Question

A square and an equilateral triangle have equal perimeters. If a diagonal of the square is
24root2 cm, then find the area of the triangle. ​

Answer 1

Answer:

256√3 cm²

Step-by-step explanation:

Diagonal of square of side length 'a' is given a√2.

Proof: let the side be 'a'. As side intersect at 90°, use Pythagoras theorem,

=> side² + side² = diagonal²

=> a² + a² = diagonal²

=> a√2 = diagonal, where 'a' is side.

So, here, if diagonal is 24√2 cm, side is 24 cm.

Perimeter(square) = perimeter(∆)

=> 4*side of square = 3(side of ∆)

=> 4*24cm = 3*side of ∆

=> 32 cm = side of ∆

Thus,

Area of ∆ = √(3)/4 * side²

= √(3)/4 * (32 cm)²

= 256√3 cm²

Answer 2

Let, the side of the square be a and side of the triangle be t.

A/Q, 4a = 3t

a = 3t/4

Diagonal of square = √2a = 24√2

Therefore, a = 24 and

t = 24*4/3 = 32

Since, area of triangle = √3/4 * t^2

= √3/4 * 1024

= 256√3 sqcm

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