CAT 2021 Slot 3 · Quant · Question 11

Work inside triangle ADC, where we know AD =10, AC =20 and the included angle \angle ADC={30}^{∘}. Let DC =x.

By the cosine rule, A{C}^{2}=A{D}^{2}+D{C}^{2}-2\,\left( AD \right)\left( DC \right)\cos{30}^{∘}:

400=100+{x}^{2}-2\left( 10 \right)x\cdot \frac{\sqrt{3}}{2}, so {x}^{2}-10\sqrt{3}\,x-300=0.

Then x=\frac{10\sqrt{3}\pm \sqrt{300+1200}}{2}=\frac{10\sqrt{3}\pm 10\sqrt{15}}{2}. A length must be positive, so x=5\left( \sqrt{3}+\sqrt{15}\right).

The area of a parallelogram with sides AD and DC and included angle {30}^{∘} is \left( AD \right)\left( DC \right)\sin{30}^{∘}=10\cdot 5\left( \sqrt{3}+\sqrt{15}\right)\cdot \frac{1}{2}=25\left( \sqrt{3}+\sqrt{15}\right) sq cm.