Question

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2 cm, then calculate the area of the triangle

Answer 1

let the side of a square be a cm

and side of a triangle be b cm

Using pythagoras theorem in ΔBCD

${(bd)}^{2} = {(bc)}^{2} + {(dc)}^{2} \\ {(12 \sqrt{2} )}^{2} = {a}^{2} + {a}^{2} \\ 288 = 2 {a}^{2} \\ \frac{288}{2} = {a}^{2} \\ \sqrt{144} = a = 12cm \\ perimetet \: of \: square = permeter \:o f \: triangle \\ 4a = 3b \\ \frac{4 \times 12}{3} = b \\ 16 cm= b \\ \: area \: of \: triangle = \frac{ \sqrt{3} }{4} {side \: of \: triangle}^{2} \\ = \frac{ \sqrt{3} }{4} {(16)}^{2} \\ = \frac{ \sqrt{3} \times 16 \times 16}{4} \\ = 16 \times 4 \times \sqrt{3} \\ = 64 \sqrt{3} {cm}^{2}$

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