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Updated on: 07 Aug 2018, 06:07
Originally posted by EgmatQuantExpert on 22 Nov 2016, 05:33.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:07, edited 4 times in total.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:07, edited 4 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
80% (02:20) correct
20%
(02:22)
wrong
based on 945
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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.
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.
Updated on: 07 Aug 2018, 06:10
Originally posted by EgmatQuantExpert on 22 Nov 2016, 11:11.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:10, edited 4 times in total.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:10, edited 4 times in total.
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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
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
22 Nov 2016, 11:14
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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.
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.
