In the following figure, the square ABCD is divided into five equal parts,

Solution:

We have a square ABCD.

We have,

$\mathrm{AB}=22 \mathrm{~cm}$

(i)We have to find the perimeter of the triangle. We have a relation as,

Area of circular region $=\frac{1}{5}($ Area of $\mathrm{ABCD})$

So,

$\pi r^{2}=\frac{1}{5}(22)^{2}$

$r=\frac{22}{\sqrt{5 \pi}}$

$=5.56$

So perimeter of the circular region,

$=2 \pi r$

$=(2) \frac{22}{7}\left(\frac{22}{\sqrt{5 \pi}}\right)$

$=34.88 \mathrm{~cm}$

(ii)We have to find the perimeter of ABEF. Let O be the centre of the circular region.

Use Pythagoras theorem to get,

$2(\mathrm{AE}+r)^{2}=22^{2}$

$\mathrm{AE}+r=15.56$

$\mathrm{AE}=(15.56-5.56) \mathrm{cm}$

$=10 \mathrm{c}$,

Similarly,

$\mathrm{BF}=10 \mathrm{~cm}$

Now length of arc EF,

$=\frac{\text { Perimeter of circular region }}{4}$

$=\frac{34.88}{4} \mathrm{~cm}$

$=8.64 \mathrm{~cm}$

So, perimeter of ABFE,

$=\mathrm{AB}+\mathrm{BF}+\mathrm{EF}+\mathrm{AE}$

$=(22+10+8.64+10) \mathrm{cm}$

$=50.64 \mathrm{~cm}$