I need to calculate the area of a polygon that looks like this:
The four lines connecting the centre of the polygon with the points of the star are 8cm long and the lines connecting the centre with the nearest point of the perimiter (which is in the big angle)are 4cm long.
I got so far:
The polygon is made out of 8 triangles.
The sides of this triangle are:
$$a=4$$
$$c=8$$
I calculated b using trigonometry:
$$b^2=a^2+c^2-2ac\cdot\cos\beta$$
Beta here is the triangle's angle that is in the center of the polygon,so it is 1/8 of 360°.
$$\beta=360°/8=45°$$
So:
$$b=\sqrt{8^2+4^2-2 \cdot 8 \cdot 4 \cdot \cos(45°)}$$
$$b=\sqrt{80-64\sqrt{1/2}}$$
$$b=\sqrt{80-32\sqrt{2}}$$
Now I used Heron's formula:
$$S=\sqrt{s \cdot (s-a) \cdot (s-b)\cdot(s-c)}$$
$$s=(a+b+c)/2$$
$$s=(4+\sqrt{80-32\sqrt{2}}+8)/2$$
$$s=6+(\sqrt{80-32\sqrt{2}})/2$$
$$S=\sqrt{(6+(\sqrt{80-32\sqrt{2}})/2) \cdot (6+(\sqrt{80-32\sqrt{2}})÷2-4) \cdot (6+(\sqrt{80-32\sqrt{2}})/2-\sqrt{80-32\sqrt{2}}) \cdot(6+(\sqrt{80-32\sqrt{2}})/2-8)}$$
After simplifying I got:
$$S=8\sqrt{2}$$
So the area of the star is 8 times bigger:
$$64\sqrt{2}$$
This seemed rather complicated,so I tried to find other ways of calculating the area.
1.
The area of the triangle is
$$S=\frac{ac\cdot \sin \beta}{2}$$
Beta is 45°
$$S=\frac{4 \cdot 8 \cdot \sin 45°}{2}$$
$$S=8\sqrt{2}$$
Then multiply by 8.
2.
Maybe it would be possible to imagine fitting this star into a larger octogan,then calculating its area using
$$S=\frac{na^2\cot\frac{\pi}{n}}{4}$$
and then cutting away a certain percent,but I didn't get far with this method.
I'd like to know if there are shorter ways of getting this area,especially if they don't include the use of trigonometry.
Best Answer
You can split one of the eight triangles by dropping a line perpendicular to the side of length 8 from the opposite vertex.
At the centre you have a 45-45-90 triangle with hypotenuse $4$, which makes its other two sides have length $\frac{4}{\sqrt{2}}=2\sqrt{2}$.
You can now look at the original triangle as having base $8$ and height $2\sqrt{2}$, which givea an area of $\frac{8\cdot2\sqrt{2}}2=8\sqrt{2}$.
The area of the star is eight times that, so $8\cdot 8\sqrt{2}=64\sqrt{2}$.
This is essentially the same as your method 1, except that I used the fact that there is an isosceles right-angled triangle (angles 45-45-90, side lengths in ratio $1$:$1$:$\sqrt{2}$) instead of explicitly using an angle.