Area of smaller region formed by chord

CAT 2024 Slot 3 · QA · Medium · Geometry

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A circular plot of land is divided into two regions by a chord of length 10310\sqrt{3} meters such that the chord subtends an angle of 120∘120^\circ at the center. Then, the area, in square meters, of the smaller region is

  1. A.

    25(4π3+3)25\left(\frac{4\pi}{3} + \sqrt{3}\right)

  2. B.

    20(4π3−3)20\left(\frac{4\pi}{3} - \sqrt{3}\right)

  3. C.

    25(4π3−3)25\left(\frac{4\pi}{3} - \sqrt{3}\right)

  4. D.

    20(4π3+3)20\left(\frac{4\pi}{3} + \sqrt{3}\right)

Answer

C

Explanation

Length of chord subtending angle θ=120∘\theta = 120^\circ at center is 2Rsin⁡(θ2)=2Rsin⁡(60∘)=R32R \sin\left(\frac{\theta}{2}\right) = 2R \sin(60^\circ) = R\sqrt{3}.

Given chord length = 103  ⟹  R3=103  ⟹  R=1010\sqrt{3} \implies R\sqrt{3} = 10\sqrt{3} \implies R = 10.

Area of smaller region = Area of sector - Area of triangle formed by chord and radii: Area of sector=120∘360∘πR2=13π(100)=100π3\text{Area of sector} = \frac{120^\circ}{360^\circ} \pi R^2 = \frac{1}{3} \pi (100) = \frac{100\pi}{3} Area of triangle=12R2sin⁡(120∘)=12(100)(32)=253\text{Area of triangle} = \frac{1}{2} R^2 \sin(120^\circ) = \frac{1}{2} (100) \left(\frac{\sqrt{3}}{2}\right) = 25\sqrt{3}

Area of smaller region=100π3−253=25(4π3−3)\text{Area of smaller region} = \frac{100\pi}{3} - 25\sqrt{3} = 25\left(\frac{4\pi}{3} - \sqrt{3}\right)

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