Triangle ABC is isosceles with AB = BC. If AB = 20 and [ABC] = \(160\), then find the perimeter of triangle ABC.
+130071 AB = BC = 20
Let the altitude = BM
And this altitude will bisect the base
Let (1/2) of the base = AM
So by the Pythagorean Theorem, the height = sqrt [ 20^2 - AM^2]
And
Area = (1/2) base * height
160 = AM * sqrt [ 20^2 - AM^2] square both sides
25600 = AM^2 [ 400 - AM^2]
25600 = 400AM^2 - AM^4
AM^4 - 400AM^2 + 25600 = 0 factor as
(AM^2 - 320) ( AM^2 - 80) = 0
So either
AM^2 - 320 = 0 or AM^2 - 80 = 0
AM^2 = 320 AM^2 = 80
AM = sqrt (320) = 8sqrt( 5 ) AM = sqrt (80) = 4sqrt (5)
We have two possible perimeters
20 + 20 + 2 (8sqrt 5) = 40 + 16sqrt (5) units
or
20 + 20 + 2 ( 4 sqrt 5) = 40 + 8 sqrt (5) units
