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

The correct option is A $\frac{41}{4},\frac{41}{5}$


$CD$ is perpendicular to $AB$
using pythagoras theorem,
$BD=\sqrt{16+y^2}DA=\sqrt{25+x^2}DC=\sqrt{41}$
In $△CDB$
$\Rightarrow (\sqrt{41}{)}^2+(\sqrt{16+y^2}{)}^2=(5+y{)}^2\Rightarrow 41+16+y^2=25+y^2+10y\Rightarrow 10y=32\Rightarrow y=\frac{16}{5}$
$\therefore BC=5+\frac{16}{5}=\frac{41}{5}$
By observation, only option A follows.