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.
5 answers
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.
that isn't right bot
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.
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.
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
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.
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.