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AP:CQ is equal to

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by sanju09 » Thu Jan 14, 2010 2:19 am
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
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by ace_gre » Thu Jan 14, 2010 12:37 pm
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
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by sanju09 » Wed Feb 17, 2010 5:22 am
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.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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