The length of each side of an equilateral triangle ABC

Question

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

A. sqrt (5)
B. sqrt (6)
C. sqrt (7)
D. sqrt (8)
E. sqrt (10)

Source : Wizako

gmatophobia · Answer

We can draw a figure as shown below, the vertical height, known as the altitude of the triangle, is same for  	riangle ABD  and  	riangle ADC

Area of  	riangle ABD = \frac{1}{2} BC * h

Area of  	riangle ADC = \frac{1}{2} DC * h

Given : Area of triangle ADC is half the area of triangle ABD

\frac{1}{2} DC * h = \frac{1}{2} * \frac{1}{2} BC * h

BD = 2 DC

Therefore BD = 2 & DC = 1

An altitude dropped on BC from point A at divides the base into two equal halves.

BO = OC = 1.5

Therefore OD = 1/2

AO =  \frac{\sqrt{3}}{2} * 3

In  	riangle AOD

AO^2 + OD^2 = AD^2

(\frac{3\sqrt{3}}{2})^2 + (\frac{1}{2})^2 = AD^2

AD^2 = \frac{28}{4} = 7

AD = \sqrt{7}

Option C

KarishmaB · Answer

Screenshot 2023-01-17 at 12.47.33 PM.png Draw the equilateral triangle with each side 3 cm. Note that the altitude AT for triangles ADB and ADC is the same. So their areas depend on the ratio of their bases only. Since area of ADB is twice the area of ADC, length of DB is twice the length of DC. Hence DB = 2 cm and DC = 1 cm. If we drop the altitude AT of the equilateral triangle, it will be the median too and hence will divide BC into 2 equal parts of 1.5 cm each. Then AD = \sqrt{AT^2 + TD^2} = \sqrt{(\frac{sqrt(3)*3}{2})^2 + (\frac{1}{2})^2} = \sqrt{7} Answer (C)

The length of each side of an equilateral triangle ABC

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