Question

a 50 foot ladder is set against the side of a house so that it reaches up 48 feet. if mila 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 42 ft.) round to the nearest tenth of a foot.

Explanation:

Step1: Find initial base - distance

The ladder, the house, and the ground form a right - triangle. Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 50\) (ladder length) and \(a = 48\) (height on the house). Let the initial base - distance be \(b_1\).
\[b_1=\sqrt{c^{2}-a^{2}}=\sqrt{50^{2}-48^{2}}=\sqrt{(50 + 48)(50 - 48)}=\sqrt{98\times2}=\sqrt{196}=14\]

Step2: Calculate new base - distance

Mila pulls the ladder 6 feet farther from the house. So the new base - distance \(b_2=b_1 + 6=14 + 6=20\)

Step3: Find new height on the house

Let the new height on the house be \(a_2\). Again, using the Pythagorean theorem \(a_2=\sqrt{c^{2}-b_2^{2}}\), with \(c = 50\) and \(b_2 = 20\)
\[a_2=\sqrt{50^{2}-20^{2}}=\sqrt{(50 + 20)(50 - 20)}=\sqrt{70\times30}=\sqrt{2100}\approx45.8\]

Answer:

45.8