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# Area of region | AMC 10B, 2016| Problem No 21

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

## Area of region- AMC-10B, 2016- Problem 21

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

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10B, 2016 Problem-21

#### Check the answer here, but try the problem first

$\pi+2$

## Try with Hints

#### First Hint

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?

#### Second Hint

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?

#### Third Hint

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 $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

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

## Area of region- AMC-10B, 2016- Problem 21

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

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10B, 2016 Problem-21

#### Check the answer here, but try the problem first

$\pi+2$

## Try with Hints

#### First Hint

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?

#### Second Hint

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?

#### Third Hint

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 $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

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