Question

Find the area of a triangle whose sides measure 13cm , 14cm and 15cm

Answer 1

Given,

=> First Side = 13 cm

=> Second Side = 14 cm

=> Third Side = 15 cm

As it is an Isosceles Triangle.

=> S = a+b+c/2 = 13+14+15/2 = 42/2 = 21 cm

$= > area = \sqrt{s(s - a)(s - b)(s - c)}$

$= > area = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$

$= > area = \sqrt{21\times 8 \times 7 \times 6}$

$= > area = \sqrt{7\times3\times2 \times 2 \times 2 \times 7 \times 3 \times 2}$

$= > area = 2 \times 2\times7 \times 3}$

$= > area = 84 \: \: {cm}^{2}$

Be Brainly

Answer 2

$\underline{\textbf{Question...}}$

To Find Area Of A Triangle Whose Sides Measures 13cm , 14cm and 15cm

$\underline{\textbf{Given...}}$

Measure Of $\implies$

$\textbf{First Side ( a ) = 13cm }$

$\textbf{Second Side ( b ) = 14cm }$

$\textbf{Third Side ( c ) = 15cm }$

So Now Let's Move For Solution ...

$\underline{\textbf{Solution..}}$

Let The Triangle = ∆ABC

By $\underline{\textbf{Heron's Formula}}$ We Have ➡️

$\boxed{\mathbf{ ar( \,\triangle \,ABC )\: = \: \sqrt{s(s - a)(s - b)(s - c)} }}$

Where

$\mathbf{S \:= \:Semi \,Perimeter = \Large{\frac{\, a \,+\, b\,+\, c}{2}}}$

That ..

$\mathbf{S\: =\: (13\:+ \:14 \:+ \:15) \:/ \:2\: \longrightarrow \: 42cm \: / \: 2 }$

$\mathbf{ \therefore\:S\: = \:21cm }$

So Now Putting Values In The Heron's Formula We Have ➡️

$\mathbf{ ar( \,\triangle \,ABC )\: = \: \sqrt{21cm(21 - 13)(21 - 14)(21 - 15)}}$

$\implies$ $\mathbf{ \: \sqrt{21\:(8)\:(7)\:(6)}}$

Now Breaking All Terms Into Smaller Parts ---

$\implies$ $\mathbf{ \sqrt{7 \:\times \:3(2\:\times\: 2 \times\: 2\:)(\:7\:)(2\: \times 3\: )}}$

$\implies$ $\mathbf{ \sqrt{7 \times 3 \times 2\times 2 \times 2 \times \: 7 \times \: 2 \times 3 }}$

$\implies$ $\mathbf{\sqrt{ {(7)}^{2}\: \times \:{(3)}^{2}\: \times\: {(2)}^{2} \:\times\: {(2)}^{2} }}$

Now Taking Out The Perfect Squares We Have ➡️

$\mathbf{ar( \triangle \: ABC \: ) \: = \: 7 \times 3 \times 2 \times 2}$

Therefore

$\mathbf{ar( \triangle \: ABC \: ) \: = \: 84{cm}^{2}}$

$\underline{\textbf{Hence The Required Answer Is...}}$

$\boxed{\mathbf{ \: 84{cm}^{2}}}$

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