Question

a 45 - foot ladder is set against the side of a house so that it reaches up 27 feet. if cooper grabs the ladder at its base and pulls it 4 feet farther from the house, how far up the side of the house will the ladder reach now? (the answer is not 23 ft.) round to the nearest tenth of a foot.

Explanation:

Step1: Find initial base - distance

We use the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $c = 45$ (length of ladder) and $a = 27$ (height on house initially). Let the base - distance be $b$.
$b=\sqrt{c^{2}-a^{2}}=\sqrt{45^{2}-27^{2}}=\sqrt{(45 + 27)(45 - 27)}=\sqrt{72\times18}=\sqrt{1296}=36$ feet.

Step2: Calculate new base - distance

The base is pulled 4 feet farther from the house. So the new base - distance $b_{new}=36 + 4=40$ feet.

Step3: Find new height on house

Again, using the Pythagorean theorem $a_{new}=\sqrt{c^{2}-b_{new}^{2}}$, with $c = 45$ and $b_{new}=40$.
$a_{new}=\sqrt{45^{2}-40^{2}}=\sqrt{(45 + 40)(45 - 40)}=\sqrt{85\times5}=\sqrt{425}\approx20.6$ feet.

Answer:

$20.6$ feet