MGMAT 25 question bank: Geometry

[GMAT 5/18]

This looks to be a very basic and I have a very basic question.

In triangle ABC to the right, if BC = 3 and AC = 4, then what is the length of segment CD?

A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3



In the answer explanations, it states the following:

To solve for the length of side CD, we can set up a proportion, based on the relationship between the similar triangles ACD and BCA:

BC/AC = CA/CD

3/4 = 4/CD

And you can solve for CD.

My question is, why was the right side of the formula CA/CD and not CD/CA? When I did the problem I used CD/CA and therefore derived the answer of 3, which according to the answer explanation, is incorrect.

Please help, thank you!

[JadranLee]

Let's think about how we know that triangles ACD and BCA are similar.

1) Let's say that <CDA is x degrees, and <DAC is y degrees. Since <ACD is 90 degrees, and the sum of all the interior angles in a triangle is 180, we know that x + y =90.

2) Now let's look at <BAC. We know that <BAC + <DAC = 90, since <BAD is labeled as a right angle. We also know that <DAC is y degrees (from step 1), and that x + y=90. Putting these facts together, we know that <BAC is x degrees.

3) We know <ACB is a right angle, since <ACD is a right angle. Since <ACB is a right angle, <BAC + <CBA = 90. Given that <BAC is x degrees, <CBA must be y degrees.

4) To summarize, <CAB has the same measure as <CDA (x degrees) , and <CBA has the same measure as <DAC (y degrees). This means that in similar triangles CAB and CAD, side BC of CAB corresponds to side CA of CAD, and side CA of CAB corresponds to side CD of CAD. Thus, as the quoted explanation says, "BC/AC = CA/CD".

Perhaps the explanation on our website would have been a bit clearer if we had labeled sides consistently - AC is just the same side as CA.

[GMAT 5/18]

Jad,

Thank you! That was a fantastic explanation! Much clearer than the MGMAT explanation (and not because of the AC/CA thing you mentioned). Your step-by-step method illustrated a great way to process these types of questions - at least I think so.

Thanks again!

[Saurabh Malpani]

Great explanation and a point to remeber in the exams.

I would like to suggest another apporach here...because I know the simmilar triagnles are always tricky I try to avoid using the rule. I just simple pythagoras principle to derive the answer.

AB^2 + AD^2 = BD^2------eq 1

AC^2 + CD^2 =AD^2----eq 2

BD= BC+CD----Eq 3

Substitue for AD from eq 2 to equation 1 , and BD from Eq 3 to eq 1 we get

AB^2 + AC^2 + CD^2 = (BC+CD)^2

AB^2 + AC^2 + CD^2 = BC^2 +CD^2 + 2 BC*CD---------eq 4

Now we know AB=5, AC=4, BC=3

we can solve eq 4.

Whenever I think of Right angle triagnle firt thing strikes me is Paythagoras

Saurabh Malpani

Let's think about how we know that triangles ACD and BCA are similar.

1) Let's say that <CDA is x degrees, and <DAC is y degrees. Since <ACD is 90 degrees, and the sum of all the interior angles in a triangle is 180, we know that x + y =90.

2) Now let's look at <BAC. We know that <BAC + <DAC = 90, since <BAD is labeled as a right angle. We also know that <DAC is y degrees (from step 1), and that x + y=90. Putting these facts together, we know that <BAC is x degrees.

3) We know <ACB is a right angle, since <ACD is a right angle. Since <ACB is a right angle, <BAC + <CBA = 90. Given that <BAC is x degrees, <CBA must be y degrees.

4) To summarize, <CAB has the same measure as <CDA (x degrees) , and <CBA has the same measure as <DAC (y degrees). This means that in similar triangles CAB and CAD, side BC of CAB corresponds to side CA of CAD, and side CA of CAB corresponds to side CD of CAD. Thus, as the quoted explanation says, "BC/AC = CA/CD".

Perhaps the explanation on our website would have been a bit clearer if we had labeled sides consistently - AC is just the same side as CA.

[Carla]

Thanks!

[Guest]

Jad,

Thank you! That was a fantastic explanation! Much clearer than the MGMAT explanation (and not because of the AC/CA thing you mentioned). Your step-by-step method illustrated a great way to process these types of questions - at least I think so.

Thanks again! :)

Wasnt that the same exact explanation that was on the website for the this problem? Lol

So I am still a bit lost from the explanation. When I first encountered this problem, I noticed that the 90 deg angle for <DCA/<BCA and the altitude that is dropped from AC. Now due to the altitude, wouldnt <CAD be 45 deg? ( 90 deg <BAD divided by 2) Thus we are left with <CDA = 45 deg also (180 - 90 - 45 = 45) - So triangle CAD is a 45-45-90 triangle. And with the same logic, triangle BCA would also be a 45-45-90 triangle.

Thus the making the triangles exactly the same making CD = 3. Can someone help me by explaining what I did wrong? and what I should change in my logic to correct the mistake? Thanks!

[StaceyKoprince]

Now due to the altitude, wouldnt <CAD be 45 deg?

Here's your mistake - that's only true if the altitude that you dropped cut the large triangle equally into two identical smaller triangles (which is what would happen if, say, AB and AD were the same length - but we weren't told that they were). Try drawing a couple of different triangles to show yourself that dropping an altitude from a 90-degree angle does not automatically chop that 90-degree angle in half - exaggerate the different in length of the two legs of the triangle you draw so that it's easy to see what I'm describing.

[MattD517]

Hello - I have read the above explanations a few times and I still do not understand how one associates ("matches up") the similar sides and angles from each similar triangle. In particular, the key seems to be in JadranLee's step 2 where we establish that <CDA = <BAC. I don't understand how we know that with the given information.

Drawing it on paper and closer to scale with AC shifted to the left (and therefore <BAC is less than <CAD), I can see how it makes sense. However, I still don't get how one would be able to recognize that the length of BC < CD and therefore be able make that assumption given the picture (which I understand is not to scale) and information provided. Please help, thanks.
-Matt

[RonPurewal]

The point is to use corresponding parts in the proportions.

If you're having trouble because these triangles are meaningless abstractions, just think about concrete example for a second.
A model car has the same shape as the real car; it's just scaled down.
Each part of the model car "corresponds" to a part on the real car. (The bumper on the model "corresponds" to the real car's bumper, and so on.)

You write the proportions in a way that accounts for this correspondence.

So, in the case of the model car, you can write, e.g.,
(length of model bumper)/(length of real bumper) = (length of model wheelbase)/(length of real wheelbase).

[RonPurewal]

The other piece of the puzzle is to figure out the actual correspondence here.

In this problem, the diagram is annoying, because all the triangles look like 45º-45º-90º triangles"”"”making it impossible to determine at a glance which parts correspond to which other parts.
So, re-draw the diagram in way that makes the correspondence obvious.
Like this:
http://geometry.freehomeworkmathhelp.co ... gles_9.gif

In that picture, it's easy to see that CD is the "short" leg of the left-hand triangle, and that AD is the corresponding "short" leg of the right-hand triangle. (These sides are labeled BC and AC, respectively, in the problem at hand.)
Also, in that picture, AD is the "long" leg of the left-hand triangle, and BD is its counterpart in the right-hand triangle. (These sides are AC and DC, respectively, in the current problem.)

[MattD517]

Thank you Ron.

[tim]

Let us know if there are any further questions on this one.

[RonPurewal]

[mistakenly posted in wrong thread -- edited]