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What is the greatest possible area of a triangular region with one ver

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What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post Updated on: 30 Sep 2019, 02:27
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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 1 and the other two vertices on the circle?

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

B. \(\frac{1}{2}\)

C. \(\frac{\pi}{4}\)

D. 1

E. \(\sqrt{2}\)

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Originally posted by slingfox on 01 Nov 2009, 22:12.
Last edited by Bunuel on 30 Sep 2019, 02:27, edited 2 times in total.
Added the OA.
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What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 06 Dec 2009, 12:47
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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?

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

B. \(\frac{1}{2}\)

C. \(\frac{\pi}{4}\)

D. 1

E. \(\sqrt{2}\)

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.
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 06 Dec 2009, 13:09
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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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New post 06 Dec 2009, 14:15
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New post 14 Mar 2010, 09:22
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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: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 23 Oct 2010, 03:38
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: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 23 Oct 2010, 06:36
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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: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 17 Oct 2013, 18:25
I solved the question the following way..

I gathered the greatest possible triangle has a 90 degree angle where 2 sides meet (each length 1, the radius)
This means the 3rd side will be \(\sqrt{2}\) (90/45/45 rule)

It's base will be \(\sqrt{2}\) and its height will be \(\sqrt{2}\)/\(2\)

So base times height over 2 looks as such-

\(\sqrt{2}*\sqrt{2}/2\) all over 2

which yields 1/2.

am I getting the right answer the wrong way?
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 17 Oct 2013, 20:56
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bscharm wrote:
I solved the question the following way..

I gathered the greatest possible triangle has a 90 degree angle where 2 sides meet (each length 1, the radius)
This means the 3rd side will be \(\sqrt{2}\) (90/45/45 rule)

It's base will be \(\sqrt{2}\) and its height will be \(\sqrt{2}\)/\(2\)

So base times height over 2 looks as such-

\(\sqrt{2}*\sqrt{2}/2\) all over 2

which yields 1/2.

am I getting the right answer the wrong way?


I think you complicated the question for no reason even though your answer and method, both are correct (though not optimum). The most important part of the question is realizing that the triangle will be a right triangle. Once you did that, you know the two perpendicular sides of the triangle are 1 and 1 (the radii of the circle). The two perpendicular sides can very well be the base and the height. So area = (1/2)*1*1 = 1/2

In fact, this is used sometimes to find the altitude of the right triangle from 90 degree angle to hypotenuse. You equate area obtained from using the perpendicular side lengths with area obtained using hypotenuse. In this question, that will be

\((1/2)*1*1 = (1/2)*\sqrt{2}*Altitude\)
You get altitude from this.

How to realize it will be a right triangle without knowing the property:
You can do that by imagining the situation in which the area will be minimum. When the two sides overlap (i.e the angle between them is 0), the area will be 0 i.e. there will be no triangle. As you keep moving the sides away from each other, the area will increase till it eventually becomes 0 again when the angle between them is 180. So the maximum area between them will be when the angle between the sides is 90.
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 25 Apr 2014, 05:51
Bunuel wrote:
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


Having some trouble figuring out why right isosceles triangle has greater area than equilateral triangle
Anyone would mind clarifying this?

Cheers!
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 27 Apr 2014, 22:39
jlgdr wrote:
Bunuel wrote:
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


Having some trouble figuring out why right isosceles triangle has greater area than equilateral triangle
Anyone would mind clarifying this?

Cheers!
J :)


Couple of ways to think about it:

Method 1:
Say base of a triangle is 1.
Area = (1/2)*base*height = (1/2)*height

Say, another side has a fixed length of 1. You start with the first figure on top left when two sides are 1 and third side is very small and keep rotating the side of length 1. The altitude keeps increasing. You get an equilateral triangle whose altitude is \(\sqrt{3}/2 * 1\) which is less than 1. Then you still keep rotating till you get the altitude as 1 (the other side). Now altitude is max so area is max. This is a right triangle.
When you rotate further still, the altitude will start decreasing again.
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Method 2:

Given in my post above.
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 28 Nov 2014, 02:15
I found this statement as a rule in a flashcard: "If you are given 2 sides of a triangle or a parallelogram, you can maximize the area by placing those two sides perpendicular." I had some concerns in appreciating this statement and I find it related to this question. Was able to understand the statement better when I solved this question.

For those who are unable to appreciate this statement from the flash card, hope this below example helps! Let's say we know that the 2 sides of a triangle are 4 and 3. The question asks when will the area of the triangle with 2 of these vertices be the maximum. Consider these 3 scenarios:
Attachment:
Triangles.png
Triangles.png [ 3.87 KiB | Viewed 2173 times ]

What do we see from the figure above? The triangle with 2 sides as perpendicular bisectors will have the greatest areas. If you knew this rule, this question would have been a cake walk! :)

Hope this helps! :-D
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 28 Apr 2015, 05:11
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Here's how this question can be solved from the first principles, by making use of only the most basic properties:

The first step is to draw a diagram to visualize the given information. Here it is:

Image

We've dropped a perpendicular OP on the base AB.

So, the area of the triangle OAB = \(\frac{1}{2}\)*AB*OP . . . (1)

Now, let's assume the angle AOP to be x degrees.

So, OP = OAcosx = 1*cosx = cosx . . . (2)
Similarly, AP = OAsinx = 1*sinx = sinx . . . (3)

Now, we know that the perpendicular drawn from the center of a circle to a chord bisects the chord.

So, perpendicular OP bisects the chord AB.
Therefore, AP = BP = AB/2 . . . (4)

Combining (3) and (4), we get:

AB = 2sinx . . . (5)

Using (2) and (5) in (1), we get:

Area of triangle OAB = \(\frac{1}{2}\)*(2sinx)*(cosx)

So, Area of triangle OAB = (sinx)*(cosx)

We have to maximize the area, and therefore, we've to maximize the value of (sinx)*(cosx)

When sinx is maximum (for x = 90 degrees), cosx = 0, and so, the product (sinx)*(cosx) = 0
When cosx is maximum (for x = 0 degrees), sinx = 0, and so, the product (sinx)*(cosx) = 0

It's easy to see that the product (sinx)*(cosx) will be maximum when sinx = cosx. This happens for x = 45 degrees. At this point, sinx = cosx = \(\frac{1}{\sqrt{2}}\)

Therefore, maximum area = (\(\frac{1}{\sqrt{2}}\))*(\(\frac{1}{\sqrt{2}}\)) = \(\frac{1}{2}\)

As you can see, in solving this question, we have used only three basic properties from the concepts of Triangles, Circles and Trigonometry respectively. All students preparing for the GMAT already know these three basic properties:

1. The formula for area of triangle
2. The property that the perpendicular drawn from the center of a circle to a chord bisects the chord
3. The values of sin x and cos x for x = 0 degrees, 45 degrees and 90 degrees


Takeaway from the discussion

The questions on GMAT do not test your knowledge of esoteric properties, but your ability to apply basic concepts from a variety of topics

Hope this helped! :)

Regards

Japinder
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Re: What is the greatest possible area of a triangular region with one ver  [#permalink]

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New post 09 Jan 2019, 09:54
slingfox wrote:
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. \(\sqrt{2}\)


Image


\(?\,\,\, = \,\,{S_{\Delta ABC}}\,\,\max\)

Let C be the center of the circle with unitary radius.

Without loss of generality, we may (and will) assume point A is one unit at the right of point C (as shown in the figure on the left).

For the last vertex (B), without loss of generality (in terms of exploring possible areas) there are only two possibilities:

(1) B is in the arc AD (figure in the middle) or (2) B is in the arc DE (figure on the right)

In BOTH cases we have:

\({S_{\Delta ABC}} = {{AC \cdot h} \over 2} = {h \over 2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = {1 \over 2}\,\,\,\,\,\,\left( {{\rm{when}}\,\,h = CD\,,\,\,{\rm{i}}{\rm{.e}}{\rm{.}},\,\,B = D} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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