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