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Area Enclosed by Arcs

CAT 2017 Slot 1 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2017 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Let ABCABC be a right-angled isosceles triangle with hypotenuse BCBC. Let BQCBQC be a semi-circle, away from AA, with diameter BCBC. Let BPCBPC be an arc of a circle centered at AA and lying between BCBC and BQCBQC. If ABAB has length 6 cm6\text{ cm} then the area, in sq. cm, of the region enclosed by BPCBPC and BQCBQC is:

  1. A.

    9π−189\pi - 18

  2. B.

    18

  3. C.

    9π9\pi

  4. D.

    9

Answer

B

Explanation

Radius AB=AC=6AB = AC = 6. Hypotenuse BC=62+62=62BC = \sqrt{6^2 + 6^2} = 6\sqrt{2}. Radius of semicircle BQC=R=32BQC = R = 3\sqrt{2}. Area of semicircle BQC=πR22=π(18)2=9πBQC = \frac{\pi R^2}{2} = \frac{\pi (18)}{2} = 9\pi. Area of sector ABPC=14π(62)=9πABPC = \frac{1}{4} \pi (6^2) = 9\pi. Area of segment BPC=Sector(ABPC)−Area(△ABC)=9π−12(6)(6)=9π−18BPC = \text{Sector}(ABPC) - \text{Area}(\triangle ABC) = 9\pi - \frac{1}{2}(6)(6) = 9\pi - 18. Area enclosed by BPCBPC and BQC=Area(BQC)−Area(BPC)=9π−(9π−18)=18BQC = \text{Area}(BQC) - \text{Area}(BPC) = 9\pi - (9\pi - 18) = 18.

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