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Q.4 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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866Alka

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

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SOLUTION

Ratio of the sides of the triangle = 12 : 17 : 25Let the common ratio be x then sides are 12x, 17x and 25xPerimeter of the triangle = 540cm12x + 17x + 25x = 540 cm⇒ 54x = 540cm⇒ x = 10Sides of triangle are,12x = 12 × 10 = 120cm17x = 17 × 10 = 170cm25x = 25 × 10 = 250cmSemi perimeter of triangle(s) = 540/2 = 270cmUsing heron's formula,Area of the triangle = √s (s-a) (s-b) (s-c) = √270(270 - 120) (270 - 170) (270 - 250)cm2 = √270 × 150 × 100 × 20 cm2 = 9000 cm2

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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)}

= 9000cm^2.

Step-by-step explanation:

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Answer 2

AnswEr :

• Perimeter = 540 cm

• Ratio of sides = 12 : 17 : 25

Let us assume that the sides be 12x, 17x & 25x.

⋆ Reference of Diagram is given below :

$\setlength{\unitlength}{1.8cm}\begin{picture}\thicklines\put(8,1){\line(1,0){3}}\put(8,1){\line(1,1){1.5}}\put(11,1){\line(-1,1){1.5}}\put(8,1){\line(1,1){1.5}}\put(8.7,1.4){\sf{Perimeter = 540cm}}\put(11,1){\line(-1,1){1.5}}\put(8.1,1.8){\sf{12x cm}}\put(10.3,1.8){\sf{17x cm}}\put(9.3,0.75){\sf{25x cm}}\end{picture}$

We know that,

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

For this, we need to know the Semi perimeter.

$\sf\longrightarrow \: S = \dfrac{Perimeter}{2}$

$\sf\longrightarrow \: S = \dfrac{540}{2}$

$\sf\longrightarrow \: { \blue{S = 270 \: cm}}$

Now,

$\sf\longrightarrow \: Perimeter = 540 \: cm \\ \\ \\ \sf\longrightarrow \: \angle1 + \angle2 + \angle3 = 540 \\ \\ \\ \sf\longrightarrow \:12x + 17x + 25x = 540 \\ \\ \\ \sf\longrightarrow \:54x = 540 \\ \\ \\ \sf\longrightarrow \:x = \dfrac{540}{54} \\ \\ \\ \sf\longrightarrow \: \large{ \boxed{ \frak{ \pink{x = 10 \: cm}}}}$

So, the required values of angles are :

• 12x = 12 × 10 = 120 cm

• 17x = 17 × 10 = 170 cm

• 25x = 25 × 10 = 250 cm

Since, we've measures of angle now ! We can calculate the required area of the triangle.

$\longrightarrow \sf \: Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \\ \longrightarrow \sf \: Area = \sqrt{270(270 - 120)(270 - 170)(270 - 250)} \\ \\ \\ \longrightarrow \sf \: Area = \sqrt{270 \times 150 \times 100 \times 20} \\ \\ \\ \longrightarrow \sf \: Area = \sqrt{27 \times 15 \times 2 \times {10}^{5} } \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg(taking \: 10 \: as \: common \bigg) \\ \\ \\ \longrightarrow \sf \: Area = \sqrt{27 \times 30 \times {10}^{5} } \\ \\ \\ \longrightarrow \sf \: Area = \sqrt{27 \times 3 \times {10}^{6} } \\ \\ \\ \longrightarrow \sf \: Area = \sqrt{81 \times {10}^{6} } \\ \\ \\ \longrightarrow \sf \: Area = 9 \times \sqrt{ {10}^{6} } \\ \\ \\ \longrightarrow \sf \: Area = 9 \times{10}^{{6}^{ \frac{1}{2} } } \\ \\ \\ \longrightarrow \sf \: Area = 9 \times{10}^{3} \\ \\ \\ \longrightarrow \sf \: Area = 9 \times1000 \\ \\ \\ \longrightarrow \sf \: \large{ \underline{ \boxed{ \frak{ \pink{Area = 9000 \: {cm}^{2} }}}}} \: \bigstar$

Therefore, Area of the required Δ is 9000 cm².