Question 539462
Find the length of the leg of an isosceles triangle whose perimeter is 64 and altitude is 8.
:
Let x = the length of one of the equal sides
then
(64-2x) = the length of the 3rd side
(32-x) = half the length of the 3rd side
:
Two right triangles are formed with the altitude, half the 3rd side and the one leg of the equal side will be the hypotenuse (x)
:
x^2 = 8^2 + (32-x)^2
x^2 = 64 + 1024 - 64x + x^2
x^2 = 1088 - 64x + x^2
x^2 - x^2 + 64x = 1088
64x = 1088
x = {{{1088/64}}}
x = 17 is the length of the equal sides
then
64 - 2(17) = 30 is the length of the 3rd side
:
:
Confirm this, find the altitude (a)
a = {{{sqrt(17^2 - 15^2)}}}
a = 8