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Explore the Back-StoryTry this beautiful Geometry Problem based on area of region from AMC 10 B, 2016. You may use sequential hints to solve the problem.

What is the area of the region enclosed by the graph of the equation $x^{2}+y^{2}=|x|+|y| ?$

,

- $\pi+\sqrt{2}$
- $\pi+2$
- $\pi+2 \sqrt{2}$
- $2 \pi+\sqrt{2}$
- $2 \pi+2 \sqrt{2}$

Geometry

Semi circle

graph

Pre College Mathematics

AMC-10B, 2016 Problem-21

$\pi+2$

The given equation is $x^{2}+y^{2}=|x|+|y|$. Expanding this equation we get four equation as mod exist here...

$x^2+y^2-x-y=0$.......................(1)

$x^2+y^2+x+y=0$....................(2)

$x^2+y^2-x+y=0$.....................(3)

$x^2+y^2+x-y=0$.....................(4)

using this four equation can you draw the figure ?

Now can you finish the problem?

now four equations can be written as $x^{2}-x+y^{2}-y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}+y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}-x+y^{2}+y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}-y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$ which represents four circles and they overlapping.....

The center of the four circles are $\left(\frac{1}{2}, \frac{1}{2}\right)$, $\left(\frac{-1}{2}, \frac{-1}{2}\right)$,$\left(\frac{1}{2}, \frac{-1}{2}\right)$,$\left(\frac{-1}{2}, \frac{1}{2}\right)$Now we have to find out the region union of the four circles.

Now can you finish the problem?

There are several ways to find the area, but note that if you connect (0,1),(1,0),(-1,0),(0,-1) to its other three respective points in the other three quadrants, you get a square of area 2 , along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU

Try this beautiful Geometry Problem based on area of region from AMC 10 B, 2016. You may use sequential hints to solve the problem.

What is the area of the region enclosed by the graph of the equation $x^{2}+y^{2}=|x|+|y| ?$

,

- $\pi+\sqrt{2}$
- $\pi+2$
- $\pi+2 \sqrt{2}$
- $2 \pi+\sqrt{2}$
- $2 \pi+2 \sqrt{2}$

Geometry

Semi circle

graph

Pre College Mathematics

AMC-10B, 2016 Problem-21

$\pi+2$

The given equation is $x^{2}+y^{2}=|x|+|y|$. Expanding this equation we get four equation as mod exist here...

$x^2+y^2-x-y=0$.......................(1)

$x^2+y^2+x+y=0$....................(2)

$x^2+y^2-x+y=0$.....................(3)

$x^2+y^2+x-y=0$.....................(4)

using this four equation can you draw the figure ?

Now can you finish the problem?

now four equations can be written as $x^{2}-x+y^{2}-y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}+y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}-x+y^{2}+y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}-y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$ which represents four circles and they overlapping.....

The center of the four circles are $\left(\frac{1}{2}, \frac{1}{2}\right)$, $\left(\frac{-1}{2}, \frac{-1}{2}\right)$,$\left(\frac{1}{2}, \frac{-1}{2}\right)$,$\left(\frac{-1}{2}, \frac{1}{2}\right)$Now we have to find out the region union of the four circles.

Now can you finish the problem?

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU

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