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10 Nov 2021, 04:15
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Two 12-foot ladders are leaned against a building. The first ladder is put on the ground 4 feet away from the building. The second ladder is placed on the ground so that its distance away from the building is equal to the height at which the ladder touches the building. What is the vertical distance between the points at which the two ladders touch the building?
A. 1
B. \(\sqrt{2}\)
C. \(2\sqrt{2}\)
D. 2
E. \(6\sqrt{2} - 4\)
A. 1
B. \(\sqrt{2}\)
C. \(2\sqrt{2}\)
D. 2
E. \(6\sqrt{2} - 4\)
PyjamaScientist
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10 Nov 2021, 06:12
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For triangle with two sides equal, let 'a' be the height of the first ladder touching the wall from the ground = 2\(a^2\) = 144 => a = \(\sqrt{72}\) = 3\(\sqrt{8}\)
Similarly, 'b' be the height of the second ladder touching the wall from the ground (at 4ft distance from the wall) => 16 + b\(^2\) = \(\sqrt{128}\) => b = 4\(\sqrt{8}\)
Positive difference between a and b = 4\(\sqrt{8}\) - 3\(\sqrt{8}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
Option C
Similarly, 'b' be the height of the second ladder touching the wall from the ground (at 4ft distance from the wall) => 16 + b\(^2\) = \(\sqrt{128}\) => b = 4\(\sqrt{8}\)
Positive difference between a and b = 4\(\sqrt{8}\) - 3\(\sqrt{8}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
Option C
28 Jun 2022, 19:03
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First ladder: 4^2 + h^2 = 12^2, h = 8√2
Second ladder: base = height, h^2 + h^2 = 12^2, h=6√2
Distance = 8√2-6√2=2√2
C
Second ladder: base = height, h^2 + h^2 = 12^2, h=6√2
Distance = 8√2-6√2=2√2
C
