In the above figure, the small circle and big

Question

https://i.imgur.com/XrqxGNd.png

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

BrentGMATPrepNow · Accepted Answer

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
 https://i.imgur.com/qxIw8NQ.png 
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...
 https://i.imgur.com/2SMxaLz.png

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...
 https://i.imgur.com/QEDwFLo.png

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
 https://i.imgur.com/Kr6WBdb.png

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 
 https://i.imgur.com/6Fp5mQd.png

So, the area of the shaded region =  (9√3)/4  +  (9√3)/4  +  3π/2  +  3π/2 
= (9√3)/2 + 3π

Answer: E

Cheers,
Brent

vaibhav1221 · Answer

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

amascarenhas · Answer

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

BrentGMATPrepNow · Answer

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

faltan · Answer

took a bit long, 3mins..

Fdambro294 · Answer

this is a great question testing Sector Areas, Symmetry, and finding angles.

Step 1: since chord AB and chord CD are parallel, fill in a Perpendicular Line from AC and BD to complete a rectangle.

Draw in the 2 diagonals of this shaded rectangle, AD and CB.

The 2 diagonals will be congruent and bisect each other (property of rectangles)

Each Diagonal will pass through the center of the picture and will equal the Diameter or the entire larger circle ———-> 6

Making each bisected segment of the diagonal ——-> 3

Step 2: the radius drawn from the center of the smaller, Inner circle (point O) to a Line Tangent to the circle will create a 90 degree angle at the point of tangency

Name the point of tangency that parallel line AB makes with the smaller circle ——-> Point X

OX will be equal to ——> radius of smaller circle ——-> 1.5

Step 3: there are TWO Pairs of Congruent, Symmetric Parts that make up the Shaded Region.

Pair 1: the triangle created at BXO and AXO will be 30-60-90 Right Triangles

We can infer this because one of the Legs (OX) = 1.5 and the hypotenuse across from the 90 degree angle at the point of tangency (point X) ——> BO = AO = 3

The leg - to - hypotenuse ratio is 1 : 2

XB will therefore be = AB = (1.5) * sqrt(3)

Since the angles at the Vertex of the Rectangle = 30 and the angle at the point of tangency = 90 ———> the will be two 60 degree angles (adding up to one bisected 120 degree angle) at the center of the figure created by AOB

Across the center, there will be a Congruent Symmetric Triangle given by COD

We can find the triangle Area of these two parts:

AB = CD = (3) * sqrt(3) = Base

OX = (3/2) = perpendicular height

Area of 2 of these triangles =

(2) * [ (1/2) * (3/2) * (3 * sqrt(3)) ]

= (9/2) * sqrt(3) ————(I)

Pair 2: the Two, Symmetric Sector Areas given by Central Angles ——-> <AOC and <BOD

From before, we found central angles —-> <AOB = <COD = 120 degrees

Since three other 2 pairs of central angles complete the circle around Point O, and they are congruent vertically opposite angles:

360 - (2) (120) = 120

120/2 = 60

You can also find that this central angle is 60 degrees by noticing that BOD triangle portion of the rectangle has all equal sides of length 3

To find the last part of the shaded area, we need to find this sector area TWICE ——-> will be sector area of Larger, Entire Circle —— which has a R = Radius = 3

= (2) * [ (60/360) * (3)^2 * (pi) ]

= (3) (pi) ———— (II)

The final shaded area is:

(I) + (II) =

(9/2) * sqrt(3) + (3) (pi)

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In the above figure, the small circle and big

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