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# 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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