If the perimeter of an isosceles triangle is 48 feet and the altitude
to the base is 12 feet, what is the area? (The method is more
important than the answer.)

We know that the area of a triangle is its base * height/2.  An
isosceles triangle has two sides of the same length and a third that
is different.  So, if the two sides have a length x and the third has
a length y, then 2x + y = 48.  By drawing the height h within the
isosceles triangle, we form two right trianges having sides h, y/2,
and a hypotenuse of x.   Using the Pythagorean Theorem, we know that
h^2 + (y/2)^2 = x^2.  h is given as 12, so after employing some
algebra, y^2 = 4x^2 - 576.  Using the perimeter formula, y^2 = (48 -
2x)^2, we then know that (48 - 2x)^2 = 4x^2 - 576.  Multiplying out
the left hand side of the equation yields 2304 - 192x + 4x^2. 
Therefore, 2304 - 192x + 4x^2 = 4x^2 - 576.  The x^2 terms cancel, so
x = 15.  Using the perimeter equation, we then know y = 18. 
Therefore, the area = 18 * 12/2 = 108 feet.  We can check by plugging
y/2 and h into the Pythagorean Theorem to solve for x.  The square
root of 81 + 144 = 15.

Sincerely,

Wonko