+887 An isosceles triangle has two sides of length $7$ and an area of $14.$ What is the product of all possible values of its perimeter?
+729 Let the base of the triangle be $b$ and the height be $h.$ Since the area of the triangle is 14, we have
\[\frac{bh}{2} = 14 \Rightarrow bh = 28.\]
By the Pythagorean Theorem, we have
\[b^2 = 7^2 - \left(\frac{b}{2}\right)^2 = 49 - \frac{b^2}{4}.\]
Multiplying both sides by 4, we get
\[4b^2 = 196 - b^2 \Rightarrow 5b^2 = 196 \Rightarrow b = \frac{4\sqrt{5}}{5}.\]
The perimeter of the triangle is $2b+7 = 2\left(\frac{4\sqrt{5}}{5}\right) + 7 = \frac{8\sqrt{5}}{5} + 7.$ The product of all possible values of the perimeter is
\[\left(\frac{8\sqrt{5}}{5} + 7\right)\left(\frac{-8\sqrt{5}}{5} + 7\right) = 49 - \frac{320}{25} = \boxed{\frac{825}{25} = 33}.\]
