CAT 2022 Slot 2 · Quant · Question 21

Triangles ABD and ADC share the same apex A and sit on the same line BC.

So they have the same height from A, and their areas are in the ratio of their bases.

The area of ADC is half that of ABD, so DC=\frac{1}{2}BD.

Together the bases make up the whole side: BD+DC=3.

Substituting, BD+\frac{BD}{2}=3, so \frac{3BD}{2}=3 and BD=2.

Then DC=1.

Now find AD using triangle ABD, where AB=3, BD=2 and the angle at B is {60}^{∘}, since ABC is equilateral.

By the cosine rule, A{D}^{2}=A{B}^{2}+B{D}^{2}-2\times AB\times BD\times \cos{60}^{∘}.

Substituting, A{D}^{2}=9+4-2\times 3\times 2\times \frac{1}{2}.

That is 13-6=7.

So AD=\sqrt{7} — the fourth option.

Shortcut. Equal heights turn the area condition straight into a base ratio, so BD:DC=2:1 needs no computation. The cosine rule then has a particularly clean form at {60}^{∘}, where the last term is simply AB\times BD.