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Diffcult geometry problem!

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by ashish1354 » Sun Nov 16, 2008 2:20 am
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6cm & 8 cm respectively (see figure in file attached). Find sides AB and AC.
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Re: Diffcult geometry problem!

by logitech » Sun Nov 16, 2008 3:01 am
ashish1354 wrote:A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6cm & 8 cm respectively (see figure in file attached). Find sides AB and AC.
This question has been around for centuries :)

https://en.wikipedia.org/wiki/Incircle

AB and AC can be found as 13 and 15 but it takes some serious calculation, if you really want to know incircles, you can take a look at the wikipedia link but I don't think that this question is within the scope of GMAT's geometry questions.
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by mals24 » Sun Nov 16, 2008 5:15 am
Agree with logitech

The calculations involve the use of trigonometry and this concept is not tested in GMAT (Thankfully!!)
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Re: Diffcult geometry problem!

by iamcste » Sun Nov 16, 2008 5:41 am
ashish1354 wrote:A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6cm & 8 cm respectively (see figure in file attached). Find sides AB and AC.

source pls...
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by stop@800 » Sun Nov 16, 2008 5:58 am
I am getting 6*sqrt(3) and 6*sqrt(5).

OA please
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answers!

by ashish1354 » Mon Nov 17, 2008 12:58 am
AC=13
AB=15

and the answer can also be calculated without trignometry :P
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source

by ashish1354 » Mon Nov 17, 2008 1:05 am
its a xth standard circles and tangents problem
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by mals24 » Mon Nov 17, 2008 4:55 am
Ok another method of solving this would be by using the Heron's formula, semi perimeter and incircle (thats what I think would be the method used in the book as well if its a CBSE book). These methods again are thankfully not required for the GMAT :)

semi perimeter s = (a+b+c)/2 = 14+x
Area of triangle = rs
= 4(14+x) = 56+4x.
Area of triangle = 1/2bh = 1/2*14*AD = 7AD

AD = AO + OD = AO +4
Area = 7(AO+4) = 7AO + 28
Equate the 2 areas: 7AO+28 = 56+4X--equ 1
AO = x^2 + 16 (pythagoras theorem).
Sustitute AO in equation 1 and youll get x = 6.79 or 7 approx

AB = 13 AC = 15

Im glad CBSE is not making the quant section of GMAT :)
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