In figure, if $ \angle \mathrm{A}=\angle \mathrm{C}, \mathrm{AB}=6 \mathrm{~cm}, \mathrm{BP}=15 \mathrm{~cm} $, $ \mathrm{AP}=12 \mathrm{~cm} $ and $ \mathrm{CP}=4 \mathrm{~cm} $, then find the lengths of $ \mathrm{PD} $ and CD.
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Given:

\( \angle \mathrm{A}=\angle \mathrm{C}, \mathrm{AB}=6 \mathrm{~cm}, \mathrm{BP}=15 \mathrm{~cm} \), \( \mathrm{AP}=12 \mathrm{~cm} \) and \( \mathrm{CP}=4 \mathrm{~cm} \).

To do:

We have to find the lengths of \( \mathrm{PD} \) and CD.

Solution:

In $\triangle \mathrm{APB}$ and $\triangle \mathrm{CPD}$,

$\angle \mathrm{A}=\angle \mathrm{C}$        (Given)

$\angle \mathrm{APS}=\angle \mathrm{CPD}$        (Vertically opposite angles)

Therefore, by AA similarity,

$\triangle A P D \sim \triangle C P D$

This implies,

$\frac{A P}{C P}=\frac{P B}{P D}=\frac{A B}{C D}$

$\frac{12}{4}=\frac{15}{P D}=\frac{6}{C D}$

Therefore,

$P D=\frac{15 \times 4}{12}$

$=5 \mathrm{~cm}$

$C D=\frac{6 \times 4}{12}$

$=2 \mathrm{~cm}$

Hence, the length of $P D$ is $5 \mathrm{~cm}$ and the length of $C D$ is $2 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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