Question

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$\huge\blue{\underline{\bf S}}}$​​
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Prove that the area of an equilateral triangle described on one side of a side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.


$\sf\red{\underline{\sf NOTE:-}}$

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Answer 1

$\color{red}\huge{\underline{\underline{GIVEN\dag}}}$

• Base of the triangle is equal to the length of square.
• Height also equals to lenght of square.

$\color{red}\huge{\underline{\underline{SOLUTION\dag}}}$

Let,the length of square be 'S';

${\boxed{\color{blue}Area\:of\:Square={side}^{2}}}$

$\therefore Area\:of\:Square={s}^{2}.........(1)$

But;

${\boxed{\color{blue}Area\:of\:Triangle=\frac{1}{2}(base)(height)}}$

We know that;

• $Base-s$
• $Height-s$

$\implies Area\:of\:Triangle=\frac{1}{2}(s)(s)$

$\therefore Area\:of\:Triangle=\frac{{s}^{2}}{2}........(2)$

From (1) & (2);

${\boxed{\color{orange}Ar.(\triangle)=\frac{Ar.(\square)}{2}}}$

Therefore area of equilateral triangle is equal to half the area of square.

HOPE THAT HELPS!!