find the base of an isoceles triangle whose area is 48 square cm and of one of the equal sides is 10

Asked by  | 16th 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  | 29th Jun, 2008, 12:13: AM

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