Question

a 51 foot ladder is set against the side of a house so that it reaches up 45 feet. if nora grabs the ladder at its base and pulls it 6 feet farther from the house, how far up the side of the house will the ladder reach now? (the answer is not 39 ft.) round to the nearest tenth of a foot.

Explanation:

Step1: Find the initial distance of the ladder base from the house

Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 51\) (ladder length) and \(b = 45\) (height on the house). Let the initial base - distance be \(a\).

$$a=\sqrt{c^{2}-b^{2}}=\sqrt{51^{2}-45^{2}}=\sqrt{(51 + 45)(51 - 45)}=\sqrt{96\times6}=\sqrt{576}=24$$

Step2: Find the new base - distance

The new base - distance \(a_{new}=24 + 6=30\)

Step3: Find the new height on the house

Using the Pythagorean theorem again, with \(c = 51\) and \(a_{new}=30\). Let the new height be \(b_{new}\)

$$b_{new}=\sqrt{c^{2}-a_{new}^{2}}=\sqrt{51^{2}-30^{2}}=\sqrt{(51 + 30)(51 - 30)}=\sqrt{81\times21}=\sqrt{1701}\approx41.2$$

Answer:

\(41.2\) feet