Question

In the given figure AB||PQ||CD, AB = x units, CD= y units and PQ = z units, then $\frac{1}{x}+\frac{1}{y}=$

• A. $\frac{2}{z}$
• B. $\frac{1}{z}$
• C. $z^2$
• D. z

Answer 1

The correct option is B

$\frac{1}{z}$

Since AB||PQ||CD, using A-A similarity $\Delta BQP\sim \Delta BDC$ and $\Delta DQP\sim \Delta DBA.$

In $\Delta DQP$ and $\Delta DBA,\frac{QD}{BD}=\frac{PQ}{AB}=\frac{z}{x}$

$\therefore \frac{QD}{BD}=\frac{z}{x}----\left(1\right)$

In $\Delta BQP$ and $\Delta BDC,\frac{BQ}{BD}=\frac{PQ}{CD}=\frac{z}{y}$

$\therefore \frac{BQ}{BD}=\frac{z}{y}$

$\Rightarrow 1-\frac{BQ}{BD}=1-\frac{z}{y}$

$\Rightarrow \frac{BD-BQ}{BD}=\frac{y-z}{y}$

$\Rightarrow \frac{QD}{BD}=\frac{y-z}{y}----\left(2\right)$

From (1)& (2), we get, $\frac{z}{x}=\frac{y-z}{y}$

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

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

Alternate Solution: ​

Since AB||PQ||CD, using A-A similarity $\Delta BQP\sim \Delta BDC$ and $\Delta DQP\sim \Delta DBA.$

In $\Delta DQP$ and $\Delta DBA,\frac{QD}{BD}=\frac{b}{a+b}=\frac{z}{x}----\left(1\right)$

In $\Delta BQP$ and $\Delta BDC,\frac{BQ}{BD}=\frac{a}{a+b}=\frac{z}{y}----\left(2\right)$

Adding 1 and 2, we get,
$\frac{a+b}{a+b}=\frac{z}{x}+\frac{z}{y}$
$\Rightarrow 1=z(\frac{1}{x}+\frac{1}{y})$
$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$