In a plane, points P and Q are 20 inches apart. If point R is randomly chosen from all the points in the plane that are 20 inches from P, what is the probability that R is closer to P than it is to Q?A. 00B. 1414C. 1313D. 1212E. 23

Answer:E. [tex]\frac{2}{3}[/tex]Step-by-step explanation:Consider a circle having centre P and radius 20 inches,So, the area of the circle,[tex]A=\pi (20)^2\text{ square inches}[/tex]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 [tex]=\frac{240^{\circ}}{360^{\circ}}\pi (20)^2[/tex][tex]=\frac{2}{3}\pi (20)^2[/tex]Hence, the probability that a point is closer to P than it is to Q = [tex]\frac{\frac{2}{3}\pi (20)^2}{\pi (20)^2}[/tex][tex]=\frac{2}{3}[/tex]i.e. OPTION E is correct.