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In the given figure, the area of the equilateral triangle is [#permalink]

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In the given figure, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

A) \(8\sqrt{2}\) B) \(24\sqrt{3}\) C) \(72\sqrt{2}\) D) \(144\sqrt{2}\) E) \(384\)

In the given figure, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

A) \(8\sqrt{2}\) B) \(24\sqrt{3}\) C) \(72\sqrt{2}\) D) \(144\sqrt{2}\) E) \(384\)

The area of equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\).

So, we are given that \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\) --> \(side=8\sqrt[4]{3}\).

The perimeter = \(9*8\sqrt[4]{3}=72\sqrt[4]{3}\) --> \(\sqrt[4]{3}\approx{\sqrt{2}}\).

Re: In the given figure, the area of the equilateral triangle is [#permalink]

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13 Feb 2014, 03:49

Hi Bunuel the question is really how can we understand the value of \(3^{1/4} = \sqrt{2}\) ??

I was stuck at this point as well. How does \(9*8*(3)^{1/4}\) convert to \(9*8*\sqrt{2}\) It may be something very small but I am not able to wrap my head around it.

Bunuel wrote:

madn800 wrote:

Buneul, how did you estimate \(fourth\sqrt{3}\)=\(\sqrt{2}\)?

Hi Bunuel the question is really how can we understand the value of \(3^{1/4} = \sqrt{2}\) ??

I was stuck at this point as well. How does \(9*8*(3)^{1/4}\) convert to \(9*8*\sqrt{2}\) It may be something very small but I am not able to wrap my head around it.

Bunuel wrote:

madn800 wrote:

Buneul, how did you estimate \(fourth\sqrt{3}\)=\(\sqrt{2}\)?

Re: In the given figure, the area of the equilateral triangle is [#permalink]

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13 Feb 2014, 09:35

Thanks Bunuel, that makes sense now, coupled with the fact that the question is asking for approximate value not exact value. Thanks!!
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Re: In the given figure, the area of the equilateral triangle is [#permalink]

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10 Jul 2015, 05:03

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Re: In the given figure, the area of the equilateral triangle is [#permalink]

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17 Nov 2016, 19:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the given figure, the area of the equilateral triangle is [#permalink]

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13 Dec 2016, 03:24

Bunuel wrote:

In the given figure, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

A) \(8\sqrt{2}\) B) \(24\sqrt{3}\) C) \(72\sqrt{2}\) D) \(144\sqrt{2}\) E) \(384\)

The area of equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\).

So, we are given that \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\) --> \(side=8\sqrt[4]{3}\).

The perimeter = \(9*8\sqrt[4]{3}=72\sqrt[4]{3}\) --> \(\sqrt[4]{3}\approx{\sqrt{2}}\).

Answer: C.

This may be really obvious but how is \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\)? Dividing by fraction is multiplication with the inverse so side^2 would be 48*4/\sqrt{3}?

In the given figure, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

A) \(8\sqrt{2}\) B) \(24\sqrt{3}\) C) \(72\sqrt{2}\) D) \(144\sqrt{2}\) E) \(384\)

The area of equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\).

So, we are given that \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\) --> \(side=8\sqrt[4]{3}\).

The perimeter = \(9*8\sqrt[4]{3}=72\sqrt[4]{3}\) --> \(\sqrt[4]{3}\approx{\sqrt{2}}\).

Answer: C.

This may be really obvious but how is \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\)? Dividing by fraction is multiplication with the inverse so side^2 would be 48*4/\sqrt{3}?

\(side^2*\frac{\sqrt{3}}{4}=48\)

\(side^2=\frac{4*48}{\sqrt{3}}\)

Multiply by \(\frac{\sqrt{3}}{\sqrt{3}}\): \(side^2=\frac{4*48}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{4*48*\sqrt{3}}{3}=64\sqrt{3}\)

This algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator.

Re: In the given figure, the area of the equilateral triangle is [#permalink]

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26 Jan 2017, 13:38

Bunuel wrote:

koenh wrote:

Bunuel wrote:

In the given figure, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

A) \(8\sqrt{2}\) B) \(24\sqrt{3}\) C) \(72\sqrt{2}\) D) \(144\sqrt{2}\) E) \(384\)

The area of equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\).

So, we are given that \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\) --> \(side=8\sqrt[4]{3}\).

The perimeter = \(9*8\sqrt[4]{3}=72\sqrt[4]{3}\) --> \(\sqrt[4]{3}\approx{\sqrt{2}}\).

Answer: C.

This may be really obvious but how is \(side^2*\frac{\sqrt{3}}{4}=48\) --> \(side^2=64\sqrt{3}\)? Dividing by fraction is multiplication with the inverse so side^2 would be 48*4/\sqrt{3}?

\(side^2*\frac{\sqrt{3}}{4}=48\)

\(side^2=\frac{4*48}{\sqrt{3}}\)

Multiply by \(\frac{\sqrt{3}}{\sqrt{3}}\): \(side^2=\frac{4*48}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{4*48*\sqrt{3}}{3}=64\sqrt{3}\)

This algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator.

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