A ball placed at point $P$ on a rectangular billiard table is shot at an angle of $45^{\circ}$ to the edge of the table. After successively bouncing off the edges of the table at $45^{\circ}$ angles, it returns to point $P$, as shown. If the ball travels $7 \mathrm{~m}$, the perimeter, in metres, of the table is closest to
(A) 7.0
(B) 7.5
(C) 8.0
(D) 8.5
(E) 9.0

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-100.jpg?height=325&width=526&top_left_y=2228&top_left_x=1190)
Label the vertices of the rectangle $A B C D$ and the points of contact of the ball with the edges in order as $P, Q, R, S, T$, and $U$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-200.jpg?height=369&width=610&top_left_y=973&top_left_x=817)

Note that the angle that each segment of the path makes with each side of the table is $45^{\circ}$.

Therefore, $P Q=\sqrt{2} A B$. Also, $Q R=\sqrt{2} A R$ and $R S=\sqrt{2} R B$, so $Q R+R S=\sqrt{2}(A R+R B)=$ $\sqrt{2} A B$.

Similarly, $S T=\sqrt{2} C D$ and $T U+U P=\sqrt{2} C D$.

Since $A B=C D$, then $P Q+Q R+R S+S T+T U+U P=4 \sqrt{2} A B$.

Since we know that the total length of the path is $7 \mathrm{~m}$, then $A B=\frac{7}{4 \sqrt{2}} \mathrm{~m}$.

Also, $P Q=\sqrt{2} \times$ horizontal distance from $P$ to $Q, Q R=\sqrt{2} Q A$ and $P U=\sqrt{2} P C$.

Therefore, $P Q+Q R+P U=\sqrt{2} A D$ (since $Q A$ plus $P C$ plus the horizontal distance from $P$ to $Q$ equals the length of the rectangle).

Similarly, $R S+S T+T U=\sqrt{2} B C=\sqrt{2} A D$.

So $P Q+Q R+R S+S T+T U+U P=2 \sqrt{2} A D$ so $A D=\frac{7}{2 \sqrt{2}}$.

Therefore, the perimeter of the table is $2 A D+2 A B=\frac{7}{\sqrt{2}}+\frac{7}{2 \sqrt{2}} \approx 7.425$, which is closest to $7.5 \mathrm{~m}$.

ANSWER: (B)