Let A B C be a triangle with C=90^ . Draw C D perpendicular to A … - Vidyadip

MCQ

Let $A B C$ be a triangle with $\angle C=90^{\circ}$. Draw $C D$ perpendicular to $A B$. Choose points $M$ and $N$ on sides $A C$ and $B C$ respectively such that $D M$ is parallel to $B C$ and $D N$ is parallel to $A C$. If $D M=5, D N=4$, then $A C$ and $B C$ are respectively equal to

✓
$\frac{41}{4}, \frac{41}{5}$
B
$\frac{39}{4}, \frac{39}{5}$
C
$\frac{38}{4}, \frac{38}{5}$
D
$\frac{37}{4}, \frac{37}{5}$

✓

Answer

Correct option: A.

$\frac{41}{4}, \frac{41}{5}$

a
(a)

Given, $A B C$ is right angled triangle

$\angle C=90^{\circ}$

$C D$ is perpendicular on $A B, D N$ and $D M$ are parallel to $A C$ and $B C$, respectively. $D N=4$ and $D M=5$

In $\triangle D M C$ and $\triangle D N B$

$\Delta D M C \sim \Delta D N B$

$\frac{D M}{D N} =\frac{M C}{N B}$

$\frac{5}{4} =\frac{4}{N B} \Rightarrow N B=\frac{16}{5}$

In $\triangle D N C$ and $\triangle D M A$

$\Delta D N C \sim \Delta D M A$

$\frac{D N}{D M} =\frac{N C}{M A}$

$\frac{4}{5} =\frac{5}{M A} \Rightarrow M A=\frac{25}{4}$

$\therefore A C=M C+A M=4+\frac{25}{4}=\frac{41}{4}$

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