Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Arcs AC, AB and BC are centered at B, C and A respectively, as shown a [#permalink]

Show Tags

11 Jun 2015, 11:15

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

81% (02:08) correct 19% (01:36) wrong based on 74 sessions

HideShow timer Statistics

Attachment:

T7908.png [ 10.93 KiB | Viewed 1387 times ]

Arcs AC, AB and BC are centered at B, C and A respectively, as shown above. If the area of equilateral triangle ABC is √3, what is the area of the shaded region?

A. \(2pi-3\sqrt{3}\) B. \(\sqrt{3} - pi\) C. 3pi D. \(\sqrt{3}+2pi\) E. 3

Official Help: Shaded region problems call for Ballparking and eliminating irrelevant answer choices. Use Ballparking and avoid unnecessary calculations. The shaded area is the target value that you seek. In your ballparking calculations you use the given value - in our case it's √3, the area of the triangle. Run on each answer using elimination criteria, from the faster elimination to the slower. Start by checking if the expression is negative - An area can't be negative. Next check if the format is correct - in this question the shaded area is the result of "subtraction with pi", don't waste time on trying to figure out what is subtracted from what. Just note that you expect a subtraction and that one of the members of the expression should include ∏. Next ballpark for estimated size and POE answers that are not in the ballpark.

After folding the shaded regions onto the triangle it is easier to guesstimate the area of the shaded region:

Attachment:

T7908B.png [ 12.31 KiB | Viewed 1380 times ]

The shaded regions add up to approximately half the area of the triangle. If you're not sure you see that, subdivide the triangle connecting the center of the circle with vertices A, B, C as shown it this figure:

Attachment:

T7908c.png [ 9.98 KiB | Viewed 1380 times ]

Therefore, your target is half of the triangle= √3/2≈0.9

Arcs AC, AB and BC are centered at B, C and A respectively, as shown above. If the area of equilateral triangle ABC is √3, what is the area of the shaded region?

Official Help: Shaded region problems call for Ballparking and eliminating irrelevant answer choices. Use Ballparking and avoid unnecessary calculations. The shaded area is the target value that you seek. In your ballparking calculations you use the given value - in our case it's √3, the area of the triangle. Run on each answer using elimination criteria, from the faster elimination to the slower. Start by checking if the expression is negative - An area can't be negative. Next check if the format is correct - in this question the shaded area is the result of "subtraction with pi", don't waste time on trying to figure out what is subtracted from what. Just note that you expect a subtraction and that one of the members of the expression should include ∏. Next ballpark for estimated size and POE answers that are not in the ballpark.

After folding the shaded regions onto the triangle it is easier to guesstimate the area of the shaded region:

Attachment:

T7908B.png

The shaded regions add up to approximately half the area of the triangle. If you're not sure you see that, subdivide the triangle connecting the center of the circle with vertices A, B, C as shown it this figure:

Attachment:

T7908c.png

Therefore, your target is half of the triangle= √3/2≈0.9

You need to give the options.

Let's focus on arc BC and its centre at A. The circle which has A as the centre and BC as an arc has radius AB - the side of equilateral triangle. The area of the equilateral triangle = \(\sqrt{3}/4 * a^2 = \sqrt{3}\) so a, the side will be 2.

Triangle ABC is equilateral so angle A is 60 degrees. Effectively, arc BC is subtending a 60 degree angle at the centre. So area of sector ABC will be 60/360 * Area of circle = \((1/6) * \pi * 2^2 = 2\pi/3\)

The shaded region area = 3* (Area of sector ABC - Area of triangle ABC) = \(3*(2\pi/3 - \sqrt{3}) = 2\pi - 3\sqrt{3}\)
_________________

Arcs AC, AB and BC are centered at B, C and A respectively, as shown above. If the area of equilateral triangle ABC is √3, what is the area of the shaded region?

Official Help: Shaded region problems call for Ballparking and eliminating irrelevant answer choices. Use Ballparking and avoid unnecessary calculations. The shaded area is the target value that you seek. In your ballparking calculations you use the given value - in our case it's √3, the area of the triangle. Run on each answer using elimination criteria, from the faster elimination to the slower. Start by checking if the expression is negative - An area can't be negative. Next check if the format is correct - in this question the shaded area is the result of "subtraction with pi", don't waste time on trying to figure out what is subtracted from what. Just note that you expect a subtraction and that one of the members of the expression should include ∏. Next ballpark for estimated size and POE answers that are not in the ballpark.

After folding the shaded regions onto the triangle it is easier to guesstimate the area of the shaded region:

Attachment:

T7908B.png

The shaded regions add up to approximately half the area of the triangle. If you're not sure you see that, subdivide the triangle connecting the center of the circle with vertices A, B, C as shown it this figure:

Attachment:

T7908c.png

Therefore, your target is half of the triangle= √3/2≈0.9

There are three shaded areas of equal area.

Area of Equilateral Triangle = \((\sqrt{3}/4) side^2 = \sqrt{3}\) i.e. Side of Triangle = 2 = Radius of arc

One shaded area = Area of sector - Area of Equilateral Triangle One shaded area = \((\frac{60}{360})(pi)R^2 - \sqrt{3}\) One shaded area = \((\frac{60}{360})(pi)2^2 - \sqrt{3}\) One shaded area = \((\frac{2}{3})(pi) - \sqrt{3}\)

Three Shaded Area = \(3*(\frac{2}{3})(pi) - \sqrt{3}\)
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Re: Arcs AC, AB and BC are centered at B, C and A respectively, as shown a [#permalink]

Show Tags

25 Oct 2017, 07:02

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Arcs AC, AB and BC are centered at B, C and A respectively, as shown a [#permalink]

Show Tags

25 Oct 2017, 11:13

reto wrote:

Attachment:

T7908.png

Arcs AC, AB and BC are centered at B, C and A respectively, as shown above. If the area of equilateral triangle ABC is √3, what is the area of the shaded region?

A. \(2pi-3\sqrt{3}\) B. \(\sqrt{3} - pi\) C. 3pi D. \(\sqrt{3}+2pi\) E. 3

1) Choose one vertex and its segment, say B

2) Area of shaded region for B: (Area of segment) - (Area of triangle)

3) Segment area - find SIDE LENGTH of equilateral triangle, which is the RADIUS of Circle B:

Find side length Area of equilateral triangle, given: \(\sqrt{3}\) Formula for that area:\(\frac{s^2\sqrt{3}}{4}\)

\(\frac{s^2\sqrt{3}}{4}\)=\(\sqrt{3}\)

\(s^2\sqrt{3}=4\sqrt{3}\)

\(s^2 =\frac{4\sqrt{3}}{\sqrt{3}}\)

\(s^2 = 4\) \(s = 2\)

So radius of Circle B = 2

3) SEGMENT AREA: Find Circle B's area. What fraction does segment area equal?

Circle area: \(\pi*r^2 = 4\pi\)

Segment area as fraction? An equilateral triangle has 3 vertices with angle measures of 60°

Fraction:\(\frac{60}{360}=\frac{1}{6}\) of circle's area Segment area: \((\frac{1}{6})(4\pi)=\frac{2}{3}\pi\)

4) Shaded area * 3 Areas: Shaded = (Segment) - (triangle) One shaded area\(:\frac{2}{3}\pi -\sqrt{3}\) There are three such areas, so total shaded area is \(3(\frac{2}{3}\pi-\sqrt{3})=2\pi-3\sqrt{3}\)