The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is
Started 1 month ago by Shashank in
Explanatory Answer
Let AB = DP = PC = x
Let the height of the parallelogram = h
Area of the parallelogram ABPD - Area of the triangle BPC = 10
xh – ½(xh) = 10
xh = 20
Area of the trapezium ABCD = xh + ½(xh) = 3/2(xh) = 30
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