In a trapezium ABCD, AB || CD, AB < CD, CD = 6 cm and distance between the parallel sides is 4 cm. If the area of ABCD is 16 cm2, then AB is
Let AB = $$x$$ cm and CD = 6 cm
Height of trapezium = $$h$$ = 4 cm
Area of trapezium = $$\frac{1}{2} \times $$ (sum of parallel sides) $$\times$$ height
=> $$\frac{1}{2} \times (x+6) \times 4=16$$
=> $$x+6=\frac{16}{2}=8$$
=> $$x=8-6=2$$ cm
=> Ans - (B)
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