Question

A man of height y is standing between two buildings with heights x and z respectively as shown in the figure. Then, which of the following statements is true about the relation of the heights?

• A. $x=y=z$
• B. $\frac{1}{x}=\frac{1}{y}=\frac{1}{z}$
• C. $\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$
• D. $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$

Answer 1

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

Given, $QB\perp AC,PA\perp ACandRC\perp AC$.
In $\Delta APC$ and $\Delta BQC$,
$\angle ACP=\angle BCQ$ [common]
$\angle PAC=\angle QBC$ [each ${90}^{\circ }$]
$\therefore \Delta APC\sim \Delta BQC$ [by AA similarity]
$\Rightarrow \frac{AP}{BQ}=\frac{AC}{BC}\Rightarrow \frac{x}{y}=\frac{AC}{BC}...\left(i\right)$
Now, in $\Delta ARC$ and $\Delta AQB$,
$\angle ACR=\angle ABQ$ [each ${90}^{\circ }$]
$\angle CAR=\angle BAQ$ [common]
$\Rightarrow \Delta ARC\sim \Delta AQB$ [by AA similarity]
$\therefore \frac{RC}{QB}=\frac{AC}{AB}$
$\Rightarrow \frac{z}{y}=\frac{AC}{AB}...\left(ii\right)$
From Eqs. (i), we get
$BC=\frac{y}{x}AC$ ...(iii)
From Eq. (ii), we get
$AB=\frac{y}{z}AC$
Adding Eqs. (iii) and (iv), we get
$AB+BC=(\frac{y}{x}+\frac{y}{z})AC\Rightarrow AC=y(\frac{1}{x}+\frac{1}{z})AC\Rightarrow \frac{1}{y}=\frac{1}{x}+\frac{1}{z}$