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In the given figure, the area of the equilateral triangle is

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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) \(384\)

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Collection of some good questions on Number System

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

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

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

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

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New post 20 Nov 2013, 02:28
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Re: In the given figure, the area of the equilateral triangle is  [#permalink]

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New post 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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New post 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 \(fourth\sqrt{3}\)=\(\sqrt{2}\)?


\(side^2=64\sqrt{3}\) --> \(side^2=8^2*\sqrt{3}\) --> \(side=\sqrt{8^2*\sqrt{3}}\)--> \(side=8\sqrt[4]{3}\).

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  [#permalink]

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New post 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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New post 13 Dec 2016, 03:24
Bunuel wrote:
Image
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}?
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Re: In the given figure, the area of the equilateral triangle is  [#permalink]

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New post 13 Dec 2016, 10:33
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koenh wrote:
Bunuel wrote:
Image
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.

Questions involving rationalization to practice:
if-x-0-then-106291.html
if-n-is-positive-which-of-the-following-is-equal-to-31236.html
consider-a-quarter-of-a-circle-of-radius-16-let-r-be-the-131083.html
in-the-diagram-not-drawn-to-scale-sector-pq-is-a-quarter-139282.html
in-the-diagram-what-is-the-value-of-x-129962.html
the-perimeter-of-a-right-isoscles-triangle-is-127049.html
which-of-the-following-is-equal-to-98531.html
if-x-is-positive-then-1-root-x-1-root-x-163491.html
1-2-sqrt3-64378.html
if-a-square-mirror-has-a-20-inch-diagonal-what-is-the-99359.html

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

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New post 26 Jan 2017, 13:38
Bunuel wrote:
koenh wrote:
Bunuel wrote:
Image
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.

Questions involving rationalization to practice:
if-x-0-then-106291.html
if-n-is-positive-which-of-the-following-is-equal-to-31236.html
consider-a-quarter-of-a-circle-of-radius-16-let-r-be-the-131083.html
in-the-diagram-not-drawn-to-scale-sector-pq-is-a-quarter-139282.html
in-the-diagram-what-is-the-value-of-x-129962.html
the-perimeter-of-a-right-isoscles-triangle-is-127049.html
which-of-the-following-is-equal-to-98531.html
if-x-is-positive-then-1-root-x-1-root-x-163491.html
1-2-sqrt3-64378.html
if-a-square-mirror-has-a-20-inch-diagonal-what-is-the-99359.html

Hope it helps.




Thanks Bunuel for this list :idea:
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Re: In the given figure, the area of the equilateral triangle is  [#permalink]

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New post 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  [#permalink]

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New post 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.

Answer: C
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