Question

$ABCD$ is a square with each side $12cm$. $P$ is a point on $BC$ such that area of $△ABP:$ area of trapezium $APCD=1:5$. Find the length of $CP$.

• A. $8cm$
• B. $9cm$
• C. $11cm$
• D. $17cm$

Answer 1

The correct option is A $8cm$

$ABCD$ is a square with each side $12cm$ square.
$P$ is a point on side $BC$.
$\frac{\text{Area of}△ABP}{\text{Area of trapezium}APCD}=\frac{1}{5}$

Let $CP=x$
$\therefore PB=12-x$
Now,
Area of $△ABP=\frac{1}{2}\times AB\times PB$
$=\frac{1}{2}\times 12\times (12-x)$
$=6(12-x)$
Area of trapezium $APCD=\frac{(AD+CD)}{2}DC$
$=\frac{(12+x)}{2}12$
$=(12+x)6$
Now,
$\frac{6(12-x)}{(12+x)6}=\frac{1}{5}$
$=>60-5x=12+x$
$=>6x=48$
$=>x=8$
$\therefore$ Length of $CP=8cm$