Question

In the adjoining figure, if AB, SC and TD are perpendicular to BD with the length indicated, then $\frac{x}{y}$ is equal to:

• A. $\frac{p}{q}$
• B. $\frac{p}{p+q}$
• C. $\frac{q}{p}$
• D. $\frac{q}{p+q}$

Answer 1

The correct option is A $\frac{p}{q}$

In $\Delta ABDand\Delta SCD,$
$\angle D$ is common to both triangles and $\angle B=\angle C$
(because each angle is right angle.)
$\therefore \frac{AB}{SC}=\frac{BD}{CD}or\frac{x}{z}=\frac{p+q}{q}$
Similarly, $\Delta TDBand\Delta SCB$ are similar.
$\therefore \frac{SC}{TD}=\frac{BC}{BD}or\frac{z}{y}=\frac{p}{p+q}\frac{x}{z}\times \frac{z}{y}=\frac{p+q}{q}\times \frac{p}{p+q}\therefore \frac{x}{y}=\frac{p}{q}$