Question

In given figure, the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22 cm, find the perimeter of the part ABEF.

Answer 1

Let the radius of the central part be $r\text{cm}$. Then,

Area of central part $=\frac{1}{5}\times$ Area of the square

$\Rightarrow \frac{22}{7}\times r^2=\frac{1}{5}\times 22\times 22$

$\Rightarrow r^2=\frac{22\times 7}{5}=\frac{154}{5}$

$\Rightarrow r=5.549\simeq 5.55\text{cm}$

Let $O$ be the centre of the central part. Clearly, $O$ is the centre of the square.
We have,

$AE=BF=OA-OE$

$=11\sqrt{2}-5.55$

$=15.51-5.55$

$=9.96\text{cm}$

$EF=\frac{1}{4}$ (Circumference of the circle) $=\frac{2\pi r}{4}$

$=\frac{\pi r}{2}$

$=\frac{1}{2}\times \frac{22}{7}\times 5.55$

$=8.72\text{cm}$

$\therefore$ Perimeter of part $ABEF=AB+AE+EF+BF=22+2\times 9.96+8.72=50.64\text{cm}$