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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 $A B C D$ are $51 \mathrm{~cm} \times 25 \mathrm{~cm}$. 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 $\overline{6}$ th part of the area of the rectangle, find the lengths QC and PD.

Answers (1)

Solution.

Given, $B C=51 \mathrm{~cm}$ and $C D=25 \mathrm{~cm}$
Area of rectangle $A B C D=(51 \times 25) \mathrm{cm}^2$
In trapezium $P Q C D$, parallel sides $Q C$ and $P D$ are in the ratio $9: 8$
Let length of $Q C=9 x$ and $P D=8 x$
We have, Area of trapezium $P Q C D=\frac{5}{6}$ th part of Area $(A B C D)$

$
\frac{1}{2} \times(\text { Sum of } \| \text { sides }) \times \text { height }=\frac{5}{6} \times \text { length } \times \text { breadth }
$

$\begin{aligned} & \frac{1}{2} \times(P D+Q C) \times C D=\frac{5}{6} \times B C \times C D \\ & \frac{1}{2} \times(8 x+9 x) \times 25=\frac{5}{6} \times 51 \times 25 \\ & \frac{1}{2} \times 17 x \times 25=\frac{5}{6} \times 51 \times 25 \\ & \frac{25}{2} \times 17 x=\frac{5}{6} \times 51 \times 25 \\ & x=\frac{5}{6} \times 51 \times 25 \times \frac{1}{17} \times \frac{2}{25}\end{aligned}$

$
\begin{aligned}
& x=5 \\
& P D=8 x \\
& =8 \times 5 \\
& =40 \mathrm{~cm} \\
& Q C=9 x \\
& =9 \times 5=45
\end{aligned}
$

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

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infoexpert21

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