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Triangle ABC has a line BC that passes through the Circle with Origin

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Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 02 Mar 2017, 07:16
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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
Triangle_ABC.png [ 15.78 KiB | Viewed 3495 times ]

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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 02 Mar 2017, 07:44
1
A)\(\sqrt{3}\) / 6

Tangent makes 90 degree angle to radius at the point of contact.
So angle <ABC = 90, <BCA = 60

tan 60 = AB/BC
\(\sqrt{3}\) = \(\frac{1}{BC}\)

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

Area of the triangle = 1/2 * base * height = 1/2 * 1 * 1/\(\sqrt{3}\) = \(\sqrt{3}\) / 6
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Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 02 Mar 2017, 08:01
Bunuel wrote:
Image
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


Hi,

GMAT does not test trigonometry,s so generally triangles given will be..
1) 45-45-90..
Here the sides are 1:1:√2
2) 30-60-90..
Here the sides are 1:√3:2

Here the tangent makes a 90° and one angle is 30..
So it is 30°-60°-90° and sides will be 1:√3:2 for BC:BA:AC or [m]\frac{BA}{BC}=\frac{√3}{1}... BC=\frac{1}{√3}\)

Area =\(\frac{1}{2}*BA*BC=\frac{1}{2}*1*\frac{1}{√3}=\frac{1}{2√3}=\frac{√3}{2√3*√3}=\frac{√3}{6}\)
A
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 02 Mar 2017, 08:54
1
1
30 - 60 - 90 triangle
\(1 - \sqrt{3} - 2\)

Area = \(\frac{1}{\sqrt{3} * 2} = \frac{\sqrt{3}}{6}\)
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 02 Mar 2017, 09:15
1
<ABC = 90°, Rule: A tangent at any point on a circle is perpendicular to the radius of the circle.

In ∆ABC,

<A = 30°, <B = 90°, <C = 60° ( Sum of interior angles of a triangle = 180°).

Hence, AB : BC : CA = √3 : 1 : 2

If AB = 1, then BC = 1/√3

Ar. ∆ABC = 1/2* AB * BC = 1/2√3 = √3/6

Option A.

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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 31 May 2017, 20:33
chetan2u wrote:
Bunuel wrote:
Image
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


Hi,

GMAT does not test trigonometry,s so generally triangles given will be..
1) 45-45-90..
Here the sides are 1:1:√2
2) 30-60-90..
Here the sides are 1:√3:2

Here the tangent makes a 90° and one angle is 30..
So it is 30°-60°-90° and sides will be 1:√3:2 for BC:BA:AC or [m]\frac{BA}{BC}=\frac{√3}{1}... BC=\frac{1}{√3}\)

Area =\(\frac{1}{2}*BA*BC=\frac{1}{2}*1*\frac{1}{√3}=\frac{1}{2√3}=\frac{√3}{2√3*√3}=\frac{√3}{6}\)
A


1/2√3=√32√3∗√3=√3612∗BA∗BC=12∗1∗1√3=12√3=√32√3∗√3=√3/6

I don't understand this part of the solution?
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 31 May 2017, 20:44
chetan2u wrote:
Bunuel wrote:
Image
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


Hi,

GMAT does not test trigonometry,s so generally triangles given will be..
1) 45-45-90..
Here the sides are 1:1:√2
2) 30-60-90..
Here the sides are 1:√3:2

Here the tangent makes a 90° and one angle is 30..
So it is 30°-60°-90° and sides will be 1:√3:2 for BC:BA:AC or [m]\frac{BA}{BC}=\frac{√3}{1}... BC=\frac{1}{√3}\)

Area =\(\frac{1}{2}*BA*BC=\frac{1}{2}*1*\frac{1}{√3}=\frac{1}{2√3}=\frac{√3}{2√3*√3}=\frac{√3}{6}\)
A


Bunuel why are we multiplying it by root 3 over root 3?
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 31 May 2017, 20:49
Nunuboy1994 wrote:
chetan2u wrote:
Bunuel wrote:
Image
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


Hi,

GMAT does not test trigonometry,s so generally triangles given will be..
1) 45-45-90..
Here the sides are 1:1:√2
2) 30-60-90..
Here the sides are 1:√3:2

Here the tangent makes a 90° and one angle is 30..
So it is 30°-60°-90° and sides will be 1:√3:2 for BC:BA:AC or [m]\frac{BA}{BC}=\frac{√3}{1}... BC=\frac{1}{√3}\)

Area =\(\frac{1}{2}*BA*BC=\frac{1}{2}*1*\frac{1}{√3}=\frac{1}{2√3}=\frac{√3}{2√3*√3}=\frac{√3}{6}\)
A


Bunuel why are we multiplying it by root 3 over root 3?


This algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator.

Questions involving rationalization to practice:
https://gmatclub.com/forum/in-the-given ... 59791.html
http://gmatclub.com/forum/if-x-0-then-106291.html
http://gmatclub.com/forum/if-n-is-posit ... 31236.html
http://gmatclub.com/forum/consider-a-qu ... 31083.html
http://gmatclub.com/forum/in-the-diagra ... 39282.html
http://gmatclub.com/forum/in-the-diagra ... 29962.html
http://gmatclub.com/forum/the-perimeter ... 27049.html
http://gmatclub.com/forum/which-of-the- ... 98531.html
http://gmatclub.com/forum/if-x-is-posit ... 63491.html
http://gmatclub.com/forum/1-2-sqrt3-64378.html
http://gmatclub.com/forum/if-a-square-m ... 99359.html

Hope it helps.
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Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 02 Jun 2017, 21:24
Nunuboy1994 we need to get answer in the form of options given. So, we need to rationalise the answer that you have got. Therefore, in order to cancel out root3 in denominator we need to multiply both numerator and denominator by root3 thus getting 3 in the denominator.
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 05 Jun 2017, 02:24
We know that angle BAC = 30 degree and BA = 1. tan 30 degree = BC/BA where tan 30 degree = 1/ sq root 3.

So, 1/ sq root 3 = BC.

Area of triangle = 1/2 * 1 * 1/ sq root 3 = 1/ 2* sq root 3 = sq root 3/ 6.
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 05 Jun 2017, 03:04
IMO A
It can be solved very easily by considering 30-60-90 triangle
As BC is 1 we can set up the equation as 1=square root (3)*x
Where x is the side opposite 30 degrees angle , there for we have x=1\square root(3)
Area = 0.5*1*(1\square root(3))
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 14 Jul 2018, 12:14
These kinds of questions check out the profundity of one's knowledge and confidence. One of the primary goals of the Standardized Tests is to check the test takers' confidence in his/her knowledge. That's why many test takers, who do not have enough confidence in their knowledge fall into the traps from these kinds of questions. One can solve the question smoothly but not reach the final conclusion.
After using 30:60:90 concept and finding the area of the triangle, the test taker should multiply both denominator and numerator with the square root of 3, then the final answer becomes apparent.
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Re: Triangle ABC has a line BC that passes through the Circle with Origin  [#permalink]

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New post 18 Aug 2018, 20:02
Bunuel wrote:
Image
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)
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Re: Triangle ABC has a line BC that passes through the Circle with Origin &nbs [#permalink] 18 Aug 2018, 20:02
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