A 16-foot ladder is placed against the side of a building, as shown in Figure 1 below. The bottom of the ladder is 8 feet from the base of the building. In order to increase the reach of the ladder against the building, the ladder is moved 4 feet closer to the base of the building, as shown in Figure 2 below. To the nearest foot, how much farther up the building does the ladder now reach? Show how you arrived at your answer.

Using the pythagorean theorem, you know that the height of the ladder up the building in the first example is [tex]\sqrt{16^2-8^2}=\sqrt{256-64}=\sqrt{192}[/tex] feet. In the second example, [tex]\sqrt{16^2-4^2}=\sqrt{256-16}=\sqrt{240}[/tex]. The difference between these two, to the nearest foot, is about 2. Hope this helps!

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swapnalimalwadeVT swapnalimalwadeVT · Answer 2 · 2022-05-07T11:37:14+02:00

The difference between these two, to the nearest foot, is about 2.

We have given the two diagram,

Using the Pythagorean theorem,

What is the Pythagorean theorem?

[tex]hypotenous^2=side^2+side^2\\[/tex]

you know that the height of the ladder up the building in the first example is

[tex]=\sqrt{16^2-8^2}\\= \sqrt{256-64} \\=\sqrt{192}[/tex]

=root(192) feet.

In the second example,

[tex]=\sqrt{16^2-4^2} \\=\sqrt{256-16} \\=\sqrt{240}[/tex]

The difference between these two, to the nearest foot, is about 2.

To learn more about the difference visit:

https://brainly.com/question/148825

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