Find the length of the sides in an Isosceles triangle, given its area and perimeter?

Question

How can I determine the length of the sides $b$, and the altitude $h$ of an isosceles triangle given its area $A= 24 \text{cm}^2$ and its perimeter $P= 16\text{cm}$?

Answer 1

Let $a$ be the length of one side and let $b$ be the length of the other two sides, and let $h$ be the altitude.

By Pythagorean Theorem we have that $$\left(\frac{1}{2}a\right)^2 + h^2 = b^2$$

We also have that $a + 2b = 16$ and $\dfrac{1}{2}ah = 24$.

Solve this system of $3$ equations.

Answer 2

Suppose such isosceles triangle exists. Denote $\displaystyle p =\frac{P}{2}$. Because\begin{gather*} P = 2p = a + 2b = 16,\\ S = \sqrt{p(p - a)(p - b)(p - b)} = 24, \end{gather*} then $\displaystyle b = 8 - \frac{a}{2}$ and$$ 24^2 = 8(8 - a)(8 - b)^2 = 8(8 - a) \left(\frac{a}{2}\right)^2 \Longrightarrow (8 - a) a^2 = 288. $$ However, by AM-GM inequality,$$ (8 - a) a^2 = 4 \cdot (8 - a) \cdot \frac{a}{2} \cdot \frac{a}{2} \leqslant 4 \cdot \left(\frac{(8 - a) + \frac{a}{2} + \frac{a}{2}}{3}\right)^3 = \frac{4 \times 8^3}{3^3} < 288, $$ a contradiction. Therefore, such isosceles triangle does not exist.