# In figure below, $\mathrm{PA}, \mathrm{QB}, \mathrm{RC}$ and $\mathrm{SD}$ are all perpendiculars to a line $l, \mathrm{AB}=6 \mathrm{~cm}$, $\mathrm{BC}=9 \mathrm{~cm}, C D=12 \mathrm{~cm}$ and $S P=36 \mathrm{~cm}$. Find $P Q, Q R$ and $R S$."

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Given:

$\mathrm{PA}, \mathrm{QB}, \mathrm{RC}$ and $\mathrm{SD}$ are all perpendiculars to a line $l, \mathrm{AB}=6 \mathrm{~cm}$, $\mathrm{BC}=9 \mathrm{~cm}, C D=12 \mathrm{~cm}$ and $S P=36 \mathrm{~cm}$.

To do:

We have to find $P Q, Q R$ and $R S$.

Solution:

The lines $P A, Q B, R C$ and $S D$ are parallel to each other.

This implies,

The lines $\mathrm{PA}, \mathrm{QB}, \mathrm{RC}$ and $\mathrm{SD}$ are perpendicular to the line $\mathrm{l}$.

According to Intercept theorem,

The ratios between the line segments created when two parallel lines are intercepted by two intersecting lines.

This implies,

$\frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{\mathrm{QR}}{ \mathrm{BC}}=\frac{\mathrm{RS}}{\mathrm{CD}}=\frac{\mathrm{PS}}{ \mathrm{AD}}$

$\mathrm{AD}=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}$

$=6+9+12$

$=27 \mathrm{~cm}$

Therefore,

$\frac{\mathrm{PQ}}{6}=\frac{\mathrm{QR}}{9}=\frac{\mathrm{RS}}{12}=\frac{36}{27}$

$\frac{\mathrm{PQ}}{6}=\frac{36}{27}$

On cross multiplication,

$\mathrm{PQ}=\frac{6\times4}{3}$

$P Q=2\times4$

$P Q=8 \mathrm{~cm}$

$\frac{\mathrm{QR}}{9}=\frac{36}{27}$

On cross multiplication,

$\mathrm{QR}=\frac{9\times4}{3}$

$\mathrm{QR}=3\times4$

$\mathrm{QR}=12 \mathrm{~cm}$

$\frac{\mathrm{RS}}{12}=\frac{36}{27}$

On cross multiplication,

$\mathrm{RS}=\frac{12\times36}{27}$

$R S=\frac{12\times4}{3}$

$R S=16 \mathrm{~cm}$

Hence, $P Q=8\ cm, Q R=12\ cm$ and $R S=16\ cm$.

Updated on 10-Oct-2022 13:28:19