CAT 2022 Question Paper (Slot 2) Question 66

Question 66

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


Area of triangle ABD is twice the area of triangle ACD

$$\angle\ ADB=\theta\ $$

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

$$BD\ =2CD$$

Therefore, BD = 2 and CD = 1

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

Applying cosine rule in triangle ADC, we get

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

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


$$AD=\sqrt{\ 7}$$

The answer is option C.

View Video Solution

Create a FREE account and get:

  • All Quant CAT Formulas and shortcuts PDF
  • 30+ CAT previous papers with solutions PDF
  • Top 500 CAT Solved Questions for Free


Boost your Prep!

Download App