Question

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

Then, $A\left(\Delta BDE\right)$ =$xA\left(\Delta ABC\right)$. Find $x$

• A. 1/4
• B. 1/2
• C. 1/8
• D. 1

Answer 1

The correct option is A 1/4

Given-

$ABC$ & $DBC$ are two equilateral triangles where D is the midpoint of BC and BD is on BC.

Solution-

$ar\left(\Delta ABC\right)=\frac{\sqrt{3}}{4}{BC}^2$

For an equilateral triangle of side $a,ar\left(\Delta \right)=\frac{\sqrt{3}}{4}{a}^2.$

And, $ar\left(\Delta BDE\right)=\frac{\sqrt{3}}{4}B{E}^2=\frac{\sqrt{3}}{4}(\frac{BC}{2}{)}^2=\frac{1}{4}{(\frac{\sqrt{3}}{4}\times BC)}^2$

$⟹\frac{ar\left(\Delta BDE\right)}{ar\left(\Delta ABC\right)}=\frac{\frac{1}{4}{(\frac{\sqrt{3}}{4}\times BC)}^2}{\frac{\sqrt{3}}{4}{BC}^2}=\frac{1}{4}$

$⟹x$ = $\frac{1}{4}$