Question

In the following figure, AB, CD and EF are perpendicular to the straight line BDF. If AB = x and, CD = z unit and EF = y unit, then

• A. $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$
• B. $\frac{1}{z}+\frac{1}{x}=\frac{1}{y}$
• C. $\frac{1}{z}+\frac{1}{y}=\frac{1}{x}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$

Let $BD=a$ and $DF=b$. In $\Delta ABF$, $CD$ is parallel to $AB$

$\Rightarrow \frac{CD}{AB}=\frac{DF}{BF}$ i.e.,

$\frac{z}{y}=\frac{b}{a+b}$

And in $\Delta BEF$, CD is parallel to EF

$\Rightarrow \frac{CD}{EF}=\frac{BD}{BF}$ i.e,, $\frac{z}{y}=\frac{a}{a+b}.$

$\therefore \frac{z}{x}+\frac{z}{y}=\frac{b}{a+b}+\frac{a}{a+b}=1$

$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$