- anonymous

will someone please go to this website i need help with problems 8-12 please http://docs.google.com/viewer?a=v&q=cache:s_Z6mEDtqnAJ:www.somersclassroom.com/files/Chapter%252011/11.5B.pdf+Answers+to+Practice+B+11.5+Geometry&hl=en&gl=us&pid=bl&srcid=ADGEEShpElU6i2Z7y85NrrTvbtv6Xn4WJVYyv7dLeO0itxvLzove7oq8uuVBixJhT8WeHarT3fU7H8VLiD-lB3r80ml3tteJXW5mtG0dWFZFEwSrLDZiHxlXHME_wCAUoHME2QVUFMpY&sig=AHIEtbRN3mHn7LH_nAJkKRNhqf_6RQ8gLw

- jamiebookeater

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- anonymous

not getting through with the link

- anonymous

the link works you just need to copy and paste the whole link, not just what is underlined

- anonymous

can someone help me with 8-12?

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## More answers

- anonymous

could you be more specific with what you are finding difficult about, for instance # 8?

- anonymous

i just dont know how to do them

- anonymous

like if i can just get the equation i can find the answers.. i just need the equation

- anonymous

well in # 8 the tringle is an equilateral triangle, so you know all the angles already. Since you are given the radius of the circle, you can use the equation A=pi*r^2

- anonymous

to find the area of the circle and then subtract the area of the triangle from the area of the circle.

- anonymous

the sides of the triangle are all 6 btw

- anonymous

so the area of the triangle is 15.59

- anonymous

so the circle area would be about 34.65?

- anonymous

one sec

- anonymous

first cut the triangle in half to make two right angle triangles. then you can use pythagorean to find the height of the triangle, which i got to be sqrt(27). finding the area of the triangle with half base times height formula, 1/2*6*sqrt(27) = 9sqrt(3) =15.59. Is that what you did?

- anonymous

yea

- anonymous

for the area of the circle did you use pi*r^2?

- anonymous

yeah and i got 50.24

- anonymous

plugging in 16 for r^2, i got the area to be 50.27

- anonymous

so subtract the area of the triangle from the area of the circle to find the area of the shaded region

- anonymous

34.68

- anonymous

super

- anonymous

would it be squared?

- anonymous

no, why would you?

- anonymous

idk.. someone helped me with number 7 and they squared it

- anonymous

might have been part of finding the area of a circle, you need to square the radius for that

- anonymous

we already did that, though, so it's fine as it is

- anonymous

okay.. will you help me with the other ones

- anonymous

well, I have a bit of hw of my own that i must do before tomorrow, but take a look at number 12... I want to try doing that one with you.

- anonymous

if you like

- anonymous

ok

- anonymous

This is a really cool and beautiful one out of the lot. do you see any way of starting to think about it?

- anonymous

doing A=pi*r^2

- anonymous

that would give you the area of the circle, it's true. but look even more closely: there are only two lines in the triangle that you do not know the length of... right?

- anonymous

yea

- anonymous

oops... I meant to say in the circle

- anonymous

and do you know the equation to find a sector a circle?

- anonymous

arc divided by 360* pi*r^2

- anonymous

A=1/2*r^2*theta

- anonymous

actually since each of these sectors is 1/4 of the circle, the equation is A=1/4*pi*r^2

- anonymous

do you see where I'm going with this?

- anonymous

so a=1/4*3.14*6^2

- anonymous

=28.26

- anonymous

well the area of the triangle is 18 right?

- anonymous

yea

- anonymous

good, and so given that you now have the area of a quarter of a circle, and the area of the triangle, you can subtract the area of the triangle away from the area of the quarter of the circle that you have.

- anonymous

sorry i skipped a step before. have you followed me up to this point or are you a bit lost?

- anonymous

So, here are the steps.
1) we found the area of a quarter of a circle.
2) we found the area of the triangle.
3) we subtract the area of the triangle from the area of the quarter of the circle in order to find the area of one of the shaded regions.

- anonymous

but since both of the triangles are the same, we also now know the area of the other shaded region. does that make sense?

- anonymous

yea

- anonymous

Awesome. I'm glad I could help :) This is my first time on this site, but this is a pretty cool idea for a site.

- anonymous

yeah

- anonymous

well I have to go do HW now, good luck!

- anonymous

thank you for your help

- anonymous

no problem

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