Question

a 52 foot ladder is set against the side of a house so that it reaches up 48 feet. if jace grabs the ladder at its base and pulls it 7 feet farther from the house, how far up the side of the house will the ladder reach now? (the answer is not 41 ft.) round to the nearest tenth of a foot.

Explanation:

Step1: Find the initial distance from the house

We can use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c\) is the length of the ladder (hypotenuse), \(a\) is the height on the house, and \(b\) is the distance from the house. Let the initial distance from the house be \(x\). So we have \(x^{2}+48^{2}=52^{2}\).
First, calculate \(48^{2}=2304\) and \(52^{2}=2704\). Then \(x^{2}=52^{2}-48^{2}=2704 - 2304=400\), so \(x = \sqrt{400}=20\) feet.

Step2: Find the new distance from the house

Jace pulls the ladder 7 feet farther from the house, so the new distance from the house is \(20 + 7=27\) feet.

Step3: Find the new height on the house

Let the new height be \(y\). Using the Pythagorean theorem again, we have \(y^{2}+27^{2}=52^{2}\).
Calculate \(27^{2}=729\) and \(52^{2}=2704\). Then \(y^{2}=52^{2}-27^{2}=2704-729 = 1975\).
So \(y=\sqrt{1975}\approx44.4\) feet.

Answer:

\(44.4\) feet