Question

sides of a triangle are in the ratio of 12:17:25
and it's perimeter is 540 cm. find it's area.
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Answer 1

$\implies\huge \sf \underline \red{Q uestion}$

sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540 cm. find its area.

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$\implies\huge \sf \underline \blue{Answer}$

$\bf{ \boxed{ \underline{ \underline{ \green{ \bf{Area = 9000 \: {cm}^{2} \: }}}}}}$

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$\implies \huge \sf \underline \orange{To \: find}$

area = ?

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$\implies \huge \sf \underline \pink{solution}$

Do you know that area of triangle :-

$\bf{ \boxed{ \underline{ \underline{ \red{ \sf{area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} \: }}}}}}$

Given area of triangle

s = semi perimeter

a,b,c = sides of the triangle

From the Question,

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{given \: perimeter = 540 \: cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{semi\: perimeter = s = \dfrac{perimeter}{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{s = \dfrac{540}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{s = 270cm}$

From the Question Given ratios of sides ,

$\: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{Given \: ratios \: of \: sides = 12 :17 : 25}$

let take sides a,b,c

a = 12x cm

b = 17 x cm

c = 25 x cm

so,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{perimeter = 540 \: cm}$

$\: \: \: \ \: \: \: \: \: \: \rm{a + b + c = 540}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{ = 12x + 17x + 25x = 540}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{ = 29x + 25x = 540}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{ = 54x = 540}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{ x = \dfrac{540}{54}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{x = 10}$

then,

$\tt \purple{a = 12x \: cm = 12 \times 10 = 120 \: cm}$

$\tt \green{b = 17x \: cm = 17 \times 10 = 170cm}$

$\tt \pink{c = 25x \: cm = 25 \times 100 = 250cm}$

Now,

$\bf{ \boxed{ \underline{ \underline{ \red{ \sf{area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} \: }}}}}}$

a = 120cm

b= 170 cm

c = 250 cm

s = 270 cm

$\: \: \sf{Area = \sqrt{270(270 - 120)(270 - 170)(270 - 250) {cm}^{2} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{270 \times 150 \times 100 \times 20 {m}^{2} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{(27 \times 15 \times 2) \times {(10)}^{5} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{(27 \times 30) \times {(10)}^{5} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{(27 \times 3) \times {(10)}^{6} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{(81) \times {(10)}^{6}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{81} \times \sqrt{ {(10)}^{6}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{ {9}^{2} \times \sqrt{ {(10)}^{6}} } }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = (9) \times {(10}^{6}) \frac{1}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = (9) \times {(10}^{3})}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 9000}$

$\rm \pink{Area = 9000 \: {cm}^{2} }$

Answer 2

Answer:

ratio of sides

12: 17:25

let them be

12x,17x, 25x respectively

perimeter of a triangle = sum of all sides

540 = 12x,17x, 25x

540 = 54x

x = 10

all sides measure

12x = 12×10 = 120

17x = 17× 10 = 170

25x= 25 × 10 = 250

it's semipetimeter = 540/2

= 270

using heron's formula area of the triangle =

root (s)(s-a)(s-b)(s-c)}

where s is the semipetimeter and a,b,c

area the sides of the triangle.

root {( 270)(270-120)(270-170)(270-250)}

= 9000cm²

hope this helps you