1. ## geometry help

I need help solving these math problems, I have no idea how to do them:

the circle one I need the are aof the shaded figure
the p-gram one, i need the area
the pentagon, I need the apothem, area, and side length

2. The area of the shaded regions in the circle.

The sides of the triangle with hypoteneuse 6 inches are the same, They are the radius of the circle:

$\displaystyle 6=\sqrt{a^{2}+a^{2}}$

$\displaystyle a=3\sqrt{2}$

The subtended angle is 90 degrees.

Area of circular segment:

$\displaystyle \frac{1}{2}(3\sqrt{2})^{2}(\frac{\pi}{2}-sin(\frac{\pi}{2}))$

That's one of them. Multiply by 2.

For the rhombus, find the area of one of the right triangles and multiply by 4.

3. Originally Posted by Zyger
I need help solving these math problems, I have no idea how to do them:

the circle one I need the are aof the shaded figure
the p-gram one, i need the area
the pentagon, I need the apothem, area, and side length
For the circle problem, you need to compute the radius first, which is $\displaystyle 3\sqrt{2}$. Then, the area of the semicircle is $\displaystyle \frac{1}{2}\pi r^2$. We subtract out the area of the triangle to obtain the area of the shaded region: $\displaystyle \frac{1}{2}\pi (3\sqrt{2})^2 - (3\sqrt{2})^2 = 9\pi - 18$.

4. may please have some more help on the others?

5. Originally Posted by Zyger
I need help solving these math problems, I have no idea how to do them:

the circle one I need the are aof the shaded figure
the p-gram one, i need the area
the pentagon, I need the apothem, area, and side length
A useful formula to know is that

For a regular n-gon of sidelength b

$\displaystyle A_{n-gon}=\frac{1}{4}nb^2\cot\bigg(\frac{\pi}{n}\bigg)$

6. Originally Posted by Zyger
I need help solving these math problems, I have no idea how to do them:

the circle one I need the are aof the shaded figure
the p-gram one, i need the area
the pentagon, I need the apothem, area, and side length

All I need now is the one with the pentagon

7. Originally Posted by Zyger
All I need now is the one with the pentagon
A general desire for the underyling is never hurtful

usign my formula $\displaystyle \frac{5}{4}\cdot{b^2}\cdot\cot\bigg(\frac{\pi}{5}\ bigg)=\frac{10(\sqrt{5}+1)\sqrt{2}x^2}{\sqrt{5-\sqrt{5}}}\approx{27.52b^2}$

So now apply it

and use the fact that

$\displaystyle A=\frac{1}{2}p\cdot{a}$

to find the apothem

8. Originally Posted by Mathstud28
A general desire for the underyling is never hurtful

usign my formula $\displaystyle \frac{5}{4}\cdot{b^2}\cdot\cot\bigg(\frac{\pi}{5}\ bigg)=\frac{10(\sqrt{5}+1)\sqrt{2}x^2}{\sqrt{5-\sqrt{5}}}\approx{27.52b^2}$

So now apply it

and use the fact that

$\displaystyle A=\frac{1}{2}p\cdot{a}$

to find the apothem
I am so sorry, but that math looks like a foreign language to me. Can you go a little slower with it please?

9. Originally Posted by Zyger
I am so sorry, but that math looks like a foreign language to me. Can you go a little slower with it please?
Of course this is a site to learn

Ok so basically we have that $\displaystyle A=\frac{1}{2}a\cdot{p}$

where a is apothem and p is permieter

Now since we know that the exterior angle of a n-gon is $\displaystyle \frac{360}{n}$ we know due to the linear pair postulate that the interior angle of an n-gon is

$\displaystyle 180-\frac{360}{n}$

so for a pentagon(5-gon)

the interior angle would be

$\displaystyle 180-\frac{360}{5}=108$

So now we know that the angle at a vertex is 108 we know that the apothem bisects it making

the angle that the apothem cuts off 44...so now we have a right triangle with angles ,90,44,46

So we need to calculate the apothem or in this case we are given it but we need to find sidelenght

so we use trig $\displaystyle \sin(46)=\frac{x}{15}\Rightarrow{x=\sin(46)\cdot{1 5}\approx{10.875}}$

and since that gives us half of our side we see that the sidelenghts are

21.75

now we go to our formula

$\displaystyle A=\frac{1}{2}a\cdot{p}$

Now since $\displaystyle P_{pentagon}=5n$ where n is the sidelength we see that $\displaystyle A=5\cdot{21.75}=107.8$

so now we see

$\displaystyle A=\frac{1}{2}\cdot{107.8}\cdot{15}=808.75$

10. SORRY POR MY POST, i HAVE TO EDIT NOW

11. Originally Posted by Mathstud28
Of course this is a site to learn

Ok so basically we have that $\displaystyle A=\frac{1}{2}a\cdot{p}$

where a is apothem and p is permieter

Now since we know that the exterior angle of a n-gon is $\displaystyle \frac{360}{n}$ we know due to the linear pair postulate that the interior angle of an n-gon is

$\displaystyle 180-\frac{360}{n}$

so for a pentagon(5-gon)

the interior angle would be

$\displaystyle 180-\frac{360}{5}=108$

So now we know that the angle at a vertex is 108 we know that the apothem bisects it making

the angle that the apothem cuts off 44...so now we have a right triangle with angles ,90,44,46

So we need to calculate the apothem or in this case we are given it but we need to find sidelenght

so we use trig $\displaystyle \sin(46)=\frac{x}{15}\Rightarrow{x=\sin(46)\cdot{1 5}\approx{10.875}}$

and since that gives us half of our side we see that the sidelenghts are

21.75

now we go to our formula

$\displaystyle A=\frac{1}{2}a\cdot{p}$

Now since $\displaystyle P_{pentagon}=5n$ where n is the sidelength we see that $\displaystyle A=5\cdot{21.75}=107.8$

so now we see

$\displaystyle A=\frac{1}{2}\cdot{107.8}\cdot{15}=808.75$

Thank you sooooooooooooooooooooooooooooooooo much