# Curved square

Can you find the area of the central part of this shape? Can you do it in more than one way?

## Problem

A square of side length 1 has an arc of radius 1 drawn from each of its corners, as in the following diagram. The arcs intersect inside the square at four points, to create the shaded region.

Image

**What is the area of the largest square that can be completely contained within the shaded region?**

Is this a good estimate of the actual shaded area?**What is the exact area of the central shaded region?**

How did that compare to your estimate?**Can you find more than one method to work out the exact area?**

Click here for a poster of this problem.

## Getting Started

You can solve this using basic properties of areas of circles, coordinate geometry and Pythagoras's theorem or simple trigonometry.

This worksheet suggests a possible method.

You can also solve it using calculus.

This worksheet suggests a possible method.

This worksheet suggests a possible method.

You can also solve it using calculus.

This worksheet suggests a possible method.

## Student Solutions

We had some really great, clear and detailed solutions for this one. It's a shame we can only write about a few of them!

Mowada and Megan used only clever coordinate geometry and the area of a sector. It can be nice to avoid tricky integrals!

Let the bottom left corner of the whole shape be the origin. We worked out the coordinates of the top point of the curved square, call this $P$. By symmetry we know that its position on the x-axis is $\frac{1}{2}$. To work out the y-value; substitute the x-value into $x^2+y^2 =1$, which is the equation of the circle the point is on.

\begin{align}

\left(\frac{1}{2}\right)^2 + y^2 &= 1\\

y^2 &= \frac{3}{4}\\

y &=\pm\frac{\sqrt 3}{2}

\end{align}

We need the positive value $y=\frac{\sqrt 3}{2}$. So the coordinates are $(\frac{1}{2},\frac{\sqrt 3}{2})$.

A very similar method can be used to find the coordinates of the right hand point of the curved square; call it $Q$. It lies on the same circle but with y coordinate equal to a half. This point has coordinates $(\frac{\sqrt 3}{2},\frac{1}{2})$.

Great work! Another way to work out the coordinates of $Q$ quickly is by symmetry; reflect the $P$ in the line $y=x$, which is the same as swapping the coordinates around.

With these points worked out; we can use Pythagoras' Theorem to find the area of the largest square inside the curved square. The largest square inside has the same corners as the curved square a side length of

\begin{equation}

\sqrt{2\left(\frac{\sqrt 3}{2}-\frac{1}{2}\right)^2}=\frac{\sqrt 3 -1}{\sqrt 2}.

\end{equation}

Then the area of the largest square in the curved square is $2-\sqrt 3$.

Well done to Brandon who found this answer. But what about the curved square? Mowada and Megan guessed the size of the angle $POQ$ made at the origin between the two points $P$ and $Q$ we've already calculated. It is indeed 30$^\circ$. This could be checked with some trigonometry if you wanted to be sure.

Since $30/360=1/12$ we have that the area of the sector $OPQ$ is $\pi/12$. Also triangles $ORP$ and $ORQ$ are identical. We calculate their area. $PR$ has length $\frac{\sqrt 3 -1}{2}$ and the 'height' to $O$ is $1/2$. The area is then $\frac{\sqrt 3 -1}{8}$. The quarter of the curved square $PQR$ is then just $\frac{\pi}{12}-\frac{\sqrt 3 -1}{4}$, so the curved square has area

\begin{equation}

\frac{\pi}{3}-\sqrt 3+1.

\end{equation}

This is brilliant deduction, and not an integral in sight. Well done! Some of you might want to practice integration however, so here is Rahul's method.

We can integrate the equation of the circle that $P$ and $Q$ lie on between $x=1/2$ and $x=\sqrt{3}/2$. From this subtract the area of the rectangle with top corners $R$ and $Q$. Using the equation of a circle we need to integrate $y=\sqrt{1-x^2}$. The substitution $x=\cos t$ so that $\frac{\mathrm{d}x}{\mathrm{d}t}=-\sin t$ will be useful.

\begin{align}

\int_{1/2}^{\sqrt 3/2}\sqrt{1-x^2}\,\mathrm{d}x &= -\int_{\pi/3}^{\pi/6}(\sin t)^2\,\mathrm{d}t\\

&= \int_{\pi/6}^{\pi/3}\frac{1}{2}\left(1-\cos(2t)\right)\,\mathrm{d}t\\

&=\left[\frac{1}{2}t-\frac{1}{4}\sin(2t)\right]_{\pi/6}^{\pi/3}\\

&=\left(\frac{\pi}{6}-\frac{\sqrt 3}{8}\right)-\left(\frac{\pi}{12}-\frac{\sqrt 3}{8}\right)\\

&=\frac{\pi}{12}

\end{align}

So subtract the area of the rectangle, which is $(\sqrt 3 -1)/4$ to get a quadrant area of

\begin{equation}

\frac{\pi}{12}-\frac{\sqrt 3 -1}{4}.

\end{equation}

Great work. This is Ben's way of doing the integration. Do you have another? All that's left is to multiply by four to get the area of the whole curved square.

\begin{equation}

\text{Area(curved square)}=\frac{\pi}{3}-\sqrt 3 +1.

\end{equation}

There are many more ways to find the answer. See if you can come up with two more! Let's compare with the straight sided square inside:

\begin{align}

\text{Area(curved square)}-\text{Area(straight square)}&=\frac{\pi}{3}-\sqrt 3 +1-2+\sqrt 3\\&=\frac{\pi}{3}-1\\&\approx 0.0472\text{ (3.s.f).}

\end{align}

Mowada and Megan used only clever coordinate geometry and the area of a sector. It can be nice to avoid tricky integrals!

Let the bottom left corner of the whole shape be the origin. We worked out the coordinates of the top point of the curved square, call this $P$. By symmetry we know that its position on the x-axis is $\frac{1}{2}$. To work out the y-value; substitute the x-value into $x^2+y^2 =1$, which is the equation of the circle the point is on.

\begin{align}

\left(\frac{1}{2}\right)^2 + y^2 &= 1\\

y^2 &= \frac{3}{4}\\

y &=\pm\frac{\sqrt 3}{2}

\end{align}

We need the positive value $y=\frac{\sqrt 3}{2}$. So the coordinates are $(\frac{1}{2},\frac{\sqrt 3}{2})$.

A very similar method can be used to find the coordinates of the right hand point of the curved square; call it $Q$. It lies on the same circle but with y coordinate equal to a half. This point has coordinates $(\frac{\sqrt 3}{2},\frac{1}{2})$.

Great work! Another way to work out the coordinates of $Q$ quickly is by symmetry; reflect the $P$ in the line $y=x$, which is the same as swapping the coordinates around.

Image

Image

With these points worked out; we can use Pythagoras' Theorem to find the area of the largest square inside the curved square. The largest square inside has the same corners as the curved square a side length of

\begin{equation}

\sqrt{2\left(\frac{\sqrt 3}{2}-\frac{1}{2}\right)^2}=\frac{\sqrt 3 -1}{\sqrt 2}.

\end{equation}

Then the area of the largest square in the curved square is $2-\sqrt 3$.

Well done to Brandon who found this answer. But what about the curved square? Mowada and Megan guessed the size of the angle $POQ$ made at the origin between the two points $P$ and $Q$ we've already calculated. It is indeed 30$^\circ$. This could be checked with some trigonometry if you wanted to be sure.

Since $30/360=1/12$ we have that the area of the sector $OPQ$ is $\pi/12$. Also triangles $ORP$ and $ORQ$ are identical. We calculate their area. $PR$ has length $\frac{\sqrt 3 -1}{2}$ and the 'height' to $O$ is $1/2$. The area is then $\frac{\sqrt 3 -1}{8}$. The quarter of the curved square $PQR$ is then just $\frac{\pi}{12}-\frac{\sqrt 3 -1}{4}$, so the curved square has area

\begin{equation}

\frac{\pi}{3}-\sqrt 3+1.

\end{equation}

This is brilliant deduction, and not an integral in sight. Well done! Some of you might want to practice integration however, so here is Rahul's method.

We can integrate the equation of the circle that $P$ and $Q$ lie on between $x=1/2$ and $x=\sqrt{3}/2$. From this subtract the area of the rectangle with top corners $R$ and $Q$. Using the equation of a circle we need to integrate $y=\sqrt{1-x^2}$. The substitution $x=\cos t$ so that $\frac{\mathrm{d}x}{\mathrm{d}t}=-\sin t$ will be useful.

\begin{align}

\int_{1/2}^{\sqrt 3/2}\sqrt{1-x^2}\,\mathrm{d}x &= -\int_{\pi/3}^{\pi/6}(\sin t)^2\,\mathrm{d}t\\

&= \int_{\pi/6}^{\pi/3}\frac{1}{2}\left(1-\cos(2t)\right)\,\mathrm{d}t\\

&=\left[\frac{1}{2}t-\frac{1}{4}\sin(2t)\right]_{\pi/6}^{\pi/3}\\

&=\left(\frac{\pi}{6}-\frac{\sqrt 3}{8}\right)-\left(\frac{\pi}{12}-\frac{\sqrt 3}{8}\right)\\

&=\frac{\pi}{12}

\end{align}

So subtract the area of the rectangle, which is $(\sqrt 3 -1)/4$ to get a quadrant area of

\begin{equation}

\frac{\pi}{12}-\frac{\sqrt 3 -1}{4}.

\end{equation}

Great work. This is Ben's way of doing the integration. Do you have another? All that's left is to multiply by four to get the area of the whole curved square.

\begin{equation}

\text{Area(curved square)}=\frac{\pi}{3}-\sqrt 3 +1.

\end{equation}

There are many more ways to find the answer. See if you can come up with two more! Let's compare with the straight sided square inside:

\begin{align}

\text{Area(curved square)}-\text{Area(straight square)}&=\frac{\pi}{3}-\sqrt 3 +1-2+\sqrt 3\\&=\frac{\pi}{3}-1\\&\approx 0.0472\text{ (3.s.f).}

\end{align}

## Teachers' Resources

### Why do this problem?

This problem draws together coordinate geometry, equations of circles, and surds, and can also be approached using integration.### Possible approach

Show the diagram (also available as a PowerPoint slide).

Give students time to study the diagram, and make notes about what they know and what they can work out. Pose the problem of finding the shaded area, and after some thinking time bring the class together to discuss possible strategies.

Two possible strategies are outlined in the following worksheets:

Sectors Method

Integration Method

You could outline the general methods to the class and give them time to solve the problem for themselves, offering the worksheet as a prompt if they get stuck.

Allow time at the end of the lesson for students to compare the different approaches.

Give students time to study the diagram, and make notes about what they know and what they can work out. Pose the problem of finding the shaded area, and after some thinking time bring the class together to discuss possible strategies.

Two possible strategies are outlined in the following worksheets:

Sectors Method

Integration Method

You could outline the general methods to the class and give them time to solve the problem for themselves, offering the worksheet as a prompt if they get stuck.

Allow time at the end of the lesson for students to compare the different approaches.

### Key questions

What information will we need to find the area?

What symmetries are present in the diagram? Does the area split up in any obvious ways?

Which would be the easiest arcs to work with? Why?

How can integration be used to find the area?

### Possible extension

Find the areas of the other parts of the diagram.

Set up a similar problem using parabolas instead of circles.

### Possible support

The worksheets suggest a suitable coordinate grid, and offer prompts for students to follow.