Question

Prove that the area of an equilateral triangle described on one side of the square is equal to half of the area of the eqilateral triangle described on one of its diagonal

Answer 1

Hey there !!

→ Given :-

A square ABCD and equilateral triangles BCE and ACF have been described on side BC and diagonal AC respectively.

→ To prove :-

➡ ar(∆BCE) = $\frac{1}{2}$ ar(∆ACF) .

→ Proof :-

Since each of the ∆BCE and ∆ACF is an equilateral triangle, so each angle of each one of them is 60°. So, the triangles are equiangular, and hence similar.

$\therefore \triangle BCE \sim \triangle ACF .$

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.


$\therefore \frac{ ar( \traingle BCE ) }{ ar( \triangle ACF ) } = \frac{ {BC}^{2} }{ {AC}^{2} } .$


$=> \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{ {BC}^{2} }{ 2 ( {BC} )^{2} } . [ \therefore AC = \sqrt{2} BC . ]$


$=> \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{1}{2} .$


▶ ar(∆BCE) = $\frac{1}{2}$ × ar(∆ACF).

✔✔ Hence, it is solved ✅✅.

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#BeBrainly.