Math Problem Statement

Two vertical poles of lengths 8 feet and 10 feet stand 15 feet apart. A cable reaches from the top of one pole to some point on the ground between the poles and then to the top of the other pole. Where should this point be located to use 25 feet of​ cable?

Solution

This problem involves a scenario where two vertical poles, one 8 feet tall and the other 10 feet tall, stand 15 feet apart. The task is to determine the point on the ground between the poles where a cable, 25 feet in length, reaches from the top of one pole, touches the ground at a point between them, and then continues to the top of the other pole.

Approach

Let the two poles be positioned along the xx-axis, with the 8-foot pole at x=0x = 0 and the 10-foot pole at x=15x = 15. The point where the cable touches the ground will be located at some point x=dx = d along the ground, where 0d150 \leq d \leq 15.

The cable can be split into two segments:

  1. The distance from the top of the 8-foot pole to the point dd on the ground is given by the distance formula:
    L1=d2+82L_1 = \sqrt{d^2 + 8^2}

  2. The distance from the top of the 10-foot pole to the same point dd on the ground is:
    L2=(15d)2+102L_2 = \sqrt{(15 - d)^2 + 10^2}

The total length of the cable is 25 feet, so we have the equation: L1+L2=25L_1 + L_2 = 25

Substituting the expressions for L1L_1 and L2L_2: d2+82+(15d)2+102=25\sqrt{d^2 + 8^2} + \sqrt{(15 - d)^2 + 10^2} = 25 This is a nonlinear equation involving dd, and solving it will give the location of the point on the ground where the cable touches.

Solving the Equation

Now, I'll solve this equation to find dd.The two possible locations where the cable could touch the ground are:

  • d1.40d \approx 1.40 feet from the base of the 8-foot pole.
  • d12.25d \approx 12.25 feet from the base of the 8-foot pole (or 2.75 feet from the base of the 10-foot pole).

These are the two points along the ground where the cable could touch while using the total length of 25 feet.

Would you like a more detailed explanation or have any other questions?

Related Questions:

  1. How is the Pythagorean theorem applied in calculating distances here?
  2. What is the significance of having two possible solutions?
  3. Can this method be applied to poles of different heights or distances apart?
  4. What would happen if the cable length was shorter or longer than 25 feet?
  5. How would the problem change if the cable did not touch the ground between the poles?

Tip: Nonlinear equations involving square roots often have multiple solutions, so always check the context to determine which solutions are valid.

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Math Problem Analysis

Mathematical Concepts

Geometry
Distance Formula
Nonlinear Equations

Formulas

L1 = √(d² + 8²)
L2 = √((15 - d)² + 10²)
L1 + L2 = 25

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 9-12