CBSE Class 9 Answered
find the base of an isoceles triangle whose area is 48 square cm and of one of the equal sides is 10
Asked by | 16 May, 2008, 12:45: PM
Expert Answer
Let the 3rd side = x
S = (x+10+10)/2 = x/2 + 10
48 = 
[(x/2 + 10) (x/2 + 10-10) (x/2 + 10-10) (x/2 + 10 - x)]
48 = (x/2 + 10)(x/2)(x/2)(-x/2 + 10)
48 = (100 - x2/4)(x2/4)
(100 - x2/4)(x2/4) = 482
400x2/16 - x4/16 = 2304
400x2 - x4 = 36864
x4-400x2 +36864 = 0
(x2-256)(x2-144) = 0
x2 = 256 or x2 = 144
x = 16 or x = 12
Answered by | 29 Jun, 2008, 12:13: AM
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