Question

$△ABC\sim △PQR,M$ is mid point of $BC$ and $N$ is mid point of $QR$. If the area of $△AC=100c{m}^2$, the area of $△PQR=144c{m}^2$ and $AM=4cm$, then find the length of $PN$

Answer 1

Using basic proportionality theorem, we can say that :

$\Rightarrow \frac{Ar\left(△ABC\right)}{Ar\left(△PQR\right)}=\frac{{\left(AM\right)}^2}{{\left(PN\right)}^2}$

$\frac{100}{144}=\frac{{\left(4\right)}^2}{{\left(PN\right)}^2}$

$\left(PN\right)=\sqrt{{\left(4\right)}^2\times \frac{144}{100}}$

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

Hence, the answer is $4.8cm$