Question

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

• A. $1$
• B. $5$
• C. $8$
• D. $10$

Answer 1

The correct option is C $8$

Given side of square $12$ cm

Then area of square = $12\times 12=144c{m}^2$

Given the area of triangle ABP $:$ area of trapezium APCD $=1:5$

Then the area of triangle ABP = $\frac{144}{6}=24c{m}^2$

Let the length of CP is $x$

Then area of trapezium APCD = $12\times \frac{12+x}{2}=6(12+x)c{m}^2$

But as per question area of triangle ABP : area of trapezium APCD = 1:5

Then

$\frac{24}{6(12+x)}=\frac{1}{5}$

$6(12+x)=24\times 5=120$

Or $6(12+x)=120$

Or $72+6x=120$

Or $6x=120-72=48$

Or $x=8cm$