anonymous
  • anonymous
A hexagon is inscribed in a circle. If the difference between the area of the circle and the area of the hexagon is 36 m2, use the formula for the area of a sector to approximate the radius r of the circle. (Round your answer to three decimal places.)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
tkhunny
  • tkhunny
Well, do it! Area of a circle? Area of inscribed hexagon? Let's see what you get.
Jack1
  • Jack1
area of polygon (from known side length) = (s^2 N ) / (4 tan (180/N) n = number of sides s = side length area of polygon (from known radius length) = ( r^2 * N * sin [360/N] ) / (2) n = number of sides r = radius of circle aka length from center of hexagon to any corner of hexagon area of circle = pi * r^2
Jack1
  • Jack1
as per attached drawing, how semicircle many segments are there that aren't part of the hexagon?
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Jack1
  • Jack1
@jlangley ?
Jack1
  • Jack1
area of a sector of a circle = [(theta)/360] * pi * r^2 |dw:1378010593857:dw|
Jack1
  • Jack1
area of the triangle part of the sector = 1/2 x base x vert ht = 1/2 * [ sin {1/2 * theta} * radius ] * [ cos {1/2 * theta} * radius ] as [ sin {1/2 * theta} * radius ] = base and [ cos {1/2 * theta} * radius ] = height
Jack1
  • Jack1
...done, now use any of the above equations to solve... even though they ask for sector theory, you can check ur answer by using the first 3 eqns
goformit100
  • goformit100
"Welcome to OpenStudy. I can answer your questions or guide you. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."
anonymous
  • anonymous
That would be great if I knew how it worked . . . A buddy is in the same class and we both had trig in high school, so we are wondering how everyone else is doing since we're having so much trouble. I managed to finally figure out how to solve using sectors only it requires knowing the relationship between the sides of a 30-60-90 triangle, which is a couple chapters down the road. What I've gotten with sector theory is this: \[A=\frac{ 1 }{ 2 } r ^{2} \theta\] \[A=\frac{ 1 }{ 2 }rh\] \[\frac{ 1 }{ 2 } r ^{2} \theta-\frac{ 1 }{ 2 }rh=36\] \[\theta=\frac {\Pi}{3}\] \[\frac{ 1 }{ 2 } r ^{2} (\frac {\Pi}{3})-\frac{ 1 }{ 2 }rh=36\] \[h=\frac{ \sqrt{3} }{ 2 }r\] \[\frac{ 1 }{ 2 } r ^{2} (\frac {\Pi}{3})-\frac{ 1 }{ 2 }r \frac{ \sqrt{3} }{ 2 }r=36\] \[\frac{ 1 }{ 2 } r ^{2} (\frac {\Pi}{3})-\frac{ 1 }{ 2 }( \frac{ \sqrt{3} }{ 2 })r ^{2}=36\] \[\frac{ 1 }{ 2 }r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})]=36\] \[6(\frac{ 1 }{ 2 }r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])=36\] \[3r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])=36\] \[r^{2}[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])=12\] \[r=\sqrt {12\div[(\frac{ \Pi }{ 3 })-(\frac {\sqrt {3}} {2})])}\] \[r \approx 8.139 \] While this works, is there a way to do it with trigonometric functions, but still apply sector theory?
Jack1
  • Jack1
|dw:1378195273896:dw|
Jack1
  • Jack1
|dw:1378195339815:dw|
Jack1
  • Jack1
|dw:1378195433942:dw|
Jack1
  • Jack1
Area of segment = area of circle/6 difference = area of segment - area of triangle therefore: sum of all differences = 36m^2 so 1 difference = 36/6 = 6 m^2 therefore: as difference = area of segment - area of triangle area of a segment of a circle = [(theta)/360] * pi * r^2 area of the triangle = 1/2 x base x vert ht = 1/2 * [ sin {1/2 * theta} * radius ] * [ cos {1/2 * theta} * radius ] so final eqn is 6 = {[(60/360)] * pi * r^2} - {1/2 * [ sin {1/2 * 60} * r ] * [ cos {1/2 * 60} * r ]} so r

Looking for something else?

Not the answer you are looking for? Search for more explanations.