Question

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

Answer 1

not getting through with the link

Answer 2

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

Answer 3

can someone help me with 8-12?

Answer 4

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

Answer 5

i just dont know how to do them

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

the sides of the triangle are all 6 btw

Answer 10

so the area of the triangle is 15.59

Answer 11

so the circle area would be about 34.65?

Answer 12

one sec

Answer 13

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?

Answer 14

yea

Answer 15

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

Answer 16

yeah and i got 50.24

Answer 17

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

Answer 18

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

Answer 19

34.68

Answer 20

super

Answer 21

would it be squared?

Answer 22

no, why would you?

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

okay.. will you help me with the other ones

Answer 27

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.

Answer 28

if you like

Answer 29

ok

Answer 30

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

Answer 31

doing A=pi*r^2

Answer 32

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?

Answer 33

yea

Answer 34

oops... I meant to say in the circle

Answer 35

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

Answer 36

arc divided by 360* pi*r^2

Answer 37

A=1/2*r^2*theta

Answer 38

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

Answer 39

do you see where I'm going with this?

Answer 40

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

Answer 41

=28.26

Answer 42

well the area of the triangle is 18 right?

Answer 43

yea

Answer 44

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.

Answer 45

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

Answer 46

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.

Answer 47

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

Answer 48

yea

Answer 49

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.

Answer 50

yeah

Answer 51

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

Answer 52

thank you for your help

Answer 53

no problem