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In the diagram to the right, triangle ABC has a right angle [#permalink]

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03 Jul 2008, 15:12

In the diagram to the right, triangle ABC has a right angle at B and a perimeter of 120. Line segment BD is perpendicular to AC and has a length of 24. AB > BC. What is the ratio of the area of triangle ABD to the area of triangle BDC?

Please see the attached diagram. Do let me know about your timings. If you are able to solve it within 2 mins. then do post your method. I was able to solve it however it took too much time around 5 mins.

Here is my quick solution. Think of 3,4,5 triangle (I guess those are popular on GMAT). AB = 40, BC = 30, AC = 50 (check that BD = (3/5)*AB = 24 -> great, we are lucky AD = (4/5)*AB = 32 -> DC = 50-32 = 18

Then ratio of areas is equal to AD/DC = 32/18 = 16/9

Here is my quick solution. Think of 3,4,5 triangle (I guess those are popular on GMAT). AB = 40, BC = 30, AC = 50 (check that BD = (3/5)*AB = 24 -> great, we are lucky AD = (4/5)*AB = 32 -> DC = 50-32 = 18

Then ratio of areas is equal to AD/DC = 32/18 = 16/9

That's nice...what would you have done if it didn't fit nicely into a 3:4:5 triangle?

maratikus wrote:

Here is my quick solution. Think of 3,4,5 triangle (I guess those are popular on GMAT). AB = 40, BC = 30, AC = 50 (check that BD = (3/5)*AB = 24 -> great, we are lucky AD = (4/5)*AB = 32 -> DC = 50-32 = 18

Then ratio of areas is equal to AD/DC = 32/18 = 16/9

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

On a separate note, and I'm not sure we'll ever really need to know this...but can you tell me the interior angles of a 3:4:5 triangle?
_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Here is my quick solution. Think of 3,4,5 triangle (I guess those are popular on GMAT). AB = 40, BC = 30, AC = 50 (check that BD = (3/5)*AB = 24 -> great, we are lucky AD = (4/5)*AB = 32 -> DC = 50-32 = 18

Then ratio of areas is equal to AD/DC = 32/18 = 16/9

Yes, you are lucky. Otherwise it is not so simple to solve

What you can always do is writing down some equations:

AB^2 + BC^2 = (AD+DC)^2 (Pythagorean theorem on triangle ABC) AB+BC+AD+AC = 120 (Perimeter of triangle ABC) AB^2 = 24^2 + AD^2 (Pythagorean theorem on triangle ABD) BC^2 = 24^2 + DC^2 (Pythagorean theorem on triangle BCD)

And you have 4 different equations with 4 unknown variables (AB,BC, AD, DC), so that it is solvable. But not easy to do...

Best way is indeed to remember that GMAT exercises are "rigged" to be simple: rectangle triangle --> think or 3:4:5 ratios.

Yes, you are lucky. Otherwise it is not so simple to solve

What you can always do is writing down some equations:

AB^2 + BC^2 = (AD+DC)^2 (Pythagorean theorem on triangle ABC) AB+BC+AD+AC = 120 (Perimeter of triangle ABC) AB^2 = 24^2 + AD^2 (Pythagorean theorem on triangle ABD) BC^2 = 24^2 + DC^2 (Pythagorean theorem on triangle BCD)

And you have 4 different equations with 4 unknown variables (AB,BC, AD, DC), so that it is solvable. But not easy to do...

Best way is indeed to remember that GMAT exercises are "rigged" to be simple: rectangle triangle --> think or 3:4:5 ratios.

This is exactly the method I followed, however as you can see there are 4 variable and few of them in quadratic form, so it took hell lot of time. It may be my bad day that I took so long however doing it in 2 mins. I found very tough.

Moreover with such a complex question, there is no way one can go back and re-check the answer. Although Maratikus method is not fool proof but it works here. Remember in GMAT each problem is designed in such a way that there will be always a shortcut by which you can finish the question within 2 mins.

Maratikus’s solution is great – and the only one that can work for <2 min time... But I wonder if the problem can be solved in other way, not involving picking numbers.

So far I found only partial simplification: instead solving 4 equations for 4 variables at once, we can implement two-stage approach to solving this problem. Basically it’s the same way to solve as others suggested:

Let a+b = x, ab=y.

Then, from P=120 and Pythagorean Theorem, we can write: (120-x)^2=x^2-2y <=> 120^2 – 240x = -2y. From ab=24c => 24(120-x)=y. So now we have system of two linear equations with big and ugly (though integer ) coefficients:

60*120 – 120x = -y 24*120 – 24x = y

x=120*84/144 =70. y=24*50=40*30

Now, returning to a and b => a+b=70, ab=40*30 => a=40, b=30, c=50 (yes, all those calculations just to prove that we indeed have 30-40-50 triangle ).

Areas ratio equals to square of a/b => 16/9.

Calculations can be done in less than 5 min… Anyway, if I see such a problem on the real exam, I’ll be dead

But I still think there should be some easier approach, completely different from the one above. Perhaps someone can find it?

Here is my quick solution. Think of 3,4,5 triangle (I guess those are popular on GMAT). AB = 40, BC = 30, AC = 50 (check that BD = (3/5)*AB = 24 -> great, we are lucky AD = (4/5)*AB = 32 -> DC = 50-32 = 18

Then ratio of areas is equal to AD/DC = 32/18 = 16/9

Correct me if Iam wrong. Ratio of Areas = square (ratio of sides)

Re: In the diagram to the right, triangle ABC has a right angle [#permalink]

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08 Aug 2014, 06:43

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Re: In the diagram to the right, triangle ABC has a right angle [#permalink]

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08 Aug 2014, 06:43

maratikus wrote:

Here is my quick solution. Think of 3,4,5 triangle (I guess those are popular on GMAT). AB = 40, BC = 30, AC = 50 (check that BD = (3/5)*AB = 24 -> great, we are lucky AD = (4/5)*AB = 32 -> DC = 50-32 = 18

Then ratio of areas is equal to AD/DC = 32/18 = 16/9

Awesome man! However, how can we be sure if the ratio is 16/9 or 9/16?

gmatclubot

Re: In the diagram to the right, triangle ABC has a right angle
[#permalink]
08 Aug 2014, 06:43

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