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}](https://tex.z-dn.net/?f=a%5E%7B2%7D%20%3Ab%5E%7B2%7D)
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} }](https://tex.z-dn.net/?f=%5Cfrac%7B50%7D%7B32%7D%20%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%28117-x%29%5E%7B2%7D%20%7D)
Cross multiply
50(117-x)^2 =32x^2
Expand the left side
50
=![32x^{2}](https://tex.z-dn.net/?f=32x%5E%7B2%7D)
Distribute the left side
684450-11700x+
=![32x^{2}](https://tex.z-dn.net/?f=32x%5E%7B2%7D)
Subtract both sides
and rewrite it ![18x^{2} -11700x+684450=0](https://tex.z-dn.net/?f=18x%5E%7B2%7D%20-11700x%2B684450%3D0)
Solve this quadratic for x.
Divide both sides of the equation by 18 to simplify.
-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