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# What is the area of the shaded region?

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Intern
Joined: 01 Mar 2014
Posts: 1

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01 Mar 2014, 20:44
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Hello everyone,
First time poster here just wanted to ask you how to solve this question quicker than the way I just did.

Provided info: Given the small circle has radius of 3, what is the area of the shaded region?

The triangle inscribed in smaller circle is split into 3 isosceles triangles with 2 sides equaling 3.
Using the formula x : x : xsqrt2
You get the side of the inscribed triangle is 3sqrt2

Once you have that you can find the area of the equilateral triangle by splitting it into 2 right triangles and using the pythagorean theorem.
Where x = height
((3sqrt2)/2)^2 + x^2 = (3sqrt2)2^2
18/4 + x^2 = 18
9/2 + x^2 = 18
subtract the 9/2
x^2 = 18 - 4.5 = 13.5
x=sqrt13.5

Then you find the area of the triangle by formula (b*h)/2
3sqrt2*sqrt13.5 = 3sqrt27 = 9sqrt3
lastly divide by 2 = 4.5sqrt3
A=4.5sqrt3

Now you know the area of the original smaller triangle.

The area of the smaller circle using A=pi r^2 = 9pi

Now given the ratio of the larger triangle to the smaller triangle 4x
You can find the area of the shaded region of the top part by using the two answers above and divide by 3.

4(4.5sqrt3) - 9pi
(18sqrt3 - 9pi)/3

6sqrt3 -3pi

Since the smaller triangle is 1/4 to the larger triangle, we know the larger circle will be 36pi and we already know the area of the larger triangle = 18sqrt3
So in knowing that we can determine the larger shaded region (at the bottom) by subtracting the area of the larger circle by the larger triangle and then divide it all by 3
(36pi - 18srqt3) / 3
which equals
12pi - 6sqrt3

6sqrt3 -3pi + 12pi - 6sqrt3

9pi

Can someone tell me if my math is correct and how can you solve this question in under ~2mins?
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Intern
Status: FTW
Joined: 01 Apr 2012
Posts: 9
Location: India
GMAT Date: 09-27-2014
GPA: 3.5
WE: Consulting (Venture Capital)

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02 Mar 2014, 01:46
Let r= Radius of smaller circle
a= side of smaller triangle
A= side of bigger triangle

Since, r=3, using the formula Radius of inscribed circle= √3a/6 (where a= side of the triangle), we get a= 6√3

Now, using the formula Radius of circumscribed circle= √3A/3 (where A= side of the triangle), we get R= 6

Now, we can see that the Area of bigger circle = Area of bigger triangle + Area of 3 big sections
=> ⊼ (6)^2 = √3 (6√3)^2 / 4 + Area of 3 Big sections
=> Area of 3 big sections = 36⊼ - 27√3
=> Area of 1 big section = 12 ⊼ - 9√3.............(i)

Now, similarly, Area of bigger triangle = Area of smaller circle + Area of 3 small sections
=>√3(6√3)^2 / 4 = ⊼ (3)^2 + Area of 3 small sections
=> Area of 3 small sections = 27√3 - 9⊼
=> Area of 1 small section = 9√3 - 3⊼...........(ii)

Now, as required by the question, the area of shaded portion = area of 1 small section + area of 1 big section
Adding (i) and (ii), we get the required area = 9⊼
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9246
Location: Pune, India

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02 Mar 2014, 23:42
joshoowa wrote:
Hello everyone,
First time poster here just wanted to ask you how to solve this question quicker than the way I just did.

Provided info: Given the small circle has radius of 3, what is the area of the shaded region?

The triangle inscribed in smaller circle is split into 3 isosceles triangles with 2 sides equaling 3.
Using the formula x : x : xsqrt2
You get the side of the inscribed triangle is 3sqrt2

Once you have that you can find the area of the equilateral triangle by splitting it into 2 right triangles and using the pythagorean theorem.
Where x = height
((3sqrt2)/2)^2 + x^2 = (3sqrt2)2^2
18/4 + x^2 = 18
9/2 + x^2 = 18
subtract the 9/2
x^2 = 18 - 4.5 = 13.5
x=sqrt13.5

Then you find the area of the triangle by formula (b*h)/2
3sqrt2*sqrt13.5 = 3sqrt27 = 9sqrt3
lastly divide by 2 = 4.5sqrt3
A=4.5sqrt3

Now you know the area of the original smaller triangle.

The area of the smaller circle using A=pi r^2 = 9pi

Now given the ratio of the larger triangle to the smaller triangle 4x
You can find the area of the shaded region of the top part by using the two answers above and divide by 3.

4(4.5sqrt3) - 9pi
(18sqrt3 - 9pi)/3

6sqrt3 -3pi

Since the smaller triangle is 1/4 to the larger triangle, we know the larger circle will be 36pi and we already know the area of the larger triangle = 18sqrt3
So in knowing that we can determine the larger shaded region (at the bottom) by subtracting the area of the larger circle by the larger triangle and then divide it all by 3
(36pi - 18srqt3) / 3
which equals
12pi - 6sqrt3

6sqrt3 -3pi + 12pi - 6sqrt3

9pi

Can someone tell me if my math is correct and how can you solve this question in under ~2mins?

It's good to understand the relations between circles and inscribed polygons. Check these two posts:
http://www.veritasprep.com/blog/2013/07 ... relations/
http://www.veritasprep.com/blog/2013/07 ... other-way/

When a circle is inscribed in an equilateral triangle, Side of the triangle = $$2\sqrt{3} *$$ Radius of the circle
Side of the larger triangle = $$2\sqrt{3} *$$ Radius of the smaller circle = $$2\sqrt{3} * 3$$ = $$6\sqrt{3}$$

When an equilateral triangle is inscribed in a circle, Side of the triangle = $$\sqrt{3} *$$ Radius of the circle
Side of the larger triangle = $$\sqrt{3} *$$Radius of the larger circle
$$6\sqrt{3} = \sqrt{3}$$ * Radius of larger circle
Radius of larger circle = 6

Area of the top shaded region = (Area of larger triangle - Area of smaller circle)/3
Area of the lower shaded region = (Area of larger circle - Area of larger triangle)/3

Total shaded area = (Area of larger circle - Area of smaller circle)/3 =$$(\pi*6^2 - \pi* 3^2)/3 = 9\pi$$
_________________
Karishma
Veritas Prep GMAT Instructor

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07 Sep 2017, 00:11
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Re: What is the area of the shaded region?   [#permalink] 07 Sep 2017, 00:11
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