Problem : A ladder 10 m long is placed between two buildings so that it will reach a window 6 m high on one building. With its foot at the same point and turned over it will reach a window 5 m high on the other building. What is the distance between the two building?
Attempt: The part "With its foot at the same point and turned over it will reach a window 5 m high on the other building." is confusing me. Below is a picture of what I have in mind. The other side of the ladder is placed 5m high above the other building.
Question: Err.. did I get the drawing right? If yes, what is the next step to solve the problem. No given angles, except maybe the 90 degrees angle in the small triangle you can form in the middle.
I tried subtracting 6 from 5 to get the opposite of the small triangle, then do Pythagorean theorem with the hypotenuse 10 which resulted to 10.05 m which is wrong.
According to the answer key: The answer is supposed to be 16.66 m. Any help would be appreciated.
Best Answer
You have misinterpreted the question. The foot of the ladder is on the ground.
Let $x$ be the distance of the foot of the ladder from the building on which the $10~\text{m}$ ladder reaches a height of $6~\text{m}$. Then, by the Pythagorean Theorem, \begin{align*} x^2 + (6~\text{m})^2 & = (10~\text{m})^2\\ x^2 + 36~\text{m}^2 & = 100~\text{m}^2\\ x^2 & = 64~\text{m}^2\\ x & = 8~\text{m} \end{align*} Let $y$ be the distance of the foot of the ladder from the building on which the $10~\text{m}$ ladder reaches a height of $5~\text{m}$. Then, by the Pythagorean Theorem, \begin{align*} y^2 + (5~\text{m})^2 & = (10~\text{m})^2\\ y^2 + 25~\text{m}^2 & = 100~\text{m}^2\\ y^2 & = 75~\text{m}^2\\ y & = 5\sqrt{3}~\text{m} \end{align*} Since the foot of the ladder is at the same point in both cases, the distance between the building is $x + y = 8~\text{m} + 5\sqrt{3}~\text{m} \approx 16.66~\text{m}$.