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

Solution

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

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

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

$$AD^2=7$$

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

The answer is option C.

Video Solution

video

Create a FREE account and get:

  • All Quant CAT Formulas and shortcuts PDF
  • 33+ CAT previous papers with video solutions PDF
  • Topic-wise Previous year CAT Solved Questions for Free

CAT Quant Questions | CAT Quantitative Ability

CAT DILR Questions | LRDI Questions For CAT

CAT Verbal Ability Questions | VARC Questions For CAT

Related Formulas With Tests

cracku

Boost your Prep!

Download App