<s>[INST] The leg of an isosceles triangle; given base and area 
The area of an isosceles triangle $\triangle ABC$ with base $AB=2c$ is $S.$ Find the leg of the triangle.

Let $AC=BC$ and $CH$ and $BP$ be the altitudes through $C$ and $B$, respectively.
I am not sure how to approach the problem. The area of the triangle is $S_{\triangle ABC}=\dfrac{AB.CH}{2}=\dfrac{AC.BP}{2}.$ What next?
 [/INST] Remember the definition of a right angle and the perpendicular bisector is that the perpendicular bisector of $AB$ is all the points $x$ so that $Ax = Bx$ and a right angle with vertex $M$ is one so that if $A,M,B$ are colinear with $AM = BM$ and $M$ between $A$ and $B$ then $\angle AMC$ is a right angle if and only if $AC = BC$.
So as the legs, $AC$ and $BC$ of an isosceles triangle are equal, then $C$ is on the perpendicular bisector of $AB$ and if $M$ is the midpoint of $AB$ then $\triangle AMC$ and $\triangle BMC$ are two congruent right triangles.
Use the Pythogorean Theorem to find $AC = BC$ via
$AC^2 = AM^2 + MC^2$
What is $AM$?

 $AB = 2c$ so $AM = \frac 12 AB = \frac 12 2c = c$.

What is $MC$?

 $MC$ is the height of $\triangle ABC$.  The area of a triangle is $\frac 12 base*height = \frac 12 AB*MC$.  The area of $\triangle ABC$ is $S$ and $AB =2c$ so $S = c*MC$.  So $MC = \frac Sc$.

So what is $AC$?

 $AC^2 = c^2 + (\frac Sc)^2$ so $AC =\sqrt{c^2 + (\frac Sc)^2}$.

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