Q16. - Geometry

Question

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

$QT=6,TR=8,PR=30.$

$PQ=?,PS=?,ST=?$

Step-by-Step Solution

Answer

The required values are,

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

Step 1. Given information.

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

Step 2. Concept used.

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

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

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

As value of , 

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

Step 3. Let S T be x and S P be y .

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

Then,

$\begin{array}{l}\frac{8}{10}=\frac{x}{20}\\ x=\frac{8\cdot 20}{10}\\ x=16\end{array}$

Same way, 

$\begin{array}{l}\frac{8}{6}=\frac{24}{y}\\ y=\frac{24\cdot 6}{8}\\ y=18\end{array}$

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