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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!

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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.

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

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03 Aug 2017, 08:10

madn800 wrote:

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

I think you need to know going into the test that root 2 is approx 1.41 and root 3 is approx 1.73 and root 5 is approx 2.2. So when you root 1.73 which is close to 2 you would get a number also close to root 2.

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) \(384\)

Since the area of the equilateral triangle is 48, we can use the following formula to determine the side:

(side^2 x √3)/4 = 48

side^2 = 192/√3

side^2 = 192/√3 x √3/√3

side^2 = 192√3/3

side^2 = 64√3

√side^2 = √(64√3)

side = (8)√(√3)

Since √3 ≈ 1.7, we have √(√3) ≈ √1.7 ≈ 1.3, and we have:

side ≈ (8)(1.3)

Since √2 is about 1.4, we have:

side ≈ 8√2

Thus, the perimeter is approximately 9 x 8√2 = 72√2.

Answer: C
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