Does the square or the circle have the greater perimeter? A surprisingly hard problem for high schoolers

Question

An exam for high school students had the following problem:

Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle?

Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square.

The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?

Answer 1

Perhaps the examiner intended the students to notice the square is determined by a $(3, 4, 5)$ triangle, because $3 + 5 = 4 + 4$ (!):

Consequently, as several others have noted, $$ \frac{\text{perimeter of the circle}}{\text{perimeter of the square}} = \frac{5 \cdot 2\pi}{4 \cdot 8} = \frac{\pi}{3.2} < 1. $$

---

For an approach less dependent on inspiration, taking the origin of the coordinate system at the center of the circle seems easier than placing the origin at the center of the square. Without loss of generality, assume the circle has unit radius:

Equating the lengths of the horizontal and vertical sides of the square in this diagram, we read off $$ x + 1 = 2y\quad\text{(or $x = 2y - 1$).} $$ Invoking the Pythagorean theorem and substituting the preceding line, \begin{align*} 0 &= x^{2} + y^{2} - 1 \\ &= (2y - 1)^{2} + y^{2} - 1 \\ &= 5y^{2} - 4y \\ &= y(5y - 4). \end{align*} Clearly $y \neq 0$, so $y = 4/5$, $x = 3/5$, and we notice the Examiner's Favorite Triangle.

Answer 2

I will let the side of the square be $2$. The center of the circle is at the intersection of the perpendicular bisectors of the chords $BC$ and $EB$. Let that point be $O$ and the foot of the perpendicular from $O$ be $F$.
Now the triangle $OEF$ is similar to $EBA.$
Therefore $$\frac{OE}{EB}= \frac{EF}{BA} \to \frac{OE}{\sqrt 5} = \frac{\sqrt 5/2}{2}\to OE = 5/4$$

The circumference of the circle is $$2\pi \frac 5 4= 5\pi/2 = 7.85 $$ which is less than $8$ the circumference of the square. They are very close.

I would like to see a proof that is not as computational as this one.

Answer 3

This is also "computational" and not purely geometic. Let the side of the square be $1$. Let $O$ be the centre of the circle and $G$ be the middle of the segement $\overline{CB}$; then $r=\vert OB\vert$ is the radius of the circle and $1-r= \vert OG\vert $. By Pythagoras we have $(1-r)^2+(\frac{1}{2})^2=\vert OG\vert^2+\vert GB\vert^2=\vert OB\vert^2=r^2$ from which one gets $r=\frac{5}{8}$. Again one gets the same as in the answer by abel. PS I also would like to see a purely geometric argument.

Answer 4

Regarding Abel's comment that he would like to see a less computational proof, consider the following. It is apparent that the length of a side of the square is $4$ small squares, and the perpendicular bisector of $EB$ crosses the horizontal centre line of the drawing $2.5$ small squares from $E$. Apologies for the inaccurate drawing.

More formally, if we call the centre of the square the origin:

EB has endpoints $(-2,0)$ and $(2,2)$ therefore it has midpoint $(0,1)$ and gradient $\frac12$.

The perpendicular bisector of EB has gradient $\frac{-1}{\mathrm{gradient\ of\ EB}}=-2$ therefore it crosses the $x$-axis at $(\frac12,0)$

Perimeter of ABCD = $16$

Perimeter of circle = $2 * \pi * 2.5 \approx 15.71$

---

---

(I liked this answer so much I wanted to illustrate step-by-step the process of finding the location of the centre of the circle:

First, the centre of the circle is defined by intersection of diameters; one is obviously the line straight across from $E$, and the other is generated by the perpendicular from the midpoint of $EB$:

Then we can locate the midpoint of $EB$ as being halfway across the square and one-quarter down from the top:

Then we can find another point on the perpendicular by turning the rectangle from E to the midpoint through a right angle:

And it's then obvious that the radius of the circle is $\frac{5}{8}$ the side of the square.

Joffan)

Answer 5

Unless you consider the Pythagorean theorem to be "a complicated geometric formula":

$$ s = \text{side length of square}\\ r = \text{radius of circle}\\ O = \text{center of circle}\\ G = \text{midpoint of BC}\\ $$

$$ r^2 = OG^2 + BG^2\\ r^2 = OG^2 + \left(\frac{s}{2}\right)^2\\ r^2 = (s - r)^2 + \frac{s^2}{4}\\ r^2 = s^2 - 2rs + r^2 + \frac{s^2}{4}\\ 0 = s^2 - 2rs + \frac{s^2}{4}\\ 2rs = \frac{5}{4} s^2\\ $$ Since $s > 0$:

$$ 2r = \frac{5}{4}s\\ r = \frac{5}{8}s\\ $$

Therefore the circumference of the circle is $2πr$; the perimeter of the square is $4s$.

$$ 2πr\stackrel{?}=4s\\ 2π \cdot \frac{5}{8}s \stackrel{?}= 4s\\ \frac{5π}{4} < 4 $$

Therefore, the circumference of the circle is less than the perimeter of the square.

Answer 6

Using the Pythagorean theorem and the Quadratic Equation

1. Use a unit circle. Radius of circle = 1, perimeter = 2 ~ 6.283.
2. Center of circle = (0, 0). Point E = (-1, 0). Point B = (x, y).
3. Perimeter of square = 4(1+x)=8y.
4. 1 + x = 2y.
5. 1 + 2x + x² = 4y².
6. x² + y² = 1.
7. 4y² = 4 - 4x².
8. 1 + 2x + x² = 4 - 4x².
9. 5x² + 2x - 3 = 0.
10. x = (-2 ± √(4+60)) / 10 = (-2 ± 8) / 10 = (-2 + 8) / 10 = 0.6
    (We can rule out x = (-2 - 8) / 10 = -1, which is Point E.)
11. y = 0.8
12. 1 + x = 1.6 = 2 * y (also a check-by-substitution)
13. Perimeter of square = 6.4 > Perimeter of circle ~ 6.283

Answer 7

Put the four corners at $(0,\pm1)$ and $(2,\pm1)$. Let the center of the circle be $(x,0)$. Then the radius of the circle is $x$ and that must be the distance from the center $(x,0)$ to the corner $(2,1)$. But that latter distance is $$ \sqrt{(2-x)^2 + 1^2}. $$ Therefore \begin{align} x & = \sqrt{(2-x)^2 + 1^2} \\ x^2 & = (2-x)^2+ 1 \\ x^2 & = 4-4x+x^2 + 1 \\ 0 & = 5-4x \\ x & = 5/4. \end{align} That's the radius, so we have $$ \text{circumference}=\frac 5 4 \cdot 2\pi \approx 7.85. $$ The perimeter of the square is $8$, so that's a bit bigger.

Answer 8

Using standard exam guesses

This happens to boil down to a 3:4:5 right triangle, so guessing can be used to solve the problem.

1. Use standard unit circle.
2. Perimeter of unit circle = 2 ~ 6.283.
3. Use standard unit circle (cartesian) coordinate system. Point E is at (-1, 0).
4. Let point O be the center of the circle, at (0, 0).
5. Let point F be the midpoint of the right side of the square.
6. Consider a few possible shapes for the right triangle OFB:

45-45-90 right triangle
Does not work, because a square based on this triangle does not reach to the left edge of the unit circle.

30-60-90 right triangle
Height BC is √3.
Width AB=EF is 1.5.
Does not work, because it is not a square.

3:4:5 right triangle
Height BC is 2 * 0.8 = 1.6
Width AB=EF is 1 + 0.6 = 1.6
Works; this is a square. Square perimeter = 6.4 > Circle perimeter = 2 ~ 6.283.

5:12:13 right triangle Don't need to evaluate, because the 3:4:5 right triangle solved the problem.

8:15:17 right triangle Don't need to evaluate, because the 3:4:5 right triangle solved the problem.

Answer 9

You can show this with a ruler-and compass construction, in the spirit of Archimedes. Or, you can get the same result with simple trig and a calculator, measuring angles in degrees only, without using calculus or knowing the value of $\pi$.

As the other answers have shown, we are comparing the perimeters of a square of side 8 and a circle radius 5.

If we circumscribe a regular polygon with $n$ sides around a circle radius $r$ so that the circle touches the mid-point of each side, each side of the polygon is length $2r \tan (180/n)^\circ$ and the circumference of the polygon is $2nr \tan (180/n)^\circ$.

It is obvious (at high school level!) that the circumference of the polygon is longer than the circumference of the circle.

So, start by circumscribing a square around the circle, and keep doubling the number of sides till something interesting happens.

$r = 5$.

Square: $n = 4$, perimeter = $2 \times 5 \times 4 \tan 45^\circ = 40$, which is a sanity check when using the calculator, of course.

Octagon: $n = 8$, perimeter = $2 \times 5 \times 8 \tan 22.5^\circ = 33.1$.

16-gon: $n = 16$, perimeter = $160 \tan 11.25^\circ < 31.9$. So the inscribed circle perimeter is $< 31.9$, which is less than the perimeter of a square with side $8$.

For extra credit, get them to check this by making a scale drawing.

Answer 10

Assume $\overline{AB}=1$.

Let $P$ be the intersection of the circle with $AB$.

Then $\overline{AP}\times\overline{AB}=\overline{AE}^2$ by power of point.

So $\overline{AP}=1/4$, and $\overline{PB}=3/4$. Hence the diameter of the circle is $\sqrt{(3/4)^2+1^2}=5/4$. The rest follows like the other proofs.

Answer 11

In the comments @Sid asked me to write this method down as an answer, even though it uses a formula.

If the students studied some Euclidean geometry, probably they know that the circumradius of a triangle with sides $a,b,c$ and area $S$ is $R=\frac{abc}{4S}$.

The circle drawn in the picture is the circumcircle of $BCE$. Let us choose units so that the square has side $2$ (to avoid denominators). Then $BE=CE=\sqrt{5}$, $BC=2$, $S_{BCE}=2$ are easy to find. Plugging them in the formula above, we get $R=\frac{5}{4}$.

So we have to compare $2\pi R = \frac{5}{2}\pi$ and $8$.

Answer 12

This calculation uses only the upper limit on the value of pi calculated by Archimedes (22/7) and simplifies the algebra somewhat by using an intermediate distance value, that of the base ZM of the triangle ZMB.

---

Define points Z as the centre of the circle and M as the midpoint of the line segment BC.

Now applying Pythagoras to the triangle ZMB in terms of a (the length of the segment ZM) and r (radius of the circle) yields (since L, the side length of the square, $ = r + a$):
$$ r^2 - a^2 = (r + a)^2 / 4 $$ and dividing by (r + a) gives $$ r - a = (r + a) /4 $$ which upon rearrangement and substitution of L back in gives $$ 3r = 5a = 5 (L - r) $$ and finally $$ 8 r = 5 L $$ Then the ratio of the circumference of the circle to that of the square is obviously $$ ratio = 2 \pi r / (4 *(8/5) r) = 10 \pi / 32 = 5 \pi / 16 $$ However, since $\pi < 22/7$ (by Archimedes), the upper limit of this ratio is $$ ratio = 5 \pi / 16 < 5 * 22 / (16 * 7) = 110/112 \approx 0.98 $$

Therefore the square has the greater circumference.

Answer 13

Here's a proof that arrives at the ratio of circle diameter to square side using similar triangles rather than the Pythagorean theorem:

Draw the diameter of the circle from $E$; let the other endpoint of that diameter be $F$ and let it cross the square at $G$. Then since the angle $\measuredangle EBF$ is a right angle, the triangles $\bigtriangleup EBG$ and $\bigtriangleup BFG$ are similar. Obviously $|EG|=2|BG|$, so $|BG|=2|FG|$ (or in other words $|FG|=\frac12|BG|$) and so $|EF|=|EG|+|FG|=2|BG|+\frac12|BG|=\frac52|BG|$. From here the proof proceeds just like the others: the perimeter of the square is $4|EG|=8|BG|$ and the circumference of the circle is $\pi|EF|=\frac{5\pi}{2}|BG|$. From $\pi\lt\frac{16}5=3.2$, we get $5\pi\lt16$ and $\frac{5\pi}{2}\lt 8$, so the perimeter of the square is longer.

Answer 14

A pure geometric solution (I think it's the same as @baum's solution and others but with details and a clarifying drawing)

Let's consider that the length of the square is 1. We can easily deduct that the square perimeter is 4.
Let (HI) be a line perpendicular to (AD), with I being its intersection with the circle).
Since (AD) is tangent to the circle, [HI] is its diameter.
$ J\hat H B =90°-J\hat B H$ (angles of a right triangle)
$J\hat B I =90°-J\hat B H$ (they both form the angle $I\hat B H=90°$ because IBH is a right triangle (inscribed in a semicircle))
therefore $ J\hat H B = J\hat B I = \theta$
$\tan(\theta)=\frac{IJ}{BJ} =2\times x$
$\tan(\theta)=\frac{BJ}{HJ} =\frac{1\over 2}{1}=\frac12$ (BJ=AH=$\frac{AD}{2}$=$\frac12$ since H is the midpoint of [AD])
$2\times x=\frac12\Rightarrow x=\frac14$
d=HI=1+$\frac14$=$\frac54$
perimeter$=2\pi r=d\pi=\frac{5\pi}{4}$

Let's compare $\frac{5\pi}{4}$ and 4
$\frac{5\pi}{4}-4=\frac{5\pi-16}{4}=\frac54(\pi-\frac{16}{5})=\frac54(\pi-3.2)<0 \text{ because } 3.2>\pi$
So $\frac{5\pi}{4}$<4
Therefore the perimeter of the square is bigger

Answer 15

This isn't a full answer, just an observation to supplement the other answers (and celebrate Pi Day).

If you construct tangents at the 20 points $(0, \pm25)$, $(\pm25, 0)$, $(\pm15, \pm20)$, $(\pm20, \pm15)$, $(\pm7, \pm24)$, $(\pm24, \pm7)$ on the circle of radius 25 centred on the origin, they intersect at the vertices $(\pm\frac{25}{7}, \pm25)$, $(\pm25, \pm\frac{25}{7})$, $(\pm\frac{125}{11}, \pm\frac{250}{11})$, $(\pm\frac{250}{11}, \pm\frac{125}{11})$, $(\pm\frac{125}{7}, \pm\frac{125}{7})$ of an irregular circumscribing icosagon, whose perimeter is $8 \times \left(\frac{25}{7} + 2 \times \frac{625}{77}\right) = \frac{12200}{77} = 158\frac{34}{77} < 160,$ as required.

Comparing this with the circumference, $50\pi$, of the inscribed circle, we get the upper bound for $\pi$: $$ \pi < \frac{244}{77} = 3\frac{13}{77} < 3.17. $$

Here, the circumscribing icosagon is shown in red, with the circle (just) visible behind it in black.

The points of tangency are marked with smaller blue circles. The 16 right-angled triangles used in construction, shown in green, are all Pythagorean, in ratio 3:4:5 or 7:24:25, with hypotenuse $\frac{625}{77}$. The other 4 edges of the icosagon are of length $\frac{50}{7}$.

There is no need to calculate any angles, but the symmetry of the figure is illuminated by knowing the sequence of angles subtended at the centre of the circle. Running counter-clockwise from the positive $x$-axis to the positive $y$-axis, it is: $\underline{\alpha\alpha\beta\beta\alpha\alpha\beta\beta\alpha\alpha}$, where $\alpha = \tan^{-1}\frac{1}{7}$, $\beta = \tan^{-1}\frac{2}{11}$, and: \begin{gather*} 2\alpha = \tan^{-1}\frac{7}{24}, \ \alpha + \beta = \tan^{-1}\frac{1}{3}, \ 2\alpha + \beta = \tan^{-1}\frac{1}{2}, \\ 2\alpha + 2\beta = \tan^{-1}\frac{3}{4}, \ 3\alpha + 2\beta = \frac{\pi}{4}, \ 4\alpha + 2\beta = \tan^{-1}\frac{4}{3}, \\ 4\alpha + 3\beta = \tan^{-1}2, \ 5\alpha + 3\beta = \tan^{-1}3, \ 4\alpha + 4\beta = \tan^{-1}\frac{24}{7}. \end{gather*}

Answer 16

You can solve this using just the quadratic formula and the equation of a circle, without bringing triangles, Pythagoras, or real geometric reasoning into it at all.

These are common enough hammers for all problems in high school, where everything turns out to be quadratic equations in disguise. Everything falls out nicely enough that it could even be what's intended.

1. Assume a unit circle centred at the origin. $P_{\mathrm{circle}} = 2\pi$.
2. $E$ is at $(-1, 0)$.
3. $A$ is at $(-1, h)$.
4. Note that $h$ is half the edge length of the square. $P_{\mathrm{square}} = 8h$.
5. $B$ is thus at $(-1 + 2h, h)$. This expression makes the quadratic formula become trivial.
6. Substitute these coordinates into $x^2 + y^2 = r^2$: $(-1 + 2h)^2 + h^2 = 1$.
7. Expand and simplify: $5h^2 - 4h = 0$.
8. So by the quadratic formula $h = {4 \pm \sqrt{(-4)^2 - 4 ~\cdot~ 5 ~\cdot~ 0} \over{2 ~\cdot~ 5}}$.
9. The discriminant simplifies almost away to get ${4 \pm \sqrt{4^2 - 0} \over 10}$, where the square root is trivial.
10. $h = {8\over10} \mathrm{~or~} {0\over 10}$.
11. $h = 0.8 \rightarrow P_{\mathrm{square}} = 8h = 6.4$.
12. $6.4 > 2\pi$.

Doing it this way also highlights the degenerate solution where the square's just a single point of mush on the perimeter of the circle, which is exactly the kind of detail I would have been expected to mention in an exam.

Answer 17

The problem is simply solved if the relative position of the center and the radius of the circle are known. The attached shows how they can be obtained.

Added:

Let the length of a side of the square be $L$. Then, the radius of the circle $= (\dfrac {5}{4}L) \times (\dfrac {1}{2})$.

$\dfrac {perimeter_{circle}}{perimeter_{square}} = ... = \dfrac {\pi}{3.2} \lt 1$.

Answer 18

Redrawn diagram above.

The $\measuredangle$CDF is determined and is given by: $$\arctan(\frac{a}{a/2}) = 63.4 ^ \circ$$ Then the $\measuredangle$FDB can be determined as:

$(180 - 2\times63.4) ^ \circ = 53.2 ^ \circ$, because $\measuredangle$CDF is equal to $\measuredangle$EDB, and the three angles involved add up to $180^ \circ$.

We then have:

$\measuredangle$FAB is 2 times that of $\measuredangle$FDB by the incribed angle theorem.
$\measuredangle$FAG is 0.5 times that of $\measuredangle$FAB.

Hence, $\measuredangle$$FAG$ $=$ $\measuredangle$$FDB$ $=53.2 ^ \circ$.

Finally, in the right angled triangle FAG, we obtain the relationship between the radius $r$ and the side a of the square as: $$\sin\space(53.2 \space^ \circ) = \frac{a/2}{r}$$ From which it follows that: $1.25 \times a=2r$. Hence, the perimeter of the circle is $\pi 2r = 3.92 a$, which is less than $4a$, the perimeter of the square.

Answer 19

A little late to the party, but a slight variation:

I drew the same diagram as Steven Stadnicki, but used it differently to find the diameter of the circle:

Define $s = |AB|$, the length of a side of the square.

Draw the diameter from $E$, intersecting the square at $G$ and the circle at $F$. Clearly $|EG|=s$ and $|BG| = s/2$.

Consider the triangles $\triangle EBG$ and $\triangle EBF$. These are both right angle triangles with the common angle at $E$, so are similar. Therefore $\dfrac{|BE|}{|EG|} = \dfrac{|EF|}{|BE|}$, so $ |EF| = \dfrac {|BE|^2}{s}$.

By Pythagorus, $|BE|^2 = s^2 + (\frac{1}{2}s)^2 = \frac{5}{4}s^2$. So $|EF| = \frac{5}{4}s$.

Then the perimeter of the square, $4s = \frac{16}{5}\frac{5}{4}s = 3.2\frac{5}{4}s$, is greater than $\pi\frac{5}{4}s$, the perimeter of the circle.

Answer 20

Another late entry (I just saw this question in the Hot Topics list), but I arrived at this solution before looking at the other answers. It roughly follows Sid's suggestion in the question; I don't think it's beyond the abilities of the high school students that the problem is aimed at.

Let the square's side length be 2, with the square centered on the origin, so the coordinates of the corners are (±1, ±1). B is (1, 1) and E is (-1, 0).

By symmetry, the centre of the circle M is on the X axis; let M be ($x$, 0).

Let $r$ be the radius of the circle.

$r = |ME| = x - -1 = x + 1$

By the Pythagorean distance formula:

$r^2 = |MB|^2 = (x - 1)^2 + (0 - 1)^2 = (x - 1)^2 + 1$

Thus $$\begin{align} (x + 1)^2 & = (x - 1)^2 + 1 \\ x^2 + 2x + 1 & = x^2 - 2x + 1 + 1 \\ 4x & = 1 \\ x & = \frac{1}{4} \\ r = x + 1 & = \frac{5}{4} \\ \end{align}$$

So the circumference $2\pi r = \frac{5\pi}{2}$. Now $\pi < \frac{22}{7}$ so the circumference is less than $\frac{5 \times 22}{2 \times 7} = \frac{55}{7}$, which is less than 8, the perimeter of the square.

No quadratic or obscure geometrical formulae required, nor the ability to magically see 3, 4, 5 triangles. :)

Answer 21

My apologies if this is already among the answers; I looked but didn't see it. (Update: Calum Gilhooley kindly informs me that it does appear among the answers, as a comment by Tebbe which simplifies Steven Stadnicki's answer.)

For convenience, let the sides of the square have length $2$. The horizontal midline of the square extends as a diameter of the circle:

By the intersecting chords theorem, we have

$$2(D-2)=1\cdot1$$

so $D=5/2$, and the inequality $\pi\lt3.2=16/5$ is enough to show

$$\pi D\lt 8$$

where $8$ is the perimeter of the square.

Answer 22

Let the circle be the unit circle. Consider the square of equal perimeter: it has side length $\pi/2$. Align this square so that the midpoint of its left side touches the leftmost point of the unit circle. Where is the upper right corner of the square? It has location $(-1+\pi/2,\pi/4)$, which is at distance $\sqrt{1-\pi+5\pi^2/16}<1$ (since $\sqrt{1-\pi+5\pi^2/16}=\sqrt{1-\pi(1-5\pi/16)}$, and $5\pi/16<5*3.2/16=1 \implies 1-5\pi/16>0 \implies 1-\pi(1-5\pi/16)<1$) from the center of the circle. Hence you must make this square bigger to get it to touch the circle. Thus the square of the original problem is bigger, and so it has bigger perimeter than the circle.

Answer 23

Here's another proof:

• Denote by $a$ the side of the square

• Take the point $F$, like in the figure. Since $\angle CBF = 90^\circ$, it follows that $CF$ is a diamenter of the circle.

• Then $\angle CEF = 90^\circ$ and triangles $\Delta DCE$ and $\Delta AEF$ are similar. This implies that $AF = a/4$ and thus $BF = 3a/4$,

• Now, apply Pythagora's theorem to find $CF = 5a/4$. This gives the radius of the circle, and the answer can be found immediately.

Answer 24

Here's another co-ordinates-free approach:

In the OP's diagram, let the centre of the circle be $O$ then clearly triangle $AOE$ is congruent to triangle $OED$ and triangle $ABO$ is congruent to triangle $CDO.$ Further, $AB, OE$ and $DC$ are all parallel. Together this implies that the area of trapezium $ABOE$ is equal to that of trapezium $CDEO.$ Clearly the area of both the trapeziums plus the isosceles triangle $BOC$ is equal to the area of the square. If the radius of the circle is $r$ so that $EO=OB=OC=r$ we then have:

$\Rightarrow \text{Ar}[ABOE] + \text{Ar}[CDEO] + \text{Ar}[BOC] = \text{Ar}[ABCD]$

$\Rightarrow 2(r+AD)\dfrac{AD}{2} + \dfrac{AD}{2}\left(\sqrt{r^2-\dfrac{AD^2}{4}}\right)= AD^2$

$\Rightarrow r =\dfrac{5AD}{8}$

$\Rightarrow \dfrac{\text{perimeter of square}}{\text{circumference of circle}} =\dfrac{4AD}{\dfrac{5\pi AD}{4}} = \dfrac{16}{5\pi}>1$