Question

In $\Delta ABC\sim \Delta PQR$, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $QR$. The area of $\Delta ABC=$ $100$ sq. cm and the area of $\Delta PQR=$ $144$ sq. cm. If $AM=4$ cm, then $PN$ is:

• A. $4.8$ cm
• B. $12$ cm
• C. $4$ cm
• D. $5.6$ cm

Answer 1

The correct option is A $4.8$ cm

$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}=\frac{100}{144}$

If triangles are similar then ratio of their areas is equal to ratio of square of their corresponding sides

$\frac{A{B}^2}{P{Q}^2}=\frac{100}{144}\frac{AB}{PQ}=\frac{10}{12}$

$AM$ and $PN$ are medians

Therefore, $\frac{AM}{PN}=\frac{AB}{PQ}$

$\Rightarrow \frac{4}{PN}=\frac{10}{12}\Rightarrow PM=4.8cm$