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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.

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. _________________

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.

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.

What does extreme point of median m mean?

Extreme point is end point. All above means that median \(m\) is drawn to side \(a\). _________________

Re: Math: Triangles [#permalink]
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.

Re: Math: Triangles [#permalink]
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

Re: Math: Triangles [#permalink]
09 Feb 2011, 00:25

1

This post received KUDOS

Bunuel wrote:

Isosceles triangletwo 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 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

Re: Math: Triangles [#permalink]
09 Feb 2011, 02:28

1

This post received KUDOS

Expert's post

tinki wrote:

Bunuel wrote:

Isosceles triangletwo 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 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. _________________

Re: Math: Triangles [#permalink]
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. _________________

D- Day December 30 2011. Hoping for the happiest new year celebrations !

Aiming for 700+

Kudo me if the post is worth it

gmatclubot

Re: Math: Triangles
[#permalink]
04 Nov 2011, 09:23

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