NO TRIGONOMETRY PLEASE

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Mon Feb 22, 2010 6:10 am

C is a point outside of a circle with centre A and radius r. CB and CD are tangents to the circle whose length is t. What is the area of triangle BCD in terms of r and t?

[spoiler]IMO r t^3/ (r^2 + t^2), NO TRIGONOMETRY PLEASE[/spoiler]

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Source: Beat The GMAT — Problem Solving | Mon Feb 22, 2010 6:10 am

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Mon Feb 22, 2010 6:16 am

sanju09 wrote:C is a point outside of a circle with centre A and radius r. CB and CD are tangents to the circle whose length is t. What is the area of triangle BCD in terms of r and t?

[spoiler]IMO r t^3/ (r^2 + t^2), NO TRIGONOMETRY PLEASE[/spoiler]

Tangents are at right angle to the line joining the radius and the point of intersection.
Also,if you join A and C then both the opposite triangles are congruent.
Hence,area of BACD = 2 x (1/2 x AD x DC) = r x t

Using pythagorus theorem,we can find AC =sqrt(r^2 + t^2).
Let the intersection pt. be E.
Hence,AE = x and EC =AC - AE =sqrt(r^2 + t^2) - (x).
Now, ED = sqrt(r^2 - x^2){In triangle AED}
Also,ED = sqrt(t^2 - AE^2)
Solving both,we get the value of x in terms of r and t.
And area of traingle = 1/2 X BD x CE = 1/2 x (2ED) x CE

Last edited by harsh.champ on Mon Feb 22, 2010 6:17 am, edited 1 time in total.

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firdaus117: Master | Next Rank: 500 Posts; Posts: 102; Joined: Sat Feb 20, 2010 5:38 am; Location: IIM Ahmedabad; Thanked: 10 times

by firdaus117 » Mon Feb 22, 2010 7:33 am

sanju09 wrote:C is a point outside of a circle with centre A and radius r. CB and CD are tangents to the circle whose length is t. What is the area of triangle BCD in terms of r and t?

[spoiler]IMO r t^3/ (r^2 + t^2), NO TRIGONOMETRY PLEASE[/spoiler]

You have got the right answer buddy.You can use properties of Similar triangles to get BE and AE.For instance,Tr. AEB is similar to Tr.ABC.Please ignore this post if similar triangles are out of purview of GMAT.

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