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:

If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached) A) 12 B) 18 C) 24 D) 30 E) 48

Attachment:

Untitled.jpg

I guess it's BE||CD (BE is parallel to CD).

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Bunuel - In this case it turned out to be a rt angled triangle so the area calc was easy.. Lets say the bigger triangle were 5 7 9 in sides and the smaller was correspondingly half in size inside of it.. In that case, what would be the easiest way to calculate the trapezium area.. I guess does GMAT require us to remember the formula of the area of a triangle (not the .5 base height) but the one with the sides? Is there an easy way to calculate the area where the sides are given for any triangle? _________________

Bunuel - In this case it turned out to be a rt angled triangle so the area calc was easy.. Lets say the bigger triangle were 5 7 9 in sides and the smaller was correspondingly half in size inside of it.. In that case, what would be the easiest way to calculate the trapezium area.. I guess does GMAT require us to remember the formula of the area of a triangle (not the .5 base height) but the one with the sides? Is there an easy way to calculate the area where the sides are given for any triangle?

If it were not right triangle we could calculate the area of triangles with Heron's Formula (as we still would knew the lengths of all sides), though I've never seen a GMAT question requiring it. _________________

If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached) A) 12 B) 18 C) 24 D) 30 E) 48

Attachment:

Untitled.jpg

I guess it's BE||CD (BE is parallel to CD).

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Hope it's clear.

Bunuel... If you say that BC = AB = 3, AE = 4 then how can BE||CD (BE is parallel to CD) true. I am not able to visualize such a figure!!!

If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached) A) 12 B) 18 C) 24 D) 30 E) 48

Attachment:

Untitled.jpg

I guess it's BE||CD (BE is parallel to CD).

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Hope it's clear.

Bunuel... If you say that BC = AB = 3, AE = 4 then how can BE||CD (BE is parallel to CD) true. I am not able to visualize such a figure!!!

Why not? Consider triangle CAD, where CA=6 and AD=8. Now draw midsegment BE of a triangle (a line segment joining the midpoints of two sides of a triangle). Properties of a midsegment:

• The midsegment is always parallel to the third side of the triangle. • The midsegment is always half the length of the third side.

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...