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. To the nearest foot, how much farther up the building does the ladder now reach?

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catbabygal catbabygal · Answer 1 · 2020-05-29T03:21:10+02:00

Answer:

Subtract 15.49 - 13.86 = 1.63 and then you simplify.

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shivishivangi1679 shivishivangi1679 · Answer 2 · 2022-06-08T14:09:09+02:00

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

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

|AC|^2 = |AB|^2 + |BC|^2

where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).

The height of the ladder up the building

In the first example

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

In the second example,

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

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

To learn more about the difference visit:

brainly.com/question/148825

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