Question

on the school playground, the slide is 7 feet west of the tire swing and 7 feet south of the monkey bars. during recess, ruth plays on the tire swing first. ruth then goes to slide down the slide before meeting up with another friend at the monkey bars. then ruth returns back to the tire swing. what is the total distance traveled by ruth? if necessary, round to the nearest tenth. feet

Explanation:

Step1: Analyze the path

Ruth's path: tire swing → slide → monkey bars → tire swing. First, distance from tire swing to slide is 7 feet (west). Then, distance from slide to monkey bars: since slide is 7 feet south of monkey bars, the horizontal (west-east) and vertical (north-south) distances between tire swing, slide, and monkey bars form a right triangle? Wait, no: tire swing to slide is 7 ft west, slide to monkey bars is 7 ft north (since slide is south of monkey bars, so monkey bars is north of slide). Wait, actually, the positions: tire swing (TS), slide (S) is 7 ft west of TS and 7 ft south of monkey bars (MB). So TS to S: 7 ft (west). S to MB: 7 ft (north, because S is south of MB). Then MB to TS: we can find this distance using Pythagoras, since TS to S is 7 west, S to MB is 7 north, so TS to MB is the hypotenuse of a right triangle with legs 7 and 7. So distance TS to MB is $\sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.9$ ft.

Now, sum the distances: TS to S (7) + S to MB (7) + MB to TS (≈9.9) = 7 + 7 + 9.9 = 23.9? Wait, no, wait: let's re-examine the path.

Wait, Ruth's path:

1. Tire swing (TS) to slide (S): 7 feet (west).

2. Slide (S) to monkey bars (MB): since slide is 7 feet south of monkey bars, so from S to MB is 7 feet north (vertical distance).

3. Monkey bars (MB) to tire swing (TS): now, the horizontal distance between TS and S is 7 ft west (so TS is 7 ft east of S), and vertical distance between MB and S is 7 ft north (so MB is 7 ft north of S). So the distance from MB to TS is the hypotenuse of a right triangle with legs 7 (east) and 7 (south, since TS is south of MB? Wait, no: S is south of MB, so MB is north of S. TS is east of S (since S is west of TS). So the coordinates: let's set TS at (0,0). Then S is at (-7, 0) (7 west). MB is at (-7, 7) (since S is 7 south of MB, so MB is 7 north of S, so y-coordinate of S is 0, so MB is at (-7, 7)). Then distance from MB (-7,7) to TS (0,0) is $\sqrt{(0 - (-7))^2 + (0 - 7)^2} = \sqrt{7^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.9$ ft.

Now, sum the three segments:

TS to S: 7 ft.

S to MB: 7 ft (from (-7,0) to (-7,7): vertical distance 7).

MB to TS: ≈9.9 ft.

Total distance: 7 + 7 + 9.9 = 23.9 ft? Wait, no, wait: wait, when going from S to MB, is the distance 7 ft? Wait, the problem says "slide is 7 feet south of the monkey bars" – so the distance between slide and monkey bars is 7 feet (vertical, north-south). Then from TS to S is 7 feet (horizontal, west-east). Then from MB to TS: the horizontal distance between TS and S is 7 (east), vertical distance between MB and S is 7 (north), so the distance from MB to TS is the hypotenuse of 7-7-√98 triangle. So that's correct.

Wait, but let's check again:

Path:

1. TS to S: 7 ft (given, west).

2. S to MB: 7 ft (since S is 7 ft south of MB, so MB is 7 ft north of S, so distance is 7 ft).

3. MB to TS: distance between MB and TS. Since TS is 7 ft east of S, and MB is 7 ft north of S, the distance between MB (north of S) and TS (east of S) is the hypotenuse of a right triangle with legs 7 (east) and 7 (south, because TS is south of MB? Wait, no: S is south of MB, so MB is north of S. TS is east of S. So the coordinates:

- Let S be at (0, 0).

- Then TS is at (7, 0) (since S is 7 ft west of TS, so TS is 7 ft east of S).

- MB is at (0, 7) (since S is 7 ft south of MB, so MB is 7 ft north of S).

Now, distance from MB (0,7) to TS (7,0) is $\sqrt{(7 - 0)^2 + (0 - 7)^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.9$ ft.

Now, sum the distances:

TS to S: 7 ft (from (7,0) to (0,0)).

S to MB: 7 ft (from (0,0) to (0,7)).

MB to TS: ≈9.9 ft (from (0,7) to (7,0)).

Total distance: 7 + 7 + 9.9 = 23.9 ft.

Wait, but let's confirm the path again. The problem says: Ruth plays on tire swing first, then goes to slide (so TS to S: 7 ft), then to monkey bars (S to MB: 7 ft north), then back to tire swing (MB to TS: the hypotenuse). So yes, that's the path.

Step2: Calculate each segment

- TS to S: 7 ft.

- S to MB: 7 ft (vertical distance, since S is south of MB).

- MB to TS: $\sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.9$ ft.

Step3: Sum the distances

Total distance = 7 + 7 + 9.9 = 23.9 feet.

Answer:

23.9