Question

In triangle ABC, D and E are points on BC such that CD = DE = EB. Find the ratio of areas of triangle ADE and triangle ABC.

• A. $\frac{1}{4}$
• B. $\frac{3}{1}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

We know a median divides a triangle into triangles of equal areas. AE is a median in triangle ADB and AD is a median in triangle AEC. Hence, we can say

A(triangle ABE) = A(triangle ADE) = A(triangle ADC)

So if area of triangle ABE = 1 Sq units (let’s say),
then area of triangle ADE = 1 Sq units and area of triangle ADC = 1 Sq units.
Area of triangle ABC = A(triangle ABE) + A(triangle ADE) + A(triangle ADC)
= 1 + 1 + 1 = 3 Sq units.
$\frac{\text{Area of triangle ADE}}{\text{Area of triangle ABC}}=\frac{1}{3}$