Question

Find the area of the triangle with sides 21 cm, 16 cm and 13 cm. Also, find the perimeter of an equilateral
triangle equal in area to this triangle.​

Answer 1

AnswEr :

☯ Given sides of the equilateral ∆ are 21 cm, 16 cm & 13 cm.

${\dag}\:\underline{\frak{Using \ Heron's \ Formula \: \: :}}$

$\star\: \small\boxed{\sf{\purple{Area \: of \: \triangle = \sqrt{s(s - a) (s - b) (s - c)}}}}$

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$:\implies\sf S_{(semi perimeter)} = \dfrac{a + b + c}{2} \\\\\\:\implies\sf s = \dfrac{21 + 16 + 13}{12} \\\\\\:\implies\sf s = \cancel\dfrac{50}{2} \\\\\\:\implies\boxed{\sf{\purple{ s = 25}}}$

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$:\implies\sf \sqrt{25(25 - 21) (25 - 16) (25 - 13)}\\\\\\:\implies\sf\sqrt{25 \times 4 \times 9 \times 12} \\\\\\:\implies\sf\sqrt{ 5 \times 5 \times 2 \times 2 \times 3 \times 3 \times 2 \times 3 \times 2} \\\\\\:\implies\sf 60 \sqrt{3}$

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$\therefore\: \underline{\sf{Here\: we \: get \: area \: is \: \bf{60 \sqrt{3}}}}$

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☯ Now, finding area of the equilateral ∆ :

$\star\: \small\boxed{\sf{\pink{ Area \: of \: equilateral \: \triangle = \dfrac{\sqrt{3}}{4} (a)^2}}}$

$:\implies\sf 60 \sqrt{3} = \dfrac{\sqrt{3}}{4}(a)^2 \qquad \quad \Bigg[ Equating \: both \: areas \Bigg] \\\\\\:\implies\sf 60 \cancel{\sqrt{3}} = \dfrac{\cancel{\sqrt{3}}}{4}(a)^2 \\\\\\:\implies\sf 60 = \dfrac{(a^2)}{4} \\\\\\:\implies\sf a^2 = 4 \times 60 \\\\\\:\implies\sf a = \sqrt{240} \\\\\\:\implies\boxed{\sf{\pink{a = 4\sqrt{15}}}}$

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☯ Perimeter of equilateral ∆ is equal to the Area of the ∆ :

$\star\: \small\boxed{\sf{\purple{Perimeter \: of \: \triangle = 3 \times Area }}}$

$:\implies\sf Area = 3 \times 4 \sqrt{15} \qquad \quad \Bigg[ Area = 4 \sqrt{15}\Bigg] \\\\\\\:\implies \boxed{\sf{\purple{ Area = 12\sqrt{15}}}}$

$\therefore\: \underline{\sf{Perimeter \: of \: equilateral \: \triangle \: is \: \bf{12\sqrt{15}}}}$

Answer 2

$\sf{Refer \ to \ the \ attachment :)}$