Question

6. (a) Find the area of the triangle whose three sides are
( (ii) 36 m, 56 m and 36 m​

Answer 1

Answer:

by Herons Formulae

$\sqrt{s \times( s - a)(s - b)(s - c)}$

here s= semi perimeter or peri/2

s= 36+56+36/2

s=72+56/2

s=128m/2

s=64cm

ar(Triangle ABC)= root 64 × (64-36)×2 × (64-56)

=>root 28,672

=>169.32m²( approx)

Answer 2

☯ Given that, measures of three sides of the triangle are 36 cm, 56 cm and 36 cm.

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Now,

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$\star\: \boxed{\sf{\blue{s_{\:(Semiperimeter)} = \dfrac{a + b + c}{2}}}}$

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$\sf{Here}\begin{cases}\sf{\:\:\:a = 36 \ cm}\\\sf{\:\:\: b = 56 \ cm}\\\sf{\:\;\:c = 36 \ cm}\end{cases}$

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$:\implies\sf s = \dfrac{36 + 56 + 36}{2} \\\\\\:\implies\sf s = \cancel\dfrac{128}{2}\\\\\\:\implies{\underline{\boxed{\sf{\pink{s = 64}}}}}$

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$\therefore\:{\underline{\sf{Hence, \ Semi- perimeter \ \triangle \ is \ \bf{64 \ cm}.}}}$⠀⠀⠀

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$\star\:\boxed{\sf{\pink{Area_{\triangle} = \sqrt{s(s - a) (s - b) (s - c)}}}}$

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$\dag\;{\underline{\frak{Substituting \ values \ in \ the \ formula,,}}}$⠀

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$:\implies\sf Area_{\triangle} = \sqrt{64(64 - 36) (64 - 56) (64 - 36)}\\\\\\:\implies\sf Area_{\triangle} = \sqrt{64 \times 28 \times 8 \times 28} \\\\\\:\implies\sf Area_{\triangle} = \sqrt{401408}\\\\\\:\implies{\underline{\boxed{\frak{\purple{ Area_{\triangle} = 448\sqrt{2}}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ area \ of \ the \ \triangle \ is \ \bf{448 \sqrt{2} \ cm^2}.}}}$⠀⠀⠀