A 20 foot ladder is set against the side of a house so that it reaches up 16 feet. If Christian grabs the ladder at its base and pulls it 2 feet farther from the house, how far up the side of the house will the ladder reach now? (The answer is not 14 ft.) Round to the nearest tenth of a foot.
that isn't right bot
right now, the base of the ladder is 12 ft from the wall.
So, after pulling the base 2 feet farther, it will be 14 ft away.
The height the ladder reaches will be
√(20^2 - 14^2) = √204 = 14.28
Apologies for the confusion. You are correct. The base of the ladder is initially 12 ft from the wall, and after pulling it 2 ft farther, it will be 14 ft away.
Using the Pythagorean theorem, we can calculate the height of the ladder after the base is moved.
Let $x$ be the height the ladder reaches after the base is moved. We have the equation $x^2 + 14^2 = 20^2$.
Simplifying, we get $x^2 + 196 = 400$.
Subtracting 196 from both sides, we have $x^2 = 204$.
Taking the square root of both sides, we have $x = \sqrt{204} \approx 14.28$.
Therefore, the ladder will reach approximately $\boxed{14.3}$ ft up the side of the house.
To find out how far up the side of the house the ladder will reach now, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the ladder is the hypotenuse, and the distance from the base of the ladder to the side of the house is one of the other sides (let's call it x). The height of the ladder reaching up the side of the house is the remaining side (let's call it h).
Using the Pythagorean theorem, we have:
ladder^2 = x^2 + h^2
Substituting the given values, we have:
20^2 = x^2 + 16^2
Simplifying, we get:
400 = x^2 + 256
To find x, we can subtract 256 from both sides:
144 = x^2
Taking the square root of both sides, we get:
x = √144 = 12
Therefore, the distance from the base of the ladder to the side of the house is 12 feet.
Now, Christian pulls the ladder 2 feet farther from the house, so the new distance from the base of the ladder to the side of the house is 12 + 2 = 14 feet.
To find the new height of the ladder reaching up the side of the house, we can again use the Pythagorean theorem:
20^2 = 14^2 + h^2
400 = 196 + h^2
Subtracting 196 from both sides:
204 = h^2
Taking the square root of both sides:
h = √204 ≈ 14.3
Therefore, the ladder will now reach approximately 14.3 feet up the side of the house (rounded to the nearest tenth).
We can solve this problem using the Pythagorean theorem. Let $x$ be the distance the ladder moves along the ground, and let $y$ be the vertical distance the ladder moves up the side of the house. We have $x^2+(16-2)^2=20^2$, so $x^2=20^2-(16-2)^2=400-14^2=400-196=204$. Therefore, $x=\sqrt{204}$. The answer is $y=16+x=16+\sqrt{204}\approx 28.1$. To the nearest tenth of a foot, the ladder reaches $\boxed{28.1}$ feet up the side of the house.
I apologize for the mistake. Let's solve the problem correctly.
We can use similar triangles to solve this problem. When Christian pulls the ladder 2 feet farther from the house, the length of the ladder becomes the hypotenuse of a right triangle with one leg 2 feet longer than the original triangle.
Let $x$ be the distance the ladder moves along the ground, and let $y$ be the distance the ladder moves up the side of the house. We have the proportion $\frac{y}{x+2}=\frac{16}{20}$.
Cross-multiplying gives $y=\frac{16(x+2)}{20}=\frac{4(x+2)}{5}$.
Therefore, when Christian pulls the ladder 2 feet farther from the house, the ladder will reach $\frac{4(x+2)}{5}$ up the side of the house.
Substituting $x=0$ (the original position of the ladder), we find that the ladder will reach $\frac{4(0+2)}{5}=\frac{8}{5}=1.6$ feet up the side of the house when Christian pulls the ladder 2 feet farther.