CAT 2017 Slot 1 QA Question & Solution
Question
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
Options
Solution
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