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If a triangle is formed using any 3 of the 8 equidistant
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02 Mar 2020, 05:31
02 Mar 2020, 05:31
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Question Stats:
46% (00:37) correct
53% (02:40) wrong based on 15 sessions
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What is the probability that a triangle formed by connecting any 3 of the 8 equidistant points on a circle is a right triangle?
A) 3/7
B) 2/7
C) 3/14
D) 2/21
E) 5/21
A) 3/7
B) 2/7
C) 3/14
D) 2/21
E) 5/21
Re: If a triangle is formed using any 3 of the 8 equidistant
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03 Mar 2020, 13:25
03 Mar 2020, 13:25
I'm getting 12/42 = 2/7, but I've been wrong before - or maybe I'm seeing double . . .
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Re: If a triangle is formed using any 3 of the 8 equidistant
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01 Feb 2021, 07:58
01 Feb 2021, 07:58
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Total ways of picking 3 points out of 8 to form a triangle = 8C3 = 56
Now, we must know that - "If one side of a triangle inscribed in a circle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle"
Each Diameter will make 6 such right triangles. And, there would be 4 different diameters which will give us different right triangles.
So, Total DIFFERENT Right triangles formed = 6 x 4 = 24
Required Probability = 24/56 = 3/7
Now, we must know that - "If one side of a triangle inscribed in a circle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle"
Each Diameter will make 6 such right triangles. And, there would be 4 different diameters which will give us different right triangles.
So, Total DIFFERENT Right triangles formed = 6 x 4 = 24
Required Probability = 24/56 = 3/7
