Q5 of 77 Page 15

In Fig. 15.78, ABCD is a trapezium in which AB=7 cm, AD=BC=5 cm, DC=x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Draw AL perpendicular to DC

And,


BM perpendicular DC


Then,


AL = BM = 4 cm


And,


LM = 7 cm


In Δ ADL, we have


AD2 = AL2+ DL2


25 = 16 + DL2


DL = 3 cm


Similarly,


MC =


=


= 3 cm


Therefore,


x = CD = CM + ML + LD


= 3 + 7 + 3


= 13 cm


Area of trapezium ABCD = (AB + CD) * AL


= (7 + 13) * 4


= 40 cm2


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