Question

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.

Answer 1

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
$\Rightarrow \frac{1}{2}\left(Sumofparallelsides\right)\times Distancebetweenparallelsides=\frac{5}{6}(BC\times CD)$
$\Rightarrow \frac{1}{2}(8x+9x)\times 25=\frac{5}{6}\times 51\times 25$
$\Rightarrow \frac{1}{2}\times 17x\times 25=\frac{5}{6}\times 51\times 25$
$\Rightarrow x=\frac{5\times 51\times 25\times 2}{25\times 17\times 6}$
$\Rightarrow x=5$
$\therefore$ Lenghts, QC = 9 $\times$ 5 = 45cm
And PD = 8 $\times$ 5 = 40cm.