Question

$$ABC$$ and $$BDE$$ are two equilateral triangles such that $$D$$ is the midpoint of $$BC$$. Ratio of the areas of triangles $$\mathrm{\Delta }ABC$$ and $\mathrm{\Delta }BDE$ is:
A) $$2:1$$
B) $$1:2$$
C) $$4:1$$
D) $$1:4$$

Answer 1

Hint: Here the given is the relation between the triangles $$\mathrm{\Delta }ABC$$ and $\mathrm{\Delta }BDE$. We have to find the ratio of the areas of given triangles. By using some triangle properties to find it.
SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.

Complete step-by-step answer:
It is given that, $$ABC$$ and $$BDE$$ are two equilateral triangles such that $$D$$ is the midpoint of $$BC$$

We need to find the ratio of $${\displaystyle \frac{\text{Area}(\mathrm{\Delta }ABC)}{\text{Area}(\mathrm{\Delta }BDE)}}$$.
Since we have to find the ratio of the areas of $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }BDE$$, we first need to prove these triangles are similar.
We know that for any equilateral triangle the sides are equal.
Since, $$ABC$$ and $$BDE$$ are two equilateral triangles such that $$D$$ is the mid-point of $$BC$$.
In $\mathrm{\Delta }ABC$,
$$\Rightarrow AB=BC=CA$$
In $\mathrm{\Delta }BDE$,
$$\Rightarrow BE=DE=BD={\displaystyle \frac{1}{2}}BC$$
Thus we have their sides would be in the same ratio.
$$\Rightarrow {\displaystyle \frac{AB}{BE}}={\displaystyle \frac{AC}{DE}}={\displaystyle \frac{BC}{BD}}$$
Hence by SSS similarity,
$$\Rightarrow \mathrm{\Delta }ABC\sim \mathrm{\Delta }BDE$$
We know that if two triangle are similar,
Ratio of areas is equal to square of ratio of its corresponding sides
$$\Rightarrow {\displaystyle \frac{\text{Area}(\mathrm{\Delta }ABC)}{\text{Area}(\mathrm{\Delta }BDE)}}={\left({\displaystyle \frac{BC}{BD}}\right)}^2$$
Since $$D$$ is the midpoint of $$BC$$
$$\Rightarrow {\displaystyle \frac{\text{Area}(\mathrm{\Delta }ABC)}{\text{Area}(\mathrm{\Delta }BDE)}}={\left({\displaystyle \frac{{\displaystyle \frac{1}{1}}BC}{{\displaystyle \frac{1}{2}}BC}}\right)}^2$$
Cancelling the common term $$BC$$,
$$\Rightarrow {\displaystyle \frac{\text{Area}(\mathrm{\Delta }ABC)}{\text{Area}(\mathrm{\Delta }BDE)}}={\left(1\times {\displaystyle \frac{2}{1}}\right)}^2$$
$$\Rightarrow {\displaystyle \frac{\text{Area}(\mathrm{\Delta }ABC)}{\text{Area}(\mathrm{\Delta }BDE)}}={\displaystyle \frac{4}{1}}$$
Hence, the areas of triangles $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }BDE$$ is: $$4:1$$ .

(C) is the correct option.

Note: If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar. The corresponding sides of similar triangles are in proportion.
We have used the following theorem.
Theorem: 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.