CAT 2002 QA Question 19 · Geometry - Triangles

Area of ∆ADC = $\frac{1}{2}$ AD × 3 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ AD × 4 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ × 3 × 4 × sin 60°

$\frac{1}{2}$ AD × 3 × sin 30° + $\frac{1}{2}$ × AD × 4 × sin 30° = $\frac{1}{2}$ × 3 × 4 × sin 60°

AD$\left[3\times \frac{1}{2}+4\times \frac{1}{2}\right]=12\times \frac{\sqrt{3}}{2}$

AD$\frac{7}{2}=\frac{12\sqrt{3}}{2}$

∴ AD = $\frac{12\sqrt{3}}{7}$

Hence, option (a).