Question

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

• \[24\sqrt{2} c m^2\]

   

• \[24\sqrt{3} c m^2\]

   

• \[48\sqrt{3} c m^2\]

   

• \[64\sqrt{3} c m^2\]

   

Answer 1

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 P₁ is given by

p₁= 3 ×  side

p = p₁

_{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`