Answer:
E.
Step-by-step explanation:
Consider a circle having centre P and radius 20 inches,
So, the area of the circle,
Also, suppose points Q and R are on the circumference,
i.e. PQ = PR = 20 inches,
If QR = 20 inches,
So, triangle PQR is an equilateral triangle,
⇒ m∠RPQ = 60°,
Now, suppose R' is another point on the circle,
Such that, ΔPQR ≅ Δ PQR',
⇒ m∠QPR' = 60°,
Thus, minor angle, m∠RPR' = m∠RPQ + m∠QPR' = 60° + 60° = 120°,
⇒ major angle, m∠RPR' = 360° - 120° = 240°,
So, the area of the circle where a point on the circumference is closer to P than it is to Q
Hence, the probability that a point is closer to P than it is to Q =
i.e. OPTION E is correct.
