CAT 2018 Slot 2 - QUANT Question 22

Question 22

From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is

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+$$\pi*24/2$$

$$\Rightarrow$$ $$88+12\pi$$


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