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Two squares are inscribed in a semi-circle, as shown below.  Find the radius of the semi-circle.

 

 Aug 18, 2023
 #1
avatar+189 
+1

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)2r2=CF2+(BC2l)2r2=52+(5l)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+(5l)24+4+4l+l2=25+2510l+l28+4l=5010l14l=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=295.3852

 Aug 19, 2023
 #2
avatar+129979 
+1

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.....

 

cool cool cool

 Aug 19, 2023
 #3
avatar+189 
+1

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.

The3Mathketeers  Aug 19, 2023

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