Triangle ABC is isosceles with AB = BC. If AC = 20 and [ABC] = 280, then find the perimeter of triangle ABC.
B
A 20 C
Let M be the midpoint of AC
area = (1/2)AC ( BM)
280 = (1/2) 20 (BM)
280 = 10 * BM
280 /10 = BM
28 = BM
Using the P Theorem
AB = sqrt ( AM^2 + BM^2) = sqrt ( 10^2 + 28^2) = sqrt (884) = sqrt (4 * 221) = 2sqrt (221)
Perimeter =
AB + BC + AC
2sqrt (221) + 2sqrt (221) + 20 =
4sqrt (221) + 20 ≈ 79.46