Chapter 10.6, Problem 23PSC

To determine

Find the value of the given segment using diagram.

Answer to Problem 23PSC

The value of $\overline{RB}=4$ .

Explanation of Solution

Given:

The given diagram is

  

  $\overline{QR}=6,\overline{AB}=13,QR\perp AB$

The segment is $\overline{RB}$ .

Calculation:

Draw a segment $\overline{PQ}$ .

  

Diameter $=\overline{AB}=13$ (given)

  $\overline{PB}\left(\text{radius}\right)=\frac{AB}{2}=\frac{13}{2}=6.5$

  $\overline{PB}\left(\text{radius}\right)=\overline{PQ}=6.5$

If $QR\perp AB$ , then $\Delta QRP$ is a right angle triangle.

  $\overline{QR}=6$ (given)

Use Pythagorean formula $\text{Base}=\sqrt{{\text{Hypotenuse}}^2-{\text{Perpendicular}}^2}$ .

  $\overline{PR}=\sqrt{\text{6}{\text{.5}}^2-{\text{6}}^2}=2.5$

  $\overline{RB}=\overline{PB}-\overline{PR}=6.5-2.5=4$

Hence the value of $\overline{RB}=4$ .