http://www.postimage.org/image.php?v=TsLg35J

[img]

http://www.postimage.org/image.php?v=TsLg35J[/img]

GMAT Forum

- vinversa
- Students
**Posts:**30**Joined:**Wed May 26, 2010 10:47 pm

- debmalya.dutta
- Students
**Posts:**47**Joined:**Wed Jan 20, 2010 9:51 am

To get the maximum area the triangle should be a right triangle should be a right triangle.

hence area is 1/2

hence area is 1/2

- RonPurewal
- ManhattanGMAT Staff
**Posts:**15904**Joined:**Tue Aug 14, 2007 8:23 am

debmalya.dutta wrote:To get the maximum area the triangle should be a right triangle should be a right triangle.

hence area is 1/2

this is actually a useful fact to memorize: if you have 2 given sides of a triangle, then you can maximize the area of the triangle by making thise two sides meet at a right angle.

(note that this will not maximize the perimeter of the triangle)

Pueden hacerle preguntas a Ron en castellano

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

- heraclesghj
- Forum Guests
**Posts:**1**Joined:**Sun Aug 08, 2010 4:12 am

i got a perfect solution.

let's suppose that the central angle is X,

then the triangle's high is sin(90-1/2x) the base is 2cos(90-1/2x)

[because the triangle is isosceles triangle,so the bisect line that both bisect the vertex angle and the base is the perpendicular line.and the perpendicular devide the triangle into two small right triangle.from each of the small right triangle.we can come to a equation that the high of the original triangle is sin(90-1/2x)(here in a right triangle, the sinx=the anglex's opposite side divide the hypotenuse, and the cosx=the other side divide the hypotenuse) each little right triangle's vertex angle is 1/2x, so the other angle is 90-1/2x,so the sin(90-1/2x) =the high of the original triangle / 1, the cos(90-1/2x)=1/2base of the original triangle / 1,so the original triangle's dimension=1/2highÂ·base=1/2Â·sin(90-1/2x)Â·2cos(90-1/2x)

and there is a formula that sin2x=2sinxcosx,so the dimension=1/2sin(180-x),because the maxsimum number of sinx=1,so the maxsimum dimension=1/2

[editor: this solution process may be valid, but you should try to come up with a solution that doesn't involve trigonometry -- remember that this test doesn't require any trigonometry, so the chances of FUTURE problems that allow this sort of reasoning are fairly small.]

let's suppose that the central angle is X,

then the triangle's high is sin(90-1/2x) the base is 2cos(90-1/2x)

[because the triangle is isosceles triangle,so the bisect line that both bisect the vertex angle and the base is the perpendicular line.and the perpendicular devide the triangle into two small right triangle.from each of the small right triangle.we can come to a equation that the high of the original triangle is sin(90-1/2x)(here in a right triangle, the sinx=the anglex's opposite side divide the hypotenuse, and the cosx=the other side divide the hypotenuse) each little right triangle's vertex angle is 1/2x, so the other angle is 90-1/2x,so the sin(90-1/2x) =the high of the original triangle / 1, the cos(90-1/2x)=1/2base of the original triangle / 1,so the original triangle's dimension=1/2highÂ·base=1/2Â·sin(90-1/2x)Â·2cos(90-1/2x)

and there is a formula that sin2x=2sinxcosx,so the dimension=1/2sin(180-x),because the maxsimum number of sinx=1,so the maxsimum dimension=1/2

[editor: this solution process may be valid, but you should try to come up with a solution that doesn't involve trigonometry -- remember that this test doesn't require any trigonometry, so the chances of FUTURE problems that allow this sort of reasoning are fairly small.]

- ya1ya2
- Students
**Posts:**7**Joined:**Thu Sep 17, 2009 8:22 am

If you remember the area of a triangle is also given by 1/2 A B sin(AB) in this case A and B are = 1, so that leaves us with maximizing sinAB which is max at 90degrees. Hence we are left with the area of 1/2

[editor: this solution process may be valid, but you should try to come up with a solution that doesn't involve trigonometry -- remember that this test doesn't require any trigonometry, so the chances of FUTURE problems that allow this sort of reasoning are fairly small.]

[editor: this solution process may be valid, but you should try to come up with a solution that doesn't involve trigonometry -- remember that this test doesn't require any trigonometry, so the chances of FUTURE problems that allow this sort of reasoning are fairly small.]

- RonPurewal
- ManhattanGMAT Staff
**Posts:**15904**Joined:**Tue Aug 14, 2007 8:23 am

i'd prefer not to discuss the merits of trigonometric solutions.

the gmat will NEVER include a problem that requires trigonometry, so we've basically got the following situation:

* if you already understand trigonometry, then trigonometric solutions (in the extremely rare cases when they're possible) will be accessible to you;

* if you don't understand trigonometry, it's definitely not worth the effort to learn it for this test.

the gmat will NEVER include a problem that requires trigonometry, so we've basically got the following situation:

* if you already understand trigonometry, then trigonometric solutions (in the extremely rare cases when they're possible) will be accessible to you;

* if you don't understand trigonometry, it's definitely not worth the effort to learn it for this test.

Pueden hacerle preguntas a Ron en castellano

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

- vjsharma25
- Students
**Posts:**31**Joined:**Sun Apr 25, 2010 2:33 pm

I very different approach to solve this question

Area of a triangle = 1/2*base*height

Now you want to maximize the area. So you have to maximize the product (base*height). Now product of two numbers is maximum when they are equal. Means base = height.

We can have maximum height of 1,as this is the radius of the circle. So base can have maximum value 1. So maximum area should be 1/2*1*1 = 1/2.

Its a bit different approach but is handy in this case.

Vjsharma25

Area of a triangle = 1/2*base*height

Now you want to maximize the area. So you have to maximize the product (base*height). Now product of two numbers is maximum when they are equal. Means base = height.

We can have maximum height of 1,as this is the radius of the circle. So base can have maximum value 1. So maximum area should be 1/2*1*1 = 1/2.

Its a bit different approach but is handy in this case.

Vjsharma25

- jnelson0612
- ManhattanGMAT Staff
**Posts:**2670**Joined:**Fri Feb 05, 2010 10:57 am

vjsharma25 wrote:I very different approach to solve this question

Area of a triangle = 1/2*base*height

Now you want to maximize the area. So you have to maximize the product (base*height). Now product of two numbers is maximum when they are equal. Means base = height.

We can have maximum height of 1,as this is the radius of the circle. So base can have maximum value 1. So maximum area should be 1/2*1*1 = 1/2.

Its a bit different approach but is handy in this case.

Vjsharma25

Thank you vsharma. This is a similar concept to the idea of making the triangle a 45-45-90 right triangle to maximize its area.

Jamie Nelson

ManhattanGMAT Instructor

ManhattanGMAT Instructor

- agha79
- Course Students
**Posts:**98**Joined:**Sun Mar 13, 2005 6:13 am

so does this mean that it is not necessary that an equilateral triangle must have the maximum area always as this a rule generally followed?

- jnelson0612
- ManhattanGMAT Staff
**Posts:**2670**Joined:**Fri Feb 05, 2010 10:57 am

agha79 wrote:so does this mean that it is not necessary that an equilateral triangle must have the maximum area always as this a rule generally followed?

Let me help you answer your own question. Draw two triangles:

1) One is a 45-45-90 triangle with sides 4, 4, 4sqrt2.

2) One is an equilateral triangle with all sides measuring 4.

Which one has the larger area?

Jamie Nelson

ManhattanGMAT Instructor

ManhattanGMAT Instructor

- agha79
- Course Students
**Posts:**98**Joined:**Sun Mar 13, 2005 6:13 am

oh yeah right 1 has larger area that is 8.

- RonPurewal
- ManhattanGMAT Staff
**Posts:**15904**Joined:**Tue Aug 14, 2007 8:23 am

agha79 wrote:oh yeah right 1 has larger area that is 8.

yep

Pueden hacerle preguntas a Ron en castellano

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

- sfbay
- Students
**Posts:**31**Joined:**Sun Apr 11, 2010 4:38 pm**Location:**San Francisco

Isosceles maximizes area

Equilateral maximizes perimeter

Equilateral maximizes perimeter

- RonPurewal
- ManhattanGMAT Staff
**Posts:**15904**Joined:**Tue Aug 14, 2007 8:23 am

munnynarang wrote:Isosceles maximizes area

true in *this* problem (with the specific restrictions given here), but not true in general.

for instance, if you have a fixed perimeter for the triangle but *no* restrictions on it, then an equilateral triangle has the greatest possible area.

Equilateral maximizes perimeter

again, dependent on the conditions of the problem.

(and i suppose you meant "minimizes" -- it's not really possible to *maximize* a perimeter in most cases; you could just make a really long, skinny triangle with a perimeter as large as you want)

Potete fare domande a Ron in italiano

On peut poser des questions à Ron en français

Voit esittää kysymyksiä Ron:lle myös suomeksi

Un bon vêtement, c'est un passeport pour le bonheur.

– Yves Saint-Laurent

- me.parashar
- Forum Guests
**Posts:**16**Joined:**Wed Dec 31, 1969 8:00 pm

thanks christiancryan. I too, answered the question wrong (marked obvious A). And the way you explained the question, I now know a very important concept of working the maxima-minima without the differential, and the beauty is that the concept can be used everywhere from number system to coordinate geometry.