Quizzes & Puzzles1 min ago
Three Overlapping Circles
18 Answers
The diagram (see link below) shows 3 identical circles, each with radius 6cm.
The centre of each circle is represented by A, B and C respectively, such that points ABC form a straight line. Calculate the blue shaded area giving your answer in terms of π
https:/
The centre of each circle is represented by A, B and C respectively, such that points ABC form a straight line. Calculate the blue shaded area giving your answer in terms of π
https:/
Answers
1 to 18 of 18
NB this approach is slightly different to the video presented by TCL. Modifying your diagram in an attempt to make the question visually more palatable. See links below. Blue shaded Area = Area of Circle B minus 4 x Sector Area (FIG 2) minus 4 x Segment Area (FIG 1) cm² Note Segment Area = Sector Area minus Equilateral Triangle Area (FIG 3)/(FIG 4) cm² In terms of π...
13:29 Mon 27th Mar 2023
Easier, I think, to work out the area of one of the segments formed by drawing a vertical straight line between the two crossing points of both circles. All four are the same size, and there's a standard formula to find their area -- which, if you don't know it, can also be derived by subtracting the area of a triangle from the area of a sector.
As it asks for the answer in terms of pi, I'll quote my answer numerically, and you can then check against that -- or I can come back to offer more hints/details.
Specifically, the shaded area, I think, comes to 21.8% of the total area of any one circle.
As it asks for the answer in terms of pi, I'll quote my answer numerically, and you can then check against that -- or I can come back to offer more hints/details.
Specifically, the shaded area, I think, comes to 21.8% of the total area of any one circle.
I hate joining in on these questions, it tests my memory for starters. I believe the angle at the centre that marks out the arc say from left circle is 120 degrees. Using the formula for area of a sector I make that 120/360 x 36π which simplifies to 12π.
The right hand one that overlaps the central circle will therefore also be 12π and those 2 together make 24π
As the are of the central circle is 6x6xπ = 36π
So blue bit is 36π - 24π = also 12π
If someone is going to point out some schoolgirl error or ludicrous assumptions by me please do it with some thoughtfulness.
The right hand one that overlaps the central circle will therefore also be 12π and those 2 together make 24π
As the are of the central circle is 6x6xπ = 36π
So blue bit is 36π - 24π = also 12π
If someone is going to point out some schoolgirl error or ludicrous assumptions by me please do it with some thoughtfulness.
This link includes a video explaining how to calculate the answer.
https:/
https:/
NB this approach is slightly different to the video presented by TCL.
Modifying your diagram in an attempt to make the question visually more palatable. See links below.
Blue shaded Area =
Area of Circle B minus 4 x Sector Area (FIG 2) minus 4 x Segment Area (FIG 1) cm²
Note Segment Area =
Sector Area minus Equilateral Triangle Area (FIG 3)/(FIG 4) cm²
In terms of π the Blue Shaded Area = 12π - (24π - 36√3) cm²
FIG 1 https:/
FIG 2 https:/
FIG 3 https:/
FIG 4 https:/
Modifying your diagram in an attempt to make the question visually more palatable. See links below.
Blue shaded Area =
Area of Circle B minus 4 x Sector Area (FIG 2) minus 4 x Segment Area (FIG 1) cm²
Note Segment Area =
Sector Area minus Equilateral Triangle Area (FIG 3)/(FIG 4) cm²
In terms of π the Blue Shaded Area = 12π - (24π - 36√3) cm²
FIG 1 https:/
FIG 2 https:/
FIG 3 https:/
FIG 4 https:/
Consider the lower half of the blue-shaded area. This consists of an equilateral triangle BXY plus a segment outside side XY less 2 equally sized segments 'bitten out' of sides BX and BY. The lower shaded area is therefore
(Triangle area) + (Segment area) - 2 x (Segment area)
= (Triangle area) - (Segment area) .... (1)
The area of one of the segments is given by the area of the sector of the circle centred on B minus the area of the equilateral triangle, so
(Segment area) = (Sector area) - (Triangle area)
Substituting for (Segment area) in equation (1) gives
(Half shaded area) = 2 x (Triangle area) - (Sector area)
doubling this to get the whole shaded area gives
(Shaded area) = 4 x (Triangle area) - 2 x (Sector area) ..... (2)
A bit of Pythagoras shows that the height of the triangle is 3√3. The base of the triangle is 6. So
(Triangle area) = (6/2).3√3 = 9√3
The sector is a sixth of the area of a complete circle of radius 6 so
(Sector area) = (π.6²)/6 = 6π
Substituting for (Triangle area) and (Sector area) in equation (2) gives
(Shaded area) = 4 x 9√3 - 2 x 6π
= 36√3 - 12π
(Triangle area) + (Segment area) - 2 x (Segment area)
= (Triangle area) - (Segment area) .... (1)
The area of one of the segments is given by the area of the sector of the circle centred on B minus the area of the equilateral triangle, so
(Segment area) = (Sector area) - (Triangle area)
Substituting for (Segment area) in equation (1) gives
(Half shaded area) = 2 x (Triangle area) - (Sector area)
doubling this to get the whole shaded area gives
(Shaded area) = 4 x (Triangle area) - 2 x (Sector area) ..... (2)
A bit of Pythagoras shows that the height of the triangle is 3√3. The base of the triangle is 6. So
(Triangle area) = (6/2).3√3 = 9√3
The sector is a sixth of the area of a complete circle of radius 6 so
(Sector area) = (π.6²)/6 = 6π
Substituting for (Triangle area) and (Sector area) in equation (2) gives
(Shaded area) = 4 x 9√3 - 2 x 6π
= 36√3 - 12π
The more I see these different solutions the more I prefer my segment method, because it seems to be the cleanest. Although possibly that's bias, and perhaps the other methods in the end just repeatedly apply the formula for different segments without necessarily saying so.
The area of a segment is generally given (for angles θ in radians) by
A = (1/2) r^2 (θ - sin θ)
where θ is the angle between the two radii bounding the segment. As with all other solutions, you still need to spot an equilateral triangle or two in order to realise that θ = 120 degrees = 2π/3 radians, but once that's done you have just
A = (1/2) r^2 ( 2π/3 - sin (2π/3))
then there are four equal segments, so multiply this by four and subtract from the area of any one circle. The total blue area B is then given by
B = πr^2 - 4A = r^2 (π - 4π/3 + 2sin (2π/3)) = r^2 (π - 4π/3 + 2(√3/2)) = r^2 (√3 - π/3)
for r = 6cm.
Eh, maybe it's much of a muchness in the end.
The area of a segment is generally given (for angles θ in radians) by
A = (1/2) r^2 (θ - sin θ)
where θ is the angle between the two radii bounding the segment. As with all other solutions, you still need to spot an equilateral triangle or two in order to realise that θ = 120 degrees = 2π/3 radians, but once that's done you have just
A = (1/2) r^2 ( 2π/3 - sin (2π/3))
then there are four equal segments, so multiply this by four and subtract from the area of any one circle. The total blue area B is then given by
B = πr^2 - 4A = r^2 (π - 4π/3 + 2sin (2π/3)) = r^2 (π - 4π/3 + 2(√3/2)) = r^2 (√3 - π/3)
for r = 6cm.
Eh, maybe it's much of a muchness in the end.
ABers FYI, my student neighbour, (step) nephew and grandson all have viewing access to Answerbank therefore they can look at ABers replies, but they are not members so they cannot write their own
posts (so I will continue to post on their behalf).
They all acknowledge there is a wealth of knowledge amongst
AB members and very grateful to you all!! Student neighbour also thinks we are all a bunch of moaning old folk LOL As if?? LOL
My grandson (doing GCSEs) wants to award BA to Zebu (the diagrams were the clincher) even though he forgot to expand the brackets by the minus sign, this would have resulted in the
final answer which has been provided by Etch!!
posts (so I will continue to post on their behalf).
They all acknowledge there is a wealth of knowledge amongst
AB members and very grateful to you all!! Student neighbour also thinks we are all a bunch of moaning old folk LOL As if?? LOL
My grandson (doing GCSEs) wants to award BA to Zebu (the diagrams were the clincher) even though he forgot to expand the brackets by the minus sign, this would have resulted in the
final answer which has been provided by Etch!!
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