Walkabout wrote:
Attachment:
observation.png
In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object?
(A) \(10-5\sqrt{3}\)
(B) \(10-5\sqrt{2}\)
(C) \(2\)
(D) \(2 \frac{1}{2}\)
(E) \(4\)
Given: VR = 10 feetLet's add a line from V to R to create the following right triangle:
![](https://i.imgur.com/3V6AhSZ.png)
Apply the Pythagorean theorem to get: \(x^2 + 5^2 = 10^2\)
Simplify: \(x^2 + 25 = 100\)
Subtract \(25\) from both sides of the equation: \(x^2 = 75\)
Solve to get: \(x = \sqrt{75}\)
To simplify \(\sqrt{75}\), we'll use the fact that
\(\sqrt{(x)(y)}=(\sqrt{x})(\sqrt{y})\)So, \(\sqrt{75}=\sqrt{(25)(3)}\)
\(=(\sqrt{25})(\sqrt{3})\)
\(=5\sqrt{3}\)Aside: Another way to find the value of x is to first recognize that, since the hypotenuse is TWICE the length of one leg, we must have a 30-60-90 right triangle on our hands, in which case x=5√3 Our diagram now looks like this:
What is the distance RS?From the diagram, we can see that \(=5\sqrt{3}+RS = 10\), which means \(RS = 10-5\sqrt{3}\)
Answer: A