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Math Expert V
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Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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11 00:00

Difficulty:   65% (hard)

Question Stats: 62% (01:57) correct 38% (02:26) wrong based on 205 sessions

### HideShow timer Statistics 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. $$24-4π$$ This question was provided by Math Revolution for the Game of Timers Competition Attachment: Untitled.png [ 3.07 KiB | Viewed 2601 times ]

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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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area of the square is 4x4=16
shaded region is approx half, 8
12-Pie is the only answer nearest to 8.
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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?

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

Looking at the figure, we can say area of shaded region <50% of area of square but not very less than 50%

Area of square=16
50% of 16 =8
So, any value less than 8 would be the area of the shaded region.

Value of π~3.14

Now replacing in choices

A. 12−π = 12-3.14=7.86
B. 12−2π =12-(~)6= 6
C. 12+π >12. Eliminate
D. 12+2π >12. Eliminate
E. 24−4π = 24- (~)12= ~12. Eliminate

B/w A &B, A is the closet. Hence A
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Originally posted by kitipriyanka on 23 Jul 2019, 08:36.
Last edited by kitipriyanka on 23 Jul 2019, 08:51, edited 1 time in total.
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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1
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

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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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

Attachments imagegmat.png [ 7.43 KiB | Viewed 2108 times ]

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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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2
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
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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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?

To answer the question you need to deduct the area of the Two semi-circles from the area of the Square.
Area of the whole square is 4*4 = 16
The area of the semi-circle is Pie/2 *R^2, R is equal to two. This is going to be multiplied by 2 as there are 2 semi-circles, but then we are counting the part of the semi-circle which is common to both twice, hence we need to deduct that.

16 - 4pie - 4 -2Pie = 12 -2pie.

Hence the answer is option B.

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

B
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GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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1
Area of the square - 4*4=16
Area of the 2 semi circles = (2*π*2*2)/2 = 4π
Area of the common segment between 2 semi circles = (Area of 4 semi circles - Area of the square)/4 <when 4 semicircles are made within the square 4 common segments similar to the one in the diagram are created and leave zero uncovered area> = ((4*π*2*2)/2 - 16)/4 = (8π - 16)/4 = 2π-4

Area of shaded region = Area of square - (area of 2 semicircle - area of the common segment) = 16-(4π-(2π-4) = 16-4π+2π-4 = 12-2π

IMO B
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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Area of square =16
area of shaded region is almost half around 8

option 1 is the closest.
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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The area of the semicircle is A = pi * r^2/2 since we have two semicircles then the total area is equal to the area of a cicle.
A = pi * r^2
Circumference is 2*pi*r
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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2

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: $$(4 - pi) + 4 + 0 + (4 - pi) = 12 - 2*pi$$

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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
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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?

See attached image below for visualization.

Shaded Area = Area of Square - Area of 2 semicircles + Area of Overlap of Semicircles

1. SQUARE
Area of Square = 4x4= 16

2. SEMICIRCLES:
Semicircle hasdiameter 4, so it has radius 2.

Area of 2 semi circles = $$Pi*2^2/2$$ + $$Pi*2^2/2$$ = 2Pi + 2Pi = 4Pi

3. OVERLAP:
The semicircles both have radius 2 and they will both pass through center of the square where they intersect. (See diagram below)

So imagine the overlap as overlap of 2 Quarter circles.

Now, isolate 1 quarter circle .... Here, Area of quarter circle - area of right triangle = half of the overlap

Half Overlap = Area of Quarter Circle - Area of Right Triangle

= $$Pi*2^2/4$$ - 1/2*2*2 = Pi - 2

Area of Overlap = 2 x half overlap = 2Pi - 4

Area of Shaded Region = 16 - 4Pi + 2Pi - 4

=12 - 2Pi

ANSWER B. $$12−2π$$

Attachment: GOT 23jul19 Geometry.jpeg [ 144.95 KiB | Viewed 1434 times ]

Originally posted by Vinit1 on 23 Jul 2019, 09:14.
Last edited by Vinit1 on 23 Jul 2019, 20:50, edited 1 time in total.
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Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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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

Area of square - (area of 2 semicircle - area of the common segment) = 16-(4π-(2π-4) = 16-4π+2π-4 = 12-2π

out of given options nearest value is IMO B 12-2pi

Originally posted by Archit3110 on 23 Jul 2019, 09:15.
Last edited by Archit3110 on 24 Jul 2019, 10:58, edited 1 time in total.
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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IMO-B

Refer attached image,
Side of square ABCD = 4 units

Now , divide the square in 4 parts by drawing lines (EF & GH) through midpoints of opposite sides.

Area of shaded region 1= 1/4 * π * r^2 = 1/4 * π * 2^2 = π ....{quarter circle}
Area of shaded region 2= 1/4 * π * r^2 = 1/4 * π * 2^2 = π ....{quarter circle}
Area of shaded region 3 = 1/4 * 4^2= 4 ....{quarter of square}

Area of region req. = Area of square - [ area of shaded region (1+2+3) ]
Req. Area = 16 - ( 4 + π + π ) = 12- 2π
Attachments WhatsApp Image 2019-07-23 at 9.33.54 PM.jpeg [ 66.06 KiB | Viewed 1971 times ]

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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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Instead of doing long calculations, I tried to approximate the area.
As both the circles are semicircles, they will intersect at the center of the square. so the top right part is slightly greater than 1/4th of the area of the square+ some more coming from the left over area after semicircles intersect
1/4th of the area of square is 4 units + approx. 1 unit from left semi circle + 1 unit from below semi circle = ~6units.

None of the answer choices have this value except B, approximating pi to 3.14.
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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

Attachments Untitled.png [ 9.44 KiB | Viewed 1900 times ]

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Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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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?

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

Sqr area=4ˆ2=16
Circle area=2ˆ2π=4π
Semi area=2π
Isosceles area=2
Segment area=Semi area/2-Iso area=2π/2-2=π-2
Shaded area = Sqr area - Circle area + 2 Segment area
Shaded area = 16 - 4π + 2(π-2) = 12 - 2π

Originally posted by exc4libur on 23 Jul 2019, 09:45.
Last edited by exc4libur on 24 Jul 2019, 09:32, edited 1 time in total.
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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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IMO correct answer is B - Explanation provided as attachment
Attachments IMG_20190723_222230.JPG [ 1.08 MiB | Viewed 1881 times ]

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Re: Two semi-circles are drawn on adjacent sides of a square with side len  [#permalink]

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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
Attachments 2019-07-23_222756 2.jpg [ 1.06 MiB | Viewed 1864 times ]

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