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# Common Mistakes in Geometry Questions - Exercise Question #3

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Joined: 04 Jan 2015
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Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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Updated on: 07 Aug 2018, 05:07
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Difficulty:

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Question Stats:

80% (02:22) correct 20% (02:17) wrong based on 386 sessions

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Common Mistakes in Geometry Questions - Exercise Question #3

ABC is a triangle inscribed in a circle and one of the sides of the triangle is the diameter of the circle with center O and radius 6 units. If angle ABC>angle BAC>angle ACB and angle OAB = $$60^o$$. Find the length of the line segment BC.

A. 6 units
B. 6$$\sqrt{3}$$ units
C. 7 units
D. 7$$\sqrt{3}$$ units
E. 12 units

Common Mistakes in Geometry are explained in detail in the following post:

Detailed solution will be posted soon.

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Originally posted by EgmatQuantExpert on 22 Nov 2016, 04:33.
Last edited by EgmatQuantExpert on 07 Aug 2018, 05:07, edited 4 times in total.
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Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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Updated on: 07 Aug 2018, 05:10
Note: This questions is related to the article on Common Errors in Geometry

Kindly go through the article once, before solving the question or going through the solution.

Official Solution

Given:

• Triangle ABC is inscribed in a circle.
• One of the sides of triangle ABC is the diameter.
• O is the centre and the radius is 6 units.
• angle ABC>angle BAC>angle ACB.
• angle OAB = $$60^o$$

Working:

ABC is inscribed in a circle, with one of the sides as the diameter.

Thus, we can conclude that ABC must be a right-angled triangle since we know that the diameter subtends an angle of 90 degrees on the circumference.

Also, angle ABC>angle BAC>angle ACB,

Therefore, we can infer angle ABC = $$90^o$$ and AC is the diameter of the circle.

From the above diagram,

We can infer that OBA an equilateral triangle and OBC is an isosceles triangle.

Hence, OB = OA = AB = 6 units.

As triangle ABC is a right-angled triangle,

We can apply Pythagoras Theorem and write:

A$$B^2$$ + B$$C^2$$ = A$$C^2$$

$$6^2$$ + B$$C^2$$ = 1$$2^2$$

B$$C^2$$ = 1$$2^2$$ - $$6^2$$

BC = $$6\sqrt{3}$$

Hence, correct option is B

Thanks,
Saquib
Quant Expert
e-GMAT

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Originally posted by EgmatQuantExpert on 22 Nov 2016, 10:11.
Last edited by EgmatQuantExpert on 07 Aug 2018, 05:10, edited 4 times in total.
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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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22 Nov 2016, 10:14
Considering triangle with AC as diameter of circle.
Since angle ABC>angle BAC>angle ACB and angle OAB = 60 deg.
angle ABC=90 deg, BAC = 60 deg and ACB = 30 deg.
length of BC = AC cos 30 deg = 12 cos 30 = 6 $$\sqrt{3}$$
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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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22 Nov 2016, 14:05
Triangle inscribed in a circle with one sides as the diameter implies a right angled triangle with the diameter as the hypotenuse

Angle ABC is the greatest therefore angle ABC = 90 degrees

This implies AC is the diameter. Given OAB is 60, therefore angle ACB is 30 degrees.

cos (30) = BC/AC => BC = 12*cos(30) = 6 (\sqrt{3})
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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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27 Nov 2016, 07:51
Hey Everyone,

The official solution has been posted. Kindly go through it and if you have any doubts feel free to post your query.

Thanks,
Saquib
Quant Expert
e-GMAT
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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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25 Aug 2018, 01:09
What if we take BC as diameter
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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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25 Aug 2018, 01:18
1
arjit5 wrote:
What if we take BC as diameter

Given that angle ABC has the highest value among all the 3 angles. As the triangle is a right-angled triangle, angle ABC = 90.

Now, if angle ABC = 90, the side opposite to that angle, AC, must be the diameter of the circle. Hence, we cannot assume BC as the diameter.

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Re: Common Mistakes in Geometry Questions - Exercise Question #3   [#permalink] 25 Aug 2018, 01:18
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# Common Mistakes in Geometry Questions - Exercise Question #3

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