Question

The area of two similar triangles are 50dm^2 and 32dm^2. The sum of their perimeters is 117dm. What us the perimeters of each of these triangle?

Answer 1

Answer:

Perimeter of one triangle is 65 dm

Perimeter of other triangle is 52 dm

Step-by-step explanation:

Please remember the concept

If sides are in the ratio of a:b

Then the area in the ratio of $a^{2} :b^{2}$

It is given sum of their perimeter is 117.

Let the small triangle has perimeter as x.

So, perimeter of big triangle is 117-x.

So, we can set up equation as

$\frac{50}{32} =\frac{x^{2} }{(117-x)^{2} }$

Cross multiply

50(117-x)^2 =32x^2

Expand the left side

50$(13689 -234x+x^{2} )$=$32x^{2}$

Distribute the left side

684450-11700x+$50x^{2}$=$32x^{2}$

Subtract both sides $32x^{2}$ and rewrite it $18x^{2} -11700x+684450=0$

Solve this quadratic for x.

Divide both sides of the equation by 18 to simplify.

$x^{2}$-650 x+38025=0

Now, if possible let's factor

Find two integers whose multiplication is 38025 but adds to -650.

-65 and -585 works.

So, we can rewrite it as

(x -65)(x-585) =0

Solve them using zero product property

x=65, x=585

So, x=65 works here.

So, perimeter of one triangle is 65 dm

Perimeter of other triangle is 117-65= 52 dm