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Re: Math: Triangles [#permalink]
18 Jun 2010, 01:14

Expert's post

bely202 wrote:

Quote:

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

Just to clarify, is P the perimeter of the circle or the triangle?

Quote:

• For an isosceles triangle with given length of equal sides right triangle (included angle) has the largest area.

Will u elaborate on this please? I'm not sure how this works.

Thanks a lot of the summary, very complete and succinct! :D

1. P is the perimeter of the triangle. 2. For instance if we have an isosceles triangle with equal sides of 1, the area will be greatest when it is a right angled triangle (max area in this case would be 1/2). _________________

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If you feel like you're under control, you're just not going fast enough. A goal without a plan is just a wish. You can go higher, you can go deeper, there are no boundaries above or beneath you.

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

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