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Re: Math: Triangles [#permalink]
A masterpiece Bunuel . Amazing summary of triangle proprties. Some Questions to have more clarity .

Are the medians equal in length in a scalene traingle ?

Does the centroid neccasarily have to be a center of circumcircle of a scalene triangle or in other words, does it mean that each triangle can be circumcircled ?
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Re: Math: Triangles [#permalink]
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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.


For Obtuse angle also, above condition is true ?
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Re: Math: Triangles [#permalink]
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gottabwise wrote:
Does the orthocenter = the centroid and if so, is it safe to say that altitudes=medians?


Generally orthocenter and centroid are not the same point. Generally altitude does not equal to median.
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Re: Math: Triangles [#permalink]
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GMATMadeeasy wrote:
A masterpiece Bunuel . Amazing summary of triangle proprties. Some Questions to have more clarity .

Are the medians equal in length in a scalene traingle ?

Does the centroid neccasarily have to be a center of circumcircle of a scalene triangle or in other words, does it mean that each triangle can be circumcircled ?


Generally medians are not equal, so in scalene triangle medians are not equal.

Centroid is not the center of the circumscribed circle. (There was a typo in the text, edited.)

As for the circumscribed triangles: yes, any triangle can be circumscribed.
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Re: Math: Triangles [#permalink]
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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
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Re: Math: Triangles [#permalink]
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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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Re: Math: Triangles [#permalink]
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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?
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Re: Math: Triangles [#permalink]
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knabi wrote:
Quote:
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 \(a\).
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Re: Math: Triangles [#permalink]
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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: \(B=2\sqrt{L^2-A^2}\)

• 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
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Re: Math: Triangles [#permalink]
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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: \(B=2\sqrt{L^2-A^2}\)

• 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.
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Re: Math: Triangles [#permalink]
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Does GMAC test trignometry? for example - could we expect such problem?

Solve the triangle below for angle x and then compute the area of the triangle.

Say a triangle ABC with angles A, B and C with opposite sides to these angles a,b, and c respectively. a = 5 inches and b = 3 inches and angle A = 75 degrees find B =?

Thanks!
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Re: Math: Triangles [#permalink]
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goutamread wrote:
Does GMAC test trignometry? for example - could we expect such problem?

Solve the triangle below for angle x and then compute the area of the triangle.

Say a triangle ABC with angles A, B and C with opposite sides to these angles a,b, and c respectively. a = 5 inches and b = 3 inches and angle A = 75 degrees find B =?

Thanks!


Trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.
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Re: Math: Triangles [#permalink]
Would someone please tell me the steps to arrive at the equation for the area of a circle inscribed in an equilateral triangle? I see the formula in the gmatclub math book but I would like to understand the logic behind the formula. Thanks
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Re: Math: Triangles [#permalink]
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Anonamy wrote:
Would someone please tell me the steps to arrive at the equation for the area of a circle inscribed in an equilateral triangle? I see the formula in the gmatclub math book but I would like to understand the logic behind the formula. Thanks


Refer to the attached image for a description of the variables.

Attachment:
2016-01-31_13-47-35.jpg
2016-01-31_13-47-35.jpg [ 11.64 KiB | Viewed 149999 times ]


You see that triangle ABC is an equilateral triangle and O is the center of the incircle such that OD= radius of the incircle and is hence perpendicular (at 90 degrees) to BC. Join O to B such that you create a right angled triangle OBD, right angled at B.

The line OB bisects\(\angle{ABC}\) such that \(\angle{OBD}\) = 30 degrees...(this is a property of equilateral triangles, something easily provable as triangles OBD and OBE are congruent).

Thus, triangle OBD is a 30-60-90 degree triangle with BD = side of triangle ABC /2 = a/2

Also, \(\frac{OD}{BD} = \frac{1}{\sqrt{3}}\) ---> \(OD = r = \frac {BD}{\sqrt{3}} = \frac{a}{2*\sqrt{3}} = \frac{a*\sqrt{3}}{6}\).

Thus, area of the incircle = \(\pi*r^2 = \pi * (\frac{a}{2*\sqrt{3}})^2 = \pi * \frac{a^2}{12}\)

Hope this helps.
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Re: Math: Triangles [#permalink]
Hi Bunuel,

Thank you for such an amazing job. I have a couple of questions:

1) Is this theory about triangle enough to solve any triangle problem on GMAT?

2)Are the following concepts are tested on GMAT?
-centroid of the triangle
-Other area formulas :
• A=P∗r/2
• A=abc/4R
-midsegment of a triangle
- congurence of triangles

Thanks
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Math: Triangles [#permalink]
Hi Bunuel,

Thanks for those links. They are helpful.

I'm having trouble understanding these formulas: A=P*r/2 and A=abc/4R. Are there any example questions with these formulas?
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Re: Math: Triangles [#permalink]
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rodri102 wrote:
Hi Bunuel,

Thanks for those links. They are helpful.

I'm having trouble understanding these formulas: A=P*r/2 and A=abc/4R. Are there any example questions with these formulas?


• \(A=\frac{P*r}{2}\)

• \(A=\frac{abc}{4R}\)

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

I was not able to find questions where you could apply those formulae and it's unlikely that you'd need those for the GMAT.
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