ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 2 :1
(b) 1:2
(c) 4:1
(d) 1:4

Given:

ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

To do: 

We have to  find the ratio of areas of $\triangle ABC$ and $\triangle BDE$.

Solution:

In $\triangle ABC$ and $\triangle ABC$,

$\angle A=\angle E$   ($\triangle ABC$ and $\triangle BDE$ are equilateral triangles)

$\angle ABC=\angle BED$   ($\triangle ABC$ and $\triangle BDE$ are equilateral triangles)

Therefore,

$\triangle ABC \sim\ \triangle BDE$        (By AA similarity)

We know that,

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Therefore,

$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=\frac{BC^2}{BD^2}$

$=\frac{(2BD)^2}{BD^2}$    ($D$ is the mid-point of $BC$)

$=\frac{4BD^2}{BD^2}$

$=\frac{4}{1}$

The ratio of areas of $\triangle ABC$ and $\triangle BDE$ is $4:1$.

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Updated on: 10-Oct-2022

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