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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

Mathematics
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not getting through with the link
the link works you just need to copy and paste the whole link, not just what is underlined
can someone help me with 8-12?

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Other answers:

could you be more specific with what you are finding difficult about, for instance # 8?
i just dont know how to do them
like if i can just get the equation i can find the answers.. i just need the equation
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
to find the area of the circle and then subtract the area of the triangle from the area of the circle.
the sides of the triangle are all 6 btw
so the area of the triangle is 15.59
so the circle area would be about 34.65?
one sec
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?
yea
for the area of the circle did you use pi*r^2?
yeah and i got 50.24
plugging in 16 for r^2, i got the area to be 50.27
so subtract the area of the triangle from the area of the circle to find the area of the shaded region
34.68
super
would it be squared?
no, why would you?
idk.. someone helped me with number 7 and they squared it
might have been part of finding the area of a circle, you need to square the radius for that
we already did that, though, so it's fine as it is
okay.. will you help me with the other ones
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.
if you like
ok
This is a really cool and beautiful one out of the lot. do you see any way of starting to think about it?
doing A=pi*r^2
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?
yea
oops... I meant to say in the circle
and do you know the equation to find a sector a circle?
arc divided by 360* pi*r^2
A=1/2*r^2*theta
actually since each of these sectors is 1/4 of the circle, the equation is A=1/4*pi*r^2
do you see where I'm going with this?
so a=1/4*3.14*6^2
=28.26
well the area of the triangle is 18 right?
yea
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.
sorry i skipped a step before. have you followed me up to this point or are you a bit lost?
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.
but since both of the triangles are the same, we also now know the area of the other shaded region. does that make sense?
yea
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.
yeah
well I have to go do HW now, good luck!
thank you for your help
no problem

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