Question

bonus: 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? show how you arrived at your answer.

Explanation:

Step1: Calculate the initial height

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 16\) (ladder length) and \(a = 8\) (initial base - distance). Let the initial height be \(h_1\).

$$h_1=\sqrt{16^{2}-8^{2}}=\sqrt{256 - 64}=\sqrt{192}\approx13.9$$

Step2: Calculate the new height

The new base - distance \(a = 4\). Let the new height be \(h_2\).

$$h_2=\sqrt{16^{2}-4^{2}}=\sqrt{256 - 16}=\sqrt{240}\approx15.5$$

Step3: Calculate the difference

Find \(h_2 - h_1\).

$$h_2 - h_1\approx15.5-13.9 = 1.6\approx2$$

Answer:

The ladder now reaches approximately \(2\) feet farther up the building.