Question 70500
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