Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

What's important to recognize about this diagram is the number of similar triangles.

When the altitude (i.e. a perpendicular from a vertex) is drawn to the hypotenuse of a right triangle, as it is here, you always get three similar triangles.

Here, altitude BD is drawn to hypotenuse AC, creating similar triangles: Triangle ABC ~ Triangle ADB ~ Triangle BDC. Again, you will always get three similar triangles right away, when you draw an altitude to a hypotenuse.

In addition, the diagram includes segment DE, which creates two more similar triangles:

Triangle ABC ~ Triangle ADB ~ Triangle BDC ~ Triangle BED ~ Triangle DEC

Since the sides of similar triangles are proportional, that implies a whole boatload proportions. So . . .

Statement #1: BE = 3

Using Pythagorean Theorem in Triangle BED, we get DE = 4. That means all five of the similar triangles are 3-4-5 triangles.

AB/BD = BD/DE

AB/5 = 5/4, and we can solve for AB. Statement #1 is sufficient.

Statement #2: DE = 4

Here, we can jump directly to AB/BD = BD/DE ---> AB/5 = 5/4 ---> solve for AB. Statement #2 is sufficient.

Answer = D.

Here's another practice question involving similar triangles:

The question at that link should be followed by a video explanation of the solution. If you are rusty on geometry, you might want to check out Magoosh ---- we have a complete video math curriculum for the GMAT, including everything you will need to know for Geometry.

Does this explanation to the question make sense? Please let me know if you have any further questions.

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4.

The same works for triangle BDC with perpendicular DE.

We know that BD=5 thus one more side and we'll be able to find any other line segment in the diagram.

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

05 Feb 2012, 17:59

Hi Bunuel - will it possible for you to explain how you make the sides proportion and which sides to make proportionally equal? Is it base to base and height to height ? I hope I am making myself clear. Also, any idea how to solve the question in the link that Mike has provided?
_________________

Hi Bunuel - will it possible for you to explain how you make the sides proportion and which sides to make proportionally equal? Is it base to base and height to height ? I hope I am making myself clear. Also, any idea how to solve the question in the link that Mike has provided?

Because both Triangles ABE & ACD share (1) a right angle, and (2) angle A, we know they are similar.

Because B is the midpoint of AC, we know 2*(AB) = AC, and thus, the ratio of the two similar triangles is 1:2. In other words, the ratio of any two corresponding sides is equal to 2.

CD/BE = 2, and if we know the length of BE, we can solve for the length of CD. Did you submit your answer and watch the video explanation to the question?

Does what I explain here make sense? Please let me know if I can answer anything else.

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

02 Sep 2012, 15:19

Hey Bunuel,

When you extend a line from the vertex of a right angle to the hypotunse, each of the new two triangles are similar to the larger one. But are the two smaller triangles similar to one another?

When you extend a line from the vertex of a right angle to the hypotunse, each of the new two triangles are similar to the larger one. But are the two smaller triangles similar to one another?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Generally if two triangles are similar to the third one, then they are naturally similar to each other.
_________________

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

03 Sep 2012, 03:42

enigma123 wrote:

In the diagram, what is the length of AB?

(1) BE = 3 (2) DE = 4

Struggling badly to solve this. Please help.

The key ingredient needed to solve this question is triangle similarity. Two similar triangles have all the angles congruent. Since the sum of the angles in a triangle is always 180, knowing that two pairs of angles in the two triangles are congruent is enough to ensure similarity. To prove that the height in the right triangle ABC determines two similar triangles, look for the congruent angles.

In the given question, even if you don't know the property that the height to the hypotenuse divides the triangle into two similar triangles (and each of them is similar to the initial right triangle), you can still prove that triangle ABD is similar to triangle BDE: Both are right angled triangles. In addition, because AB is perpendicular to BC and so is DE, AB is parallel to DE, so angle ABD is congruent to angle BDE. Two pairs of congruent angles means the two triangles are similar.

Knowing all the sides in the triangle BDE, then writing the proportions and because BD belongs to both triangles, you can find AB.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

03 Sep 2012, 09:46

Bunuel wrote:

alphabeta1234 wrote:

Hey Bunuel,

When you extend a line from the vertex of a right angle to the hypotunse, each of the new two triangles are similar to the larger one. But are the two smaller triangles similar to one another?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Generally if two triangles are similar to the third one, then they are naturally similar to each other.

Hey Bunuel,

Thank you very much for your response. I guess then as corrolary to the first question. Since all the angles are the same in all three triangles (let say angles x*,y*, and z*, where z*=90), Is there an easier way to identify which angle is referring to which side in the triangle within the equation you stated, AB/AC=AD/AB=BD/BC? Or are you mentally rotating the triangles making sure angle y* is opposite to AB, angle x* is opposite to AC, and checking it with angle y* opposite to AD, and angle x* is opposite to AB. Are you mentally rotating and checking that the all the sides are in conjunction with (opposite to x*)/(opposite to y*). Or is there an easier, more systematic method your using to ensure that all those sides are in fact in the same proportion.

On another side note, unrelated to my question above: I also noticed that in your equation that each ratio is of sides of the same triangle equaling the ratio of the other triangle, opposite degree sides. So (Side 1, Triangle A)/(Side 2, Triangle A)=(Side 1 Triangle B)/(Side 2, Triangle B). I always imagined that if Triangle A (sides X,Y,Z) and Triangle B(sides M N P), where triangle A is bigger than B, then their similarity defined the following relationship:

X=Mc , where c is a some positive constant>1 and sides X and M are opposite to the same angle Y=Nc Z=Pc

Then the relationship would be defined as c=X/M=Y/N=Z/P= (side 1 Triangle A)/(side 1 Triangle B)=(side 2 Triangle A)/(side 2 Triangle B)=(side 3 Triangle A)/(side 3 Triangle B) I just noticed that this relationship can always be manipulated to get c=X/M=Y/N ==> new constant=M*c/Y=X/Y=M/N=(Side 1 Triangle A)/(Side 2 Triangle A)=(Side 1 Triangle B)/(Side 2 Triangle B), which i think is very interesting. Same sides of the triangle are also in proportion to the same sides of the other triangle and another new constant the related the two triangles is embedded within the relationship of the triangles own sides!!! (eek, hope that made sense)

I also learned from you that if you had two similar triangles then (Area Triangle A)/(Area Triangle B)=(Side 1 Triangle A)^2/(Side 2 Triangle B)^2.

Are there any other key relationships that are important about similar triangles?

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

26 Apr 2014, 18:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

24 May 2014, 13:02

Bunuel wrote:

enigma123 wrote:

In the diagram, what is the length of AB?

(1) BE = 3 (2) DE = 4

Struggling badly to solve this. Please help.

Attachment:

Triangle2.GIF

In the diagram, what is the length of AB?

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4.

The same works for triangle BDC with perpendicular DE.

We know that BD=5 thus one more side and we'll be able to find any other line segment in the diagram.

(1) BE = 3. Sufficient. (2) DE = 4. Sufficient.

Answer: D.

Hi Bunuel,

I understand that a line perpendicular to a hyp will always create TWO similar triangles. That being said, line BD creates ADB and BDC. Correct?

Now, we have another segment DE, since it's perpendicular to BC, it will create BED and DEC. Correct?

Are BED and DEC similar to ADB and BDC or are the two groups standalone? The reason why I ask is because they are both segmented off another hypotenuse?

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4.

The same works for triangle BDC with perpendicular DE.

We know that BD=5 thus one more side and we'll be able to find any other line segment in the diagram.

(1) BE = 3. Sufficient. (2) DE = 4. Sufficient.

Answer: D.

Hi Bunuel,

I understand that a line perpendicular to a hyp will always create TWO similar triangles. That being said, line BD creates ADB and BDC. Correct?

Now, we have another segment DE, since it's perpendicular to BC, it will create BED and DEC. Correct?

Are BED and DEC similar to ADB and BDC or are the two groups standalone? The reason why I ask is because they are both segmented off another hypotenuse?

Angles in ABC are identical to those in ABD and BDC. Angles in BDC are identical to those in BDE and DEC.

Obviously, angles in all those triangles are identical.
_________________

Re: In the diagram, what is the length of AB? [#permalink]

Show Tags

09 Apr 2015, 13:30

Hi Bunuel,

Can we do this by using isosceles triangle property for triangle ABD in which angle ABD is 45 ( BD is dropping a perpendicular ) so angle A is 45 hence AD is also 5 and we can calculate AB via Pythagoras and same can be done for remaining angles as well.

Can we do this by using isosceles triangle property for triangle ABD in which angle ABD is 45 ( BD is dropping a perpendicular ) so angle A is 45 hence AD is also 5 and we can calculate AB via Pythagoras and same can be done for remaining angles as well.

BD is not a bisector of angle B, so ABD is NOT 45 degrees. BD would be a bisector if AB were equal to BC in this case the perpendicular from B will coincide with the bisector from B.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...