CAT 2022 Slot 2 QA Question & Solution

GeometryMedium

Question

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is

Options

$\sqrt{8}$

$\sqrt{6}$

$\sqrt{7}$

$\sqrt{5}$

Solution

Area of triangle ABD is twice the area of triangle ACD

$\angle ADB = \theta$

$\frac{1}{2} (AD)(BD)\sin\theta = 2\left(\frac{1}{2} (AD)(CD)\sin(180^\circ - \theta)\right)$

$BD = 2CD$

Therefore, BD = 2 and CD = 1

$\angle ABC = \angle ACB = 60^\circ$

Applying cosine rule in triangle ADC, we get

$\cos(\angle ACD) = \frac{AC^2 + CD^2 - AD^2}{2(AC)(CD)}$

$\frac{1}{2} = \frac{9 + 1 - AD^2}{6}$

$AD^2 = 7$

$AD = \sqrt{7}$

The answer is option C.