The area of the ✽ in a ⬡

This geometry problem comes from a recent math test.

The question is the following:

We have a regular hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the hexagon in it creating a flower-shaped-like object. Find the area of the flower.
a flower in a hexagon

I tried creating an equilateral triangle by connecting the point in the center to the vertices of the boundary but was unable to proceed.

Would be thankful if you could help me out.

  • 2
    $\begingroup$ This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs. $\endgroup$ Mar 3, 2019 at 11:01
  • $\begingroup$ Note that you are editing the post after it is closed. That willl automatically send your question to the reopen queue (and some others would review that and cast their reopen vote when they see fit). If you want your question to have a better chance to be reopened, try to make a more substantial edit, in particular respond to the accusation that this post is missing context (see the yellow box) by (e.g.) describe what you have tried and where you are stuck. $\endgroup$ Apr 11, 2019 at 20:52
  • $\begingroup$ @DAVO 12 times difference between sector and equilateral triangle is $12( \pi 1^2/6- \sqrt3/2. 1^2) =2 \pi -3 \sqrt3 $ $\endgroup$
    – Narasimham
    Apr 11, 2019 at 20:55

5 Answers 5


Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.

The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $\frac{\pi}{3}$ (radian measure) of a circle of radius $1$.

The area of one such segment is $\frac 12 r^2(\theta - \sin\theta) = \frac 12(\frac{\pi}{3} - \frac{\sqrt{3}}{2})$.

There are $12$ such segments, yielding the total area of the "flower" as $2\pi - 3\sqrt 3$.

  • $\begingroup$ Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs. $\endgroup$
    – Trebor
    Mar 3, 2019 at 3:08
  • 1
    $\begingroup$ @Deepak Welcome! Anytime!!!! You deserve more than that. $\endgroup$ Mar 3, 2019 at 7:20
  • 1
    $\begingroup$ @Deepak Thank you a lot for the answer. :) And yea I just realized those aren't ellipses :D $\endgroup$
    – DAVO
    Mar 3, 2019 at 14:32
  • 1
    $\begingroup$ @Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you. $\endgroup$ Mar 3, 2019 at 14:52
  • 1
    $\begingroup$ @DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393] $\endgroup$ Mar 3, 2019 at 14:59

Here's a solution that doesn't require trigonometry. The hexagon is made up of 6 "petals" and 6 "wedges" (the scooped out triangles between the petals). Call the area of a petal $p$, and the area of a wedge $w$.

What is the area enclosed by two adjacent sides of the hexagon and the arc drawn from the vertex between them? (For example, the region EFAGE). It's one third of a unit circle, and contains three petals and two wedges:

$3p + 2w = {\frac 1 3} \pi r^2 = \frac \pi 3$

What is the area of one triangle (one sixth of the hexagon)? The altitude of the triangle is $\sqrt{1 - {\frac 1 4}} = \frac{\sqrt 3}{2}$, so the area is $\frac{\sqrt 3}{4}$. The triangle is made up of two half-petals and one wedge:

$p + w = \frac{\sqrt 3}{4}$

Solving the two simultaneous equations:

$2p + 2w = \frac{\sqrt 3}{2}$

$p = \frac{\pi}{3} - \frac{\sqrt 3}{2}$

And the area of the flower is $6p = 2\pi - 3\sqrt 3$.


enter image description here

Shaded blue area is

(Area of pie)-(Area of rhombus)=$\pi\cdot 1^2\cdot\frac{120}{360}-1^2\cdot\sin(120)$=$ \frac{\pi}{3}-\frac{\sqrt{3}}{2}$

And we have 6 of them, and so total area of flower is

$ 6(\frac{\pi}{3}-\frac{\sqrt{3}}{2})=2\pi-3\sqrt{3}$


[This answer was posted for the original version of the problem, which had essentially no information about the figure other than the picture itself]

We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $\approx 3.1$. (3) $\approx 4.7$. (4) $\approx 2.5$. (5) $\approx 3.3$. (6) $\approx 1.1$.

If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.

Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $\frac{\sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $\frac32\sqrt{3}\approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2\pi-3\sqrt{3}$.

And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.


Here's another way of looking at it. Imagine taking two circles of radius 1 and cutting them into thirds. Arrange them so that the six 120° angles you have (one from each wedge) form the corners of the hexagon.

Every point in one of the "petals" is covered by three of these wedges. Every point in the hexagon that's not in one of the petals is covered by two of these wedges. The total area covered by the six wedges is therefore twice the area of the total hexagon plus the area of the petals; or, in other words, the area of the petals is the area of two unit circles minus twice the area of the hexagon.

The area of the two circles is $2\pi$; the area of the hexagon is $3\sqrt{3}/2$. The area of the petals is therefore $2 \pi - 3 \sqrt{3} \approx 1.087...$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .