Question

find the area of a triangle whose sides are in ratio 5:12:13 and its perimeter is 60cm​

Answer 1

AnswER :

$\bf{\Large{\underline{\sf{Given\::}}}}$

Triangle whose sides are in the ratio 5:12:13 and its perimeter is 60 cm.

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The area of triangle.

$\bf{\Large{\underline{\tt{\pink{Explanation\::}}}}}$

Let the ratio be R.

$\bf{We\:have}\begin{cases}\sf{1st\:side\:of\:\triangle\:=\:5R}\\ \sf{2nd\:side\:of\:\triangle\:=\:12R}\\ \sf{3rd\:side\:of\:\triangle\:=\:13R}\end{cases}}$

A/q

$|\implies\sf{Perimeter\:of\:\triangle\:=\:60cm}\\\\\\|\implies\sf{5R+12R+13R\:=\:60}\\\\\\|\implies\sf{30R\:=\:60}\\\\\\|\implies\sf{R\:=\:\cancel{\dfrac{60}{30} }}\\\\\\|\implies\sf{\pink{R\:=\:2\:cm}}$

So,

• 1st triangle = (5*2)cm = 10 cm
• 2nd triangle = (12*2)cm = 24 cm
• 3rd triangle = (13*2)cm = 26 cm

$\bigstar\rm{\large{\underline{Using\:Heron's\:Formula\::}}}}$

$|\hookrightarrow\sf{Semi-perimeter\:(S)\:=\:\dfrac{A+B+C}{2} }\\\\\\|\hookrightarrow\sf{Semi-perimeter\:(S)\:=\:\dfrac{10cm+24cm+26cm}{2} }\\\\\\|\hookrightarrow\sf{Semi-perimeter\:(S)\:=\:\cancel{\dfrac{60cm}{2} }}\\\\\\|\hookrightarrow\sf{\pink{Semi-perimeter\:(S)\:=\:30cm}}$

Now,

$|\hookrightarrow\sf{Area\:of\:triangle\:=\:\sqrt{S(S-A)(S-B)(S-C)} }\\\\\\|\hookrightarrow\sf{Area\:of\:triangle\:=\:\sqrt{30(30-10)(30-24)(30-26)} }\\\\\\|\hookrightarrow\sf{Area\:of\:triangle\:=\:\sqrt{30(20)(6)(4)} }\\\\\\|\hookrightarrow\sf{Area\:of\:triangle\:=\:\sqrt{2*3*5*2*2*5*2*3*2*2}} \\\\\\|\hookrightarrow\sf{Area\:of\:triangle\:=\:(2*2*2*3*5)cm^{2} }\\\\\\|\hookrightarrow\sf{\pink{Area\:of\:triangle\:=\:120cm^{2} }}$

Thus,

The area of triangle is 120 cm².

Answer 2

Let the sides be =>

• a =5x
• b =12x
• c =13x

Perimeter =60cm (given)

So,

5x+12x+13x =60

=> 30x=60

=> x=2cm

NOW,

a=5x=5×2=10cm

b=12x=2×2=24cm

c=13x=13×2=26cm

$\mathbf \red{area \: of \: right \: traingle}$

1/2×base×height

=>1/2×24×10

=>120 cm²