Question

The dimensions of a rectangle ABCD are $51cm\times 25cm$. 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.

• A. QC = 81cm, PD = 72cm
• B. QC = 36cm, PD = 32cm
• C. QC = 45cm, PD = 40cm
• D. QC = 54cm, PD = 48cm

Answer 1

The correct option is C

QC = 45cm, PD = 40cm

ABCD is a rectangle in which AB = 51cm and BC = 25 cm

Since parallel sides QC and PD are in the ratio 9:8, so let $QC=9x$ and $PD=8x$.

Area of a trapezium = $\frac{1}{2}$ x sum of parallel sides x distance between parallel sides

Now area of trapezium PQCD

$=\frac{1}{2}\times (9x+8x)\times 25c{m}^2$

$=\frac{1}{2}\times 17x\times 25$

Area of rectangle = length x breadth

Area of rectangle ABCD

$=BC\times CD=51\times 25$

It is given that area of trapezium PQCD

$=\frac{5}{6}\times \text{Area of rectangle ABCD}$.

$\therefore \frac{1}{2}\times 17x\times 25=\frac{5}{6}\times 51\times 25$

$\Rightarrow x=\frac{5}{6}\times 51\times 25\times 2\times \frac{1}{17\times 25}=5$

Hence the length $QC=9x=9\times 5=45cm$. and the length $PD=8x=8\times 5=40cm$.