The dimensions of a rectangle ABCD are 51 cm×25 cm. A trapezium with its parallel sides QC and PD in the ratio 9:8, is cut of from the rectangle as shown in figure. If the area of the trapezium PQCD is 56 the area of the rectangle, find the lengths QC and PD.
QC = 45cm, PD = 40cm
ABCD is a rectangle in which AB = 51cm and BC = 25 cm
Since parallel sides QC and PD are in the ratio 9:8, so let QC=9x and PD=8x.
Now area of trapezium PQCD
=12×(9x+8x)×25 cm2
=12×17x×25
Area of rectangle ABCD
=BC×CD=51×25
It is given that area of trapezium PQCD
=56×Area of rectangle ABCD.
∴12×17x×25=56×51×25
⇒x=56×51×25×2×117×25=5
Hence the length QC=9x=9×5=45cm. and the length PD=8x=8×5=40cm.