Question

In the given figure, PA, QB and RC are perpendiculars to AC.

Determine the value of $\frac{1}{x}+\frac{1}{z}$.

• A. y
• B. $\frac{1}{xz}$
• C. $\frac{1}{y}$
• D. $\frac{1}{xyz}$

Answer 1

The correct option is C

$\frac{1}{y}$

In $\Delta$PAC and $\Delta$QBC

$\angle PAC=\angle QBC={90}^{\circ }$

$\angle PCA=\angle QCB$ ......(common)

So, $\Delta PAC\sim \Delta QBC$ ........(AA criterion)

Therefore, $\frac{z}{y}=\frac{AC}{BC}$

or $\frac{y}{z}=\frac{BC}{AC}$ ____(1)

and $\frac{y}{x}=\frac{AB}{AC}$___(2)

Adding equations (1) and (2), we get

$\frac{y}{z}+\frac{y}{x}=\frac{BC}{AC}+\frac{AB}{AC}$

$y(\frac{1}{z}+\frac{1}{x})=\frac{BC+AB}{AC}$

$y(\frac{1}{z}+\frac{1}{x})=\frac{AC}{AC}$ = 1

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