Question

In $△ABC$ and $△DEF,$ it is given that $AB=5$ cm, $BC=4$ cm, $CA=4.2$ cm, $DE=10$ cm, $EF=8$ cm and $FD=8.4$ cm. If $AL$ is perpendicular to $BC$ and $DM$ is perpendicular to $EF,$ then what is the ratio of $AL$ to $DM$?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $1$

Answer 1

Answer: (A)

$\frac{1}{2}$

Given, $AB=5$ cm, $BC=4$ cm, $CA=4.2$ cm, $DE=10$ cm, $EF=8$ cm and $FD=8.4$ cm

Now, $\frac{AB}{DE}=\frac{5}{10}=\frac{1}{2}$

$\frac{BC}{EF}=\frac{4}{8}=\frac{1}{2}$

$\frac{CA}{FD}=\frac{4.2}{8.4}=\frac{1}{2}$

Since, $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$

Thus. $△ABC\sim △DEF$ ....... (SSS rule)

Now, in similar triangles the altitudes are in the ratio of corresponding sides.

$\frac{AL}{DM}=\frac{AB}{DE}=\frac{1}{2}$