## Isosceles Triangle

Isosceles triangle $ABC$ has a base of length $8$ units. $\overline{BD}$ is the altitude to the base, and $E$ is the midpoint of side $\overline{BC}$. The triangular area $BDE$ is three square units. Find the perimeter of triangle $ABC$.

Source: NCTM Mathematics Teacher 2008

SOLUTION

Let $F$ be the midpoint of $\overline{AB}$. Triangle $FBE$ is an isosceles triangle and its area is one fourth the area of triangle $ABC$ because its area equals the area of triangle $BDE$.
Area of $ABC=4$ times area of $BDE$
$\dfrac{1}{2}AC\cdot BD=4\cdot 3$
$\dfrac{1}{2}8\cdot BD=12$
$BD=3$
Altitude $\overline{BD}$ in an isosceles triangle is also a perpendicular bisector.
$BC^2=BD^2+DC^2$ by the Pythagorean theorem
$=3^2+4^2$
$=25$
$BC=5$
Perimeter of triangle $ABC$
$8+5+5=18$

Answer: $18$ length units