\(A\) and \(B\) are the endpoints of the longest line that

This topic has expert replies
Moderator
Posts: 2237
Joined: Sun Oct 29, 2017 2:08 pm
Followed by:2 members

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Princeton Review

\(A\) and \(B\) are the endpoints of the longest line that can be drawn in a circle with center \(X\). If \(C\) is a point of the circle such that \(AC=AX=3\), what is the perimeter of triangle \(ABC\)?

A. \(\frac{9}{2}\)
B. \(9\)
C. \(6+3\sqrt{3}\)
D. \(9+3\sqrt{3}\)
E. \(9\sqrt{3}\)

OA D

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2621
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Tue May 07, 2019 5:39 am
The longest line you can draw in a circle is a diameter, so AB is a diameter of the circle. If X is the center of the circle, then AX is a radius. The question tells us AX is 3, so the radius of the circle is 3.

If you draw a diagram, and look at triangle ACX, two of the sides of that triangle, AX and CX, are radii, so have length 3. The third side is AC, which the question tells us also has length 3. So triangle ACX is equilateral, and all of its angles are 60 degrees.

If you now look at triangle ABC, it shares the angle at point A with triangle ACX, so triangle ABC has a 60 degree angle. We're also connecting a diameter AB to a point C on the circle, which always creates a 90 degree angle at that point. So triangle ABC has a 60 and a 90 degree angle, and must be a 30-60-90 triangle. We already know its hypotenuse is the diameter, so is 6, and its short side is AC which we know is 3, and then using either Pythagoras or 30-60-90 triangle ratios (the sides are in a 1 to √3 to 2 ratio), we find that the length of the missing side, BC, is 3√3. Adding the sides we get the perimeter of 6+3+3√3 = 9 + 3√3.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 7223
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Thu May 09, 2019 5:09 pm
AAPL wrote:Princeton Review

\(A\) and \(B\) are the endpoints of the longest line that can be drawn in a circle with center \(X\). If \(C\) is a point of the circle such that \(AC=AX=3\), what is the perimeter of triangle \(ABC\)?

A. \(\frac{9}{2}\)
B. \(9\)
C. \(6+3\sqrt{3}\)
D. \(9+3\sqrt{3}\)
E. \(9\sqrt{3}\)

OA D
We see that AB is a diameter of the circle, and AX is a radius. Since AX = 3, AB = 6. Furthermore, since triangle ABC contains the diameter AB as a side, it must be a right triangle, and AB is its hypotenuse.

We know AC = 3 and AB = 6, and we need to determine BC. Since AB is twice AC, triangle ABC must be a 30-60-90 right triangle. Therefore, BC = 3√3, and the perimeter of triangle ABC is AC + AB + BC = 3 + 6 + 3√3 = 9 + 3√3.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage