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

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

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03 Jul 2008, 16:12
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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.
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Triangle_Jumble.doc [26 KiB]

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Director
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Re: Can you solve it in 2 Minutes [#permalink]

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03 Jul 2008, 17:03
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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

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Director
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Re: Can you solve it in 2 Minutes [#permalink]

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03 Jul 2008, 17:28
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

That was rocking +1 to you.

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Re: Can you solve it in 2 Minutes [#permalink]

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03 Jul 2008, 18:04
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

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Re: Can you solve it in 2 Minutes [#permalink]

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03 Jul 2008, 18:06
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?
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Director
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Re: Can you solve it in 2 Minutes [#permalink]

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03 Jul 2008, 18:30
jallenmorris wrote:
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?

one of them is 90 degrees, another one asin(3/5), the last one 90-asin(3/5) //nasty angles

c>b>a, the question is asking about b^2/a^2

a+b+c = 120
ab=24*c
a^2+b^2 = c^2

you can solve equations ...

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 03:27
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

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.

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 04:22
the perimeter of the triangle is an integer. So, It is reasonable to assume that the lengths of all sides are also integers...
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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 05:05
walker wrote:
the perimeter of the triangle is an integer. So, It is reasonable to assume that the lengths of all sides are also integers...

Maybe (this is not necessarily the case but why not?). But it doesn't give you more information about the ratio 3:4:5.

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 06:58
Oski wrote:
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.

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 08:10
Someone has done some work on http://www.physicsforums.com/archive/in ... 60382.html and mentions that the mesure of the smallest angle in a 3-4-5 triangle is 36.87 deg

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 08:31
hi0parag wrote:
Someone has done some work on http://www.physicsforums.com/archive/in ... 60382.html and mentions that the mesure of the smallest angle in a 3-4-5 triangle is 36.87 deg

Just have to find which angle X satisfies cos(X)=4/5

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 12:35
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?

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 20:31
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

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

I think the final answer is (16/9)^2.

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Re: Can you solve it in 2 Minutes [#permalink]

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04 Jul 2008, 20:33
Other sequence for Right angled triangles:

5:12:13
8:15:17

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Re: Can you solve it in 2 Minutes [#permalink]

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05 Jul 2008, 21:22
I could nt understand the axiom/rule behind.
BD = (3/5) AB
OR BD = (4/5) BC
then the calculation of AD on the same axiom...
pls some one explain??

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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, 07:43
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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, 07:43
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?

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

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25 Aug 2017, 19:22
Thanks a lot for your useful information

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Re: In the diagram to the right, triangle ABC has a right angle   [#permalink] 25 Aug 2017, 19:22
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