Question

Please Help!
A circle is circumscribed about a hexagon. Determine the area of the hexagon if the area outside the hexagon but inside the circle is 15 sq. cm.

Answer 1

Let radius of circle = r ; Area of circle = (pi)r^2
You can find the area of the triangle OAB in terms of r.
Area of OAB = 1/2 * AB * OC = BC * OC = rcos(30) * rsin(30) = r^2*sqrt(3)/4
Area of hexagon = 6 * area of OAB = 6 * r^2*sqrt(3)/4 = 3/2 * sqrt(3) * r^2

Difference between the areas of hexagon and circle =
(pi)r^2 - 3/2 * sqrt(3) * r^2 = 15
solve for r. put it in the area of hexagon and find its area.

Answer 2

??????????????????????????????????

Answer 3

um, can you use the equation button below sir? I can't really understand the flow of the solution.

Answer 4

@ranga

Answer 5

If the math part is not clear you can still follow the description part and find the areas.

Answer 6

the ^ symbol means the next character is an exponent.

Answer 7

[Area of OAB = 1/2 * AB * OC = BC * OC ] this part sir, how did [1/2 * AB * OC] became [BC * OC]?

Answer 8

* means multiplication.
1/2 * AB = BC (BC is half of AB)

Answer 9

so the scope of 1/2 is only the AB excluding the OC? another question sir if I may, what about [rcos(30) * rsin(30) = r^2*sqrt(3)/4]?

Answer 10

it's done by multiplying the coordinates of sin and cos of 30?

Answer 11

how did you end up in rsin(30) and rcos(30)?

Answer 12

For example, if you have 1/2 * 6 * 8 you can associate the 1/2 with 6 or 8 and you will get the same answer. 1/2 * 6 * 8 = 3 * 8 = 24 or 1/2 * 6 * 8 = 6 * 4 = 24

Answer 13

is it "SOH CAH TOA"? sir, i really apologize if I'm asking too much.

Answer 14

ok sir I get the 1/2 part now.

Answer 15

Look up trig tables. sin(30) = 1/2 ; cos(30) = sqrt(3) / 2

Answer 16

sine = opposite / hypotenuse ; cosine = adjacent / hypotenuse

Answer 17

yes sir I know that trigo part but the transition in [BC * OC = rcos(30) * rsin(30)] is what I'm confused at.

Answer 18

sin(30) = BC / OB = BC / r
Therefore, BC = r * sin(30)

Answer 19

Angle BOC = 30 degrees (half of 60 degrees)

Answer 20

aaaah. I understand it now sir :D

Answer 21

alright.

Answer 22

now all I need is to solve for r and get the area of the hexagon.

Answer 23

yes. But I will be logging off now but hopefully someone will be here to help if needed.

Answer 24

Thank you sir. @ranga

Answer 25

you are welcome.

Answer 26

I can't solve for r. too complex.

Answer 27

The area of region between the circle and the hexagon is given by ..Area of the circle(pir^2)-(area of the hexagon(1/2 r^2sin60x6 (because hexagon forms six identical triangles)

there, 15=pi(3.142) x r^2 -(0.5 x r^2sin60 x6)

15=3.142r^2 -2.598r^2
15=0.544r^2
15/0.544 =r^2
27.58=r^2
square root both sides to get radius

Please verify your email address. We've sent you an email with a verification link inside.
OpenStudy

Home
Invite
omarani1336
Settings
Log out

SmartScore

← 96 members online

0 replying
2 viewing

omarani1336

I am a little bit confused of how to obtain ratios from the unit cycle plz help

a few moments ago

Type your reply

You have an open question:

I am a little bit confused of how to obtain ratios from the unit cycle plz help

Asked a few moments ago
No Medals Yet

lowcard2: help?!!!!!
3 viewing
No Medals Yet
ECEstudent9405: Please Help! A circle is circumscribed about a hexagon. Determine the area of the hexagon if the area outside the hexagon but inside the ci…
1 replying right now...
1 viewing
No Medals Yet
Ratedover: Hey! I know this is math but I've been trying English but I have't had any luck with it. Could anyone give me some feedback on my essay? It …
2 viewing
No Medals Yet
stupidinmath: Two regular pentagons are constructed one on the hypotenuse and one on a leg of an isosceles right triangle. What is the ratio of the areas …
1 viewing
No Medals Yet
Geometry
omarani1336: I am a little bit confused of how to obtain ratios from the unit cycle plz help
2 viewing
No Medals Yet
Abbles: Algebra homework help?
updated a few moments ago
No Medals Yet
brendafn: change to exponential form y=log base(6) X
2 viewing
2 Medals
Yttrium: Stats question: I just wanna verify if I got the correct answer. "Three coins, consisting two 10$, three 5$, and five 1$, are to be selecte…
1 viewing
No Medals Yet
stupidinmath: Two regular pentagons are constructed one on the hypotenuse and one on a leg of an isosceles right triangle. What is the ratio of the areas …
1 viewing
1 Medal
NotTim: Problem 6. 6. (2 pts) Construct a 98% confidence interval for the population slope b. b = 34, Se = 7; SSx = 45;n = 11 …
1 viewing
No Medals Yet
No more questions. Working hard or hardly working?

Privacy Policy
Terms and Conditions
Code of Conduct

Mathematics
Search Subjects

95 Online

58 Online

53 Online

47 Online

46 Online

46 Online

41 Online

39 Online

39 Online

27 Online
Loading more…

Create a New Subject
Enter Subject Name

Answer 28

then when you get the radius you can simply find the area of the hexagon by using the formula 1/2 r^2sin60 x6

0.5x(5.3)^2sin60 x 6=72.98 sq cm

Answer 29

hey are you getting my response/