Bunuel wrote:

Triangle ABC has a line BC that passes through the Circle with Origin O. Line BA is tangent to the circle at point B, and line BA has length 1. What is the area of Triangle ABC?

A. \(\frac{\sqrt{3}}{6}\)

B. \(\frac{\sqrt{3}}{3}\)

C. 1

D. \(\frac{\sqrt{3}}{2}\)

E. \(\sqrt{3}\)

Attachment:

Triangle_ABC.png

Responding to a pm:

A tangent is perpendicular to the radius at the point of contact. So angle ABC is 90 degrees. Since angle A is 30 degrees, angle C must be 60.

So we have a 30-60-90 triangle in which the sides are in the ratio \(1:\sqrt{3}:2 = BC : AB : AC\)

So if \(BC = x\),

\(AB = \sqrt{3}x\)

\(AC = 2x\)

Given, \(AB = 1 = \sqrt{3}x\)

So \(x = 1/\sqrt{3}\)

So \(BC = 1/\sqrt{3}\)

Area of triangle ABC \(= (\frac{1}{2})*AB*BC = (\frac{1}{2})*1*\frac{1}{\sqrt{3}} = \frac{1}{2\sqrt{3}}\)

Note that none of the options have \(\sqrt{3}\) is the denominator. So the denominator has been rationalised.

Multiply and divide the fraction by \(\sqrt{3}\) to get

Area = \(\frac{1*\sqrt{3}}{2\sqrt{3} * \sqrt{3}} = \frac{\sqrt{3}}{6}\)

Answer (A)

_________________

Karishma

Veritas Prep GMAT Instructor

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