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If points A and B are randomly placed on the circumference o [#permalink]
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Expert's post 00:00

Question Stats: 26% (01:38) correct 73% (01:18) wrong based on 60 sessions
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. $$\frac{3}{4}$$
[Reveal] Spoiler: OA

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Re: If points A and B are randomly placed on the circumference o [#permalink]
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In trigonometry, we use Law of Cosines to find out the angle of a triangle when 3 sides are known. Here, two sides of triangle are 2,2 and third side should be less than 2. If, in a triangle, two equal sides (lines) are at 120 degrees, the sum of two sides (or the third side) will be same as the other two side. So in our question the angle should be above 120 degree. So probability of 120 degree in a 360 degree circle is
120/360 =1/3 where the length of chord will be less than 2.
But for the chord being greater than 2, the probability is
1 - 1/3
2/3
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Re: If points A and B are randomly placed on the circumference o [#permalink]
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Expert's post
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. 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. There's also ANOTHER location for point B that creates another chord that is EXACTLY 2 cm long. 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. 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) Since these 2 triangles have sides of equal length, they are equilateral triangles, which means each interior angle is 60 degrees. The 2 central angles (from the equilateral triangles) add to 120 degrees.
This means the remaining central angle must be 240 degrees. 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

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Re: If points A and B are randomly placed on the circumference o [#permalink]
Sir, If you solved this problem in another way it will be easy for me to absorve it.
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Re: If points A and B are randomly placed on the circumference o [#permalink]
Expert's post
Hi Mithun

I do not think to solve such problem with algebra (which is the alternative method) rather than visualization will be easier _________________

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Re: If points A and B are randomly placed on the circumference o [#permalink]
Hey how can we take the ratio of upper limits to find the probability ?

Last edited by skythelimit on 07 Apr 2020, 12:58, edited 1 time in total. GRE Instructor Joined: 10 Apr 2015
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Re: If points A and B are randomly placed on the circumference o [#permalink]
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Expert's post
skythelimit wrote:
Hey how can me take the ratio of upper limits to find the probability ?

I'm not how that strategy will help. It's certainly beyond the scope of the GRE.
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Re: If points A and B are randomly placed on the circumference o [#permalink]
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My approach.

calculating the circumference: 4π

Following this I try to determine in how many pieces I would need to "slice up" to get the equivalent 2 cm chords to 4π.

following this I go 4π/x =2-->yielding back 6,28-->rounding it up I assume it is 6, since 4π/6 =2,0933-->meaning I would need 6 pieces

Now since we know that it is random where the two fall, we can make our own starting point of some sort.

Following upon this I imagine we have point 0. Looking counterclockwise and clockwise we get the 2cm chord distance twice.

Since we know that we have "two pieces" out of the 6 that yield back 2cm-->we divide 2/6 =1/3 the probability that we get 2cm

Since we are looking for the opposite, the probability of not getting a chord that is less than 2cm, we have 2/3 probability Re: If points A and B are randomly placed on the circumference o   [#permalink] 15 Apr 2020, 04:57
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