Question

Four circular cardboard pieces of radii $7\mathrm{cm}$ are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

Answer 1

Step 1: Construction

Given, four circles are placed such that each piece touches the other two pieces.

Join the centers of adjacent circles

We need to find the area shaded in grey.

Step 2: Define the problem

Given, the radii of the four circles $r=7\mathrm{cm}$ each

We can observe that the area of the grey-shaded area is the difference between the area of $\square ABCD$ and the area of the four quarter circles at the vertex of the square.

Thus, area of the grey-shaded region $=$ area of the square $-$ area of the four sectors.

Step 3: Calculate the area of the square

The side of the square is formed by joining the radii of the circles.

Thus side length of the square, $s=r\times 2=7\times 2=14\mathrm{cm}$

Area of the square $A_1=s^2={14}^2=196{\mathrm{cm}}^2$

Step 4: Calculate the area of the four sectors

$ABCD$ is a square. Thus, $\angle A=\angle B=\angle C=\angle D=\frac{\mathrm{\pi }}{2}$.

Thus, the sectors subtended at the four corners have an angle of $\frac{\mathrm{\pi }}{2}$.

We know that the area of a sector$=\frac{1}{2}{r}^2\theta$

where, $r$ and $\theta$ are the radius and angle of the sector.

here, $r=7\mathrm{cm}$ and $\theta =\frac{\mathrm{\pi }}{2}$

Thus, area of one sector$=\frac{1}{2}\times 7^2\times \frac{\mathrm{\pi }}{2}=\frac{49·{\displaystyle \frac{22}{7}}}{4}=38.5{\mathrm{cm}}^2$

Thus, the combined area of all the four sectors, $A_2=4\times 38.5=154{\mathrm{cm}}^2$.

Step 5: Calculate the area of the grey-shaded region

Thus, the area of the grey-shaded region, $A=A_1-A_2$

$$\begin{gathered} =196-154 \\ =42{\mathrm{cm}}^2 \end{gathered}$$

Therefore, the area of the grey-shaded region is $42{\mathrm{cm}}^2$.