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In the figure below, four semicircles are drawn, each centered at the
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02 Nov 2019, 03:45
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85% (hard)
Question Stats:
53% (02:24) correct 47% (02:08) wrong based on 64 sessions
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In the figure below, four semicircles are drawn, each centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
A. \(8\pi\) B. \(32 - 8\pi\) C. \(16 - 8\pi\) D. \(8\pi - 32\) E. \(8\pi - 16\)
Re: In the figure below, four semicircles are drawn, each centered at the
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02 Nov 2019, 03:55
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Area of the shaded region = 8*[(area of quadrant)- (area of right angle triangle)] = \(8*[{\frac{(pi*4)}{4}}- {\frac{1}{2}*2*2}]\) = 8*(pi-2)
Bunuel wrote:
In the figure below, four semicircles are drawn, each centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
A. \(8\pi\) B. \(32 - 8\pi\) C. \(16 - 8\pi\) D. \(8\pi - 32\) E. \(8\pi - 16\)
Re: In the figure below, four semicircles are drawn, each centered at the
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02 Nov 2019, 03:58
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total area of the square = 16 and total area of the shaded region should be slightly less than 16 see answer options A. \(8\pi\) 25 ; not possible B. \(32 - 8\pi\) ; 7 ; too less C. \(16 - 8\pi\) ; -ve value D. \(8\pi - 32\) ; -ve value E. \(8\pi - 16\) ; 9 slightly less than 16 IMO E;
Bunuel wrote:
In the figure below, four semicircles are drawn, each centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
A. \(8\pi\) B. \(32 - 8\pi\) C. \(16 - 8\pi\) D. \(8\pi - 32\) E. \(8\pi - 16\)
Re: In the figure below, four semicircles are drawn, each centered at the
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02 Nov 2019, 04:38
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Finding the shaded area is equal to the area of the square less the area not shaded. There are 4 "not shaded" regions.
Since we are given that AB = 4: Area Square = \(4*4=16\).
We can find the area of 2 "not shaded" regions by calculating the area of the square less two semi-circles (one circle): Area Circle = \(\pi r^2 = pi (2)^2 = 4\pi\)
Therefore, the area of 2 "not-shaded" regions is: \(16-4\pi\),
and the area of 4 "not-shaded" regions is: \(2* (16-4\pi) = 32-8\pi\).
Since we know the area of the "pedals" is the area of the square less 4 "not shaded" regions: \(16- (32-8\pi) = -16+8\pi = 8pi -16.\)
In the figure below, four semicircles are drawn, each centered at the
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Updated on: 02 Nov 2019, 11:20
Step Approach:
- Divide the Square in 4 equal sections(Squares) comprising of one shaded region(that will be a shaded region in square of side 2) - area of triangle \(\frac{1}{2}∗2∗2=2\) - area of circular region (it is exactly 1/4th of full circle) = \(\frac{1}{4}∗π∗r^2\) => radius is 2 => \(\frac{1}{4}∗π∗2^2 = π\) - to get the area of shaded region we'll minus the area of triangle from entire circular region and multiple by 2 to cover full shaded region area. -> \(2(π−2)=(2π−4)\) - we have 4 similar squares with shaded region, so we'll multiply the above result by 4 to get area covered by all 4 shaded regions: \(4*(2π−4) = 8π−16\)
In the figure below, four semicircles are drawn, each centered at the
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02 Nov 2019, 06:54
Bunuel wrote:
In the figure below, four semicircles are drawn, each centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
A. \(8\pi\) B. \(32 - 8\pi\) C. \(16 - 8\pi\) D. \(8\pi - 32\) E. \(8\pi - 16\)
Another way..
use of figure to simplify the question
Mark the areas as shown from a to h.. Now the area of semi-circle AB = a+b+c the area of semi-circle BC = c+e+h the area of semi-circle CD = f+g+h the area of semi-circle DA = a+d+e Area of all arcs =a+b+c+c+e+h+f+g+h+a+d+e= area of 4 semi-circles of radius 4/2 or 2 = \(2*\pi*r^2=2*\pi*2^2=8*\pi\)....(i)
Now area of square = a+b+c+d+e+f+g+h=4*4=16...(ii)
a+b+c+c+e+h+f+g+h+a+d+e-( a+b+c+d+e+f+g+h)=\(8\pi - 16\)......\(a+c+f+h=8\pi-16\) And a+c+f+h is nothing but area of shaded region
E
Attachments
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Re: In the figure below, four semicircles are drawn, each centered at the
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02 Nov 2019, 11:00
Bunuel wrote:
In the figure below, four semicircles are drawn, each centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
A. \(8\pi\) B. \(32 - 8\pi\) C. \(16 - 8\pi\) D. \(8\pi - 32\) E. \(8\pi - 16\)
Area of Square ABCD = 4 * 4 = 16 = Area of 4 petal regions + Area of non shaded regions So, Area of non shaded regions = 16 - Area of 4 petal regions
Area of semicircles with radius 2 = 4 * \(\pi\) \(2 * \frac{2}{2}\) = 8 * \(\pi\) (since there are 4 semicircles) Here the area of semicircles include the shaded petals 's area twice as every petal is included in area of two semicircles. Thus, 8 * \(\pi\) = Area of non - shaded regions + 2 * Area of Petal regions = 16 - Area of 4 petal regions + 2 * Area of 4 petal regions (from equation in green) = 16 + Area of 4 petal regions
Hence Area of 4 petal regions = 8 * \(\pi\) -16
IMO Answer E.
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53% (02:24) correct 
