Question

Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540 cm. Find its area.​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

For a triangle :-

Given :-

$\sf{ratio \: of \: the \: sides = 12 \: : \: 17 \: : \: 25}$

$\sf{perimeter (p) = 540 \: cm}$

To Find :-

$\sf{Area \: of \: this \: triangle = ?}$

Using Formulas :-

$\sf{perimeter \: of \: triangle = Sum \: of \: all \: sides}$

$\sf{For \: Scalene \: triangle-}$

By Heron's Formula :-

$\sf{Area =\sqrt{S(S-a)(S-b)(S-c)}}$

$\sf{here \: , \: S=Semi-perimeter=\dfrac{a+b+c}{2}}$

$\sf{a \: , b \: and \: c \: are \: the \: sides \: of \: the \: triangle}$

Solution :-

$\sf{Let \: the \: sides \: are \: 12x \: , \: 17x \: and \: 25x.}$

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

$\sf{(12x + 17x + 25x) = 540 \: cm}$

$\sf{54x = 540}$

$\sf{x = 10}$

$\sf{So \: , \: the \: sides \: are -}$

$\sf{a = 12(10) = 120 \: cm}$

$\sf{b = 17(10) = 170 \: cm}$

$\sf{c = 25(10) = 250 \: cm}$

Now ,

$\sf{S =\dfrac{a + b + c}{2} =\dfrac{540}{2} = 270 \: cm}$

So ,

$\sf{Area \: of \: the \: triangle -}$

$\sf{\sqrt{270(270-120)(270-170)(270-250)}}$

$\sf{\sqrt{270\times 150\times 100\times 20}}$

$\sf{\sqrt{27\times 15\times 100\times 100\times 20}}$

$\sf{\sqrt{9\times (3\times 15)\times 10000\times 20}}$

$\sf{\sqrt{9\times 45\times 20\times 10000}}$

$\sf{\sqrt{9\times 900\times 10000}}$

$\sf{\sqrt{(3^{2}\times 30^{2}\times 100^{2})}}$

$\sf{\sqrt{(3\times 30\times 100)^{2}}}$

$\sf{\sqrt{(9000)^{2}}}$

$\sf{9000 \: cm^{2}}$

Result :-

$\sf{Area \: of \: the \: triangle \: is \: 9000 \: cm^{2}}$

Extra important Formulas :-

$\star$$\sf{Area \: of \: a \: right \: angled \: triangle =\dfrac{1}{2}\times base\times height}$

$\star$$\sf{Area \: of \: an \: isosceles \: triangle =\dfrac{b}{4}\sqrt{(4a^{2} - b^{2})}}$

Answer 2

$\sf \large \color{grey} \: QUESTION$

Sides of a triangle are in the ratio of 12 : 17 : 25 and it's perimeter is 540 cm . Find its area .

$\\ \sf \large \color{purple} \: GIVEN$

• Sides of triangle are in ratio as 12 : 17 : 25.
• It's perimeter is 540 cm

$\sf \large \color{red} \: REQUIRED \: TO \: FIND$

Area

$\\ \sf \large \color{pink} \: SOLUTION$

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

Here, s is the semi perimeter .

and a , b , c are the sides of the triangle.

Given Perimeter = 540 cm

$\sf \: semi \: perimeter = s = \dfrac{perimeter}{2} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s = \frac{540}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf s = 270 \: cm$

Given Ratio sides is 12 : 17 : 25

Let sides be a = 12x cm , b = 17x cm , c = 25x cm

Where x is any number.

Now,

Perimeter = 540 cm

a + b + c = 540

12x + 17x + 25x = rep

54 x = 540

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

x = 10

So,.

a = 12x cm = 12 × 10 = 120 cm

b = 17x cm = 17 × 10 = 170 cm

c = 25x cm = 25 × 10 = 250 cm

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

Putting a = 120 cm , b = 170 cm , c = 250 cm and S = 270 cm

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

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

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

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

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

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

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

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

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

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

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

$\bf \color{red} \sf \: thus \: area = 9000 {cm}^{2}$