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Area of Sector and Triangle in Circle

CAT 2019 Slot 1 · Quantitative Ability · Easy · Geometry

This is an easy Quantitative Ability question from the CAT 2019 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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In a circle with centre O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC=ODOC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

  1. A.

    (π4)12(\frac{\pi}{4})^{\frac{1}{2}}

  2. B.

    (π6)12(\frac{\pi}{6})^{\frac{1}{2}}

  3. C.

    (π43)12(\frac{\pi}{4\sqrt{3}})^{\frac{1}{2}}

  4. D.

    (π33)12(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}

Answer

D

Explanation

Area of region R (sector of 60°) = 60360⋅π(1)2=π6\frac{60}{360} \cdot \pi (1)^2 = \frac{\pi}{6}. Area of triangle OCD = 12⋅OC⋅OD⋅sin⁡(60∘)=12⋅x2⋅32=x234\frac{1}{2} \cdot OC \cdot OD \cdot \sin(60^\circ) = \frac{1}{2} \cdot x^2 \cdot \frac{\sqrt{3}}{2} = \frac{x^2 \sqrt{3}}{4}. Given Area(OCD) = 12Area(R)\frac{1}{2} \text{Area}(R): x234=12⋅π6=π12\frac{x^2 \sqrt{3}}{4} = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12} x2=4π123=π33x^2 = \frac{4\pi}{12\sqrt{3}} = \frac{\pi}{3\sqrt{3}} x=(π33)12x = \left(\frac{\pi}{3\sqrt{3}}\right)^{\frac{1}{2}}.

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