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}$