Question

The perimeter of a triangle is 144 m. If one of its sides is 48 m and the remaining (4)

two sides are in the ratio 3 : 5, find the area of the triangle.​

Answer 1

Given:

Perimeter of a triangle = 144 m

One side = 48 m

The ratio of other two sides = 3 : 5

let the actual side be 3x and 5x

We know, the perimeter = a + b + c

144 = 48 + 3x + 5x

144 - 48 = 8x

96 = 8x

96 / 8 = x

12 = x

So now the other two sides are:

3x => 3 × 12

=> 36 m

5x => 5 × 12

=> 60 m

Semi - perimeter is half of the perimeter i.e,

s = 144/2

= 72

So the area of a ️= under root s× (s - a) (s - b)(s - c)

just put the values...

for the area refer to the attachment

hope this helped you dear...

Answer 2

Step-by-step explanation:

We have,

•First side of the triangle,a = 48 m

•Let, second side of the triangle be,b=3x

•Therefore, the third side of the triangle,c=5x

The perimeter of the triangle = 144 m

$\sf{\implies{a+b+c=144 \:m}}$

$\sf{\implies{48+3x+5x=144}}$

$\sf{\implies{8x=144-48}}$

$\sf{\implies{8x=96}}$

$\sf{\implies{x=\cancel{\frac{96}{8}}}}$

$\sf{\implies{\boxed{\sf{x=12}}}}$

Now substituting the value of x in eqn (i) and (ii),we get,

second side,$\sf{b=3\times{12} = \boxed{\sf{36 \:m}}}$

and third side,$\sf{c=5\times{12} = \boxed{\sf{60 \:m}}}$

Now,,

semiperimeter of the triangle,

$\sf{s=\frac{a+b+c}{2}}$

$\sf{\implies{s=\frac{144}{2}}}$

$\implies{\boxed{\sf{s=72\:m}}}$

Therefore the area of the triangle is given by,

$\sf{ar(∆)=\sqrt{s(s-a)(s-b)(s-c)}}$

$\implies{\sf{ar(∆)=\sqrt{72(72-48)(72-36)(72-60)}}}$

$\implies{\sf{ar(∆)=\sqrt{72\times{24}\times{36}\times{12}}}}$

$\implies{\boxed{\sf{ar(∆)=864\:m^{2}}}}$

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HOPE IT HELPS ❣️❣️❣️