Two squares are inscribed in a semi-circle, as shown below. Find the radius of the semi-circle.
I had to stare at this one a long time to solve this one! I have created a diagram below with points labeled so that it is easier to follow along.
In the diagram above, notice that OF and OE are both radii of the semicircle. These lengths I have labeled as r. I have labeled L as the length of segment OB. We can now apply Pythagorean's Theorem twice to find the radius of this semicircle.
r2=AE2+(AB+l)2r2=22+(2+l)2 | r2=CF2+(BC2−l)2r2=52+(5−l)2 |
Now, we have expressions for both r and L, so we can solve for r, and we are finished. Probably the best way is to solve for L first since that one is easier to solve.
22+(2+l)2=52+(5−l)24+4+4l+l2=25+25−10l+l28+4l=50−10l14l=42l=3
Now that we have solved for L, we can find the radius.
r2=22+(2+l)2r2=22+(2+3)2r2=4+25r2=29r=√29 or r=−√29
Since we are dealing with the radius, which is a length, we should reject r=−√29. Therefore, r=√29≈5.3852
I tried a similar procedure as you, 3Mathketeers and I got the same answer.....but notice that a square of 5 will not fit inside the semi-circle [a 4 x 5 rectangle will !! ]
In short, I believe that this is an impossible problem as set up by the poster.....
I apologize for the confusion. As you discovered, the diagram I created is intentionally NOT drawn to scale so that it looks as similar as possible to the original diagram. I figured that as long as points E and F were on the circle, then I could still go ahead and find the radius anyway.