Question

In triangle ABC; angle A = 90^o, side AB = x cm, AC = (x + 5) cm and area = 150 cm². Find the sides of the triangle.

Answer 1

The area of the triangle is given as 150 sq.cm
`1/2 xx x xx ( x + 5 ) = 150`

x² + 5x - 300 = 0
( x + 20 )( x - 15 ) = 0
x = 15

Hence AB = 15cm, AC = 20 cm and
BC = `sqrt( 15^2 + 20^2 )` = 25 cm.