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The area of an equilateral triangle is greater than

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The area of an equilateral triangle is greater than [#permalink] New post 17 Sep 2017, 03:57
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Question Stats:

80% (00:50) correct 20% (00:38) wrong based on 10 sessions



The area of an equilateral triangle is greater than \(25 \sqrt{3}\) but less than \(36 \sqrt{3}\)

Quantity A
Quantity B
The length of one of the sides of the triangle
9


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The area of an equilateral triangle is greater than [#permalink] New post 21 Sep 2017, 07:38
The area of an equilateral triangle is \(\frac{b*h}{2}\), where b one of the sides we are looking for to compute column A. Then h, since half equilateral triangle is a 30-60-90 triangle, is equal to \(b/2*sqrt(3)\). If we take the minimum value that the area can take, we can write this equation \(\frac{b*\frac{b}{2}*sqrt(3)}{2}=25sqrt(3)\) that leads to a value of b equal to 10. If 10 is the length of the side needed to get the minimal area, this means that to get all the other area values, which are greater by definition, the side must be greater than 10. Thus we are sure that column A is greater than column B, Answer A!
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Re: The area of an equilateral triangle is greater than [#permalink] New post 22 Sep 2017, 10:12
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IlCreatore wrote:
The area of an equilateral triangle is \(\frac{b*h}{2}\), where b one of the sides we are looking for to compute column A. Then h, since half equilateral triangle is a 30-60-90 triangle, is equal to \(b/2*sqrt(3)\). If we take the minimum value that the area can take, we can write this equation \(\frac{b*\frac{b}{2}*sqrt(3)}{2}=25sqrt(3)\) that leads to a value of b equal to 10. If 10 is the length of the side needed to get the minimal area, this means that to get all the other area values, which are greater by definition, the side must be greater than 10. Thus we are sure that column A is greater than column B, Answer A!



Just want to prove the area of Equilateral triangle first -

Image

In the figure above, the sides of an equilateral triangle are equal to “a” units.
We know that the area of Triangle is given by;
A = \(\frac{1}{2} * base * height\)




To find the height, consider Triangle ABC,
Applying Pythagoras Theorem we know,

\(AB^2\) = \(AD^2\)+\(BD^2\)

\(a^2\) = \(h^2\) + \(\frac{a}{2} ^2\)

\(h^2\) = \(a^2\) - \(\frac{a^2}{4}\)

\(h^2\) = \(\frac{3a^2}{4}\)

h = \(\sqrt{3}\frac{a}{2}\)

Thus, we can calculate area by the basic equation,

Area = \(\frac{1}{2} * base * height\) = \(\frac{1}{2} * a*\sqrt{3}\frac{a}{2}\)

Therefore, Area of Equilateral Triangle = \(\sqrt{3}\frac{a^2}{4}\)

Now back to ques we can use -

\(25 sqrt3\) < \(\sqrt{3}\frac{a^2}{4}\) < \(36 sqrt3\)

solving we get - 10<a<12 (since denominator 4 = \(2^2\) which is always positive)

Therefore a>9. so option A is correct
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Re: The area of an equilateral triangle is greater than   [#permalink] 22 Sep 2017, 10:12
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