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Find the area of the square.

 

 May 6, 2020
 #1
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The area of the square is  324 u²   indecision

 May 6, 2020
 #2
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The area of the square is  324 u²  

 

I made a boo-boo  cheeky       it's not 324 u² 

Guest May 6, 2020
 #4
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The area of the square is       A = 256 u²  indecision

 

Dragan  May 6, 2020
edited by Dragan  May 6, 2020
edited by Dragan  May 6, 2020
edited by Dragan  May 6, 2020
edited by Dragan  May 6, 2020
edited by Dragan  May 7, 2020
 #3
avatar+29978 
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As follows:

 

 May 6, 2020
 #5
avatar+678 
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This is real math! smiley

Dragan  May 6, 2020
 #6
avatar+24909 
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Find the area of the square.

 

\(\text{Let the side of the square $=a$} \\ \text{Let the area of the square $=a^2$} \\ \text{Let the radius of the circle $=r$} \\ \text{Let the center of the circle $(x_c,y_c)=(-r,\ 0)$} \)

 

circle:

\(\begin{array}{|rcll|} \hline \mathbf{\text{circle:}} \\ \hline \mathbf{(x+r)^2+y^2} &=& \mathbf{r^2} \quad | \quad x=-\dfrac{a}{2} \\\\ \left(r-\dfrac{a}{2}\right)^2+y^2 &=& r^2 \\\\ r^2 -2\cdot \dfrac{ar}{2}+\dfrac{a^2}{4} +y^2 &=& r^2 \\\\ r^2 - ar +\dfrac{a^2}{4} +y^2 &=& r^2 \\ y^2 &=& ar -\dfrac{a^2}{4} & \boxed{2r = a-6 \\ r=\dfrac{a-6}{2}} \\ y^2 &=& a\left(\dfrac{a-6}{2}\right) -\dfrac{a^2}{4} \\\\ y^2 &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} & \boxed{4+y = \dfrac{a}{2}\\y=\dfrac{a}{2}-4\\y=\dfrac{a-8}{2} } \\ \left(\dfrac{a-8}{2}\right)^2 &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} \\\\ \dfrac{(a-8)^2}{4} &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} \\\\ (a-8)^2 &=& 2a(a-6) -a^2 \\ a^2-16a+64 &=& 2a^2-12a-a^2 \\ -16a+64 &=& -12a \\ 64 &=& 4a \quad | \quad :4 \\ 16 &=& a \\ \mathbf{a} &=& \mathbf{16} \\ \hline \end{array}\)

 

The area of the square is \(a^2 = 16^2 = \mathbf{256}\)

 

laugh

 May 6, 2020

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