Question

5. The sides of a triangle are 16cm , 30cm,34cm. Its
area is (in cm?)
(a) 225
(b) 22573
(c) 225v2
(d) 240​

Answer 1

Given that,

• Sides of the triangle are 16 cm, 30 cm, 34 cm.

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☯ $\underline{\boldsymbol{Using \: Heron's \: Formula \: :}}$

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

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$:\implies\sf S_{\:(Semiperimeter)} = \dfrac{a + b + c}{2} \\\\\\:\implies\sf S_{\:(Semiperimeter)} = \dfrac{16 + 30 + 34}{2} \\\\\\:\implies\sf S_{\:(Semiperimeter)} = \cancel\dfrac{80}{2}\\\\\\:\implies{\underline{\boxed{\frak{\purple{ S_{\:(Semiperimeter)} = 40 \: cm}}}}}$

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$\dag\;{\underline{\frak{Substituting \: Values \; in \ formula,}}}$

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$:\implies\sf Area_{\triangle} = \sqrt{40 (40 - 16) (40 - 30) (40 - 34)} \\\\\\:\implies\sf Area_{\triangle} = \sqrt{ 40 \times 24 \times 10 \times 6} \\\\\\:\implies\sf Area_{\triangle} = \sqrt{4 \times 10 \times 4 \times 6 \times 10 \times 6} \\\\\\:\implies\sf Area_{\triangle} = \sqrt{4 \times 6 \times 10}\\\\\\:\implies{\underline{\boxed{\frak{\purple{ Area_{\triangle} = 240 \: cm^2}}}}} \: \bigstar$

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$\therefore\:{\underline{\sf{The\: area \: of \ \triangle \:is\: \bf{ Option \: (d) \: 240\: cm^2.}}}}$⠀⠀⠀

Answer 2

Answer:

Given :-

Sides of the triangle are 16 cm, 30 cm, 34 cm.

To Find :-

Area

Solution :-

By using Herons formula

$\frak \red{S = \: \dfrac{a + b + c}{2} }$

$\sf S = \: \dfrac{16 + 30 + 34}{2}$

$\sf \: S = \cancel \dfrac{80}{2}$

$\frak \pink{S = \: 40 \: cm}$

Finding area

By using Herons formula

$\frak \red{A = \sqrt{s(s - a)(s - b)(s - c)}}$

$\sf \: A \: = \sqrt{40(40 - 16)(40 - 30)(40 - 34)}$

$\sf \: A \: = \sqrt{40(24)(10)(6)}$

$\sf \: A \: = \sqrt{4 \times 6 \times 10}$

$\sf \: A \: = 240 {cm}^{2}$

Hence :-

Option D is correct