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Length of Angle Bisector in a Triangle

CAT 2002 Slot 1 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2002 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 ΔABC\Delta ABC, the internal bisector of ∠A\angle A meets BCBC at DD. If AB=4AB = 4, AC=3AC = 3 and ∠A=60∘\angle A = 60^\circ, then the length of ADAD is

  1. A.

    2\sqrt{3}

  2. B.

    \frac{12\sqrt{3}}{7}

  3. C.

    \frac{15\sqrt{3}}{8}

  4. D.

    \frac{6\sqrt{3}}{7}

Answer

B

Explanation

Using the area of ΔABC\Delta ABC: Area(ΔABC)=Area(ΔABD)+Area(ΔADC)\text{Area}(\Delta ABC) = \text{Area}(\Delta ABD) + \text{Area}(\Delta ADC) 12⋅AB⋅AC⋅sin⁡(60∘)=12⋅AB⋅AD⋅sin⁡(30∘)+12⋅AC⋅AD⋅sin⁡(30∘)\frac{1}{2} \cdot AB \cdot AC \cdot \sin(60^\circ) = \frac{1}{2} \cdot AB \cdot AD \cdot \sin(30^\circ) + \frac{1}{2} \cdot AC \cdot AD \cdot \sin(30^\circ) 4⋅3⋅32=4⋅AD⋅12+3⋅AD⋅124 \cdot 3 \cdot \frac{\sqrt{3}}{2} = 4 \cdot AD \cdot \frac{1}{2} + 3 \cdot AD \cdot \frac{1}{2} 63=72AD  ⟹  AD=12376\sqrt{3} = \frac{7}{2} AD \implies AD = \frac{12\sqrt{3}}{7}.

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