# Find the ratio of the area inside the square but outside the circle to the area of the square in the figure.

All I'm worried about it is (a); for now. Okay, so let's start off like this: I know what the question is asking, sort of. I know it wants the ratio of the inside corner pieces of the square to the square itself. But what??? I obviously know that $A_{circle}=\pi r^2$ and $A_{square}=wh$ and also that the piece they are looking for is $A_{circle}-A_{square}$ but the ratio? I don't understand that. Can someone explain to me what they mean by that and stay posted for I may have questions about (b)-(e). Thanks.

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They want $\frac{A_{square}-A_{circle}}{A_{square}}$ –  Karolis Juodelė Aug 10 '12 at 8:45
$\dfrac{wh-\pi r^2}{wh}$? –  Austin Broussard Aug 10 '12 at 8:46
I think that you're just looking for the definition of a ratio. it just measure proportionally how much larger or smaller is one quantity when compared to another. For example, Profit Margin is the amount of net income generated per dollar of revenue, so the formula would be $$\frac{\text{net income}}{\text{total revenue}}$$. Similarly, problem (a) is asking for the proportion of the area of the square that is cover by its corners. –  John Joy Aug 20 '14 at 18:06

If $A_0$ is the total area of the four corner pieces, and $A_1$ is the area of the square, they want $\frac{A_0}{A_1}$: that fraction is by definition the ratio of $A_0$ to $A_1$. Note also that for the square you have $w=h=2r$.

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You misread, $A_1$ is the square. –  Karolis Juodelė Aug 10 '12 at 8:48
$w=h=2r\equiv A_1=2r^2$? –  Austin Broussard Aug 10 '12 at 8:49
@Karolis: Miswrote, actually; thanks. Fixed. –  Brian M. Scott Aug 10 '12 at 8:51
Thus, $\dfrac{4r^2-\pi r^2}{4r^2}$? –  Austin Broussard Aug 10 '12 at 8:53
@Austin: That’s how I’d leave it; some might prefer to leave it as $\frac{4-\pi}4$. –  Brian M. Scott Aug 10 '12 at 9:00

The side length of the square is $2r$. Therefore, the area of the square is $(2r)^2 = 4r^2$. The area of the circle inside the square is $\pi r^2$. Therefore, the area inside of the square outside of the circle is $A_{square} -A_{circle}= 4r^2 - \pi r^2 = r^2(4-\pi).$ Thus, the ratio of the area inside of the square but outside of the circle to the area of the square is $r^2(4-\pi)/(4r^2)=\frac{4-\pi}{4}$

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Answer for Q b) (4+pi)r I.e.,( perimeter of semicircle with radius r is pi.r)+ (peri meter of rectangle with sides r & 2r is r+r+2r+2r)- (length of rectangle is 2r)(because we need peri meter of the given diagram) So [pi.r+6r-2r] =[ pi.r+4r]= (4+pi)r

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For Q a)

The answer is $(4-\pi):4$ (or) $(4-\pi)r^2:4r^2$

First we have to find out the area of corner pieces. That we will get by subtracting the area of circle from the area of square. I.e.,area of square of side $2r$ is $4r^2$ ( bz. They gave half of the side is $r$) And area of circle with radius $r$ is $\pi r^2$ So area of corner pieces is $4r^2-\pi r^2 = (4-\pi) r^2$ Finally the ratio of area of corner pieces to square is

$$(4-\pi)r^2:4r^2$$ (or) $$(4-pi):4$$

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While your answer might be correct, please edit it and use MathJax to typset the formulas (it's LaTeX, essentially). As it's currently written, your answer is very hard to understand. –  TZakrevskiy Feb 21 '14 at 11:15