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What's the area of the main square? (I think the attached picture defines the problem clearly.)

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Not really an important comment, but the bigger circle is irrelevant. –  barto Jan 31 at 14:17
2  
@barto It serves to establish that the two interior lines bisect the square exactly. –  Harald Hanche-Olsen Jan 31 at 14:23

4 Answers 4

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Let $A$ denote the leftmost point (shown) on the small circle and $C$ the rightmost point shown. Let $B$ denote the "top" point shown on the small circle. Let $O$ be the center of the square.

Now let $x$ be the length of $AO$. Then the length of $BO$ is $2+x$ and the length of $OC$ is $6+x$.

An elementary theorem from geometry tells us $\angle ABC$ is $90^\circ$. It follows that triangles $\triangle ABC$ and $\triangle AOB$ are similar. By similar triangles we have $$ {6+2x\over \sqrt {x^2+(2+x)^2}}= {{\sqrt {x^2+(2+x)^2}}\over x} $$ whence $x=2$.

The side length of the square is thus $16$.

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How do you know that BO is (2 + x)? –  user124432 Jan 31 at 17:03
    
@user124432 $6+x = 4+BO$ (look at the upper and left parts of the "+"). –  David Mitra Jan 31 at 17:05
    
indeed, thank you, your solution is the easiest to follow –  user124432 Jan 31 at 17:11

Let the square be of side $2a$. Clearly $a > 6$. Then with the origin at the centre of the square, the inside circle can be expressed as $(x-b)^2+y^2=(a-b)^2$

So we know that the intercept of this circle with the positive $Y$ axis must be $a-4$ and the intercept with the $-X$ axis is $6-a$. Hence $$b^2+(a-4)^2=(a-b)^2, \qquad (6-a-b)^2=(a-b)^2$$

$$\implies a \in \{3, 8\} \implies a = 8$$

So the area required is $256$.

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Inside the small circle are two intersecting chords. Call the lengths of their segments $h$, $h$, $\ell$ and $L$. I hope it's obvious the two $h$'s correspond to the two pieces of the vertical chord. Clearly $L=h+4=6+\ell$, since all three are the radius of the large circle. By the Intersecting Chords Theorem,

$$ h^2=\ell L=(h-2)(h+4)=h^2+2h-8$$

from which $h=4$ follows easily. This implies the large circle has radius $8$, which means the square has sides of length $16$, hence area $256$.

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Let the point at the center of the inner circle be $R$;

Let the point at the center of the square be $O$;

Let the point labeled at the "top" of the circle (near the number 4) be $A$;

Let $r$ = radius of small circle;

Let $2L$ = length of the side of the square;

Let $x$ = the distance from the $O$ to $R$;

We can write a few equations immediately.

1) Measuring the center horizontal line in two different ways and equating, we see $6+2r=r+x+L$. This gives: \begin{equation} x=(r-L)+6. \end{equation}

2) From right triangle $OAR$ we can write: \begin{equation} (L-4)^2 + x^2 = r^2 \end{equation}

3) From the horizontal line again we get: \begin{equation} 2r+6=2L \implies r=L-3 \end{equation}

Plugging equation (3) into (1) we get that $x=3$. Plugging equation (3) and $x=3$ into equation (2) gives:

\begin{equation} (r-1)^2+3^2=r^2 \end{equation}

which gives $r=5$. Therefore, using (3) again, $2L=2(r+3)=16$ and $Area=256$.

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