Question

7. 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.
368
ICSE MATHEMATICS​

Answer 1

Please find the attachment

Answer 2

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 ∆ :

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

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

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