CAT 2018

Slot 2

QA

Question & Solution

Area of the semicircle with AB as a diameter = $\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$

$\Rightarrow$ $\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$ = $72*\pi$

$\Rightarrow$ $AB = 24 cm$

Given that area of the rectangle ABCD = 768 sq.cm

$\Rightarrow$ AB*BC = 768

$\Rightarrow$ BC = 32 cm

We can see that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

$\Rightarrow$ 32+24+32+$\pi*24/2$

$\Rightarrow$ $88+12\pi$