bek_gmat wrote:
A circle is inscribed in EQUILATERAL triangle ABC such that point D lies on the circle and on line segment AC, point E lies on the circle and on line segment AB, and point F lies on the circle and on line segment BC. If line segment AB = 6, what is the area of the figure created by line segments AD, AE, and minor arc DE?
A 3√3- 9π/4
B 3√3- π
C 6√3- π
D 9√3- 3π
E It cannot be determined from the information given.
The question is missing a key word:
equilateral. I've amended the question to reflect its original wording.
Here is a drawing:
Region ADE = (triangle - circle)/3.
Triangle:
The formula for the area of an equilateral triangle is A=(s^2)/4 * √3.
A = (6^2)/4 * √3 = 9√3.
Circle:
The sides of a 30-60-90 triangle are proportioned x : x√3 : 2x.
In the 30-60-90 triangle shown above, x√3=3, so x=√3.
x=√3 is also the radius of the circle.
Area = πr^2 = π(√3^2) = 3π.
Region ADE:
Area = (triangle - circle)/3 = (9√3 - 3π)/3 = 3√3 - π.
The correct answer is
B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
