SOLUTION: Find the legs of an isosolese triangle if the perimeter is 32 and the altitude to the base is 8.

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Question 70500: Find the legs of an isosolese triangle if the perimeter is 32 and the altitude to the base is 8.
Found 2 solutions by checkley75, venugopalramana:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
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


Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Find the legs of an isosolese triangle if the perimeter is 32 and the altitude to the base is 8.
LET THE EQUAL LEGS BE = X
LET THE BASE BE = B
PERIMETER=2X+B=32.....B=32-2X=2(16-X)
ALTITUDE TO BASE = SQRT[X^2-(B/2)^2]=8
X^2-B^2/4=64
X^2-[2(16-X)]^2/4=64
X^2-256+32X-X^2=64
32X=64+256=320
X=10
HENCE EQUAL LEG = 10
BASE =32-20=12