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The diagram shows a British 50 pence coin.

enter image description here

The seven arcs $AB$, $BC$, . . . , $FG$, $GA$ are of equal length and each arc is formed from the circle of radius a having its centre at the vertex diametrically opposite the mid-point of the arc. Show that the area of the face of the coin is

$$\frac{a^2}{2}(\pi-7\tan\frac{\pi}{14})$$

How can i prove it?

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closed as off-topic by heropup, Martin R, Matthew Conroy, 3SAT, Dan Feb 19 '16 at 19:21

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  • 2
    $\begingroup$ What have you tried so far? A small hint: Start with the $7$ sectors and work out the area you counted multiple times. $\endgroup$ – AlexR Feb 19 '16 at 18:19
  • $\begingroup$ This is a curve of constant width. $\endgroup$ – Lucian Feb 19 '16 at 18:41
  • $\begingroup$ Wonder if area for $n$ sided loonie is $ a^2/2 ( \pi - n \tan \pi/( 2 n) $ $\endgroup$ – Narasimham Feb 19 '16 at 19:06
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Taking O to be the centre, let $AO=OE=r$ and we have $\angle AOB=\frac {2\pi}{7}, \angle AOE=\frac {6\pi}{7}, \angle AEO=\frac{\pi}{14}$

Using the Sine Rule in $\triangle AOE$, we have$$\frac{a}{\sin\left(\frac{6\pi}{7}\right)}=\frac{r}{\sin\left(\frac{\pi}{14}\right)}$$ $$\Rightarrow r=\frac{a}{2\cos\left(\frac{\pi}{14}\right)}$$

The area of $\triangle AOE$ is $$\frac 12 r^2\sin\left(\frac{6\pi}{7}\right)$$

This simplifies to $$\frac{a^2}{4}\tan\left(\frac{\pi}{14}\right)$$

Hence the required area is$$7\times\left[\frac{a^2}{2}\times\frac {\pi}{7}-2\times\frac{a^2}{4}\tan\left(\frac{\pi}{14}\right)\right]$$

Hence the answer.

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Hint: the shape is the union of a regular heptagon and seven circular segments.

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Ok, i solved so:

Fixing the point $O$(centre) is equidistant from each of three vertices $A$, $B$ and $E$. I started looking for the area of the sector $AOB$ by calculating the area of AEB and subtracting the areas of the two triangles $OBE$ and $OAE$. The required area is 7 times this.

So i need angle $\angle AEB$. And we know that $\angle AOB = \frac{2\pi}{7}$ and hence $\angle BOE = \frac{2\pi-\frac{2\pi}{7}}{2}= \frac{6\pi}{7}$ So, $$\angle AEB=2\angle OEB = 2\times\frac{\pi-\frac{6\pi}{7}}{2} = \frac{\pi}{7}$$ Now $ABE$ is a sector of a circle of radius $a$, so its area is $$\pi a^2 \times \frac{\pi/7}{2\pi}=\frac{\pi a^2}{14}$$ The area of $\angle OBE$ is $\frac{1}{2}BE \times$height, i.e. $$\frac{1}{2}a \times \frac{a}{2}\tan \angle OBE=\frac{a^2}{4}tan\frac{\pi}{14}$$ The area of the coin is therefore, $$\color{red}{7 \times(\frac{\pi a^2}{14}-2 \times \frac{a^2}{4}\tan \frac{\pi}{14})}$$ Thanks for your help

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