Question

In ΔABC, D and E are the mid- points of AB and AC respectively. Find the ratio of the areas ΔADE and ΔABC.

• A. 1 : 5
• B. 2 : 5
• C. 2 : 3
• D. 1 : 4

Answer 1

The correct option is D 1 : 4

Since D and E are the midpoints of AB and AC respectively.
We can say,
DE || BC
[By converse of mid-point theorem]

Also, $DE=\left(\frac{1}{2}\right)BC$

In ΔADE and ΔABC,
∠ADE = ∠B (Corresponding angles)
∠DAE = ∠BAC (common)
Thus, ΔADE ~ ΔABC (AA Similarity)

Now, we know that,
The ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides, so

$\frac{Ar\left(\Delta ADE\right)}{Ar\left(\Delta ABC\right)}$ $=\frac{A{D}^2}{A{B}^2}$
$\frac{Ar\left(\Delta ADE\right)}{Ar\left(\Delta ABC\right)}$ $=\frac{1^2}{2^2}$
$\frac{Ar\left(\Delta ADE\right)}{Ar\left(\Delta ABC\right)}$ $=\frac{1}{4}$

Therefore, the ratio of the areas ΔADE and ΔABC is 1:4