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  • The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the Figure.

The dimensions of a rectangle ABCD are 51 cm×25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9:8, is cut 5 off from the rectangle as shown in the Figure. If the area of the trapezium PQCD is 6― th part of the area of the rectangle, find the lengths QC and PD.

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Solution.

Given, BC=51 cm and CD=25 cm
Area of rectangle ABCD=(51×25)cm2
In trapezium PQCD, parallel sides QC and PD are in the ratio 9:8
Let length of QC=9x and PD=8x
We have, Area of trapezium PQCD=56 th part of Area (ABCD)

12×( Sum of ‖ sides )× height =56× length × breadth 

12×(PD+QC)×CD=56×BC×CD12×(8x+9x)×25=56×51×2512×17x×25=56×51×25252×17x=56×51×25x=56×51×25×117×225

x=5PD=8x=8×5=40 cmQC=9x=9×5=45

The lengths of PD and QC are 40 cm and 45 cm respectively.

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