# 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 7 . What are the maximum and minimum possible areas of triangle B?

Apr 27, 2018

There's a possible third side of around $11.7$ in triangle A. If that scaled to seven we'd get a minimal area of $\frac{735}{97 + 12 \sqrt{11}}$.

If the side length $4$ scaled to $7$ we'd get a maximal area of $\frac{735}{16.}$

#### Explanation:

This is perhaps a trickier problem than it first appears. Anybody know how to find the third side, which we seem to need for this problem? Normal trig usual makes us calculate the angles, making an approximation where none is required.

It's not really taught in school, but the easiest way is Archimedes' Theorem, a modern form of Heron's Theorem. Let's call A's area $A$ and relate it to A's sides $a , b$ and $c .$

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$

$c$ only appears once, so that's our unknown. Let's solve for it.

${\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2} = 4 {a}^{2} {b}^{2} - 16 {A}^{2}$

${c}^{2} = {a}^{2} + {b}^{2} \setminus \pm \sqrt{4 {a}^{2} {b}^{2} - 16 {A}^{2}}$

We have $A = 15 , a = 4 , b = 9.$

${c}^{2} = {4}^{2} + {9}^{2} \setminus \pm \sqrt{4 \left({4}^{2}\right) \left({9}^{2}\right) - 16 {\left(15\right)}^{2}} = 97 \setminus \pm \sqrt{1584}$

$c = \sqrt{97 \setminus \pm 12 \sqrt{11}}$

$c \approx 11.696 \mathmr{and} 7.563$

That's two different values for $c$, each of which should give rise to a triangle of area $15$. The plus sign one is of interest to us because it's larger than the other two sides.

For maximal area, maximal scaling, that means the smallest side scales to $7$, for a scale factor of $\frac{7}{4}$ so a new area (which is proportional to the square of the scale factor) of ${\left(\frac{7}{4}\right)}^{2} \left(15\right) = \frac{735}{16}$

For minimal area the largest side scales to $7$ for a new area of

$15 {\left(\frac{7}{\sqrt{97 + 12 \sqrt{11}}}\right)}^{2} = \frac{735}{97 + 12 \sqrt{11}}$