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 10 months 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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