A 45 foot ladder is placed against a house, reaching 27 feet up the wall. The ladder is then pulled 4 feet further away from the house. We want to find how far up the side of the house the ladder now reaches, rounded to the nearest tenth of a foot.

1. Problem Description

A 45 foot ladder is placed against a house, reaching 27 feet up the wall. The ladder is then pulled 4 feet further away from the house. We want to find how far up the side of the house the ladder now reaches, rounded to the nearest tenth of a foot.

2. Solution Steps

First, we can use the Pythagorean theorem to find the initial distance of the base of the ladder from the house. Let $a$ be the initial distance, $b$ be the height the ladder reaches up the wall (27 feet), and $c$ be the length of the ladder (45 feet).

$$a^2 + b^2 = c^2$$

$$a^2 + 27^2 = 45^2$$

$$a^2 + 729 = 2025$$

$$a^2 = 2025 - 729$$

$$a^2 = 1296$$

$$a = \sqrt{1296}$$

$$a = 36 \text{ feet}$$

So, initially, the base of the ladder is 36 feet from the house.

Now, the ladder is pulled 4 feet further from the house, so the new distance of the base of the ladder from the house is $36 + 4 = 40$ feet. Let $h$ be the new height the ladder reaches up the wall. We can again use the Pythagorean theorem, with $a = 40$, $c = 45$, and $b = h$.

$$a^2 + h^2 = c^2$$

$$40^2 + h^2 = 45^2$$

$$1600 + h^2 = 2025$$

$$h^2 = 2025 - 1600$$

$$h^2 = 425$$

$$h = \sqrt{425}$$

$$h \approx 20.6155$$

Rounding to the nearest tenth of a foot, we get $h \approx 20.6$ feet.

3. Final Answer

20.6