Math: Triangles : GMAT Quantitative Section
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 03:15

### 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 June
Open Detailed Calendar

# Math: Triangles

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36586
Followers: 7088

Kudos [?]: 93304 [122] , given: 10557

### Show Tags

22 Nov 2009, 22:51
122
KUDOS
Expert's post
144
This post was
BOOKMARKED
TRIANGLES

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, walker

--------------------------------------------------------
Get The Official GMAT Club's App - GMAT TOOLKIT 2.
The only app you need to get 700+ score!

[iOS App] [Android App]

--------------------------------------------------------

Triangle A closed figure consisting of three line segments linked end-to-end. A 3-sided polygon.

Vertex The vertex (plural: vertices) is a corner of the triangle. Every triangle has three vertices.

Base The base of a triangle can be any one of the three sides, usually the one drawn at the bottom.

• You can pick any side you like to be the base.
• Commonly used as a reference side for calculating the area of the triangle.
• In an isosceles triangle, the base is usually taken to be the unequal side.

Altitude The altitude of a triangle is the perpendicular from the base to the opposite vertex. (The base may need to be extended).

• Since there are three possible bases, there are also three possible altitudes.
• The three altitudes intersect at a single point, called the orthocenter of the triangle.

Median The median of a triangle is a line from a vertex to the midpoint of the opposite side.

The three medians intersect at a single point, called the centroid of the triangle.
• Each median divides the triangle into two smaller triangles which have the same area.
• Because there are three vertices, there are of course three possible medians.
• No matter what shape the triangle, all three always intersect at a single point. This point is called the centroid of the triangle.
• The three medians divide the triangle into six smaller triangles of equal area.
• The centroid (point where they meet) is the center of gravity of the triangle
Two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.
• $$m=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$, where $$a$$, $$b$$ and $$c$$ are the sides of the triangle and $$a$$ is the side of the triangle whose midpoint is the extreme point of median $$m$$.

Area The number of square units it takes to exactly fill the interior of a triangle.

Usually called "half of base times height", the area of a triangle is given by the formula below.
• $$A=\frac{hb}{2}$$

Other formula:
• $$A=\frac{P*r}{2}$$

• $$A=\frac{abc}{4R}$$

Where $$b$$ is the length of the base, $$a$$ and $$c$$ the other sides; $$h$$ is the length of the corresponding altitude; $$R$$ is the Radius of circumscribed circle; $$r$$ is the radius of inscribed circle; P is the perimeter

• Heron's or Hero's Formula for calculating the area $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $$a,b,c$$ are the three sides of the triangle and $$s = \frac{a+b+c}{2}$$ which is the semi perimeter of the triangle.

Perimeter The distance around the triangle. The sum of its sides.

• For a given perimeter equilateral triangle has the largest area.
• For a given area equilateral triangle has the smallest perimeter.

Relationship of the Sides of a Triangle

• The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.

Interior angles The three angles on the inside of the triangle at each vertex.

• The interior angles of a triangle always add up to 180°
• Because the interior angles always add to 180°, every angle must be less than 180°
• The bisectors of the three interior angles meet at a point, called the incenter, which is the center of the incircle of the triangle.

Exterior angles The angle between a side of a triangle and the extension of an adjacent side.

• An exterior angle of a triangle is equal to the sum of the opposite interior angles.
• If the equivalent angle is taken at each vertex, the exterior angles always add to 360° In fact, this is true for any convex polygon, not just triangles.

Midsegment of a Triangle A line segment joining the midpoints of two sides of a triangle

• A triangle has 3 possible midsegments.
The midsegment is always parallel to the third side of the triangle.
• The midsegment is always half the length of the third side.

• A triangle has three possible midsegments, depending on which pair of sides is initially joined.

Relationship of sides to interior angles in a triangle

• The shortest side is always opposite the smallest interior angle
• The longest side is always opposite the largest interior angle

Angle bisector An angle bisector divides the angle into two angles with equal measures.

• An angle only has one bisector.
• Each point of an angle bisector is equidistant from the sides of the angle.
The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC: $$\frac{BD}{DC}=\frac{AB}{AC}$$
The incenter is the point where the angle bisectors intersect. The incenter is also the center of the triangle's incircle - the largest circle that will fit inside the triangle.

Similar Triangles Triangles in which the three angles are identical.

• It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
• In similar triangles, the sides of the triangles are in some proportion to one another. For example, a triangle with lengths 3, 4, and 5 has the same angle measures as a triangle with lengths 6, 8, and 10. The two triangles are similar, and all of the sides of the larger triangle are twice the size of the corresponding legs on the smaller triangle.
• If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$

Congruence of triangles Two triangles are congruent if their corresponding sides are equal in length and their corresponding angles are equal in size.

1. SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

2. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.

3. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.

So, knowing SAS or ASA is sufficient to determine unknown angles or sides.

NOTE IMPORTANT EXCEPTION:
The SSA condition (Side-Side-Angle) which specifies two sides and a non-included angle (also known as ASS, or Angle-Side-Side) does not always prove congruence, even when the equal angles are opposite equal sides.

Specifically, SSA does not prove congruence when the angle is acute and the opposite side is shorter than the known adjacent side but longer than the sine of the angle times the adjacent side. This is the ambiguous case. In all other cases with corresponding equalities, SSA proves congruence.

The SSA condition proves congruence if the angle is obtuse or right. In the case of the right angle (also known as the HL (Hypotenuse-Leg) condition or the RHS (Right-angle-Hypotenuse-Side) condition), we can calculate the third side and fall back on SSS.

To establish congruence, it is also necessary to check that the equal angles are opposite equal sides.

So, knowing two sides and non-included angle is NOT sufficient to calculate unknown side and angles.

Angle-Angle-Angle
AAA (Angle-Angle-Angle) says nothing about the size of the two triangles and hence proves only similarity and not congruence.

So, knowing three angles is NOT sufficient to determine lengths of the sides.

Scalene triangle all sides and angles are different from one another

• All properties mentioned above can be applied to the scalene triangle, if not mentioned the special cases (equilateral, etc)

Equilateral triangle all sides have the same length.

• An equilateral triangle is also a regular polygon with all angles measuring 60°.
• The area is $$A=a^2*\frac{\sqrt{3}}{4}$$

• The perimeter is $$P=3a$$
• The radius of the circumscribed circle is $$R=a*\frac{\sqrt{3}}{3}$$
• The radius of the inscribed circle is
$$r=a*\frac{\sqrt{3}}{6}$$
• And the altitude is $$h=a*\frac{\sqrt{3}}{2}$$ (Where $$a$$ is the length of a side.)
• For any point P within an equilateral triangle, the sum of the perpendiculars to the three sides is equal to the altitude of the triangle.
• For a given perimeter equilateral triangle has the largest area.
• For a given area equilateral triangle has the smallest perimeter.
• With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle.

Isosceles triangle two sides are equal in length.

• An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length.
• For an isosceles triangle with given length of equal sides right triangle (included angle) has the largest area.
• To find the base given the leg and altitude, use the formula: $$B=2\sqrt{L^2-A^2}$$

• To find the leg length given the base and altitude, use the formula: $$L=\sqrt{A^2+(\frac{B}{2})^2}$$

• To find the altitude given the base and leg, use the formula: $$A=\sqrt{L^2-(\frac{B}{2})^2}$$ (Where: L is the length of a leg; A is the altitude; B is the length of the base)

Right triangle A triangle where one of its interior angles is a right angle (90 degrees)

• Hypotenuse: the side opposite the right angle. This will always be the longest side of a right triangle.
• The two sides that are not the hypotenuse. They are the two sides making up the right angle itself.
Theorem by Pythagoras defines the relationship between the three sides of a right triangle: $$a^2 + b^2 = c^2$$, where $$c$$ is the length of the hypotenuse and $$a$$, $$b$$ are the lengths of the the other two sides.
• In a right triangle, the midpoint of the hypotenuse is equidistant from the three polygon vertices
• A right triangle can also be isosceles if the two sides that include the right angle are equal in length (AB and AC in the figure above)
• Right triangle with a given hypotenuse has the largest area when it's an isosceles triangle.
• A right triangle can never be equilateral, since the hypotenuse (the side opposite the right angle) is always longer than the other two sides.

• Any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.
• A Pythagorean triple consists of three positive integers $$a$$, $$b$$, and $$c$$, such that $$a^2 + b^2 = c^2$$. Such a triple is commonly written $$(a, b, c)$$, and a well-known example is $$(3, 4, 5)$$. If $$(a, b, c)$$ is a Pythagorean triple, then so is $$(ka, kb, kc)$$ for any positive integer $$k$$. There are 16 primitive Pythagorean triples with c ≤ 100:
(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

• A right triangle where the angles are 45°, 45°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio $$1 : 1 : \sqrt{2}$$. With the $$\sqrt{2}$$ being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.
• Area of a 45-45-90 triangle. As you see from the figure above, two 45-45-90 triangles together make a square, so the area of one of them is half the area of the square. As a formula $$A=\frac{S^2}{2}$$. Where S is the length of either short side.

Right triangle inscribed in circle:

$$R = \frac{AC}{2}$$

• If M is the midpoint of the hypotenuse, then $$BM = \frac{1}{2}AC$$. One can also say that point B is located on the circle with diameter $$AC$$. Conversely, if B is any point of the circle with diameter $$AC$$ (except A or C themselves) then angle B in triangle ABC is a right angle.
• A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

Circle inscribed in right triangle: $$r = \frac{ab}{a+b+c}=\frac{a+b-c}{2}$$

Note that in picture above the right angle is C.

• Given a right triangle, draw the altitude from the right angle.

Then the triangles ABC, CHB and CHA are similar. Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Official GMAC Books:

The Official Guide, 12th Edition: DT #19; DT #28; PS #48; PS #152; PS #205; PS #209; PS #229; DS #20; DS #56; DS #74; DS #109; DS #140; DS #144; DS #149; DS #157; DS #160; DS #173;
The Official Guide, Quantitative 2th Edition: PS #44; PS #71; PS #85; PS #145; PS #157; PS #162; DS #19; DS #65; DS #88; DS #91; DS #123;
The Official Guide, 11th Edition: DT #19; DT #28; PS #45; PS #152; PS #158; PS #226; PS #248; DS #27; DS #32; DS #51; DS #66; DS #108; DS #113; DS #124; DS #136; DS #152;

Generated from [GMAT ToolKit 2]

--------------------------------------------------------
Get The Official GMAT Club's App - GMAT TOOLKIT 2.
The only app you need to get 700+ score!

[iOS App] [Android App]

--------------------------------------------------------

[Reveal] Spoiler: Images

 Attachment: Math_Tri_1.png [ 4.73 KiB | Viewed 261206 times ] Attachment: Math_Tri_Al.png [ 5.9 KiB | Viewed 261495 times ] Attachment: Math_Tri_Median.png [ 6.66 KiB | Viewed 261425 times ] Attachment: Math_Tri_Exterior.png [ 5.43 KiB | Viewed 260667 times ] Attachment: Math_Tri_Midsegment.png [ 4.74 KiB | Viewed 260855 times ] Attachment: Math_Tri_Bisector.png [ 5.96 KiB | Viewed 260735 times ] Attachment: Math_Tri_Equilateral.png [ 3.46 KiB | Viewed 260790 times ] Attachment: Math_Tri_Isosceles.png [ 3.63 KiB | Viewed 260636 times ] Attachment: Math_Tri_Right.png [ 3.67 KiB | Viewed 260201 times ] Attachment: Math_Tri_Right3060.png [ 3.64 KiB | Viewed 281639 times ] Attachment: Math_Tri_Right4545.png [ 2.5 KiB | Viewed 269934 times ] Attachment: Math_Tri_inscribed.png [ 6.47 KiB | Viewed 261067 times ] Attachment: Math_Tri_circumscribed.png [ 8.24 KiB | Viewed 260347 times ] Attachment: Math_Tri_Similar.png [ 4.62 KiB | Viewed 261828 times ] Attachment: Math_icon_triangles.png [ 4.03 KiB | Viewed 257499 times ]

_________________

Last edited by walker on 25 Jan 2015, 08:29, edited 5 times in total.
Manager
Joined: 29 Jul 2009
Posts: 123
Location: France
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 102 [2] , given: 15

### Show Tags

23 Nov 2009, 01:57
2
KUDOS
1
This post was
BOOKMARKED
Wow!!! +1 - really good.
_________________
Retired Moderator
Status: The last round
Joined: 18 Jun 2009
Posts: 1310
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 79

Kudos [?]: 1004 [0], given: 157

### Show Tags

23 Nov 2009, 02:06
Woo!! Woo! Woo!!

So here comes our master, to theoretical posts

God bless you!!

Wonderful team, wonderful site. +2
_________________
Manager
Joined: 24 Jul 2009
Posts: 73
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 124 [0], given: 124

### Show Tags

23 Nov 2009, 05:13
Bunuel, You Rock!!!!
+10.
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14455
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3724

Kudos [?]: 23012 [0], given: 4514

### Show Tags

23 Nov 2009, 12:19
WWWWWWWWWOOOOOW ++++
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 18

Kudos [?]: 164 [0], given: 22

### Show Tags

23 Nov 2009, 15:28
I totally needed this.
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
Followers: 103

Kudos [?]: 1281 [4] , given: 18

### Show Tags

23 Nov 2009, 16:23
4
KUDOS
1
This post was
BOOKMARKED
Wow Bunuel... You never cease to amaze me!

Great Post. Really enjoyed going through it. Cant imagine how much time and effort it must have taken you to compile and format it.

Kudos to you.

Ps. Noticed a couple of typos you might've overlooked.

Bunuel wrote:
• $$m=\sqrt{\frac{2b^2+c^2-a^2}{4}}$$, where $$a$$, $$b$$ and $$c$$ are the sides of the triangle and $$a$$ is the side of the triangle whose midpoint is the extreme point of median $$m$$.

The formula should be $$m=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$ where $$m$$ is the length of the median.

Bunuel wrote:
• The radius of the circumscribed circle is $$R=a*\frac{\sqrt{3}}{4}$$

The radius of the circumscribed circle for an equilateral triangle should be : $$R=a*\frac{\sqrt{3}}{3}$$

Also, was just wondering, since you have included a number of formulae for calculating area of a triangle, might as well include this one : $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $$a,b,c$$ are the three sides of the triangle and $$s = \frac{a+b+c}{2}$$ which is the semi perimeter of the triangle.

Once again.. Great post.

Keep 'em coming!

Cheers.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 18

Kudos [?]: 164 [0], given: 22

### Show Tags

23 Nov 2009, 16:39
sriharimurthy, Bunuel,

All of this is from S Chand Intermediate First year Mathematics Textbook, Andhra Pradesh Intermediate Public Examination Board, India?
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Math Expert
Joined: 02 Sep 2009
Posts: 36586
Followers: 7088

Kudos [?]: 93304 [8] , given: 10557

### Show Tags

23 Nov 2009, 17:03
8
KUDOS
Expert's post
sriharimurthy wrote:
Wow Bunuel... You never cease to amaze me!

Great Post. Really enjoyed going through it. Cant imagine how much time and effort it must have taken you to compile and format it.

Kudos to you.

Ps. Noticed a couple of typos you might've overlooked.

Bunuel wrote:
• $$m=\sqrt{\frac{2b^2+c^2-a^2}{4}}$$, where $$a$$, $$b$$ and $$c$$ are the sides of the triangle and $$a$$ is the side of the triangle whose midpoint is the extreme point of median $$m$$.

The formula should be $$m=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$ where $$m$$ is the length of the median.

Bunuel wrote:
• The radius of the circumscribed circle is $$R=a*\frac{\sqrt{3}}{4}$$

The radius of the circumscribed circle for an equilateral triangle should be : $$R=a*\frac{\sqrt{3}}{3}$$

Also, was just wondering, since you have included a number of formulae for calculating area of a triangle, might as well include this one : $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $$a,b,c$$ are the three sides of the triangle and $$s = \frac{a+b+c}{2}$$ which is the semi perimeter of the triangle.

Once again.. Great post.

Keep 'em coming!

Cheers.

Thank you very much for spotting the typos. +1.

Quite a long post so no wonder I missed some of them. I'll try to avoid them in the future.

BTW if someone has any other properties of triangles, which might be useful for GMAT, he/she can post here and I'll add them to the original post.

BarneyStinson wrote:
sriharimurthy, Bunuel,

All of this is from S Chand Intermediate First year Mathematics Textbook, Andhra Pradesh Intermediate Public Examination Board, India?

Sorry, I don't even know which resource you are referring to. These properties are form various sources: GMAT problems, geometry sites, etc.
_________________
Manager
Joined: 07 Jul 2009
Posts: 230
Followers: 3

Kudos [?]: 99 [0], given: 13

### Show Tags

23 Nov 2009, 17:21
As always Awesome post Bunuel!!!! Many Kudos. You have taken this forum to a new level. Thank you very much for your time and efforts.
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 548

Kudos [?]: 3559 [0], given: 360

### Show Tags

23 Nov 2009, 18:14
Excellent job!!!

+1
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Joined: 25 Jul 2009
Posts: 32
Location: Lahore, Pakistan
Schools: going to Melbourne Business School
Followers: 2

Kudos [?]: 9 [0], given: 0

### Show Tags

24 Nov 2009, 02:46
touche, great post !~!
Senior Manager
Joined: 18 Aug 2009
Posts: 328
Followers: 8

Kudos [?]: 296 [0], given: 13

### Show Tags

24 Nov 2009, 05:29
Bunuel

might I suggest you to create another post with subject: "All you want to know about Quantitative" and provide links to your various posts in it? That way we all can bookmark just one page and still can have all the information from you? Also in this way if you add any new post, we will not miss it.

You might also want to create a master thread containing links to all your posts of different sets of problems. That weay we can have a central repository to different types of problems eg. Inequality, Combinatorics, etc.
Manager
Joined: 28 Jul 2009
Posts: 127
Followers: 2

Kudos [?]: 25 [0], given: 12

### Show Tags

24 Nov 2009, 08:06
This will help a lot .. another thanks ..
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 18

Kudos [?]: 140 [0], given: 69

### Show Tags

24 Nov 2009, 09:29
I wish somebody made a pdf compilation of Bunuel's posts...
Intern
Joined: 31 Aug 2009
Posts: 19
Followers: 1

Kudos [?]: 20 [0], given: 3

### Show Tags

24 Nov 2009, 09:49
awesome guide thanks!
so there are other guides like this? I'm taking the GMAT tomorrow, I would love to have a quick but thorough review. can someone post the other guides? Thanks
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 18

Kudos [?]: 140 [0], given: 69

### Show Tags

24 Nov 2009, 10:06
manifestdestiny wrote:
awesome guide thanks!
so there are other guides like this? I'm taking the GMAT tomorrow, I would love to have a quick but thorough review. can someone post the other guides? Thanks

Check out sticky threads in Math and Verbal forums. Pretty much everything important is collected up there.

and Good luck for tomorrow!
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 68

Kudos [?]: 735 [0], given: 19

### Show Tags

24 Nov 2009, 14:48
+1

I would give +10 for this million dollar resources if it were allowed.

You wont find this much materials free of cost anywhere in the world.

You are highly resourceful.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Intern
Joined: 16 Nov 2009
Posts: 5
Followers: 1

Kudos [?]: 2 [1] , given: 0

### Show Tags

24 Nov 2009, 21:42
1
KUDOS
Thanks Bunuel, wonderpost. concise, crisp and containing all the key properties of triangle. It would have taken me ages to assimilate all those in on single piece of paper. Thanks a lot!! +2!!!!!
Intern
Joined: 28 Oct 2009
Posts: 6
Location: Bangalore
Followers: 1

Kudos [?]: 1 [1] , given: 4

### Show Tags

24 Nov 2009, 22:18
1
KUDOS
thanks for the Information.People like you make the internet a better place . Printing this out now. Request to the moderator-you could put all the links of subject info like this one into one page.Is there a forum page like that?
Re: Math: Triangles   [#permalink] 24 Nov 2009, 22:18

Go to page    1   2   3   4   5    Next  [ 94 posts ]

Similar topics Replies Last post
Similar
Topics:
Maths 1 23 Sep 2011, 13:34
2 Triangles 2 13 Jul 2011, 15:31
2 Triangle 1 16 Apr 2011, 23:11
Triangles 0 31 May 2010, 19:25
1 GEOMETRY - TRIANGLES 11 17 Sep 2008, 22:02
Display posts from previous: Sort by