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What is the greatest possible area of a triangular region

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What is the greatest possible area of a triangular region [#permalink] New post 01 Nov 2009, 21:12
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What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius one and the other two vertices on the circle?

A. \frac{\sqrt{3}}{4}

B. \frac{1}{2}

C. \frac{\pi}{4}

D. 1

E. \sqrt{2}
[Reveal] Spoiler: OA

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Last edited by Bunuel on 17 Oct 2013, 07:49, edited 1 time in total.
Added the OA.
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Re: Maximum Area of Inscribed Triangle [#permalink] New post 02 Nov 2009, 01:36
gmattokyo wrote:
I'd go with B. 1/2
right triangle. a rough sketch shows that taking one of the sides either left or right seems to be reducing the area.

right triangle area =1/2x1x1 (base=height=radius)=1/2


The logic just striked me... area=1/2xbasexheight.
In this case, if you keep the base is constant=radius. Height is at its maximum when it is right triangle.

is that the OA?
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Re: Maximum Area of Inscribed Triangle [#permalink] New post 03 Nov 2009, 16:58
So I came across this question in my test and got it wrong..I assumed the equilateral triangle has the greatest area and marked root3/4 :(
Now i see the logic..any triangle drawn by the above specifications will have two legs as the radius..we have to maximise the area so the third leg should be the largest.

However,is this some kind of a theoram/fact that we should be knowing?That to get the largest area of a triangle,the triangle has to be a right angle and not an equilateral one?
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Re: Maximum Area of Inscribed Triangle [#permalink] New post 03 Nov 2009, 17:53
tejal777 wrote:
So I came across this question in my test and got it wrong..I assumed the equilateral triangle has the greatest area and marked root3/4 :(
Now i see the logic..any triangle drawn by the above specifications will have two legs as the radius..we have to maximise the area so the third leg should be the largest.

However,is this some kind of a theoram/fact that we should be knowing?That to get the largest area of a triangle,the triangle has to be a right angle and not an equilateral one?


Yes, if the bases are the same. In this case 1 would be the base (radius) and a 45-45-90 maximizes area

Try using any number for the base, for example 4

45-45-90 = 1/2(4)(4) = 8

60-60-60 = 1/2(4)(2 sqrt(3)) = 4 sqrt(3)
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Re: GMAT Prep Triangle/Circle [#permalink] New post 06 Dec 2009, 11:47
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What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius one and the other two vertices on the circle?

Clearly two sides of the triangle will be equal to the radius of 1.

Now, fix one of the sides horizontally and consider it to be the base of the triangle.

area=\frac{1}{2}*base*height=\frac{1}{2}*1*height=\frac{height}{2}.

So, to maximize the area we need to maximize the height. If you visualize it, you'll see that the height will be maximized when it's also equals to the radius thus coincides with the second side (just rotate the other side to see). which means to maximize the area we should have the right triangle with right angle at the center.

area=\frac{1}{2}*1*1=\frac{1}{2}.

Answer: B.

You can also refer to other solutions:
triangular-region-65317.html
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Re: GMAT Prep Triangle/Circle [#permalink] New post 06 Dec 2009, 12:09
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Adding onto what Bunuel said, there is an important property about isosceles triangles that will help you understand and solve this question.

First though, let us see how this particular triangle must be isosceles.

If one vertex is at the centre of the circle and the other two are on the diameter, then the triangle must be isosceles since two of its sides will be = radius of circle = 1.

Now for an isosceles triangle, the area will be maximum when it is a right angled triangle. One way of proving this is through differentiation. However, since that is well out of GMAT scope, I will provide you with an easier approach.

An isosceles triangle can be considered as one half of a rhombus with side lengths 'b'. Now a rhombus of greatest area is a square, half of which is a right angled isosceles triangle. Thus for an isosceles triangle, the area will be greatest when it is a right angled triangle.

[Note to Bunuel : I think this one might have been missed in the post on triangles?]

Now for the right angled triangle in our case, b = 1 and h = 1

Thus area of triangle = \frac{1}{2}*b*h = \frac{1}{2}

Answer : B

Note : I believe the mistake you might have made is considered the base to be = 2 (or the diameter of the circle) and height to be 1. This can only be possible if all three vertices lie on the circle not when one is at the centre.
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Re: GMAT Prep Triangle/Circle [#permalink] New post 06 Dec 2009, 13:15
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sriharimurthy wrote:
[Note to Bunuel : I think this one might have been missed in the post on triangles?]


This is a useful property, thank you. +1.

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

And vise-versa:

Right triangle with a given hypotenuse has the largest area when it's an isosceles triangle.
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Re: GMAT Prep Triangle/Circle [#permalink] New post 10 Dec 2009, 19:52
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Has anyone got a diagram of the trangle in circle for this question? I'm unable to visualize the diagram from the question


fig attached
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Re: Maximum Area - No clues [#permalink] New post 14 Mar 2010, 01:57
mustdoit wrote:
What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

a. rt3/4
b. 1/2
c. Pi/4
d. 1
e. rt2


OA:
[Reveal] Spoiler:
B


let the vertex at Centre be A and B and C are vertices of trianle on the circle
so length of side AB and AC will be equal to radius of circle =1.In this case the maximum area will be obtained for a right angled isosceles traiangle

1/2* AB* AC = 1/2 *1*1 = 1/2
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Re: Maximum Area - No clues [#permalink] New post 14 Mar 2010, 02:20
mustdoit wrote:
What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

a. rt3/4
b. 1/2
c. Pi/4
d. 1
e. rt2


OA:
[Reveal] Spoiler:
B


Let say b is the third side's length and a is the equal sides' lenght.
then the area of triangle by hero's formula will be b * sqrt(4 a^2 - b^2)/4
putting value of a
=> Area = b * sqrt(4 - b^2)/4
now to get maximum value of Area we have to take derivative of Area in terms of the third side.
For maximum Area its square will also be maximum, that's why squaring both the sides
=> Area^2 = b^2 * (4 - b^2)/16
=> Taking derivative both the sides
=> d(Area^2)/db = (8b-4b^3)/16
equate RHS to 0 to get value of b for which Area is maximum
(8b - 4b^3)/16 = 0
=>2b-b^3 = 0
=>b (2-b^2) = 0
b = 0, |b| = sqrt 2
now b cannot be negative so
b = 0, b = sqrt 2
for these two values sqrt 2 will give the maximum area and put this value in
Area = b * sqrt(4 - b^2)/4

Area = (sqrt 2 * sqrt 2) / 4 = 1/2 hence b is the answer.
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Re: Maximum Area - No clues [#permalink] New post 14 Mar 2010, 08:22
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Can I see it this way?

If you know what is function sin, it has a range from -1 to 1:

Since area of triangle = 1/2 x (side a x side b x sin C), where C is the angle in between side a and b.
The area would be at its maximum when C equals 90 degrees, i.e. sin C = 1.

In this case, we can take side a and side b the radii and C 90 degrees:
1/2 x 1 x 1 x 1 = 1/2

Hope this helps.
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Re: Maximum Area of Inscribed Triangle [#permalink] New post 06 Apr 2010, 16:49
I got this right on my test, does my thought process make sense?

I know that for a set perimeter of a quadrilateral a square will maximize area, so if you have 16 feet of fence to enclose a garden and want to maximize the area of the garden you would build a square fence around the garden.

EX: Perimeter= 16 Area of square=16
Ex: Perimeter of a rectangle with width of 2 and length of 6=16 Area of the rectangle= 12

So for this problem I thought that a 45-45-90 triangle is half of a square therefore this triangle must maxmize the area with given base.

Sorry if this is confusing, but is this mathmatically correct?
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Re: GMAT PREP (PS) [#permalink] New post 06 May 2010, 12:13
With one of the vertices at the centre, the two sides of the traingle could be perpendicular to each other (2 radii) and the third side joining the two vertices will be the hypotenuse. Hence, the area will be 1/2 * 1 *1 = 1/2 !
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Re: PS question: need help [#permalink] New post 23 Oct 2010, 02:38
satishreddy wrote:
ps question

Trignometry based solution

Note that such a triangle is always isosceles, with two sides=1 (the radius of the circle).
Let the third side be b (the base) and the height be h.
If you imagine the angle subtended at the centre by the thrid side, and let this angle be x.

The base would be given by 2*sin(x/2) and the height by cos(x/2); where x is a number between 0 and 180

The area is therefore, sin(z)*cos(z), where z is between 0 and 90.
We can simplify this further as sin(z)*\sqrt{1-sin^2(z)}, with z between 0 and 90, for which range sin(z) is between 0 and 1.

So the answer is maxima of the function f(y)=y*\sqrt{1-y^2} with y between 0 and 1.
This is equivalent to finding the point which will maximize the square of this function g(y)=y^2(1-y^2) which is easy to do taking the first derivative, g'(y)=2y-4y^3, which gives the point as y=\frac{1}{\sqrt{2}}.

If we plug it into f(y), the answer is area = 0.5 .. Hence answer is (b)

Basically the solution above proves that for an isosceles triangle, when the length of the equal sides is fixed, the area is maximum when the triangle is a right angled triangle (y=sin(x/2)=\frac{1}{\sqrt{2}} means x=90). This is a result you will most liekly see being quoted on alternate solutions.
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Re: Maximum Area - No clues [#permalink] New post 23 Oct 2010, 05:36
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Interesting Question!
As CalvinHobbes suggested, the easiest way to deal with it might be through the area formula:
Area = (1/2)abSinQ
a and b are the lengths of two sides of the triangle and Q is the included angle between sides a and b.
(It is anyway good to remember this area formula if you are a little comfortable with trigonometry because it could turn your otherwise tricky question into a simple application.)

If we want to maximize area, we need to maximize Sin Q since a and b are already 1.
Maximum value of Sin Q is 1 which happens when Q = 90 degrees.

Therefore, maximum area of the triangle will be (1/2).1.1.1 = (1/2)
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Re: Maximum Area - No clues [#permalink] New post 11 Jun 2011, 16:59
Area is maximum in an isosceles triangle when angle between two same sides is 90.

Maximum area = 1/2 (r)(r) = (1/2) (r^2) = 1/2

Answer is B.
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some tough PS questions from GMATprep [#permalink] New post 20 Mar 2012, 10:28
Hi guys,

I took GMATprep test couple of days back and was faced with these tricky questions in PS. Can someone please answer how to solve them?

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PS - Greatest possible area of a triangular region [#permalink] New post 25 Jul 2012, 14:57
What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other 2 vertices on the circle?

A. sqrt (3) / 4
B. 1/2
C. (pi) / 4
D. 1
E Sqrrt(2)
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Re: PS - Greatest possible area of a triangular region [#permalink] New post 25 Jul 2012, 15:01
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GODSPEED wrote:
What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other 2 vertices on the circle?

A. sqrt (3) / 4
B. 1/2
C. (pi) / 4
D. 1
E Sqrrt(2)


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Re: What is the greatest possible area of a triangular region [#permalink] New post 17 Oct 2013, 07:48
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