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circular measure question(difficult)

Here is the question attached
circular measure.JPG33.7KB
Reply 1
Avatar for lil_jack
lil_jack
OP
10
Help anyone:s-smilie::s-smilie:
Construct a sector from the point C, since they're all radii, all lengths will be 2r, subtract the area of the triangle to find the area of the segment then multiply by 2.
circle.png

So you have to work out the area of the sector DCAB, subtract the area of the triangle DCB and multiply it by 2 to get the full area of the overlap.
Reply 4
Avatar for lil_jack
lil_jack
OP
10
Original post by NotNotBatman
Construct a sector from the point C, since they're all radii, all lengths will be 2r, subtract the area of the triangle to find the area of the segment then multiply by 2.


Thanks alot man :biggrin: i really appreciate it, but still having a problem in simplifying the answer as exact as in the question :/
Original post by lil_jack
Thanks alot man :biggrin: i really appreciate it, but still having a problem in simplifying the answer as exact as in the question :/


What have you got so far?
Reply 6
Avatar for JN17
JN17
10
I remember this question and I remember another method, using the above persons diagram, find sector ADC, multiply by four and subtract two of triangle ADC. If you look at symmetry between the top and bottom halves, they are two overlapping sectors for which the area is the area of both sectors minus any overlap (the central triangle they both share).
Reply 7
Avatar for lil_jack
lil_jack
OP
10
Original post by NotNotBatman
What have you got so far?

area of sector= 1/2*4r^2*2/3pie
area of triangle=1/2*2r*2r*sin(120)
and then i minus them from each other,,trying to simplify it but gets it wrong
Original post by lil_jack
area of sector= 1/2*4r^2*2/3pie
area of triangle=1/2*2r*2r*sin(120)
and then i minus them from each other,,trying to simplify it but gets it wrong


You're right so far, simplifying that;

Area of sector = 43πr2 \frac{4}{3} \pi r^2
Area of triangle = 3r2 \sqrt{3} r^2

Remember when you subtract them, that's only half the area you need and make sure you're calculator is in the right mode (degrees or radians).
Reply 9
Avatar for lil_jack
lil_jack
OP
10
Original post by NotNotBatman
You're right so far, simplifying that;

Area of sector = 43πr2 \frac{4}{3} \pi r^2
Area of triangle = 3r2 \sqrt{3} r^2

Remember when you subtract them, that's only half the area you need and make sure you're calculator is in the right mode (degrees or radians).

thanks alot for your time :smile:
Original post by JN17
I remember this question and I remember another method, using the above persons diagram, find sector ADC, multiply by four and subtract two of triangle ADC. If you look at symmetry between the top and bottom halves, they are two overlapping sectors for which the area is the area of both sectors minus any overlap (the central triangle they both share).


Good method; another way which I think is the easiest, is to work out the area of sectors DCB and ACB and subtract twice the triangle.
Original post by lil_jack
thanks alot for your time :smile:


No problem :smile:
How did you get the angle CDB as 2π/3
How did you get the angle CDB as 2π/3
Original post by GrannyMcPoppins
How did you get the angle CDB as 2π/3


Good Q. Split the triangle into two by drawing a horiz line until it meets its vert base. Pick one: angle @/2, adj side = radius/2, hyp = radius.

So cos(@/2) = adj/hyp = 1/2, @/2 = 60°, @ = 120°
(edited 6 years ago)

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