Q. 90 - Precalculus Enhanced with Graphing Utilities

Question

Geometry: Find formulas for the base b and one of the equal sides l of an isosceles triangle in terms of its altitude h and perimeter P.

Step-by-Step Solution

Answer

We have $l=\frac{4h^2+P^2}{4P}$ and $b=\frac{P^2-4h^2}{2P}$.

Step 1. Given information

For an isosceles triangle,

$h=\frac{\sqrt{4l^2-b^2}}{2}$

 and $P=2l+b$.

Step 2. Solve the system of equations.

From second equation,

$$\begin{gathered} P=2l+b \\ P-2l=b \\ b=P-2l \end{gathered}$$

Substitute this in first equation,

$$\begin{gathered} h=\frac{\sqrt{4l^2-{(P-2l)}^2}}{2} \\ {h}^2=\frac{4l^2-(P^2-4Pl+4l^2)}{4} \\ 4h^2=4l^2-P^2+4Pl-4l^2 \\ 4h^2=4Pl-P^2 \\ 4h^2+P^2=4Pl \\ \frac{4h^2+P^2}{4P}=l \end{gathered}$$

Step 3. Substitute l = 4 h 2 + P 2 4 P in b = P - 2 l .

We get

$$\begin{gathered} b=P-\frac{4h^2+P^2}{4P}\times 2 \\ =P-\frac{4h^2+P^2}{2P} \\ =\frac{P^2-4h^2}{2P} \end{gathered}$$