ABCD is a trapezium in which AB = (3x+2) cm. DC = (x+3) cm and AD= (x-1) cm.?

Is 'AD' the height of the trapezium? [Unless AD is the height, the end result may not be possible to show].

Area of a trapezium = (1/2) x (height) x (Sum of the two parallel sides)

==> Area here = (1/2) x AD x (AB + CD)

==> (1/2)(x-1)(3x + 2 + x + 3) = 10

==> (x - 1)(4x + 5) = 20

Expanding and simplifying, 4x^2 + x - 25 = 0
 
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