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What is the greatest possible area of a triangular region

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

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New post 28 Nov 2014, 01:15
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 1122 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  [#permalink]

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New post 28 Apr 2015, 04:11
1
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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What is the greatest possible area of a triangular region with o  [#permalink]

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New post 26 Oct 2015, 02:34
The height of the given triangle within circumference of the circle will be either one or less than that. Since Radi is 1. Therefore, to maximise area height as well as base have to be as high as possible. Since its given one of the points is at the centre, base=1 & max height=1; Hence answer=1\2.

Hope this helps.
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Re: What is the greatest possible area of a triangular region  [#permalink]

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New post 17 Jan 2016, 02:27
If you are given two sides of a triangle or parallelogram, you can maximize the area by placing those two sides perpendicular to each other.

1*1/2=1/2
Answer B
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Re: What is the greatest possible area of a triangular region  [#permalink]

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New post 09 Jan 2019, 08:54
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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Re: What is the greatest possible area of a triangular region &nbs [#permalink] 09 Jan 2019, 08:54

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