Question

Can anyone plz answer this fast .​

Answer 1

$\underline{\underline{\tt{\pink{To \: Prove:-}}}}$

• The area of equilateral triangle described one side of a square is equal to half area of the equilateral triangle described one of its diagonals.

$\underline{\underline{\tt{\orange{Proof:-}}}}$

➝ Let side of square be a .

From the figure ,

In ∆ABD :

$\hookrightarrow \tt BD = \sqrt{ a^2 + a^2 } \\ \\ \\ = \tt\sqrt{2 {a}^{2} } \\ \\ \\ \tt \: = \sqrt{2} a.$

Hence,

◉ Diagonal = a√2

Now,

$\tt \: \hookrightarrow ar. \triangle \: APB = \sqrt{3}/4 \times a^2 ....(1)$

$\tt \hookrightarrow Ar.\triangle \: BQD = \frac{ \sqrt{3} }{4} \times (a \sqrt{2} ) {}^{2} \\ \\ \\ \tt \: = \frac{ \sqrt{3} }{4} \times 2 {a}^{2} \\ \\ \\ \tt \: = \frac{ \sqrt{3} }{2} {a}^{2} ....(2)$

Therefore,

From eqn (1)&(2) we get :

$\bigstar \boxed{\pink{\tt{Area \: of \: \triangle APB = 1/2 \times Area\: of\: \triangle BQD }}}$

Hence proved !

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