Chapter 11.6, Problem 90AYU

To determine

To find: The formulas for the base $b$ and one of the equal sides $l$ of isosceles triangle in terms of its altitude $h$ and perimeter $P$.

Answer to Problem 90AYU

Solution:

$l = \frac{P^2 + 4h^2}{4P}$

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

Explanation of Solution

Given:

$length = l; base = b; altitude = h; perimeter = P.$

Formula Used:

$\text{Perimeter} \text{of} \text{an} \text{isosceles} \text{triangle} = P = 2l + b$

Calculation:

$P = 2l + b$   -----(Eq 1)

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

$4h^2 + b^2 - 4l^2 = 0$

Substituting Eq 1 in the above equation,

$\begin{array}{c}4h^2 + {(P - 2l)}^2 - 4l^2 = 0\\ 4h^2 + P^2 - 4Pl = 0\\ 4Pl = 4 h^2 + P^2\end{array}$

$l = \frac{P^2 + 4h^2}{4P}$

From Eq 1, $b = P - 2l$

$\begin{array}{c}b = P - 2\left(\frac{4 h^2 + P^2}{4P}\right)\\ = \frac{2P^2 - 4h^2 - P^2}{2P}\\ b = \frac{P^2 - 4h^2}{2P}\end{array}$

Therefore,

$\begin{array}{c}l = \frac{P^2 + 4h^2}{4P}\\ b = \frac{P^2 - 4h^2}{2P}\end{array}$