Question

Triangle A has an area of `15` and two sides of lengths `4` and `9`. Triangle B is similar to triangle A and has a side of length `12`. What are the maximum and minimum possible areas of triangle B?

Answer 1

135 and `~~15.8`, respectively.

Explanation:

The tricky thing in this problem is that we do not know which of the tree sides of the original triangle corresponds to the one of length 12 in the similar triangle.

We know that the area of a triangle can be calculated from Heron's formula

`A = sqrt{s(s-a)(s-b)(s-x)}`

For our triangle we have `a=4` and `b=9` and so `s={13+c}/2`, `s-a = {5+c}/2`, `s-b={c-5}/2` and `s-c = {13-c}/2`. Thus

`15^2 = {13+c}/2 xx {5+c}/2 xx {c-5}/2 xx {13-c}/2`

This leads to a quadratic equation in `c^2` :

`c^4 - 194 c^2 + 7825 = 0`

which leads to either `c ~~ 11.7` or `c ~~ 7.5`

So the maximum and minimum possible value for the sides of our original triangle are 11.7 and 4, respectively. Thus the maximum and minimum possible value of the scaling factor are `12/4=3` and `12/11.7~~ 1.03`. Since area scales as square of length, the maximum and minimum possible values of the area of the similar triangle are `15 xx 3^2 = 135` and `15 xx 1.03^2 ~~ 15.8`, respectively.