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1. Three squares are drawn inside a circle as shown. The area of the circle is 60pi square inches. How many square inches are in the area of one square?

 

2. Two circles inside a square are externally tangent to each other and are tangent to certain sides of the square as shown. The perimeter of the square is \(2 \sqrt{2}\) . What is the sum of the circumferences of the two circles?

 

 Feb 19, 2021
 #1
avatar+36915 
+2

Here is  A  way to solve first one....may not be THE BEST way....

     circle radius = sqrt 60

draw a line from origin to corne of square that touches circle     this line forms a   arc tan (1/3) = 18.435  degree angle

     sqrt 60 cos 18.435   = length of 3 boxes = 7.34      so each square is side length   7.34/3

              then total area of  1 square   =     (7.34/3  *  7.34/3) =  6 in2    

 Feb 19, 2021
edited by ElectricPavlov  Feb 19, 2021
 #2
avatar+45 
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Wow! Thank you for answering very fast.

Penguin  Feb 19, 2021
 #3
avatar+505 
+2

ElectricPavlov nice solution!

Solution for 2:

Since the perimeter of the square is \(2\sqrt{2}\), the side length of one side of the square is \(\frac{\sqrt{2}}{2}\). The diagonal of that square is then equal to \(\frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1\)

Let x be the radius of one circle, and let y be the radius of the other circle. 

Notice that the diagonal is equal to \(x\sqrt{2} + x + y\sqrt{2} + y\). Therefore, 

\(x\sqrt{2} + x + y\sqrt{2} + y = 1\\ x(\sqrt{2}+1)+ y(\sqrt{2}+1)=1\\ x+y=\frac{1}{\sqrt{2}+1} = \sqrt{2}-1\)

Since the sum of the circumference of the circle is 2 times the radius times pi, just multiply 2pi to get the final answer:

\(\boxed{2\sqrt{2}\pi-2\pi}\)

 Feb 19, 2021
 #4
avatar+45 
0

Thank you for your efforts and response, but the answer is wrong

Penguin  Feb 19, 2021
 #5
avatar+505 
+3

I'm really sorry that I got it wrong. Could you tell me the answer so I can see where I made a mistake?

Edit: It seems like you mistyped the problem according to this:

https://web2.0calc.com/questions/help_12091

 

The solution would have been:

\(x(\sqrt{2}+1)+y(\sqrt{2}+1)=\frac{1}{2}(\sqrt{2}+1)\\ x+y=\frac{1}{2}\)

Since the sum of the circumference of the circle is 2 times the radius times pi, just multiply 2pi to get the final answer:

\(\frac{1}{2} \cdot 2 \cdot \pi = \boxed{\pi}\)

textot  Feb 19, 2021
edited by textot  Feb 20, 2021
 #6
avatar+45 
+1

Oh my bad. I'm really sorry for making you think you were wrong.

Penguin  Feb 20, 2021

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