+45 Triangle $ABC$ is isosceles with $AB = BC.$ If $AC = 20$ and $[ABC] = 150,$ then find the perimeter of triangle $ABC.$
+1459 Since △ABC is isosceles, let AB=AC=x. Then the area of △ABC is [[ABC] = \frac{1}{2} \cdot x \cdot h,]where h is the altitude drawn from A to BC.
Setting the area equal to 150, we get [\frac{1}{2} \cdot x \cdot h = 150.]
Then xh=300. However, by the Pythagorean Theorem, [x^2 = (h + 8)^2 = h^2 + 16h + 64,]so x2−h2=16h+64.
Substituting 300 for xh, we get [x^2 - 300 = 16h + 64.]Then x2−16h−364=0.
By Vieta's formulas, the sum of the roots of this quadratic is 16, so x=20.
Hence, the perimeter is x+x+8=48.
