Question

The sides of triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.​

Answer 1

SolutioN :-

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Let the sides of triangle be 13x , 14x and 15x

Then , the perimeter of triangle = 13x + 14x + 15x

But , it is given that perimeter = 84

So , 13x + 14x + 15x = 84

$\small{ \sf \: \implies \: 42x = 84}$

$\small{ \sf \: \implies \:x = \frac{84}{42} }$

$\small{ \sf \: \implies \:x = 2 }$

$\boxed{ \mathrm{x = 2}}$

Let the first side be a , second side b and third side be c of a triangle.

Then , by putting the value of x

• a = 13x = 13 × 2 = 26
• b = 14x = 14 × 2 = 28
• c = 15x = 15 × 2 = 30

Now ,

By using Heron's formula , we have

$\boxed{\sf \small \: Area\:of\:\triangle = \sqrt{s(s - a)(s - b)(s - c)} }$

Where ,

• $\sf {s\:=\:\frac{perimeter}{2} }$
• a, b, c denotes the three sides of triangle.

By putting their values ,

$\small\sf{Area \: of \: \triangle = \sqrt{ \frac{84}{2}( \frac{84}{2} - 26)( \frac{84}{2} - 28) ( \frac{84}{2} - 30 )} }$

$\small \sf{ \implies \: Area \: of \: \triangle = \sqrt{42(42 - 26)(42 - 28)(42 - 30)} }$

$\small \sf{ \implies \: Area \:of \: \triangle = \sqrt{42(16)(14)(12)} }$

$\small \sf{ \implies \: Area \: of \: \triangle = \sqrt{42 \times 16 \times 14 \times 12} }$

$\small \sf{ \implies \: Area \: of \: \triangle = \sqrt{112896} }$

$\small \sf{ \implies \: Area \: of \: \triangle = {336 \: cm}^{2} }$

Therefore, area of triangle = 336m²

Answer 2

Correct Question-:

The sides of a triangle are in ratio 13:14:15 . If its perimeter is 84cm , Then Find the area of triangle .

AnswEr-:

$\boxed{\sf { \:The\:Area\:\:of\:Triangle \:is\:336cm^{2}\:}}$

EXPLANATION-:

• $\frak{Given \: -:} \begin{cases} \sf{The\:\:sides\:of\:triangle\:are\:in\:ratio\:= \frak{ 13:14:15}} & \\\\ \sf{The\:Perimeter \:of\:triangle\:is\:= \frak{84\: cm}}\end{cases}$

• $\frak{To \:Find\: -:} \begin{cases} \sf{The\:area\:of\:triangle \:.}\end{cases}$

$\sf{\dag{\underline {Solution\:of\:Question\: -:}}}$

• $\frak{Let's \:Assume \: -:} \begin{cases} \sf{The\:First\:side\:of\:triangle\:be\:= \frak{ 13x\: cm}} & \\\\ \sf{The\:Second \:side\:of\:triangle\:is\:= \frak{14x\: cm}}& \\\\ \sf{The\:Third \:side\:of\:triangle\:is\:= \frak{15x\: cm}}\end{cases}$

• $\underline{\boxed{\star{\sf{\red{ Perimeter_{(Triangle)}  \: = \: Side _{1}+ Side _{2} + Side_{3}}}}}}$

• $\frak{Here \:\: -:} \begin{cases} \sf{Side_{1}\:or\:First\:Side\:= \frak{13x \: cm}} & \\\\ \sf{Side_{2}\:or\:Second\:Side\:= \frak{14x\: cm}}& \\\\ \sf{Side_{3}\:Third \:side\:= \frak{15x\: cm}}& \\\\ \sf{The\:Perimeter \:of\:triangle\:is\:= \frak{84\: cm}}\end{cases}$

$\sf{\underline {Now,}}$

• $\longrightarrow{\sf { \:13x + 14x + 15x = 84 \:}}$

• $\longrightarrow{\sf { \:27x + 15x = 56 \:}}$

• $\longrightarrow{\sf { \:42x = 84 \:}}$

• $\longrightarrow{\sf { \:x = \dfrac{\cancel {84}}{\cancel {42}}\:}}$

• $\longrightarrow{\sf { \:x = 2\:}}$

Therefore,

• $\boxed{\sf { \:x = 2\:}}$

Now ,

• $\frak{Putting \:x =2\: -:} \begin{cases} \sf{Side_{1}\:or\:First\:Side\:= \frak{13x \:=\:13 \times 2=26 cm}} & \\\\ \sf{Side_{2}\:or\:Second\:Side\:= \frak{14x\:=14 \times 2\:=28 cm}}& \\\\ \sf{Side_{3}\:Third \:side\:= \frak{15x =15\times 2 = 30\: cm}}\end{cases}$

Therefore ,

• $\boxed{\sf { \:The\:three\:sides\:of\:Triangle \:are\:26cm\:,28cm\;and\:30cm\:}}$

Now ,

• By using Heron's Formula

• $\boxed{\sf{\red{Area\:= \sqrt{ S(s-a)(s-b)(s-c)}}}}$

• Here -:

• S = Semi Perimeter = $\sf{\dfrac{Perimeter}{2}}$ = $\sf{\dfrac{84}{2} = 42 cm}$

• A = First Side = 26 cm

• B = Second Side = 28 cm

• C = Third Side = 30 cm

Now , By Putting known Values in Formula of Area -:

• $\longrightarrow{\sf{ \sqrt{ 42(42-26)(42-28)(42-30)}}}$

• $\longrightarrow{\sf{ \sqrt{ 42(16)(14)(12)}}}$

• $\longrightarrow{\sf{ \sqrt{ 42\times 16 \times 14 \times 12 }}}$

• $\longrightarrow{\sf{ \sqrt{ 112896}}}$

• $\longrightarrow{\sf{ Area\: \triangle = 336cm^{2}}}$

Hence -:

• $\boxed{\sf { \:The\:Area\:\:of\:Triangle \:is\:336cm^{2}\:}}$

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