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A square and an equilateral triangle have equal perimeters. If the diagonal of the square is \[12\sqrt{2}\] cm, then area of the triangle is

#### Options

- \[24\sqrt{2} c m^2\]
- \[24\sqrt{3} c m^2\]
- \[48\sqrt{3} c m^2\]
- \[64\sqrt{3} c m^2\]

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#### Solution

It is given the perimeter of a square ABCD is equal to the perimeter of triangle PQR.

The measure of the diagonal of the square is given `12 sqrt(2)` cm.We are asked to find the area of the triangle

In square ABCD, we assume that the adjacent sides of square be *a.*

Since, it is a square then* a= b *

By using Pythagorean Theorem

`a^2 +b^2 = (12sqrt(2))^2`

`a^2 +a^2 = 288`

`2a^2 = 288`

`a^2 = 288/2`

` a = sqrt(144)`

a = 12 cm

Therefore, side of the square is 12 cm.

Perimeter of the square ABCD say *P* is given by

p = 4 × side

Side = 12 cm

p = 4 × 12

p = 48 cm

Perimeter of the equilateral triangle PQR say *P*_{1} is given by

p_{1}= 3 × side

p = p_{1}

_{p = 3 × side }

_{48 = 3 × side }

_{side = `48/3`}

_{side = 16 cm }

The side of equilateral triangle PQR is equal to 16 cm.

Area of an equilateral triangle say *A,* having each side *a* cm is given by

`A = sqrt(3)/4 a^2`

Area of the given equilateral triangle having each equal side equal to 4 cm is given by

*a* = 16 cm

`A = sqrt(3)/4 (16)^2 `

`A = sqrt(3)/4 xx 256`

`A=64 sqrt(3) cm^2`

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