GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Feb 2019, 10:21

### GMAT Club Daily Prep

#### 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

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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Valentine's day SALE is on! 25% off.

February 18, 2019

February 18, 2019

10:00 PM PST

11:00 PM PST

We don’t care what your relationship status this year - we love you just the way you are. AND we want you to crush the GMAT!

# Given a triangle ABC, where G is the intersection of point of medians

Author Message
TAGS:

### Hide Tags

Intern
Joined: 16 Apr 2017
Posts: 9
Location: Armenia
GMAT 1: 740 Q51 V39
GPA: 3.97
Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

15 Jan 2019, 03:57
1
1
5
00:00

Difficulty:

85% (hard)

Question Stats:

38% (02:24) correct 62% (02:08) wrong based on 59 sessions

### HideShow timer Statistics

Given a triangle ABC, where G is the intersection of point of medians, what fraction of the triangle ABC is shaded?

A. $$\frac{1}{4}$$

B. $$\frac{1}{12}$$

C. $$\frac{1}{6}$$

D. $$\frac{1}{8}$$

E. $$\frac{1}{9}$$

Attachments

Triangle2.png [ 23.55 KiB | Viewed 776 times ]

Intern
Joined: 10 May 2018
Posts: 7
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

15 Jan 2019, 08:07
Intern
Joined: 27 Nov 2018
Posts: 20
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

15 Jan 2019, 15:05
1
1
From the question, we can deduct that ADE &ABC are similar triangles, so DE=1/2BC, DE//BC

Since DE//BC, DGE&ABC are similar triangles
Since DE=1/2BC, then height of the two triangles (from G) also is 1:2, which makes the height of the trapezoid DECB 3
Since height of triangle ADE=1/2ABC, which makes height of ADE 3 as well.
So, the height of ABC=3+3=6

Since 2DE=BC, then, Area of ABC = (6x2)times of DGE.

Intern
Joined: 16 Jan 2019
Posts: 1
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

16 Jan 2019, 08:41
1
2asc wrote:
From the question, we can deduct that ADE &ABC are similar triangles, so DE=1/2BC, DE//BC

Since DE//BC, DGE&ABC are similar triangles
Since DE=1/2BC, then height of the two triangles (from G) also is 1:2, which makes the height of the trapezoid DECB 3
Since height of triangle ADE=1/2ABC, which makes height of ADE 3 as well.
So, the height of ABC=3+3=6

Since 2DE=BC, then, Area of ABC = (6x2)times of DGE.

2asc I dont understand how you get the height from G to base.
Intern
Joined: 10 May 2018
Posts: 7
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

16 Jan 2019, 10:29
1
I have solved it like this .
We can deduce that ADE and ABC are similar triangles.(SAS property) AD/AB= 1/2 ,AE/Ac=1/2 and angle A is common .
Therefore DE/BC=1/2.
Also BC !! DE (D and E are mid points so angle EDG = angle GCB and angle DEG = angle GBC so the third angles are also same and hence both the triangles DGE are GBC are similar .For a centroid the areas of all the six trinagles are equal so area of BGC will be (1/6 of area of ABC + 1/6 of area of ABC i.e 2/6 of area of ABC .
Also as DE/BC=1/2 so the area of DBC/Area of BGC = 1/4

So the area of DBC = 1/4 area pf BGC
= 1/4*2/6 area of ABC
==> Area of DBC/Area of ABC = 1/12
Intern
Joined: 27 Nov 2018
Posts: 20
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

16 Jan 2019, 11:37
1
1
rezalotif
Because DE//BC, it makes DGE & BGC similar triangles.
From earlier we established that ADE and ABC are similar triangles, where the sides of ABC is twice of ADE, which means DE=1/2BC
Now look at DE & BC in terms of triangle DGE and BGC. We established DGE&BGC are similar triangles and DE=1/2BC, then the height of the two triangles should reflect the same ratio.
Suppose height from G to DE is a, and G to BC is b, then DE/BC=a/b=1/2.
Suppose a is 1, then b is 2, a+b=3=half of height of triangle ABC
So height of ABC is 3x2=6 times of height of DGE

Hope this helps!
CEO
Joined: 11 Sep 2015
Posts: 3431
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

16 Jan 2019, 14:46
Top Contributor
Yuridias wrote:
Given a triangle ABC, where G is the intersection of point of medians, what fraction of the triangle ABC is shaded?

A. $$\frac{1}{4}$$

B. $$\frac{1}{12}$$

C. $$\frac{1}{6}$$

D. $$\frac{1}{8}$$

E. $$\frac{1}{9}$$

Since the question does not mention anything about the kind of triangle we have (isosceles, scalene, etc), we can assume that we can use ANY triangle.
So, let's use a nice equilateral triangle with sides of length 2

Area of an equilateral triangle = (√3)(side²/4)
So, area = (√3)(2²/4)
= (√3)(4/4)
= √3

Now let's draw our medians

We can also add some angles

We need to find the area of the RED triangle in the middle

So, let's flip that red triangle like so...

If we draw an altitude from the top vertex, the line divides the triangle into two 30-60-90 special right triangles.

When we compare the RED 30-60-90 triangles with the base 30-60-90 triangle we can calculate the height (h)
Since we're comparing SIMILAR triangles, we can write: 0.5/√3 = h/1
Cross multiply to get: (0.5)(1) = (√3)(h)
Simplify: 1/2 = (√3)(h)
Solve: h = 1/(2√3)

We have the following

Area of triangle = (base)(height)/2
= (1)[1/(2√3)]/2
= 1/(4√3)

What fraction of the triangle ABC is shaded?
Fraction = 1/(4√3)/√3
= [1/(4√3)](1/√3)
= 1/12

Cheers,
Brent
_________________

Test confidently with gmatprepnow.com

CEO
Joined: 11 Sep 2015
Posts: 3431
Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

16 Jan 2019, 14:49
Top Contributor
Yuridias wrote:
Given a triangle ABC, where G is the intersection of point of medians, what fraction of the triangle ABC is shaded?

A. $$\frac{1}{4}$$

B. $$\frac{1}{12}$$

C. $$\frac{1}{6}$$

D. $$\frac{1}{8}$$

E. $$\frac{1}{9}$$

IMPORTANT: the diagrams in GMAT Problem Solving questions (like the one above) are DRAWN TO SCALE unless stated otherwise.
So, even if we have no idea where to begin, we can certainly eliminate 2 or 3 answers by simply "eyeballing" the diagram.

RELATED VIDEO FROM OUR COURSE

_________________

Test confidently with gmatprepnow.com

Intern
Joined: 18 Oct 2018
Posts: 1
Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

### Show Tags

22 Jan 2019, 11:08
I don't understand why we can just call it an equilateral triangle.

EDIT:

To anyone else confused. The fact the that the question is only asking what fraction of the triangle is shaded means we can pic any triangle we want because the proportions created by this scenario remain the same for any triangle. Choosing an equilateral triangle then just makes the math much simpler.
Given a triangle ABC, where G is the intersection of point of medians   [#permalink] 22 Jan 2019, 11:08
Display posts from previous: Sort by