To solve this problem, we need to establish coordinates for each dancer based on the given information.
Let's place Belle at the origin of our coordinate system:
- Belle's position: \( (0, 0) \)
Since Isabella is 2 feet directly left of Belle, we place Isabella at:
- Isabella's position: \( (-2, 0) \)
Dalton is positioned 4 feet straight in front of Isabella, meaning he is directly behind her along the same vertical line (the y-axis). Therefore, since Isabella is at \((-2, 0)\), we find Dalton’s position by moving 4 feet upwards (in the positive y-direction):
- Dalton's position: \( (-2, 4) \)
Now we have the positions of the dancers clearly defined:
- Belle: \( (0, 0) \)
- Isabella: \( (-2, 0) \)
- Dalton: \( (-2, 4) \)
Now, let's calculate the distances Isabella travels during her movements.
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From Isabella to Belle:
- Isabella moves from \((-2, 0)\) to \( (0, 0)\).
- The distance is calculated as follows: \[ \text{Distance} = \sqrt{(0 - (-2))^2 + (0 - 0)^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \text{ feet} \]
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From Belle to Dalton:
- Isabella moves from \( (0, 0) \) to \( (-2, 4) \).
- The distance is calculated as follows: \[ \text{Distance} = \sqrt{((-2) - 0)^2 + (4 - 0)^2} = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \text{ feet} \]
- Approximating \( \sqrt{5} \approx 2.236 \): \[ 2\sqrt{5} \approx 2 \times 2.236 = 4.472 \text{ feet} \]
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From Dalton back to Isabella's original position:
- Isabella moves from \( (-2, 4) \) back to \( (-2, 0) \).
- The distance is: \[ \text{Distance} = \sqrt{((-2) - (-2))^2 + (0 - 4)^2} = \sqrt{(0)^2 + (-4)^2} = \sqrt{16} = 4 \text{ feet} \]
Finally, we sum up all the distances that Isabella travels: \[ \text{Total distance} = 2 + 2\sqrt{5} + 4 \] Substituting \( 2\sqrt{5} \approx 4.472 \): \[ \text{Total distance} \approx 2 + 4.472 + 4 = 10.472 \text{ feet} \]
Rounding to the nearest tenth, we find: \[ \text{Total distance} \approx 10.5 \text{ feet} \]
Thus, the total distance Isabella traveled is 10.5 feet.