Question

In figure 3, APB and CQD are semi-circles of diameter 7 cm each, while ARC and BSD are semi - circles of diameter 14 cm each. Find the perimeter of the shaded region. $[Use\pi =\frac{22}{7}]$

OR
Find the area of a quadrant of a circle, where the circumference of circle is 44 cm.$[Use\pi =\frac{22}{7}]$

Answer 1

Perimeter of the shaded region = Length of APB + Length of ARC + Length CQD + Length of DSB
Now, Perimeter of APB $=\frac{1}{2}\times 2\pi \left(\frac{7}{2}\right)cm=\frac{22}{7}\times \frac{7}{2}cm=11cm$
Perimeter of ARC $=\frac{1}{2}\times 2\pi \left(7cm\right)=\frac{22}{7}\times 7cm=22cm$
Perimeter of CQD $=\frac{1}{2}\times 2\pi \left(\frac{7}{2}cm\right)cm=\frac{22}{7}\times \frac{7}{2}cm=11cm$
perimeter of DSB $=\frac{1}{2}\times 2\pi \left(7cm\left(\frac{7}{2}\right)cm\right)=\frac{22}{7}\times 7cm=22cm$
Thus, perimeter of the shaded region = 11 cm + 22 cm + 11 cm = 66 cm
OR
Let the radius of the circle be r.
It is given that perimeter of the circle is 44 cm.
$\therefore 2\pi r=44cm$
$\Rightarrow 2\times \frac{22}{7}\times r=44cm$
$\Rightarrow$ r = 7 cm
Area of a quadrant of a circle
$=\frac{1}{4}\times \pi r^2=\frac{1}{4}\times \frac{22}{7}\times (7cm{)}^2$
$=\frac{1}{4}\times \frac{22}{7}\times 49c{m}^2=38.5c{m}^2$
Thus, the area of a quadrant of the given circle is $38.5c{m}^2$.