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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+177 
+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.

 

\(r^2 = {\rm AE}^2 + ({\rm AB} + l)^2 \\ r^2 = 2^2 + (2 + l)^2\)\(r^2 = {\rm CF}^2 + ({\rm BC}^2 - l)^2 \\ r^2 = 5^2 + (5 - l)^2\)

 

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.

 

\(2^2 + (2 + l)^2 = 5^2 + (5 - l)^2 \\ 4 + 4 + 4l + l^2 = 25 + 25 - 10l + l^2 \\ 8 + 4l = 50 - 10l \\ 14l = 42 \\ l = 3\)

 

Now that we have solved for L, we can find the radius.

 

\(r^2 = 2^2 + (2 + l)^2 \\ r^2 = 2^2 + (2 + 3)^2 \\ r^2 = 4 + 25 \\ r^2 = 29 \\ r = \sqrt{29} \text{ or } r = -\sqrt{29}\)

 

Since we are dealing with the radius, which is a length, we should reject \(r = -\sqrt{29}\). Therefore, \(r = \sqrt{29} \approx 5.3852\)

 Aug 19, 2023
 #2
avatar+128818 
+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+177 
+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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