If an equilateral triangle is inscribed in a larger circle and is

Question

Equilateral Triangle.jpg

IanStewart · Accepted Answer

If we find the area of the big circle, and subtract the area of the triangle, we'll get the total area of the three equal sections outside of the triangle and inside the big circle. We only want the area of one of those sections, so if we do that subtraction, we'll need to divide by 3. So we want to find:

(Area of big circle - Area of triangle)/3

Now if you look at the small triangle at the top right of the diagram, the one with the dotted lines as two of its edges, notice that the line of length m in that triangle must meet the big equilateral triangle at 90 degrees, because the edge of the equilateral is tangent to the circle (and tangent lines are always perpendicular to radii at the tangent point). So that dotted-line triangle is a right triangle, and since we're bisecting the 60 degree angle at the top of the diagram, it must be a 30-60-90 triangle, and its sides are in a 1 to √3 to 2 ratio. Its shortest side is m, so its longest side, the vertical dotted line, is 2m. But that's the radius of the big circle, so the area of the big circle is π (2m)^2 = 4πm^2

Notice also that the third side of that small triangle, the downwards sloping side on the right, has length √3m. But that's half the length of a side of the equilateral triangle, so each side of the equilateral has length 2√3m. We need the area of that triangle, so we need to find a height (unless you know an equilateral triangle formula). If we take the horizontal side at the bottom as our base, then the vertical height is just made up of one radius of the small circle, of length m, plus the radius of the big circle which we found above was 2m. So the height of the triangle is 3m. Since our base is just the length of a side in the equilateral triangle, our base is 2√3m, and the area of the equilateral triangle is bh/2 = (2√3m)(3m)/2 = 3√3m^2.

Finally,

(Area of big circle - Area of triangle)/3 = (4πm^2 - 3√3m^2)/3 = (4π - 3√3)m^2/3

CareerGeek · Answer

Let ABC are the vertices with vertices A on top.

Property of equilateral triangle is centroid = center of incircle = Center of circumcircle

Centroid divides Median in ratio 2:1
So, if radius of incircle = m
Radius of circumcircle = 2m
—> OA = OB = OC = 2m

Required shaded area = area of sector OBC - area of triangle OBC.

OBC is isosceles triangle with 120-30-30 angles (sides ratio 3^1/2:1:1) . If OB = OC = 2m
—> BC = 2*3^1/2*m

Area of sector = 120/360*pi*(2m)^2 
= 4/3*pi*m^2

Area of triangle = 1/2*m*2*3^1/2*m
= sqrt(3)*m^2

Required Area = (4/3*pi - sqrt(3))*m^2
= (4*pi - 3*sqrt(3))*m^2/3

IMO Option A

Posted from my mobile device

Archit3110 · Answer

indu1954 
please see attached solution

EMPOWERgmatRichC · Answer

Hi indu1954,

This question can be approached with a mix of TESTing VALUES and Geometry rules. To start, we're going to focus on the Equilateral triangle:

1) We can split that triangle into 3 equal pieces (break the center of the circle into three 120-degree angles) and you'll have 3 equal triangles with angles of 30/30/120. That central 120-degree angle that 'points down' will create exactly 1/3 of the circle (and that piece contains the shaded area that we're interested in.)

2) Let's TEST M = 1. When the radius of the little circle is 1, then the 'height' of the 30/30/120 triangle will equal 1 and the 'base' will be 2√3. You can calculate that by breaking the 30/30/120 triangle into two 30/60/90 triangles.

The area of that 30/30/120 triangle is --> (1/2)(base)(height) = (1/2)(2√3)(1) = √3

You also now have the radius of the big circle (since the radius of the big circle equals the hypotenuse of the 30/60/90 triangles). That larger radius is 2.

3) We can now find the area of the BIG circle --> Area = π(R^2) = π(2^2) = 4π. One third of that circle is 4π/3.

4) To find the area of the shaded region, we subtract the area of the triangle from the area of 1/3 of the big circle:

4π/3 - √3 =
4π/3 - 3√3/3 =
(4π - 3√3)/3

Since M=1, it won't be difficult to find the one answer that matches...

A

GMAT assassins aren't born, they're made,
Rich

KarishmaB · Answer

Responding to a pm: My first thought is this: the arc of the area of the shaded region subtends 60 degree angle so the central angle it subtends must be 120 degrees. Shaded Area = (1/3)*Area of big circle - (1/3)*Area of triangle Side of equilateral triangle = 2*sqrt(3)*Radius of incircle = 2*sqrt(3)*m Area of triangle = \frac{\sqrt{3}}{4} * side^2 = \frac{\sqrt{3}}{4} * ^2 = 3*\sqrt{3}*m^2 Side of equilateral triangle = sqrt(3)*Radius of circumcircle 2*sqrt(3)*m = sqrt(3)*Radius of circumcircle Radius of circumcircle = 2m Area of big circle = \pi*radius^2 = \pi*(2m)^2 Shaded Region = (1/3)* Answer (A) Check these posts for discussion on the given relations: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/0 ... relations/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/0 ... other-way/

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If an equilateral triangle is inscribed in a larger circle and is

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If an equilateral triangle is inscribed in a larger circle and is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards