Prove that the area of an equilateral triangle described on one side

Solution:

ABCD is a square having sides of length = a.



Then the diagonal $\mathrm{BD}=\mathrm{a} \sqrt{2}$.

We construct equilateral $\Delta \mathrm{s}$ PAB and QBD

$\Rightarrow \Delta \mathrm{PAB} \sim \Delta \mathrm{QBD}$ (Equilateral triangles are similar)

$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{PAB})}{\operatorname{ar}(\Delta \mathrm{OBD})}=\frac{A B^{2}}{B D^{2}}=\frac{a^{2}}{(a \sqrt{2})^{2}}=\frac{1}{2}$

$\Rightarrow$ ar $(\Delta \mathrm{PAB})=\frac{1}{2}$ ar $(\Delta \mathrm{QBD})$

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