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# In the above figure, the small circle and big circle have di

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In the above figure, the small circle and big circle have di [#permalink]  03 Jul 2018, 07:23
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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
[Reveal] Spoiler: OA

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Re: In the above figure, the small circle and big circle have di [#permalink]  03 Jul 2018, 22:41
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GreenlightTestPrep 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

Here,

plz refer to the diagram attached,

First of all let us joined the AO , CO , BO, DO as they are all radii of the bigger circle = 3 (since radius = dia/2 = 6/2 = 3)

Let us join AC and BD, which are = 3 (since the diameter of the smaller circle = 3)

we find that △ AOC and △BOD are equilateral △ as all sides are equal and so all angles in the △'s are 60°

Now let us consider the sector AOC, we have the central angle = 60°

We know,

$$\frac{Central angle}{360} = \frac{Sector Area AOC}{Circle Area}$$

or $$\frac{60}{360}= \frac{(Sector Area)}{(9\pi)}$$

or Sector Area AOC = $$\frac{(3\pi)}{2}$$

Similarly Sector Area BOD = $$\frac{(3\pi)}{2}$$

Now we need the area of the △ AOB and △COD, in which we need the base of both the △'s i.e. AB and CD

Now let us consider the right angled △ BCD

we have BD =3 , CB = 6 , angle BDC =90° , angle DBC = 60° and angle BCD = 30° (angle BDC = 90° because the line AB is parallel to line CD and both are inscribed within a circle)

SO the triangle is a 30-60-90 triangle and the sides are in the ratio of $$1: \sqrt{3} : 2$$

or we can write as $$3 : 3\sqrt{3} : 6$$ ; such that the side $$CD = 3\sqrt{3}$$

Now the Area of the △COD,

$$\frac{1}{2} * base * altitude = \frac{1}{2} * CD * \frac{3}{2}$$ (Altitude = 1.5 i.e. the radius of the smaller circle)

= $${\frac{1}{2} * 3\sqrt{3} * \frac{3}{2}}$$

i.e. Area of the △COD = $$\frac{9\sqrt3}{4}$$

Similarly in △ AOB the area = $$\frac{9\sqrt3}{4}$$

Now the area of the shaded region = Area of the sector AOC + Area of the sector BOD + Area of the △ AOB + Area of the △ COD = $$\frac{(3\pi)}{2} + \frac{(3\pi)}{2} + \frac{9\sqrt3}{4} + \frac{9\sqrt3}{4}$$ = $$3\pi + \frac{9\sqrt3}{2}$$
Attachments

FIG 1jpg.jpg [ 21.5 KiB | Viewed 663 times ]

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Re: In the above figure, the small circle and big circle have di [#permalink]  05 Jul 2018, 06:20
Expert's post
GreenlightTestPrep 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

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
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Brent Hanneson – Creator of greenlighttestprep.com

Re: In the above figure, the small circle and big circle have di   [#permalink] 05 Jul 2018, 06:20
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