Question

pls answer this using heron's formula..
pls give me the answer as step-by-step explanation

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Answer 1

Answer:

a=18cm b=10cm and let the third side be 'c'

a+b+c=perimeter

18+10+c=42

28+c=42

c=42-28

c=14cm

s=a+b+c/2

s=18+10+14/2

s=42/2

s=21cm

Area of triangle=√s(s-a) (s-b) (s-c)

=√21(21-18) (21-10) (21-14) cm^2

√21×3×11×7 cm^2

√7×3×3×11×7 cm^2

7×3×√11 cm^2

21√11 cm^2

Answer 2

Answer:

Given that : Measures of three sides of the triangle are 18 cm and 10 cm  and Perimeter of Triangle is 42 cm.

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Now,

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⠀⠀⠀⠀⠀Finding Third Side of Triangle :

$\star\: \boxed{\sf{\blue{P_{\:(perimeter)} = a + b + c}}}$

$\sf{Here}\begin{cases}\sf{\:\:\:a = 18 \ cm}\\\sf{\:\:\: b = 10 \ cm}\\\sf{\:\;\:c = ?? \ cm}\\\sf{\:\:\:Perimeter=42cm} \end{cases}$

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$:\implies\sf 42 = 10 + 18 + c \\\\\\:\implies\sf 42 = 28 + c \\\\\\:\implies\sf c = 42 - 28\\\\\\:\implies{\underline{\boxed{\sf{\pink{c = 14\:cm}}}}}$

$\therefore\:{\underline{\sf{Hence, \ Third\:Side\:of\: \ \triangle \ is \ \bf{14 \ cm}.}}}$⠀

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⠀⠀⠀⠀⠀Finding Semi-Perimeter of Triangle :

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

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$\sf{Here}\begin{cases}\sf{\:\:\:a = 18 \ cm}\\\sf{\:\:\: b = 10 \ cm}\\\sf{\:\;\:c = 14 \ cm}\end{cases}$

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$:\implies\sf s = \dfrac{18 + 10 + 14}{2} \\\\\\:\implies\sf s = \cancel\dfrac{42}{2}\\\\\\:\implies{\underline{\boxed{\sf{\pink{s = 21 \:cm}}}}}$

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$\therefore\:{\underline{\sf{Hence, \ Semi- perimeter \ \triangle \ is \ \bf{21 \ cm}.}}}$⠀

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⠀⠀⠀⠀⠀Finding Area of Triangle :

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

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$\begin{gathered}\dag\;{\underline{\frak{Substituting \ values \ in \ the \ formula,,}}}\\ \\\end{gathered}$

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$:\implies\sf Area_{\triangle} = \sqrt{21(21 - 18) (21 - 10) (21 - 14)}\\\\\\:\implies\sf Area_{\triangle} = \sqrt{21 \times 3 \times 11 \times 7} \\\\\\:\implies\sf Area_{\triangle} = \sqrt{4851}\\\\\\:\implies\sf Area_{\triangle} = \sqrt{21\times 21 \times 11}\\\\\\:\implies\sf Area_{\triangle} = \sqrt{21^{2} \times 11}\\\\\\:\implies{\underline{\boxed{\frak{\purple{ Area_{\triangle} = 21\sqrt{11}}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ area \ of \ the \ \triangle \ is \ \bf{21 \sqrt{11} \ cm^2}.}}}$

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