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 350,000 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:

In the link below, I found explanation showing that a triangle formed by connecting midpoints of the triangle divides the area of this bigger triangle into 4 equal areas: http://mathworld.wolfram.com/MedialTriangle.html

M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that \(S_{MON}=\frac14*S_{ABC}\)

1. Let's consider vertex A: M and O are midpoints of AB and AC. In other words, all linear sizes of MAO triangle is smaller by 2 times than all linear sizes of BAC. Therefore,\(S_{MAO}=\frac14*S_{ABC}\)

2. Applying the same reasoning for each vertex we will get: \(S_{MON}=S_{ABC} - (S_{MAO}+S_{MBN}+S_{NCO}) = S_{ABC} - (\frac14*S_{ABC}+\frac14*S_{ABC}+\frac14*S_{ABC}) =\frac14*S_{ABC}\) _________________

M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that \(S_{MON}=\frac14*S_{ABC}\)

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral. AB, BC, and OC can each be a different length, as can MN, ON, OM.

M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that \(S_{MON}=\frac14*S_{ABC}\)

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral. AB, BC, and OC can each be a different length, as can MN, ON, OM.

Mid point therom says that a triangle made from connecting mid-points of the sides of a given triangle is 1/4 of the original triangle. In that case, it is not required to be equilateral. _________________

for any triangle, ABC if you join mid points and divide it into 4 parts, the area will be divided into 4 triangles with each having an area equal to 1/4 of triangle ABC.

Lets prove is with simple method, visualize the triangle ABC on co-ordinate plane

A = (0,0) B = (x,0) C = (a,b)

Area = xb/2 (Note that area does not depend on a)

M = (x/2, 0) N = ((x+a)/2, b/2) O = (a/2, b/2)

base of triangle, NO = x/2 Height = b/2 Area of MNO = xb/8

Similarly triangle AMO base AM = x/2 height = b/2 Area = xb/8

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral. AB, BC, and OC can each be a different length, as can MN, ON, OM.

This problem test similarity. ABC and MAO (as other small triangles) are similar triangles: the same angle A and the same relation between AB/AC=AM/AO. _________________

MN, NO and OM each are midsegments of triangle ABC (midsegment is a line segment joining the midpoints of two sides of a triangle). Important property of a midsegment: the midsegment is always half the length of the third side. So, \(MN=\frac{AC}{2}\), \(NO=\frac{AB}{2}\) and \(OM=\frac{BC}{2}\)

Next, since each side of triangle MNO is half of the side of triangle ABC then these triangles are similar (the ratio of all the sides are the same). Important property of similar triangles: if two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).

Since the sides of two similar triangles MNO and ABC are in the ratio 1:2 then then their areas are in the ratio 1:4 --> (area of MNO)=(area of ABC)/4.

So, in order to find the area of MNO we should find the area of ABC.

(1) The area of ABC is \(\frac{\sqrt{3}}{4}\). Sufficient. (2) ABC is an equilateral triangle with height \(\frac{sqrt3}{2}\) --> we can find the area of equilateral triangle with given altitude. Sufficient.

Re: DS Geometry (m08q22) [#permalink]
28 Aug 2012, 09:52

Based on the prompt, We know that the each side of triangle MON is going to be exactly HALF the length of the corresponding sides of triangle ABC. Which means that these two are SIMILAR triangles, and we will be able to figure out the area of MON if we are given the area of ABC. based on the ratio s^2:s^2 (s=side).

1) SUFFICIENT. 2) if ABC is an equilateral triangle, its height is the same from any base, and also cuts the triangle ABC in half to make it 2 right triangles with sides ratio of x:x^(1/2):2x. Knowing the side of the longest leg, we are able to calculate the rest of the sides and hence the area of ABC. SUFFICIENT.