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Math Expert V
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Posts: 60727
In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 53% (02:24) correct 47% (02:08) wrong based on 64 sessions

### HideShow timer Statistics 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

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Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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2
= 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. $$8\pi - 16$$

Are You Up For the Challenge: 700 Level Questions

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The attachment Paper_II_21_77.gif is no longer available

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Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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2
total area of the square = 16
and total area of the shaded region should be slightly less than 16
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. $$8\pi - 16$$

Are You Up For the Challenge: 700 Level Questions

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Paper_II_21_77.gif
Intern  S
Joined: 14 Nov 2018
Posts: 39
Location: Austria
Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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2
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:
$$16- (32-8\pi) = -16+8\pi = 8pi -16.$$

E. $$8\pi - 16$$
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In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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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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Originally posted by fauji on 02 Nov 2019, 06:04.
Last edited by fauji on 02 Nov 2019, 11:20, edited 3 times in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 8327
In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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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
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Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

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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. $$8\pi - 16$$

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

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GMATPREP1 590(Q48,V23) March 6, 2019
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GMATPREPSoft1 680(Q48,V35) June 26, 2019 Re: In the figure below, four semicircles are drawn, each centered at the   [#permalink] 02 Nov 2019, 11:00
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# In the figure below, four semicircles are drawn, each centered at the  