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Re: Math: Triangles [#permalink]
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07 Jul 2010, 22:33
Wow, thanks for this summary!



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Re: Math: Triangles [#permalink]
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22 Jul 2010, 02:25
Under Equilateral triangles its been mentioned that "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.' Could you please help me out with this? I am not able to comprehend. Also are the perpendicular bisectors and the altitudes the same in case of an equilateral triangles. Thanks Bunuel!!! I am hunting and tracking down every post that your have posted in this forum. All of them that I have read till now have been extremely clear, precise and are tuned to the GMAT. Thanks again!



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Re: Math: Triangles [#permalink]
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22 Jul 2010, 03:06
sanram2205 wrote: Under Equilateral triangles its been mentioned that "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.' Could you please help me out with this? I am not able to comprehend. Also are the perpendicular bisectors and the altitudes the same in case of an equilateral triangles. Thanks Bunuel!!! I am hunting and tracking down every post that your have posted in this forum. All of them that I have read till now have been extremely clear, precise and are tuned to the GMAT. Thanks again! You won't need first one for GMAT. As for the second question: in equilateral triangle angle bisectors, medians and altitudes (heights) are the same and equal in length.
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Re: Math: Triangles [#permalink]
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24 Jul 2010, 06:01
Quote: m=\sqrt{\frac{2b^2+2c^2a^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. What does extreme point of median m mean?



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Re: Math: Triangles [#permalink]
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24 Jul 2010, 07:30



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Re: Math: Triangles [#permalink]
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30 Aug 2010, 13:22
amazing! one day GMATclub might get to publish its own book : )



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31 Aug 2010, 22:03
This is awesome...thanks!!!!
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Re: Math: Triangles [#permalink]
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10 Oct 2010, 22:14
gmatdelhi wrote: Under Insoceles triangle section: To find the base given the leg and altitude, use the formula:...
How do you derive these formulae? What's the logic behind them??? The explanation is based on simple fact that  In an Isosceles triangle the Altitude (coming from the vertex holding equal sides to the base) is Same as its Median. ie., Altitude ( which forms 90 degrees with base cuts the base in 2 equal parts).. Therefore Applying Phythogras theorem for L is hypotenuse A is side B/2 is another side Therefore L square = A Square + (B/2) Square All the 3 formulas shown in the page are same , and derived from this only. Hope this clarifies. Regards, Sridhar.



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Re: Math: Triangles [#permalink]
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16 Oct 2010, 08:48
Another useful property for triangles,
If the sum of any two angles of a triangle equals the third angle then the triangle must be a right triangle i.e. If the angles of a triangle are A, B and C, Then
=> A + B = C
But for any triangle A + B + C = 180
=> 2C = 180 or C = 90



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Re: Math: Triangles [#permalink]
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10 Nov 2010, 13:36
Quote: In similar triangles, the sides of the triangles are in some proportion to one another. Hey Bunuel, a very small correction. Above statement should be  In similar triangles, the sides of the triangles are in same proportion to one another.



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Re: Math: Triangles [#permalink]
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06 Jan 2011, 10:55
Wow thank you so much!! This will be the great GMAT math book since sliced bread
Medians woah!!!! What I would do in a GMAt exam if I knew all these facts, I would beat the GMAT!
Writing in two day sand this is my final brush up of all my concepts ... great job ...



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Re: Math: Triangles [#permalink]
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11 Jan 2011, 11:18
Excellent post. I went through all this triangle formulas and it is very very helpful......



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Re: Math: Triangles [#permalink]
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04 Feb 2011, 13:13
A very useful collection !! Thank you!  +



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Re: Math: Triangles [#permalink]
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05 Feb 2011, 13:29
Material is Great !!! + KUDO FROM ME



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Re: Math: Triangles [#permalink]
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09 Feb 2011, 00:25
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Bunuel wrote: 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: \(B=2\sqrt{L^2A^2}\) • To find the leg length given the base and altitude, use the formula: \(L=\sqrt{A^2+(\frac{B}{2})^2}\) • To find the leg length given the base and altitude, 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) hello Bunuel you said: To find the leg length given the base and altitude, use the formula: A=\sqrt{L^2(\frac{B}{2})^2} i believe you were supposed to say Altitude instead. ? right? my question: is there specific formula for isosceles Area? somebody mentioned but i was no able to find the one. thanks



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Re: Math: Triangles [#permalink]
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09 Feb 2011, 02:28
tinki wrote: Bunuel wrote: 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: \(B=2\sqrt{L^2A^2}\) • To find the leg length given the base and altitude, use the formula: \(L=\sqrt{A^2+(\frac{B}{2})^2}\) • To find the leg length given the base and altitude, 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) hello Bunuel you said: To find the leg length given the base and altitude, use the formula: A=\sqrt{L^2(\frac{B}{2})^2} i believe you were supposed to say Altitude instead. ? right? my question: is there specific formula for isosceles Area? somebody mentioned but i was no able to find the one. thanks First of all: do not fully quote such big texts. Do as I edited know: quote only the specific part you are referring to. Next, formula indicates Altitude=... so yes there was a typo. Thanks for spotting. Edited. As for your question: you won't need any other formula for the area of an isosceles triangle but area=1/2*base*height. The are of isosceles right triangle is area=leg^2/2.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Math: Triangles [#permalink]
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22 Apr 2011, 06:11
Hi Bunuel, Can you please explain the last point in the median section?
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Re: Math: Triangles [#permalink]
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22 Jul 2011, 21:59
Thanks Bunuel ... kudos...



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Re: Math: Triangles [#permalink]
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20 Aug 2011, 06:05
Bunuel wrote: 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)
Shouldn't that be CB and AC? Great material. Thanks for the compilation.



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Re: Math: Triangles [#permalink]
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04 Nov 2011, 09:23
Bunnel great post. But i have one qstn. Isnt it too much information about triangles from GMAT perspective. Or is it necessary to grab all the concepts put up.
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