Common Mistakes in Geometry Questions - Exercise Question #3

Hi, how were you able to assume that ABO is 60º?
Am I missing something?

Thanks!

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

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.

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})

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

What if we take BC as diameter

EgmatQuantExpert are you sure that this is a 600-700 lvl question? It takes quite some time to comprehend the given data. It feels like 700 lvl question.

judging from the question and what is given, I doubt that the answer is that clear - splitting up the 90-angle in 30 and 60 will lead to a shape that cannot be constructed (the angles, one side being the diameter and the line going from B to O)

the diagram given above looks impossible if drawn to scale

I think the Key takeaways are this:


1) If an Inscribed Triangle inside a Circle has 1 Side = Diameter ----- then the Angle at the Vertex Opposite the Diameter = 90 Degrees

2)Since in any Right Triangle the 90 Degree Angle will be Across from the Longest Side and we are Given that Angle ABC is the Largest, the Vertex of the Inscribed Triangle across from the Diameter must be B.

2) If Angle OAB = 60 Degrees, then the Last Angle must be 30 Degrees.

Furthermore, because we know Vertex B is across from the Diameter and is the Vertex with the 90 Degree Angle, Side BC must be OPPOSITE Angle OAB = 60 Degrees


Using the 30 - 60 - 90 Right Triangle Ratio:

Side Opp. 90 Degrees / Side Opposite 60 Degrees = 2x / x * sqrt(3) = 12 / BC

Solving for Side BC, you get:

BC = 6 * sqrt(3)

-B-

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