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DavisAshley
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will someone please go to this website i need help with problems 812 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=ADGEEShpElU6i2Z7y85NrrTvbtv6Xn4WJVYyv7dLeO0itxvLzove7oq8uuVBixJhT8WeHarT3fU7H8VLiDlB3r80ml3tteJXW5mtG0dWFZFEwSrLDZiHxlXHME_wCAUoHME2QVUFMpY&sig=AHIEtbRN3mHn7LH_nAJkKRNhqf_6RQ8gLw
 3 years ago
 3 years ago
DavisAshley Group Title
will someone please go to this website i need help with problems 812 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=ADGEEShpElU6i2Z7y85NrrTvbtv6Xn4WJVYyv7dLeO0itxvLzove7oq8uuVBixJhT8WeHarT3fU7H8VLiDlB3r80ml3tteJXW5mtG0dWFZFEwSrLDZiHxlXHME_wCAUoHME2QVUFMpY&sig=AHIEtbRN3mHn7LH_nAJkKRNhqf_6RQ8gLw
 3 years ago
 3 years ago

This Question is Closed

beens Group TitleBest ResponseYou've already chosen the best response.0
not getting through with the link
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
the link works you just need to copy and paste the whole link, not just what is underlined
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
can someone help me with 812?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
could you be more specific with what you are finding difficult about, for instance # 8?
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
i just dont know how to do them
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
like if i can just get the equation i can find the answers.. i just need the equation
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
to find the area of the circle and then subtract the area of the triangle from the area of the circle.
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
the sides of the triangle are all 6 btw
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
so the area of the triangle is 15.59
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
so the circle area would be about 34.65?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
one sec
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
for the area of the circle did you use pi*r^2?
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
yeah and i got 50.24
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
plugging in 16 for r^2, i got the area to be 50.27
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
so subtract the area of the triangle from the area of the circle to find the area of the shaded region
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
super
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
would it be squared?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
no, why would you?
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
idk.. someone helped me with number 7 and they squared it
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
might have been part of finding the area of a circle, you need to square the radius for that
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
we already did that, though, so it's fine as it is
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
okay.. will you help me with the other ones
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
if you like
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
This is a really cool and beautiful one out of the lot. do you see any way of starting to think about it?
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
doing A=pi*r^2
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
oops... I meant to say in the circle
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
and do you know the equation to find a sector a circle?
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
arc divided by 360* pi*r^2
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
A=1/2*r^2*theta
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
actually since each of these sectors is 1/4 of the circle, the equation is A=1/4*pi*r^2
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
do you see where I'm going with this?
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
so a=1/4*3.14*6^2
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
well the area of the triangle is 18 right?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
sorry i skipped a step before. have you followed me up to this point or are you a bit lost?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
but since both of the triangles are the same, we also now know the area of the other shaded region. does that make sense?
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
well I have to go do HW now, good luck!
 3 years ago

DavisAshley Group TitleBest ResponseYou've already chosen the best response.0
thank you for your help
 3 years ago

generallyclay Group TitleBest ResponseYou've already chosen the best response.0
no problem
 3 years ago
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