The dimensions of a rectangle ABCD are 51 cm 25 cm. A trapezium $P Q C D$ with its parallel sides QC and PD in the ratio $9 : 8$ , is cut off from the rectangle as shown in the figure. If the area of the trapezium PQCD is $\frac{5}{6}$ th part of the area of the rectangle, find the length QC.

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Solution

Area of trapezium = Sum of opposite sides $\times \frac{H e i g h t}{2}$
Given:
$P D = \frac{8}{9} Q C$
Area of trapezium = $(P D + Q C) \frac{D C}{2}$ = $\frac{17 \times 25}{18} Q C$
Area of Rectangle = $25 \times 51$
Now given that $\frac{A r e a o f T r a p e z i u m}{A r e a o f T r i a n g l e} = \frac{5}{6}$
Therefore, $\frac{\frac{17 \times 25}{18} Q C}{25 \times 51} = \frac{5}{6}$
Therefore, $L e n g t h o f Q C = 45$

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Q. Question 7
The dimensions of a rectangle ABCD are 51cm x 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. If the area of the trapezium PQCD is

\frac{5}{6}

th part of the area of the rectangle. Find the lengths QC and PD.

The dimensions of a rectangle ABCD are $51 c m \times 25 c m$ . 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 $\frac{5}{6}$ the area of the rectangle, find the lengths QC and PD.