Question

Question 9
Find the area of the shaded region in figure.

Answer 1

Join GH and FE



Here, Breadth of the rectangle = 12 m

$\therefore$ Breadth of the inner rectangle EFGH = 12 – ( 4 + 4 ) = 4 cm

Which is equal to the diameter of the semi - circle, d = 4m

$\therefore$ Radius of semi – circle, r = 2m

$\therefore$ Length of inner rectangle EFGH = 26 - ( 5 +5 ) = 16 m

$\therefore$ Area of two semi - circles
$=2\left(\frac{\pi r^2}{2}\right)=2\times \pi \frac{(2{)}^2}{2}=4\pi m$

Now, area of inner rectangle EFGH
$=EH\times FG=16\times 4=64m^2$

And area of outer rectangle ABCD
$=26\times 12=312m^2$

$\therefore$ Area of shaded region
= Area of outer rectangle - ( Area of two semi - circles + Area of inner rectangle)

$=312-(64+4\pi )=(248-4\pi )m^2$