Two semi-circles are drawn on adjacent sides of a square with side len

Question

https://gmatclub.com/forum/download/file.php?id=49179 Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region? A. 12-π B. 12-2π C. 12+π D. 12+2π E. 24-4π Untitled.png

ancored · Accepted Answer

Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?

I divided the square into 4 small squares.

Square #1 (I quadrant) - its area is 2*2=4 and we need to count it.
Squares #2 and #4 (II quadrant and III quadrant) if we put them together then we can see semicircle and we need to calculate eternal area 2*4 - (Pi*2^2)/2=8-2Pi
Square #3 we don't need to count this area is the answer.

Thus, 4+8-2Pi= 12-2Pi

The answer is B

globaldesi · Answer

Consider only one semi-circle
Its area will 
 \frac{\pi}{2} *2^2 
area of the triangle will  \frac{1}{2} *2\sqrt{2}*2\sqrt{2} 
thus area of those two leaves will be
=  \frac{\pi}{2} *2^2  -  \frac{1}{2} *2\sqrt{2}*2\sqrt{2} 
=  2\pi - 4

Area of two semicircles without overall =  4\pi  -  2\pi + 4 
=  2\pi +4  ----- eq 1

Total area of square =  4^2 = 16 
subtract eq1 from the square area to get shaded part =  16 - 2\pi +4 
=  12 - 2\pi

= B answer

coloredflash · Answer

I wasn't sure how to solve for the section between the two circles which I knew needed to be added back in, so I approximated. The area of the square would be 16. The shaded section is approximately only slightly less than half the area, which would be 8.
π ~= 3

A. 12−π
12 - 3 = 9 - This is more than half

B. 12−2π
12 - 2*3 = 6 - This seems to be the closest answer

C. 12+π
12 + 3 = 16 - This is basically the size of the square so cannot be the answer

D. 12+2π
12+2*3 = 18 - This is larger than the area of the square so cannot be the answer

E. 24−4π
24 - 4*3 = 12 - This is too big

RusskiyLev · Answer

https://gmatclub.com/forum/download/file.php?id=49179

Quote:

Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?

This question is a bit tricky. We have a square with side length of 4 and two semi-circles in it. We need to find the area of space not occupied by semicircles. Let us first of all, divide our square into 4 equal pieces and numerate them 1 to 4 from upper and left one to lower and right one. Let us look at these small circles one by one. First of all, the area of such a small circle is \(S=(a/2)^2=(4/2)^2=4\)
Also, we might notice that the diameter of a semicircle is equal to the side of a bigger square. Thus, \(r=d/2=a/2=2\)
The area of semicircle is \(S=(pi*r^2)/2=(pi*2^2)/2=2*pi\)

Now, let us have a closer look at these squares:
1) Upper left square.
We can see that there is a half of semicircle in this small square. The area of the half of semicircle is \(S=2*pi/2=pi\)
So the area of the shaded region will be \(area of small square - area of half of semicircle = 4 - pi\)

2) Upper right square
This small square does not have its area covered by a semicircle, thus its all area is shaded and is equal to 4

3) Lower left square
This square is covered by 2 halves of semicircles and we can notice that it does not have any area shaded. Thus, its shaded area is 0.

4) Lower right square
Here, as in the first example, the shaded area is equal to \(area of small square - area of half of semicircle = 4 - pi\)

Adding all these areas together we get: \((4 - pi) + 4 + 0 + (4 - pi) = 12 - 2*pi\)

Answer: B

hiranmay · Answer

Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?

Solution :
If we draw 4 semi-cycles on each one of the sides, then all the semi-cycles will meet/intersect in the center(crossing point of the square's diagonals). 
So (the area of the square) = 4 * (each semi-cycle area) - 4 * (area of the each eye-shaped region)
=> 4^2= (4 * π*((4/2)^2)/2) - 4 * (area of the each eye-shaped region)
=> 4= (π*(2^2)/2) - (area of the each eye-shaped region)
=> (area of the each eye-shaped region) = 2π -4

So finally,
the area of the shaded region = (the area of the square) - 2 * (each semi-cycle area) + (area of the each eye-shaped region) = 4^2 -2 * π*((4/2)^2)/2 + (area of the each eye-shaped region) 
the area of the shaded region = 16 -4π + (2π -4) = 12 -2π

A. 12−π
B. 12−2π -->  correct 
C. 12+π
D. 12+2π
E. 24−4π

Vinit1 · Answer

Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?

A. 12−π12−π
B. 12−2π12−2π
C. 12+π12+π
D. 12+2π12+2π
E. 24−4π

so area of square is 16

and the area of the two semicircle - oval part has to be subtracted from 16 = shaded area
16 - (area of two semicircle - oval part) = shaded area

16 - area of circle + oval part = shaded area

area of circle = π * (4/2) ^2

so shaded area = 16 - 4π + oval part ...............1

first method:
so shaded area = 16 - 4 (3.14) + 0val part
this oval part is definitely infact a semicircle is 2 ovals plus a area almost equal to it. less than half the area of the semicircle. which is 2 pie

so it is less than 16 - 2( 3.14) = 9.78

so option c, d, e are greater than this value ..
remaining only a and b

a) 12-3.14 and b) 12 - 6.28
a is close to 9.78 so this can be rejected to get the option as B

second method.
this is lengthy

the square can be divided into figure as mentioned
4 oval parts and 4 white parts

2 white parts is area of square - area of circle = 16-.4π

so 4 white = 32 -8π
so 4 ovals = 16 -4 white area = 16 -32 + 8π = -16 + 8π

so 1 oval = -4 + 2π
so shaded area = 16 - 4π + oval part
16-4π + -4 + 2π

12-2π is the answer

Attachments

gmatclub upload.png [ 112.47 KiB | Viewed 22625 times ]

firas92 · Answer

Since the two semi circles are congruent, they intersect at the top of the arc.

We can divide this into 3 regions

I. A quarter circle with radius 2 (Area =  \frac{\pi*2*2}{4} = \pi )
II. A square with side 2 (Area =  2*2 = 4 )
III. Another quarter circle with radius 2 (Area =  \frac{\pi*2*2}{4} = \pi )

Total area =  4*4 = 16

Shaded area = Total Area - I - II - II

Therefore shaded area =  16 - \pi - 4 - \pi = 12-2\pi

Answer is  (B)

Kinshook · Answer

Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?
A. 12−π
B. 12−2π
C. 12+π
D. 12+2π
E. 24−4π

Please see the solution in image below

IMO B

ccheryn · Answer

Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?

A. 12−π12−π
B. 12−2π12−2π
C. 12+π12+π
D. 12+2π12+2π
E. 24−4π

so area of square is 16

and the area of the two semicircle - oval part has to be subtracted from 16 = shaded area
16 - (area of two semicircle - oval part) = shaded area

16 - area of circle + oval part = shaded area

area of circle = π * (4/2) ^2

so shaded area = 16 - 4π + oval part ...............1

first method:  
so shaded area = 16 - 4 (3.14) + 0val part 
this oval part is definitely infact a semicircle is 2 ovals plus a area almost equal to it. less than half the area of the semicircle. which is 2 pie

so it is less than 16 - 2( 3.14) = 9.78

so option c, d, e are greater than this value ..
remaining only a and b

a) 12-3.14 and b) 12 - 6.28 
a is close to 9.78 so this can be rejected to get the option as B

second method.  
this is lengthy

the square can be divided into figure as mentioned 
4 oval parts and 4 white parts

2 white parts is area of square - area of circle = 16-.4π

so 4 white = 32 -8π
so 4 ovals = 16 -4 white area = 16 -32 + 8π = -16 + 8π

so 1 oval = -4 + 2π
so shaded area = 16 - 4π + oval part
16-4π + -4 + 2π

12-2π is the answer

JonShukhrat · Answer

One of the   easiest   ways to figure out the area of the yellow region will require us to slightly change the diagram. In the below picture we can see that the yellow area didn’t change after a small modification. The point  O  is the cross point of the diagonals of the bigger square  DCBE  because both  FO  and  AO  are radiuses.

Now the bigger square  DCBE  contains a small square  FOAE  and two quarter-circles  FDO  and  AOB .

The yellow region    = (square DCBE) – (square FOAE) – (quarter-circle FDO) – (quarter-circle AOB)

The area of the square  DCBE =4*4=16

The area of the square  FOAE =2*2=4

The area of the quarter-circle  FDO = \frac{P*r^2}{4}=\frac{4P}{4}=P

The area of the quarter-circle  AOB  is also  =P

So  The yellow region   =16-4-P-P=12-2P

Hence  B

freedom128 · Answer

As shown in the attached figure, the total shaded area= (Area I) + (Area IIa)

Area of semi circle  = \frac{1}{2}*\pi*r^2 = \frac{1}{2}*\pi*2^2 = 2\pi 
Area of quarter circle  = \frac{1}{4}*\pi*r^2 = \frac{1}{4}*\pi*2^2 = \pi

Area I   = (2 * 4) -  Area of quarter circle  = 8 - \pi

Area IIa + Area IIb  = (2 * 4) -  Area of semi circle  = 8 - 2\pi  
 Area IIa = Area IIb  --->  Area IIa   = \frac{1}{2}* (8 - 2\pi) = 4 - \pi

Therefore,  the total shaded area= (Area I) + (Area IIa)   = (8 - \pi) + (4 - \pi) = 12 -2\pi

ANSWER IS (B)

Sayon · Answer

Consider only the attached portion of the figure given (Figure A):
  
 Finding the area of the portion STQUS: 
We can do that by finding the area of STQS and multiply it by 2.
Area of STQS = Area of sector (SRQTS) – Area of Right Triangle (SRQ)
Area of STQS = ( \frac{1}{4}  *  \pi  *  2^2 ) – ( \frac{1}{2} * 2 * 2 )
Area of STQS =  \pi – 2

Therefore, Area of the portion STQUS:   2 * (\pi -2)  =  2\pi – 4  ->  (a)

Area of shaded portion = Area of the Square –  
 Area of shaded portion = 4 * 4 –   
Area of shaded portion =  16 –   
Area of shaded portion =  16 – 2 \pi – 4 
 Area of shaded portion =  12 – 2 \pi 
 
 Answer B

mira93 · Answer

Please find the attached below

manish1708 · Answer

Image
Two semi-circles are drawn on adjacent sides of a square with side length 4 as shown above. What is the area of the shaded region?

Area of the main square = 4x4 = 16.
Now we have 2 semi-circles drawn on adjacent sides of a square. Therefore diameter of circle = 4 and radius = 2.
Also, two semi-circles will intersect at the intersection of the diagonals of the square (since radius = 2 and square side length = 4).
From the intersection point, we can draw two lines - one perpendicular to the base of the square and similarly one to the left side of the square.
So, we will have non-shaded area as
1) half of semicircle at the left top of the square. ---> Area = π2^2/4 = π 
2) half of semicircle at the bottom right of the square. ---> Area = π2^2/4 = π 
3) square with side length =2 at the bottom left of the main square. ---> Area = 2x2 = 4

Therefore total area of the shaded region will be 
Area of the main square (side length = 4) - Area of non shaded region (sum of areas calculated above)
--> 16 - (π +π +4) =  12 - 2π  
 
A. 12−π 
B. 12−2π 
C. 12+π
D. 12+2π 
E. 24−4π

Answer Choice:  B

shridhar786 · Answer

area of the shaded part 
= 12-2∏
B is the answer

QuantMadeEasy · Answer

Posted from my mobile device

Abhi077 · Answer

Solution:

The area of the shaded region will be Area of square - areas of two semicircles + the area of leaf formed in between(since it has been double counted)

The focus here must be finding out the area of the leaf, rest all is pretty simple.

If we consider one semi circle, and divide it by two we'll get a quarter circle

i.e since we know that side of a square which is 4 is equal to diameter of the semicircle, radis is two

Area of circle with radius 2 will be  \pi  r^2  i.e  \pi  X 4, therefore area of semi circle will be 2 \pi

area of the other semicircle also will be 2 \pi . total area of two semi circle will be 4  \pi.

we know the area of square will be 4^2  i.e 16

Now the area of leaf in one semicircle shall be computed as in the following diagram:

We divide the semicircle in the quarter circle and subtract from that the area of triangle which will be 2.

Area of leaf will be \pi  - 2
Area of the leaf in the other semi circle too will be the same

Therefore total area of leaf will be 2 \pi  - 4

Total area of shaded region will be

16 - 4 \pi  + ( 2\pi  -4)

= 16-  2\pi  - 4
= 12 - 2 \pi

Hence the answer is  B

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Two semi-circles are drawn on adjacent sides of a square with side len

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Two semi-circles are drawn on adjacent sides of a square with side len

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards