Carcass wrote:
If points A and B are randomly placed on the circumference of a circle with a radius of 2 cm, what is the probability that the length of chord AB is greater than 2 cm?
A. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)
We'll begin by arbitrarily placing point A somewhere on the circumference.
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So, we want to know the probability that a randomly-placed point B will yield a chord AB that is at least 2 cm long.
So, let's first find a location for point B that creates a chord that is EXACTLY 2 cm long.
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There's also ANOTHER location for point B that creates another chord that is EXACTLY 2 cm long.
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IMPORTANT: For chord AB to be
greater than or equal to 2 cm, point B must be placed somewhere along the red portion of the circle's circumference.
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So, the question really boils down to, "What is the probability that point B is randomly placed somewhere on the red line?"
To determine this probability, notice that the 2 cm chords are the same length as the circle's radius (2 cm)
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Since these 2 triangles have sides of equal length, they are equilateral triangles, which means each interior angle is 60 degrees.
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The 2 central angles (from the equilateral triangles) add to 120 degrees.
This means the remaining central angle must be 240 degrees.
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This tells us that the red portion of the circle represents 240/360 of the entire circle.
So, P(point B is randomly placed somewhere on the red line) = 240/360 = 2/3
Answer: D