Question

A wire when bent in the form of an equilateral triangle encloses an area of 36root3 find the area enclosed by the same wire when went form a square

Answer 1

The wire's length will remain same unless it is cut down

Given area of the triangle =36√3

so, √3/4 a² = 36√3

=> a² = 36√3 * 4/√3

=>a² = 36*4

=>a= √144=12

so one side of the triangle is 12

so the total length of the wire is 36cm

A square has 4 sides

so each side will be 36/4 = 9cm

Area of the square = 9² = 81cm²

hope you got it Mark as brainliest

Answer 2

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Here is the Solution....

Area of equilateral∆ = $\frac{\sqrt{3}a^2}{4}$

$36\sqrt{3} = \frac{\sqrt{3}a^2}{4}$

$\frac{36\sqrt{3} \times 4}{\sqrt{3}} = a^2$

$36 \times 4 = a^2\\144 = a^2\\ \sqrt{144} = a\\a = 12$

a = 12.

Now, perimeter of ∆ = 3 x 12

= 36

Perimeter of ∆ = perimeter of square

36 = 4 x side

36/4 = side

9 = side

Each side of square = 9

Area is square = (side)²

= 9²

= 81

${\bold{\boxed{Answer\:is\:81sq\:units}}}$

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I Hope This Will Help You ❣️