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:

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe [#permalink]

Show Tags

04 Jun 2015, 06:24

I found a nice way; since the lines are AC and DB are perpendicular, this means that the triangles are proportionate to each other and that they are right triangles. Since we know sides AM and DM we can use pythagorem theorem to solve for AD

1.2 = 6/5 1.8 = 8/5

6/5*6/5 + 8/5*8/5 = 4 which is a perfect square and solves for 2 So AD is 2, we know that CB is proportionate to AD. CB is 1 1/2 x more than AD So CM must be 1 1/2 times more than DM. 6/5 * 3/2 = 18/10 = 1.8 Therefore answer C

Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe [#permalink]

Show Tags

14 Dec 2016, 06:43

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.
_________________

If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe [#permalink]

Show Tags

22 Dec 2016, 17:45

Could some confirm this for me, I couldn't make out the theory that GMATInsight was using in the blue ink, so I don't want to make this assumption if it only holds true here:

If two chords intersect in a circle, are the "vertical" arcs proportional?

I.e. for this example, given the chords AC and DB, are arcs AD and BC similar/proportionate to arcs CD and AB? I guess my question is: is this a theory, or does this proportional relationship happen to be the case because perhaps these chords are perpendicular (or some other relationship that I'm missing)?

Making this assumption, we could the ratio of the hypotenuses are equal to the ratio of the other angles/sides in a "reflective" manner.

EDIT: Another thought/question: Going off that theory, if these were NOT right triangles and they had provided an area for one of the triangles, along the lengths of two related sides, could we calculate the area of the other triangle? Like in this example, if the area of the smaller triangle were X, and they told us CB=3 and AD = 2, could we use that ratio of 3:2 to calculate the area of the larger triangle? I.e. would the area always equal X*(3/2)^2?

Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe [#permalink]

Show Tags

22 Dec 2016, 21:37

arven wrote:

Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here?

This is because the ratio of sides you took is not proper. You can use the below method. Triangles ADM and BCM are similar. Just take any two points at a time for each of the triangle(A,D from first triangle and corresponding B,C from second triangle and so on..) So, AD/BC = DM/CM = AM/BM

Check , the proportion should be DM/AM = CM/BM. But in your case, you have reversed the LHS to AM/DM. Hope this helps.

Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe [#permalink]

Show Tags

23 Dec 2016, 06:55

shashank1tripathi wrote:

arven wrote:

Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here?

This is because the ratio of sides you took is not proper. You can use the below method. Triangles ADM and BCM are similar. Just take any two points at a time for each of the triangle(A,D from first triangle and corresponding B,C from second triangle and so on..) So, AD/BC = DM/CM = AM/BM

Check , the proportion should be DM/AM = CM/BM. But in your case, you have reversed the LHS to AM/DM. Hope this helps.

I thought I just have to take Base1/Base2=height1/height2

So as I understand do I have to take the sides that are perpendicular to each other?
_________________

Re: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpe [#permalink]

Show Tags

23 Dec 2016, 09:34

arven wrote:

shashank1tripathi wrote:

arven wrote:

Ι took the two similar triangles ADM and BCM: AM/DM=CM/BM. Why am I wrong here?

This is because the ratio of sides you took is not proper. You can use the below method. Triangles ADM and BCM are similar. Just take any two points at a time for each of the triangle(A,D from first triangle and corresponding B,C from second triangle and so on..) So, AD/BC = DM/CM = AM/BM

Check , the proportion should be DM/AM = CM/BM. But in your case, you have reversed the LHS to AM/DM. Hope this helps.

I thought I just have to take Base1/Base2=height1/height2

So as I understand do I have to take the sides that are perpendicular to each other?

I found out the theory in use: When two chords intersect, the product of one chords parts = the product of the other chord's parts. So AM*MC=DM*MB; Therefore, AM/DM = MB/MC; which basically says 2 sides of these triangles are similar. That, along with the similar vertical angle of 90 tells us the triangles are similar. If AM and DM are two parts of a 3-4-5, then we know MB and MC have to be parts of a 3-4-5. The previous ratio shows which is the 3 and which is the 4 (Numerators are "4s")

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...