# Find the sides of the triangle given its area and the sides of the similiar triangle

The triangle $\triangle ABC$ is similar to the triangle $\triangle A'B'C'$. The sides of the triangle $\triangle ABC$ are $a$, $b$ and $c$ and the sides of the triangle $\triangle A'B'C$ are $a'$, $b'$ and $c'$. The area of the triangle $\triangle ABC$ is $A=63$. The sides of the triangle $\triangle A'B'C'$ are $a'=45,b'=40,c'=13$. How can we find the sides $a,b,c$?

Unfortunately, I couldn't crack this problem altough I did try out a few approaches (all of them eventually led me to a dead end). But, if someone of you needs the area of the triangle $\triangle A'B'C'$ it is $A'=18\sqrt{861}$ (well, at least I think so).

• Could you show your workings on just one approach (regardless of your "dead end")? Jun 20, 2018 at 23:28

If the areas of two similar triangles have ratio X then corresponding side lengths will have ratio Sqrt(X). So if you are correct in your calculation of the area of triangle A'B'C', you should now have a solution for the side lengths of triangle ABC.

1) How to find the area of a triangle from its sides:

If a triangle has sides $a,b,c$ then perimeter is $a + b + c$ and the semi-perimeter is $p = \frac {a+b+c}2$ then the area of the triangle is $\sqrt{p(p-a)(p-b)(p-c)}$.

So... $a′=45,b′=40,c′=13$ so $p = \frac {45 + 40 + 13}2 = 49$.

So Area $\triangle A'B'C' = \sqrt{49(49-45)(49-40)(49-13)} =\sqrt{49*4*9*36} = 7*2*3*6=252$.

2) Areas are proportion to the squares of the proportions of the sides.

If $\frac a{a'} = \frac b{b'} = c{c'} = k$ then $\frac {Area \triangle ABC}{Area \triangle A'B'C' }= k^2$

So $\frac {Area \triangle ABC}{Area \triangle A'B'C' }= \frac {63}{252} = \frac {7*9}{7*2*3*6}= \frac 1{4}$.

So $k = \frac 12$

So $a = ka'=\frac 1245=\frac {45}2 ; b = kb'=\frac 1240=20; c = kc'=\frac 1213=\frac {13}2$

Hint:

If $w$ is the scale factor of the corresponding sides, then $w^2$ is the scale factor for the areas.

Since you know one of the two areas, and can find the other (Heron's formula), you can find $w^2$, hence you can find $w$.

By the way, for the other triangle, check your area calculation (I get $252$).

Your area calculation is wrong. Try it again with Heron's formula. Cross-check by plugging your values here. I get 252. Thus, $\triangle A'B'C'$ is 4 times the size of $\triangle ABC$. Thus, $a, b, c$ are half of $a', b', c'$.