Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Given: 

An equilateral triangle described on one side of the square.

To do: 

We have to prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let the side of the square $\mathrm{ABCD}$ be $a$.

This implies,

$\mathrm{AC}^{2} =\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$=a^{2}+a^{2}$

$=2 a^{2}$

$\mathrm{AC}=\sqrt{2} a$

$\angle PAD=\angle QAC=60^o$

$\angle PDA=\angle QCA=60^o$

Therefore, by AA similarity,

$\Delta \mathrm{PAD} \sim \Delta \mathrm{QAC}$

This implies,

$\frac{\text { ar } \triangle \mathrm{PAD}}{\operatorname{ar} \Delta \mathrm{QAC}}=\frac{\mathrm{AD}^{2}}{\mathrm{AC}^{2}}=\frac{a^{2}}{(\sqrt{2} a)^{2}}$

$=\frac{1}{2}$

$ar(\triangle \mathrm{PAD})=\frac{1}{2}(ar \Delta \mathrm{QAC})$

Hence proved.

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Updated on: 10-Oct-2022

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