cat 2022 Question 21

Question

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

Accepted Answer

Area of triangle ABD is twice the area of triangle ACD

$\angle ADB = heta$

$\frac{1}{2}(AD)(BD)\sin heta = 2\left(\frac{1}{2}(AD)(CD)\sin(180 - heta) ight)$

$BD = 2CD$

Therefore, BD = 2 and CD = 1

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

Applying cosine rule in triangle ADC, we get

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

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

$AD^2 = 7$

Question 21.

Question Explanation