linear solve plzzzzzzzzzzzz
Asked by
| 4th Jul, 2009,
11:13: AM
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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