Question 1197946: There is a circle of diameter 24 cm. It is folded along vertical chord AB so that point D on the circumference coincides with the centre C. The area of the shaded region, in cm^2 is:
a) 192π - 72√3
b) 24π - 18√3
c) 48π - 18√3
d) 192π - 144√3
e) 48π - 36√3
I have tried to draw out some angles and such to find ratios to no avail.
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
It depends on where the shaded region is located. Is it to the left of the mirror line? Or to the right? Unfortunately you haven't specified.
But luckily there are answer choices to pick from which will help narrow down the correct region.
This is probably what the diagram looks like
I'm placing point A as the center, rather than C. But feel free to swap those letters if you prefer.
The colored regions aren't necessarily what is shaded in the diagram of your textbook.
They're shaded that way to break the figure into manageable pieces.
d = 24 = diameter
r = d/2 = 24/2 = 12 = radius
The radius goes from A to B, and also from A to D.
This explains how AB = 12 and AD = 12.
The dashed vertical mirror line will have point D reflect over it to land on point A, which is the center of the circle.
Because of this mirror symmetry, we know that AE = ED
Furthermore, AE = AD/2 = 12/2 = 6
Focus on triangle AEB.
This is a 30-60-90 triangle because of the fact hypotenuse AB = 12 is exactly double that of the short leg AE = 6.
This makes the long leg BE = 6*sqrt(3).
Recall that
longLeg = shortLeg*sqrt(3)
which applies to 30-60-90 triangles exclusively.
Since we know we're dealing with a 30-60-90 triangle, we know that angle EAB is 60 degrees.
This doubles to 2*60 = 120 degrees to represent angle BAC.
This doubling process is valid because triangles EAB and EAC are congruent, which in turn makes angle EAB = angle EAC.
The portion of the circle not shaded red (including the blue region) is 120/360 = 1/3 of the circle.
Therefore, the red shaded region is the remaining 2/3 of the circle.
The area of the full circle of radius r = 12 is:
A = pi*r^2
A = pi*12^2
A = pi*144
A = 144pi
The red shaded area, shaped like pacman, is 2/3 of that full area.
red shaded area = (2/3)*(144pi) = 96pi
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Now let's find the area of triangle EAB shaded in blue.
area = 0.5*base*height
area = 0.5*AE*BE
area = 0.5*6*6*sqrt(3)
area = 18*sqrt(3)
Double this result to get the area of triangle CAB.
2*18*sqrt(3) = 36*sqrt(3)
again this doubling process is valid because triangles EAB and EAC are congruent.
Then combine this with the 96pi found earlier to get the area of everything to the left of the dashed mirror line.
96pi+36*sqrt(3)
This is not listed as one of the answer choices, so we'll see if we can find the area of the region to the right of the mirror line.
We subtract from 144pi, which was the area of the full circle
144pi - (96pi+36*sqrt(3))
144pi - 96pi - 36*sqrt(3)
48pi - 36*sqrt(3)
This is listed as one of the answer choices. So this is likely the correct answer.
Of course this is assuming the shaded region in your textbook is to the right of the mirror line.
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Answer: choice E) 48pi - 36*sqrt(3)
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Nice problem -- except we don't know what the shaded region is.
The diagram below should help YOU find your answer, since YOU know what the shaded region is.
The problem says a given circle with diameter 24 (radius 12) is folded along a vertical chord so that the point D on the circumference of the circle coincides with the center C of the circle. We can picture that by drawing a second circle with radius 12 centered at D.
The picture of the two circles before circle C is folded along vertical chord AB then looks like this:
You should be able to calculate the area of any region of the figure using these....
Each of the equilateral triangles ACD and BCD has an area of (side squared times sqrt(3))/4 = 36*sqrt(3).
Each of the four 30-60-90 triangles formed by the chord AB and the two equilateral triangles has an area of half the area of each equilateral triangle: 18*sqrt(3).
Each of the four segments of a circle CAD, CBD, DCA, and DCB has an area one-sixth of the area of each circle: (144pi)/6 = 24pi.
Each of the four regions between a side of one of the equilateral triangles and one of the circles has an area (circular segment of circle, minus equilateral triangle) of 24pi-36*sqrt(3).
It seems the most likely shaded region in your problem is the area of the folded-over portion of circle C; in that case, using the work shown you should get answer choice e.
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