Question

The dimensions of a rectangle $ABCD$ are $51cm\times 25cm$. A trapezium $PQCD$ having its parallel side $QC$ and $PD$ in the ratio $9:8$ is cut off from the rectangle as shown in the given figure. If the area of the trapezium $PQCD$ $=\frac{5}{6}th$ part of the rectangle, find the lengths of $QC$ and $PD$.

Answer 1

Area of rectangle $ABCD=51cm\times 25cm=1275{cm}^2$
Given, the sides $QC$ and $PD$ of the trapezium $PQCD$ are in the ratio $9:8$.

Let the sides be $9x$ cm and $8x$ cm.
Area of trapezium $PQCD=\frac{1}{2}\times (QC+PD)\times height$

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

Again, it is given that area of trapezium $PQCD=\frac{5}{6}$ of area of rectangle

$\Rightarrow \frac{1}{2}\times (9x+8x)\times 25$$=\frac{5}{6}\times 1275$

$\Rightarrow 17x=85\Rightarrow x=5\therefore QC=9x=9\times 5cm=45cm$

And $PD=8x=8\times 5cm=40cm$