circle is same.

88 = 2 $\pi$ $\times$ r
     =  2 $\frac{22}{7}\times r$

r = 14 cm.

Area of track = Area of Field - Area inside the track
The area of the track =106 x 80 —106 x 60 + $2\times \frac{1}{2}\times \frac{22}{7}\times ({40}^2-{30}^2)$ 
=106 x 20 + $\frac{22}{7}$ $\times$  (70) $\times$ (10)
=2120 +700 x $\frac{22}{7}$ = 2120 + 2200
= 4320 $m^2$

As the wiper sweeps, area covered by each wiper is indicated in brown color below.  

Area cleaned by each wiper =  Area of sector AOB – Area of sector COD   

Area of the sector of angle  Ɵ =  $\left(\frac{\theta }{{360}^{\circ }}\right)\times \pi \times r^2$

Area of sector AOB - Area of sector COD =  $\frac{120}{360}\times \frac{22}{7}\times {35}^2-\left(\frac{120}{360}\right)\times \frac{22}{7}\times \left(7^2\right)$  = 1232 $c{m}^2$

So, the total area cleaned by both the wipers = 2 × 1232 cm² = 2464 $c{m}^2$

Given: radius for sector OAC = 14 $cm$ and angle subtended =  ${40}^{\circ }$ and 
radius for sector OBD = 7 $cm$ and angle subtended =  ${40}^{\circ }$
Area of Sector = $\frac{x^{\circ }}{{360}^{\circ }}\times \pi r^2$
Required area = [Area of sector OAC – Area of sector OBD]
=$\frac{40}{360}\times \frac{22}{7}\times {14}^2–\frac{40}{360}\times \frac{22}{7}\times 7^2$
= 68.42 - 17.1
= $51.32c{m}^2$
$\therefore$ Area of shaded region = $51.32c{m}^2$

Area of rectangle = $22\times 14=308c{m}^2$
Area of semicircle
= $\frac{1}{2}\times \frac{22}{7}\times (7{)}^2=77c{m}^2$
Required area = [Area of rectangle - Area of semicircle]
= 308 – 77 = $231c{m}^2$