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Q16.

Expert-verifiedFound in: Page 337

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

**$\overline{SR}$ is tangent to $\odot P$ and $\odot Q$.**

**$QT=6,\text{}TR=8,\text{}PR=30.$**

**$PQ=?,\text{}PS=?,\text{}ST=?$**

The required values are,

$PQ=20,PS=18\text{and}ST=16$

$\odot P$ Any $\overline{SR}$ is tangent to $\odot P$ and $\odot Q$. Also $QT=6,TR=8\text{and}PR=30$.

Applying Pythagoras in triangle $\Delta RTQ$ to find the value of $PQ$.

Here, $\Delta RTQ$ is right angle triangle,

$\begin{array}{l}Q{R}^{2}={6}^{2}+{8}^{2}\\ Q{R}^{2}=36+64\\ QR\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\end{array}$

As value of ,

$\begin{array}{l}PQ=30-10\\ PQ=20\end{array}$

Here, $\overline{TQ}\left|\right|\overline{SP}$

Then,

$\begin{array}{l}\frac{8}{10}=\frac{x}{20}\\ x\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8\cdot \text{\hspace{0.33em}}20}{10}\\ x\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16\end{array}$

Same way,

$\begin{array}{l}\frac{8}{6}=\frac{24}{y}\\ y\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{24\cdot \text{\hspace{0.33em}}6}{8}\\ y\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=18\end{array}$

Therefore, the values are, $PQ=20,PS=18\text{and}ST=16$.

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