Question

. In ∆ABC, DE//BC and 3DE=BC then find the ratio of areas of ∆ADE and ∆ABC​

Answer 1

Given:

In Δ ABC,

DE // BC

3DE = BC

To find:

The ratio of areas of ∆ADE and ∆ABC

Solution:

We have,

$3DE = BC$

⇒ $\bold{\frac{DE}{BC} = \frac{1}{3}}$ ....... (i)

In ΔADE and ΔABC, we have

∠A = ∠A ...... [common angle]

∠ADE = ∠ABC ....... [corresponding angle]

∴ ΔADE ~ ΔABC ...... [By AA similarity]

We know that the ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

∴ $\frac{Area(\triangle ADE)}{Area(\triangle ABC)} = \bigg(\frac{DE}{BC} \bigg )^2$

substituting from (i)

$\implies \frac{Area(\triangle ADE)}{Area (\triangle ABC)} = \bigg(\frac{1}{3} \bigg )^2$

$\implies \bold{\frac{Area(\triangle ADE)}{Area (\triangle ABC)} = \frac{1}{9}}$

Thus, the ratio of areas of ∆ADE and ∆ABC​ is 1 : 9.

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