In a triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, then AP:CQ is equal to (A) 4:1 (B) 3:1 (C) 1:1 (D) 3:2 (E) 3:8

Hi, Here is my approach. I enjoyed solving this ABC is a right angled triangle with B=90. (based on Pythagoras theorem) Now use area of triangle 0.5*BC *AC= 0.5*AC * BD BD = 4.8 Now since AB is perpendicular to BC, I re-drew the triangle so that AB coincides with Y axis and BC with X axis and B(0,0).This made it easier. We already know that A(0,6), B(0,0) and C(8,0). Radius of circle = 4.8 ==> P(0,4.8) and Q(4.8,0) AP = 6-4.8 = 1.2 CQ = 8-4.8=3.2 AP/CQ = 3:8 IMO E

ace_gre wrote:Hi, Here is my approach. I enjoyed solving this ABC is a right angled triangle with B=90. (based on Pythagoras theorem) Now use area of triangle 0.5*BC *AC= 0.5*AC * BD BD = 4.8 Now since AB is perpendicular to BC, I re-drew the triangle so that AB coincides with Y axis and BC with X axis and B(0,0).This made it easier. We already know that A(0,6), B(0,0) and C(8,0). Radius of circle = 4.8 ==> P(0,4.8) and Q(4.8,0) AP = 6-4.8 = 1.2 CQ = 8-4.8=3.2 AP/CQ = 3:8 IMO E A better explanation could need a germane portrayal to the situation that could be easily understood by the newbie as well. Do we have visual artists too? I expect it from the outlook that BTG will have features that would facilitate us to canvas our mind out while posting a reply, wherever and whenever it’s mandatory.