## Two Isoceles Triangles

Two noncongruent triangles have the same area but different perimeters. Triangle A has sides lengths of 8, 8 and 6 cm. Triangle B has two sides with length of 8 cm. Find the length of the third side of Triangle B. Round your answer to the nearest hundredth of a cm.
Source: mathcontest.olemiss.edu 10/2/2006

SOLUTION

In isosceles $\triangle ADE$ , altitude $\overline{AC}$ is also the perpendicular bisector of base $\overline{DE}$.
In right $\triangle ACD$,
$AC^2+CD^2=AD^2$
$AC^2+3^2=8^2$
$AC^2+9=64$
$AC^2=64-9=55$
$AC=\sqrt{55}$
Area of $\triangle ADE=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times DE\times AC$
$=\frac{1}{2}\times 6\times \sqrt{55}$
$=3\sqrt{55}$

In isosceles $\triangle BFG$, altitude $\overline{BE}$ is also the perpendicular bisector of base $\overline{FG}$.
Let $FG=2x$.
Area of $\triangle BFG=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times FG\times BE$
$=\frac{1}{2}\times 2x\times BE$
$=x\times BE$
Since the two triangles have the same area
$3\sqrt{55}=x\times BE$
$BE=\frac{3\sqrt{55}}{x}$

In right $\triangle BEF$,
$BE^2+EF^2=BF^2$
$\left (\frac{3\sqrt{55}}{x}\right )^2+x^2=8^2$
$\frac{495}{x^2}+x^2=64$
$495+x^4=64x^2$        multiply by $x^2$
$x^4-64x^2+495=0$        (1)

Let $u=x^2$. Then, Eq.(1) has the following quadratic form
$u^2-64u+495=0$
$\left (u-9\right )\left (u-55\right )=0$
factoring the equation
There are two solutions
$u=9$ and $u=55$
If $u=9$, then $x^2=9$ which implies that $x=3$. In this case,
$FG=2x=6$. We reject this solution because the two triangles are noncongruent.
If $u=55$, then $x^2=55$ which implies that $x=\sqrt{55}$. In this case,
$FG=2x$
$=2\sqrt{55}=14.83$

Answer: $14.83$