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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

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. \(8\pi - 16\)

Are You Up For the Challenge: 700 Level Questions


Attachment:
Paper_II_21_77.gif
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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: \(4*(2π−4) = 8π−16\)

IMO Option E
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Bunuel
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
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Bunuel

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. \(8\pi - 16\)

Are You Up For the Challenge: 700 Level Questions


Attachment:
Paper_II_21_77.gif

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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it took me a lot of time until i got the idea to solve

if we can find the area of unshaded regions, we can subtract it from the area of square to get the answer.

if we look only two adjecent semi circles, and if we subtract the area of these adjecent semicircles, it will give us the area of two unshaded region. similarly, after again subtracting the area of two adjecent semicircles of another direction from the area of square, we will the area of remaining two unshaded portion.
Thus our quation for finding out the area of unshaded region is
(16 - 4pi)*2 = 32 - 8Pi

16 is the area of square and 4pi is the area of circle (two adjecent semicircle equal one circle).
ad we have multiplied this by two because, we have to find the areas of remaining two unshaded regions.
16 - 32 + 8pi
which is 8pi - 16
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A different approach:
If you look closely at the figure, you can see that if you add the areas of all the 4 semi circles, it comes out to be the entire area of the square + those shaded regions. How?

Label semicircles from top-right-bottom-left as S1-S2-S3-S4
If you add S1 and S2, you will notice that the shaded region joining vertex B of the square occurs in both so technically it is being counted twice when adding S1 and S2.

Similarly, once you add all 4 semicircles (S1+S2+S3+S4), you will see that each shaded part is being counted twice and you will also notice that apart from the counting twice part, all 4 semicircles also cover the entire square so we can safely say that

Area of all 4 semicircles = Area of square + Area of shaded regions (since they are counted twice)
4* ((Pie * r^2)/2) = 4^2 + A
A = 16 - 8Pie
E
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