Question

In the figure, $PQ\parallel BC$ and AP : PB = 1 : 2. Find $\frac{ar\left(\Delta APQ\right)}{ar\left(\Delta ABC\right)}$.

• A. 1 : 4
• B. 4 : 1
• C. 1 : 9
• D. 2 : 9

Answer 1

Answer: (C)

1 : 9

If $AP:PB=1:2$ then $AP:AB$ will be $1:3$.

So, in similar triangles, the ratio of the area of $\frac{\Delta APQ}{\Delta ABC}={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$