Bunuel wrote:
The corners of an equilateral triangle are rounded to form semi-circles of radius 1. If the length of a side of the triangle before rounding is 10, what is the area of the resulting figure?
A. \(25\sqrt{3} + 3\)
B. \(25\sqrt{3} + \frac{3\pi}{2}\)
C. \(25\sqrt{3}\)
D. \(22\sqrt{3} + 3\)
E. \(22\sqrt{3} + \frac{3\pi}{2}\)
Area of the resulting figure = Area of Bigger equilateral - (Area cut out)
Area cut out from one corner = (1/3)*(Area of Equilateral triangle with side 2√3 - Area of Circle with radius 1) .
[The inscribed circle with radius 1 will give us the side of equilateral as 2√3 using 30-60-90 property)Area cut out from one corner \(= (1/3)*[(√3/4)* 2√3)^2 - π1^2) = √3-(π/3)\) .
Total Area cut out \(= 3*[√3-(π/3)] = 3√3 - π\)
Area of the resulting figure = \((√3/4)* 10)^2 - (3√3 - π) = 22√3 +π\)
I think answer must be option E but highlighted part is still a bit confusing. Taking semicircle out from corners will involve different calculations.
Attachments
File comment: www.GMATinsight.com
8.png [ 5.81 KiB | Viewed 518 times ]
_________________
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
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
close
39% (02:29) correct 
