The area of the flower

So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end and create an equilateral triangle but still was unable to find the solution.

Would be thankful if you could help me out.

The question is the following:

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

Here are the possible answers ↓

1. $pi$

3. $3pi/2$

4. $4sqrt2-pi$

5. $pi/2+sqrt3$

6. $2pi-3sqrt3$

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 $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.

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

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 $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.

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 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 $6p = 2pi - 3sqrt 3$.