linear solve plzzzzzzzzzzzz

Asked by  | 4th Jul, 2009, 11:13: AM

Expert Answer:

Let the altitude of the triangle be 'a' and the base of the triangle be 'b'.

According to the question,

a = b - (5/2)

  = (2b - 5)/2

Area of a Trinagle = (1/2) x Base x Altitude

                                 = (1/2) x b x a

On substitution,

22 = (1/2) x b x (2b - 5)/2

      = (1/4) (2b2 - 5b)

22 x 4 = 2b2 - 5b

Therefore, 2b2 - 5b - 88 = 0

On solving the above equation, we get

2b2 - 5b - 88 = 0

2b2 - 16b + 11b - 88 = 0

2b (b - 8) + 11 ( b - 8) = 0

(2b + 11) ( b - 8) = 0

So, base (b)  = 8 cm                     or                  Base (b)= -11/2 cm (rejected, because length cannot be negative)

Therefore, altitude (a) = (2b - 5)/2 

                                        = 11 /2 = 5.5cm

 

Answered by  | 4th Jul, 2009, 04:47: PM

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