Question

Prove that the area ofequilateral triangle onscribed on diagonal of a square is double of the triangle onscribed on the side of rhe square

Answer 1

Answer:

Step-by-step explanation:

let side of square be a

length of diagonal =√2a

area of eq Δ=√3*a²/4

=√3a²/4

area of triangle on eqΔ = √3*(√2a)²/4

=√3a²/2

this 2 triangles are similar (all eqΔ are similar)

Δside~Δdiagonal

arΔside/arΔdiagonal=(√3a²/4)/(√3a²/2)

=1/2

hence proved area of Δdiagonal is double then Δside

Answer 2

Refer the attachment for figure. In figure, AEC is an equilateral triangle onscribed on the diagonal AC and BFC is an equilateral triangle onscribed on the side BC of square ABCD.

Let the side of square be of length 'a'

So, by Pythagoras Theorem, diagonal would be

$\sqrt{(a^2 + a^2)}$

$\implies \: \sqrt{2 {a}^{2} } \\ = a \sqrt{2}$

So, we know that area of equilateral triangle =

$\frac{ \sqrt{3} }{4} {side}^{2}$

For triangle AEC, area =

$\frac{ \sqrt{3} }{4} \times (a \sqrt{2} ) {}^{2} \\ \\ \implies \: \frac{ \sqrt{3} }{4} \times 2{a}^{2} \\ \\ \implies \: \frac{ \sqrt{3} }{2} {a}^{2}$

Area of triangle BFC, area =

$\frac{ \sqrt{3} }{4} \times {a}^{2} \\ \\ = \frac{ \sqrt{3} }{4} {a}^{2}$

So, comparing their areas,

$\frac{ar( \triangle AEC)}{ar(\triangle BFC )} = \frac{ \frac{ \sqrt{3} }{2} {a}^{2} }{ \frac{ \sqrt{3} }{4} {a}^{2} } \\ \\ \implies \: \frac{ar( \triangle AEC)}{ar(\triangle BFC )} \: = \frac{2}{1} \\ \\ \implies\! ar( \triangle AEC) = 2ar(\triangle BFC )$

Hence, area of equilateral triangle onscribed on diagonal of a square is double of the triangle onscribed on the side of rhe square