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Q24.

Expert-verified
Found in: Page 581

Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

A rolling ball that does not have much spin will bounce off a wall so that the two angles that the path forms with the wall are congruent. Thus, to roll the ball from B off the wall shown and into hole H, you need to aim the ball so that $\angle 1\cong \angle 2$.Let $H\text{'}$ be the image of H by reflection in the wall. $\stackrel{\to }{BH\text{'}}$ Intersects the wall at P. Why is $\angle 1\cong \angle 3$? Why is $\angle 3\cong \angle 2$? Why is $\angle 1\cong \angle 2$? You can conclude that if you aim for $H\text{'}$, the ball will roll to H.Show that the distance travelled by the ball equals the distance H.

1. $\angle 1\cong \angle 2,\angle 1\cong \angle 3\text{and}\angle 3\cong \angle 2$.
2. The distance travelled by the ball equals the distance $BH\text{'}$.
See the step by step solution

a.Step 1. Given Information.

$H\text{'}$ Is the image of H by reflection in the wall and $\overline{BH\text{'}}$ intersects the wall at P.

Step 2. Proof.

$\angle 1\cong \angle 3\text{}\left[\text{vertically opposite angles}\right]$

In triangles $PXH\text{and}PXH\text{'}$.

$\begin{array}{c}PX=PX\text{}\left[\text{common side}\right]\\ m\angle X=m\angle X\text{}\left[\text{each equals to 90}°\right]\\ HX=HX\text{'}\text{}\left[\text{perpendicular bisector}\right]\end{array}$

By side angle side congruency

$\begin{array}{c}\Delta PXH\cong \Delta PXH\text{'}\\ \angle 2\cong \angle 3\end{array}$

Since $\angle 1\cong \angle 3\text{and}\angle 2\cong \angle 3$ therefore $\angle 1\cong \angle 2$

Step 3. Conclusion.

Hence, the congruency of angles is proved.

b.Step 1. Given Information.

$H\text{'}$ Is the image of H by reflection in the wall and $\overline{BH\text{'}}$ intersects the wall at P.

Step 2. Proof.

$\begin{array}{c}\Delta PXH\cong \Delta PXH\text{'}\\ PH=PH\text{'}\end{array}$

Distance travelled by the ball is,

$\begin{array}{c}BP+PH=BP+PH\text{'}\\ =BH\text{'}\end{array}$

Step 3. Conclusion.

Hence, it is proved that the distance travelled by the ball equals the distance $BH\text{'}$.