Question

$\Delta ABC$ and $\Delta BDE$ are two equilateral triangles, such that D is the midpoint of BC. The ratio of the areas of triangles ABC and BDE is:

• A. $2:1$
• B. $1:2$
• C. $4:1$
• D. $1:4$

Answer 1

The correct option is C $4:1$

Solution for both cases

Given: $△ABC$ and $△EBD$ are equilateral triangles.

So, measures of all the angles of both triangles is ${60}^o$.

Hence, $△EBD\sim △ABC$ ...AA test of similarity

$\therefore \frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}$ ...C.S.S.T

But $BD=\frac{1}{2}BC$ ...[Since, D is midpoint of BC]

$\therefore \frac{BD}{BC}=\frac{1}{2}$

By theorem on ratio of areas of similar triangles,

$\frac{A\left(△EBD\right)}{A\left(△ABC\right)}={\left(\frac{BD}{BC}\right)}^2$

$\therefore \frac{A\left(△EBD\right)}{A\left(△ABC\right)}=\frac{1}{4}$