CAT 2017 [slot 1] Question with solution 17

Question 17:
Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is
  1. $9\pi-18$
  2. 18
  3. $9\pi$
Option:
Explanation:

The image of the figure is as shown.

AB = AC = 6cm. Thus, BC = $\sqrt{6^2 + 6^2}$ = 6?2 cm

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

Area of semicircle BQC

Diameter BC = 6?2cm

Radius = 6?2/2 = 3?2 cm

Area = $\pi r^2 $/2 = $ \pi$ × $(3 \sqrt{2})^2 $/2 = 9$\pi$

Area of quadrant BPC

Area = $\pi r^2$/4 = $\pi×(6)^2$/4 = 9$\pi$

Area of triangle ABC

Area = 1/2 × 6 × 6 = 18

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

= 9$\pi$ - 9$\pi$ + 18 = 18

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