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CAT 2022 Slot 2 — QA Question 22

Basics of TrianglesEasy

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

Answer & solution

  • √7

  • B

    √6

  • C

    √5

  • D

    √8

Solution

Easy

Triangles ABDABD and ACDACD share the same height from AA, so their areas are in the ratio of their bases BD:DCBD:DC. That fixes DD on BCBC. Drop the altitude from AA to the midpoint EE of BCBC and use Pythagoras in △AED\triangle AED.

A B C D E AE AD
1

Locate D. Same height ⇒\Rightarrow area ratio = base ratio:

[ACD][ABD]=DCBD=12 ⇒ BD=2, DC=1(BD+DC=3)\dfrac{[ACD]}{[ABD]}=\dfrac{DC}{BD}=\dfrac12\ \Rightarrow\ BD=2,\ DC=1\quad(BD+DC=3)
2

Altitude and offset. EE is the midpoint, BE=1.5BE=1.5, so ED=BD−BE=0.5ED=BD-BE=0.5. Height of the equilateral triangle:

AE=32×3=332AE=\dfrac{\sqrt3}{2}\times 3=\dfrac{3\sqrt3}{2}
3

Pythagoras in △AED\triangle AED:

AD2=AE2+ED2=(332)2+(12)2AD2=274+14=7 ⇒ AD=7\begin{aligned} &AD^2=AE^2+ED^2=\left(\dfrac{3\sqrt3}{2}\right)^2+\left(\dfrac12\right)^2\\ &AD^2=\dfrac{27}{4}+\dfrac14=7\ \Rightarrow\ AD=\sqrt7 \end{aligned}
AD=7 cmAD=\mathbf{\sqrt7}\text{ cm}
CAT 2022 Slot 2 QA Q22: 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 t — Solution | TheCATExam