Question 70500
2X+Y=32
Y=32-2X 
THE EQUAL SIDES ARE X AND THE BASE IS Y
THEREFORE A TRIANGLE IS FORMED BY 1/2 BASE (32-2X)/2 AND THE OTHER SIDE IS THE ALTITUDE OF 8.
THEREFORE WE HAVE A TRIANGLE WITH SIDES 8, (32-2X)/2 & X.
USING THE PATAGOREAN THEROM OF A^2+B^2=C^2 
8^2+(32-2X)/2)^2=X^2
64+(16-X)^2=X^2
64+256-32X+X^2=X^2 CANCELLING OUT THE X^2 WE GET
-32X=-256-64
-32X=-320
X=-320/-32
X=10 FOR THE TWO EQUAL LEGS THUS 
2X+Y=32
2*10+Y=32
Y=32-20
Y=12 FOR THE BASE