Question

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC then ar(ΔABC) : ar(BDE) = _________.

Answer 1

Given:
ABC and BDE are two equilateral triangles
D is the mid-point of BC


$$\begin{gathered} \mathrm{ar}\left(∆ABC\right)=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(BC\right)}^2...\left(1\right) \\ \mathrm{ar}\left(∆BDE\right)=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(BD\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(\frac{1}{2}BC\right)}^2\left(\because D\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}BC\right) \\ =\frac{\sqrt{3}}{4}\times \frac{1}{4}\times {\left(BC\right)}^2 \\ =\frac{\sqrt{3}}{16}\times {\left(BC\right)}^2...\left(2\right) \\ \mathrm{Dividing}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{\mathrm{ar}\left(∆ABC\right)}{\mathrm{ar}\left(∆BDE\right)}=\frac{{\displaystyle \frac{\sqrt{3}}{4}}{\left(BC\right)}^2}{{\displaystyle \frac{\sqrt{3}}{16}}{\left(BC\right)}^2} \\ =\frac{4}{1} \end{gathered}$$


Hence, ar(ΔABC) : ar(ΔBDE) = 4 : 1.