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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AcademicMathematicsNCERTClass 10

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 \( R S=16\ cm \).

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

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