the length of side of a triangle are in ratio 2:3:4 and it perimeter i

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

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

Answer»

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 :

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;

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