+0

0
55
6

Find the area of the square.

May 6, 2020

#1
0

The area of the square is  324 u²

May 6, 2020
#2
0

The area of the square is  324 u²

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

Guest May 6, 2020
#4
+678
0

The area of the square is       A = 256 u²

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
+29978
+3

As follows:

May 6, 2020
#5
+678
0

This is real math!

Dragan  May 6, 2020
#6
+24909
+1

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 $$a^2 = 16^2 = \mathbf{256}$$

May 6, 2020