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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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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 ninesided 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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Originally posted by TirthankarP on 15 Sep 2013, 10:08.
Last edited by Bunuel on 15 Sep 2013, 12:31, edited 3 times in total.
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Re: In the given figure, the area of the equilateral triangle is
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15 Sep 2013, 12:29




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Re: In the given figure, the area of the equilateral triangle is
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19 Nov 2013, 22:18
Buneul, how did you estimate \(fourth\sqrt{3}\)=\(\sqrt{2}\)?



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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 \(fourth\sqrt{3}\)=\(\sqrt{2}\)? \(side^2=64\sqrt{3}\) > \(side^2=8^2*\sqrt{3}\) > \(side=\sqrt{8^2*\sqrt{3}}\)> \(side=8\sqrt[4]{3}\). Hope it's clear.
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13 Feb 2014, 06:50
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 \(fourth\sqrt{3}\)=\(\sqrt{2}\)? \(side^2=64\sqrt{3}\) > \(side^2=8^2*\sqrt{3}\) > \(side=\sqrt{8^2*\sqrt{3}}\)> \(side=8\sqrt[4]{3}\). Hope it's clear. The trick here is that any positive integer root from a number more than 1 will be more than 1. For example: \(\sqrt[1000]{2}>1\). So, we know that \(1<\sqrt[3]{3}<2\). We also know that \(\sqrt{2}\approx {1.4}\). So, \(1<(\sqrt[4]{3}\approx{\sqrt{2}})<2\). Does this make sense?
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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!!
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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 ninesided 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}?



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Re: In the given figure, the area of the equilateral triangle is
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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 ninesided 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. Questions involving rationalization to practice: ifx0then106291.htmlifnispositivewhichofthefollowingisequalto31236.htmlconsideraquarterofacircleofradius16letrbethe131083.htmlinthediagramnotdrawntoscalesectorpqisaquarter139282.htmlinthediagramwhatisthevalueofx129962.htmltheperimeterofarightisosclestriangleis127049.htmlwhichofthefollowingisequalto98531.htmlifxispositivethen1rootx1rootx163491.html12sqrt364378.htmlifasquaremirrorhasa20inchdiagonalwhatisthe99359.htmlHope it helps. Thanks Bunuel for this list



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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.



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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 ninesided 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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