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. \(\frac{3}{4}\)

The probability that the length of chord AB > 2 cm is 2/3 ⇒ D

Step-by-step explanation:

* Lets explain how to solve the problem

- Points A and B are randomly placed on the circumference of a

 circle with a radius of 2 cm

- We need to find the probability that the length of chord AB is

 greater than 2 cm

- Probability = required length of arc / circumference

- Assume that the length of the chord AB is 2 and the center of

 the circle is labeled M

- The radius of the circle is 2 and AB = 2

- In Δ AMB

∵ MA= MB = 2 radii

∵ AB = 2

∴ Δ AMB is an equilateral triangle

∴ m∠AMB = π/3

∵ The length of an arc = r Ф , where r is the radius of the circle and

  Ф is the central angle subtended by this arc

∵ r = 2 , Ф = π/3

∴ The length of arc AB = 2 × π/3 = 2π/3

- The arc AB is in the half of the circle then there are another arc

  with the same length in the other half of the circle

- The length of the arc we must excluded from the length of the

  circle to have the part of length of AB is greater than 2 is

   2(2π/3) = 4π/3

∴ The length of the excluded arc is 4π/3

∵ The length of the circle is 2πr

∵ r = 2

∴ The length of the circle = 2π(2) = 4π

- The required arc = length circle - length excluded arc

∵ The length of the excluded arc = 4π/3

∵ The length of the circle = 4π

∴ The required arc = 4π - 4π/3 = 8π/3

- Probability = required length of arc / circumference

∴ Probability = (8π/3)/(4π) = 2/3

∴ The probability that the length of chord AB > 2 cm is 2/3

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

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step-by-step explanation: