GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 28 Jan 2020, 15:17

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# In the figure below, four semicircles are drawn, each centered at the

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60727
In the figure below, four semicircles are drawn, each centered at the  [#permalink]

### Show Tags

02 Nov 2019, 03:45
00:00

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

Attachment:

Paper_II_21_77.gif [ 2.48 KiB | Viewed 1332 times ]

_________________
VP
Joined: 19 Oct 2018
Posts: 1297
Location: India
Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

### Show Tags

02 Nov 2019, 03:55
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

Attachment:
The attachment Paper_II_21_77.gif is no longer available

Attachments

Paper_II_21_77.gif [ 3.21 KiB | Viewed 1105 times ]

GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5748
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

### Show Tags

02 Nov 2019, 03:58
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

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

### Show Tags

02 Nov 2019, 04:38
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. $$8\pi - 16$$
Manager
Status: So far only Dreams i have!!
Joined: 05 Jan 2015
Posts: 228
WE: Consulting (Consulting)
In the figure below, four semicircles are drawn, each centered at the  [#permalink]

### Show Tags

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: $$4*(2π−4) = 8π−16$$

IMO Option E
Attachments

Untitled.png [ 17.98 KiB | Viewed 906 times ]

_________________
Unable to Give Up!

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
Joined: 02 Aug 2009
Posts: 8327
In the figure below, four semicircles are drawn, each centered at the  [#permalink]

### Show Tags

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

Paper_II_21_77.gif [ 3.53 KiB | Viewed 977 times ]

_________________
Director
Joined: 07 Mar 2019
Posts: 609
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: In the figure below, four semicircles are drawn, each centered at the  [#permalink]

### Show Tags

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. $$8\pi - 16$$

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

_________________
Ephemeral Epiphany..!

GMATPREP1 590(Q48,V23) March 6, 2019
GMATPREP2 610(Q44,V29) June 10, 2019
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
Display posts from previous: Sort by