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

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 $P= 16\text{cm}$? • Does the triangle have that all 3 sides are equal (such a triangle is called equilateral)? – He Is Alive gamme Jan 11 '18 at 8:23
• No. An Isosceles triangle has two equal sides and one base different. – ferret96 Jan 11 '18 at 8:26

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.
• Solving the system we notice that $a,h<0$, i.e such triangle does not exist. – Galc127 Jan 11 '18 at 8:56
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.