## Equal Triangles

Two different triangles have the same area but different perimeters. Triangle $A$ has side lengths of $10,10$ and $6$ cm. Triangle $B$ has two sides with length of $10$ cm. Find the length of the third side of triangle $B$. Round your answer to the nearest tenth of a cm.
Source: mathcontest.olemiss.edu 9/9/2013

SOLUTION

Since triangle $A$ is isosceles, altitude $h_A$ is the perpendicular bisector of the base.
$h_A^2+3^2=10^2$
$h_A^2=91$
$h_A=\sqrt{91}$
Area of triangle $A$
$\frac{1}{2}\times 6\times \sqrt{91}=3\sqrt{91}$

Since triangle $B$ is isosceles, altitude $h_B$ is the perpendicular bisector of the base.
$h_B^2+x^2=10^2\qquad \left (1\right )$
Area of triangle $B$ = area of triangle $A$
$\frac{1}{2}\times 2x\times h_B=3\sqrt{91}$
$xh_B=3\sqrt{91}$
$x^2h_B^2=819$
$h_B^2=\frac{819}{x^2}$
Substitute the value of $h_B^2$ into Eq. $\left (1\right )$
$\frac{819}{x^2}+x^2=10^2$
$819+x^4=100x^2$
$x^4-100x^2+819=0$
The equation is quadratic in $x^2$
$x^2=\frac{100\pm\sqrt{6724}}{2}$
$=\frac{100\pm 82}{2}$
First solution
$x^2=\frac{100+82}{2}=91$
$x=\sqrt{91}$
Length of third side
$2x=2\sqrt{91}=19.1$ cm

Second solution
$x^2=\frac{100-82}{2}=9$
$x=3$
Length of third side
$2x=2\times 3=6$ cm which is the length of the base of triangle $A$.

Answer: $19.1$ cm