Math, asked by haariskhan1978, 9 months ago

the length of side of a triangle are in ratio 2:3:4 and it perimeter is 144 cm find the area of the triangle and the height corresponding to longest side​

Answers

Answered by savitafamily289
17

Answer:

Step-by-step explanation:

Ratios are as follows

2:3:4

Let's take the common multiple between their ratios as x

Therefore the equation will be

2x+3x+4x

Perimeter given was 144cm

Let's put it in the equation we made earlier

2x+3x+4x=144

9x=144

X=144/9

X=16

Now we got x so now we can find all numbers

2x= 2×16=32

3x=3×16=48

4x=4×16=64

Answered by Anonymous
38

Solution :

\bf{\red{\underline{\underline{\bf{Given\::}}}}}

The length of side of a triangle are in the ratio 2:3:4 and it perimeter is 144 cm.

\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}

The area of the triangle and the height corresponding to longest side.

\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}

Let the ratio be r

The sides of the triangle :

  • 2r
  • 3r
  • 4r

We know that formula of the perimeter of triangle :

\mapsto\sf{\orange{Perimeter\:of\:triangle=Side+Side+Side}}\\\\\mapsto\sf{144cm=2r+3r+4r}\\\\\mapsto\sf{144cm=9r}\\\\\mapsto\sf{r=\cancel{\dfrac{144}{9}} }cm\\\\\mapsto\sf{\pink{r=16\:cm}}

Now;

  • 1st side = 2(16) = 32 cm
  • 2nd side = 3(16) = 48 cm
  • 3rd side = 4(16) = 64 cm

\dag\bf{\underline{\underline{\bf{Using\:Heron's\:Formula\::}}}}}

\mapsto\sf{\orange{Sem-perimeter=\dfrac{a+b+c}{2} }}\\\\\\\mapsto\sf{Semi-perimeter=\dfrac{32cm+48cm+64cm}{2} }\\\\\\\mapsto\sf{Semi-perimeter=\cancel{\dfrac{144}{2}} cm}\\\\\\\mapsto\sf{\pink{Semi-perimeter=72cm}}

So;

\mapsto\sf{Area\:_{triangle}=\sqrt{s(s-a)(s-b)(s-c)} }\\\\\mapsto\sf{Area\:_{triangle}=\sqrt{72(72-32)(72-48)(72-64)}}\\ \\\mapsto\sf{Area\:_{triangle}=\sqrt{72(40)(24)(8)} }\\\\\mapsto\sf{Area\:_{traingle}=\sqrt{552960} cm^{2} }\\\\\mapsto\sf{\pink{Area\:_{triangle}=743.61\:cm^{2} }}

Now;

\leadsto\sf{\orange{Area\:of\:triangle=\frac{1}{2} \times base\times height}}\\\\\\\leadsto\sf{743.61cm^{2} =\dfrac{1}{\cancel{2}} \times \cancel{64}cm\times h}\\\\\\\leadsto\sf{743.61\:cm^{2} =32\times h}\\\\\\\leadsto\sf{h=\cancel{\dfrac{743.61cm^{2} }{32cm} }}\\\\\\\leadsto\sf{\pink{h=23.23\:cm}}

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