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In the given figure, the area of the equilateral triangle is
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Updated on: 15 Sep 2013, 12:31
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68% (02:51) correct 32% (03:08) wrong based on 359 sessions
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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) \(384\)
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15 Sep 2013, 12:29
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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) \(384\)
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 = \(9*8\sqrt[4]{3}=72\sqrt[4]{3}\) --> \(\sqrt[4]{3}\approx{\sqrt{2}}\).
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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 \(fourth\sqrt{3}\)=\(\sqrt{2}\)?
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13 Feb 2014, 06:50
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gmatprav wrote:
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
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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!!
_________________
Re: In the given figure, the area of the equilateral triangle is
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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}?
Re: In the given figure, the area of the equilateral triangle is
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13 Dec 2016, 10:33
1
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
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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
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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.
Re: In the given figure, the area of the equilateral triangle is
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09 Aug 2017, 12:44
TirthankarP wrote:
Attachment:
Q4_Quant.png
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.
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