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7th Grade Math · Question of the Day

7th Grade Math Question of the Day

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Friday, October 9, 2026

A string is wrapped around a cylindrical can exactly 33 times. If the string length is 37.6837.68 cm and the can height is 1010 cm, what is the diameter of the can?

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Question of the Day

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A string is wrapped around a cylindrical can exactly 33 times. If the string length is 37.6837.68 cm and the can height is 1010 cm, what is the diameter of the can?

  1. 88 cm
  2. 1212 cm
  3. 22 cm
  4. 44 cm (correct answer)

Explanation: When a string wraps around a cylinder, it creates a helical (spiral) path. To find the cylinder's diameter, you need to "unwrap" this path and use the Pythagorean theorem. Picture unrolling the cylindrical can into a flat rectangle. The height remains 1010 cm, but the width becomes the circumference of the base circle. Since the string wraps around exactly 33 times, it travels a horizontal distance of 3×circumference3 \times \text{circumference} while moving vertically 1010 cm. The string's path forms the hypotenuse of a right triangle where:

  • One leg = height = 1010 cm
  • Other leg = 3×circumference=3×πd3 \times \text{circumference} = 3 \times \pi d (where dd is diameter)
  • Hypotenuse = string length = 37.6837.68 cm
Using the Pythagorean theorem: 102+(3πd)2=37.68210^2 + (3\pi d)^2 = 37.68^2 Solving: 100+9π2d2=1419.78100 + 9\pi^2 d^2 = 1419.78
9π2d2=1319.789\pi^2 d^2 = 1319.78
d2=1319.789π2≈16d^2 = \frac{1319.78}{9\pi^2} \approx 16
d=4d = 4 cm Choice A (88 cm) gives a circumference too large, making the horizontal distance too long. Choice B (1212 cm) makes this error even worse. Choice C (22 cm) creates too small a circumference, making the horizontal distance too short to reach the required string length of 37.6837.68 cm. Study tip: When you see string wrapping problems, always visualize "unwrapping" the cylinder into a rectangle and apply the Pythagorean theorem to the resulting right triangle.