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GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
In the above figure, the small circle and big  [#permalink]

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Top Contributor
15 00:00

Difficulty:   95% (hard)

Question Stats: 47% (03:08) correct 53% (03:18) wrong based on 58 sessions

### HideShow timer Statistics In the above figure, the small circle and big circle have diameters of 3 and 6 respectively. If AB||CD, and both circles share the same center, what is the area of the shaded region?

A) 9/2 + 2π
B) 9/2 + 3π
C) 3√3 + 3π
D) 6√3 + 2π
E) (9/2)√3 + 3π

*kudos for all correct solutions

_________________
If you enjoy my solutions, you'll love my GMAT prep course. Originally posted by BrentGMATPrepNow on 03 Jul 2018, 07:25.
Last edited by BrentGMATPrepNow on 03 Jul 2018, 08:57, edited 1 time in total.
Manager  G
Joined: 19 Nov 2017
Posts: 249
Location: India
Schools: ISB'21
GMAT 1: 670 Q49 V32
GPA: 4
Re: In the above figure, the small circle and big  [#permalink]

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1
GMATPrepNow wrote: In the above figure, the small circle and big circle have diameters of 3 and 6 respectively. If AB||CD, and both circles share the same center, what is the area of the shaded region?

A) 3 + 2π
B) 3 + 3π
C) 2√3 + 3π
D) 4√3 + 2π
E) 3√3 + 3π

*kudos for all correct solutions

Consider the figure.
Angle AOC = 60
The area of sector AC is
(60/360)*9π = 3π/2 (1)
area of bigger circle is 9π

Area of sector BD is also same as AC = 3π/2 (2)

To find are a of remaining shaded region,
Draw a perpendicular from O on line CD this divides 120 measure into 60 - 60, making it a 30-60-90 triangle.
area of COD is
(1/2)*(3/2)*(3*√3) = (9√3)/4 (3)

Area of AOB is also = 9√3/4 (4)

Adding 1,2,3, and 4, We get

3π + (9√3)/2

Option E was the closest to this solution. I checked twice but could not find an error in my solution. Please check for a typo or help me find a mistake in my method/calculations.

Regards,
V
Attachments Q700.png [ 10.29 KiB | Viewed 2072 times ]

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Vaibhav

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Joined: 30 May 2017
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Re: In the above figure, the small circle and big  [#permalink]

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1
By approximation, the area would be similar to that of a rectangle with sides 3*6 =18

considering that pi =3.14 and sq root(3) = 1.7, the closest solution would be E
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
Re: In the above figure, the small circle and big  [#permalink]

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Top Contributor
Both of you are correct.
As you were solving the question, I changed the answer choices after realizing I made a silly error in my calculations.
Kudos for all!!

Cheers,
Brent
_________________
If you enjoy my solutions, you'll love my GMAT prep course. GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
Re: In the above figure, the small circle and big  [#permalink]

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Top Contributor
3
GMATPrepNow wrote: In the above figure, the small circle and big circle have diameters of 3 and 6 respectively. If AB||CD, and both circles share the same center, what is the area of the shaded region?

A) 9/2 + 2π
B) 9/2 + 3π
C) 3√3 + 3π
D) 6√3 + 2π
E) (9/2)√3 + 3π

*kudos for all correct solutions

If we recognize that the hypotenuse of the blue right triangle below is twice the length of one side, we can see that this is a 30-60-90 SPECIAL TRIANGLE In the base triangle (on the right), the side opposite the 30° has length 1.
So, the magnification factor of the triangle in the question is 3/2 (i.e., the triangle in the diagram is 3/2 times the size of the base triangle)
So, the length of the 3rd side = (3/2)(√3) = (3√3)/2

Let's add this to our diagram... From here, we can calculate the area of the shaded triangle.
The length of the base = (3√3)/2 + (3√3)/2 = 3√3

Area = (base)(height)/2
= (3√3)(3/2)/2
= (9√3)/4
Let's add this to our diagram... Important: we earlier learned that the triangle in the first image is a 30-60-90 special triangle.
In fact, there are four such special triangles in the diagram.
So, let's add the 30° angles Let's now find the area EACH sector.
Area of sector = (central angle/360°)(π)(radius²)
= (60°/360°)(π)(3²)
= (1/6)(π)(9)
= 3π/2 So, the area of the shaded region = (9√3)/4 + (9√3)/4 + 3π/2 + 3π/2
= (9√3)/2 + 3π

Cheers,
Brent
_________________
If you enjoy my solutions, you'll love my GMAT prep course. Manager  B
Joined: 12 Mar 2018
Posts: 105
Re: In the above figure, the small circle and big  [#permalink]

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took a bit long, 3mins..
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Posts: 15605
Re: In the above figure, the small circle and big  [#permalink]

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_________________ Re: In the above figure, the small circle and big   [#permalink] 02 Apr 2020, 16:35

# In the above figure, the small circle and big  