SajjadAhmad
A and B are the end points of the longest line that can be drawn in a circle with center X. If C is a point on the circle such that AC = AX = 3, what is the perimeter of triangle ABC?
(A) 9/2
(B) 9
(C) 6 + 3 \(\sqrt{3}\)
(D) 9 + 3 \(\sqrt{3}\)
(E) 9 \(\sqrt{3}\)
Dear
SajjadAhmad,
I'm happy to respond.
First, "
the longest line that can be drawn in a circle" = the diameter. We know AB is a diameter.
X is the center, so AX & BC are radii. AC is a chord that is equal in length to the radius, and XC is another radius, so ACX is an equilateral triangle, with three 60 degree angles. Let's look at a diagram:
Attachment:
circle X with triangle ABC.png [ 77.94 KiB | Viewed 5138 times ]
The next piece we need is that any triangle with three vertices on a circle that has a diameter a side is a right triangle. Another way to say that is that any angle that intersects an arc of a semicircle is a right angle. See the "special case" in this blog article:
Inscribed and Circumscribed Circles and Polygons on the GMATThus, ABC is a 30-60-90 triangle. The ratios in this triangle should be familiar. See this blog if they are not:
The GMAT’s Favorite TrianglesWe know AC = 3.
We know AB = AX + BX = 6
The third side is the side opposite the 60 degree angle in the 30-60-90 triangle. This is \(\sqrt{3}\) times the shortest side. Thus:
BC = \(3\sqrt{3}\)
perimeter of ABC = AC + AB + BC = \(9 + 3\sqrt{3}\)
OA =
(D) Does all this make sense?
Mike