Question

Let $ABC$ be a triangle with $\angle C=90^{\circ }$. Draw $CD$ perpendicular to $AB$. Choose points $M$ and $N$ on sides $AC$ and $BC$ respectively such that $DM$ is parallel to $BC$ and $DN$ is parallel to $AC$. If $DM=5,DN=4$, then $AC$ and $BC$ are respectively equal to

• A. $\frac{41}{4},\frac{41}{5}$
• B. $\frac{39}{4},\frac{39}{5}$
• C. $\frac{38}{4},\frac{38}{5}$
• D. $\frac{37}{4},\frac{37}{5}$

Answer 1

Answer: (A)

Given, $ABC$ is right angled triangle
$\angle C=90^{\circ }$
$CD$ is perpendicular on $AB,DN$ and $DM$
are parallel to $AC$ and $BC$, respectively.
$DN=4$ and $DM=5$

In $\Delta DMC$ and $\Delta DNB$,
$\Delta DMC\sim \Delta DNB$
$\therefore \frac{DM}{DN}=\frac{MC}{NB}$
$\Rightarrow \frac{5}{4}=\frac{4}{NB}$
$\Rightarrow NB=\frac{16}{5}$
$\therefore BC=CN+NB$
$=5+\frac{16}{5}=\frac{41}{5}$
In $\Delta DNC$ and $\Delta DMA$,
$\Delta DNC\sim \Delta DMA$
$\Rightarrow \frac{DN}{DM}=\frac{NC}{MA}$
$\Rightarrow \frac{4}{5}=\frac{5}{MA}$
$\Rightarrow MA=\frac{25}{4}$
$\therefore AC=MC+AM$
$=4+\frac{25}{4}=\frac{41}{4}$