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.

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Solution

We have dimensions of a rectangle ABCD as 51cm and 25cm. also, in trapezium PQCD, parallel sides QC and PD are in the ratio 9:8.
i.e., QC : PD = 9 : 8
let, length QC = 9x
and PD = 8x

According to the question,
Area of trapezium PQCD = $\frac{5}{6}$ area of rectangle ABCD
$\Rightarrow \frac{1}{2} (S u m o f p a r a l l e l s i d e s) \times D i s t a n c e b e t w e e n p a r a l l e l s i d e s = \frac{5}{6} (B C \times C D)$
$\Rightarrow \frac{1}{2} (8 x + 9 x) \times 25 = \frac{5}{6} \times 51 \times 25$
$\Rightarrow \frac{1}{2} \times 17 x \times 25 = \frac{5}{6} \times 51 \times 25$
$\Rightarrow x = \frac{5 \times 51 \times 25 \times 2}{25 \times 17 \times 6}$
$\Rightarrow x = 5$
$∴$ Lenghts, QC = 9 $\times$ 5 = 45cm
And PD = 8 $\times$ 5 = 40cm.

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