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Common Mistakes in Geometry Questions - Exercise Question #3

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Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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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:


Common Mistakes in Geometry Questions



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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New post 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.

Image

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


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Saquib
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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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New post 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 \(\sqrt{3}\)
Answer B.
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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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New post 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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Re: Common Mistakes in Geometry Questions - Exercise Question #3  [#permalink]

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New post 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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New post 25 Aug 2018, 01:18
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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.

Hope this answers your query. :-)

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