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A 10-foot ladder leans against a wall so that it is 6 feet high at the top. The ladder is moved so that the base of the ladder travels toward the wall twice the distance that the top of the ladder moves up. How much higher is the top of the ladder now? (Hint: Let 8−2x be the distance from the base of the ladder to the wall.)

2 Answers

3 votes

Answer:

Explanation:

The Pythagorean Theorem holds for both positions.

The first configuration is an 6-8-10 right triangle.

The second configuration using the Pythagorean Theorem:

10^2 = (8-2x)^2 + (6 + x)^2

100 = 64 - 32x + 4x^2 + 36 + 12x + x^2

100 = 100 - 20x + 5x^2

5x^2 - 20x = 0

5x(x - 4) = 0

x = 4 feet (we throw out the x = 0 solution as extraneous)

User Tal Pressman
by
6.1k points
3 votes

Answer:

Explanation:

The Pythagorean Theorem holds for both positions.

The first configuration is an 6-8-10 right triangle.

The second configuration using the Pythagorean Theorem:

10^2 = (8-2x)^2 + (6 + x)^2

100 = 64 - 32x + 4x^2 + 36 + 12x + x^2

100 = 100 - 20x + 5x^2

5x^2 - 20x = 0

5x(x - 4) = 0

x = 4 feet (we throw out the x = 0 solution as extraneous)

A 10-foot ladder leans against a wall so that it is 6 feet high at the top. The ladder-example-1
User Eugene B
by
6.9k points
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