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CAT 2022 Slot 2 Paper

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Question : 66 of 66

Marks: +1, -0

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 thearea of triangle $ABD$ . Then the length of $AD$ , in cm, is

$\sqrt{8}$
$\sqrt{6}$
$\sqrt{7}$
$\sqrt{}$

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-\theta)) \right)

BD = 2 CD

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.

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