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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

Explanation:

Area of ∆ACD is half of area of ∆ABD.
Since their height is same, ratio of their areas will be same as the ratio of their bases.
⇒ BD = 2CD
⇒ BD = 2 cm and CD = 1 cm.

Taking E as the midpoint of BC, BE = 3/2 = 1.5
⇒ ED = 2 – 1.5 = 0.5 cm.

Also, AE = height of the equilateral triangle = $\frac{\sqrt{3}}{2}$ × 3 = $\frac{3 \sqrt{3}}{2}$

In ∆AED
⇒ AD² = AE² + ED²
⇒ AD² = ${(\frac{3 \sqrt{3}}{2})}^{2}$ + (0.5)²

⇒ AD² = $\frac{27}{4}$ + $\frac{1}{4}$ = 7

⇒ AD = $\sqrt{7}$

Hence, option (a).

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Doubts

J said (2023-07-25 08:22:20)

how did BE = 3/2

Reply from Admin:

E is the mid-point of side BC, hence BE = half of BC = 3/2 = 1.5

Also, in an equilateral triangle median is also the altitude of the triangle.

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