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TirthankarP
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In the given figure, the area of the equilateral triangle is [#permalink] New post  Updated on: 15 Sep 2013, 11:31
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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 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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Originally posted by TirthankarP on 15 Sep 2013, 09:08.
Last edited by Bunuel on 15 Sep 2013, 11:31, edited 3 times in total.
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Re: In the given figure, the area of the equilateral triangle is [#permalink] New post  15 Sep 2013, 11:29
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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] New post  19 Nov 2013, 21:18
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] New post  20 Nov 2013, 01:28
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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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Re: In the given figure, the area of the equilateral triangle is [#permalink] New post  13 Feb 2014, 02:49
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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Re: In the given figure, the area of the equilateral triangle is [#permalink] New post  13 Feb 2014, 05:50
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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] New post  13 Feb 2014, 08:35
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] New post  13 Feb 2014, 09:12
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gmatprav wrote:
Thanks Bunuel, that makes sense now, coupled with the fact that the question is asking for approximate value not exact value. Thanks!!


Yes, the fact that we need only the approximate value is the key here.
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Re: In the given figure, the area of the equilateral triangle is [#permalink] New post  13 Dec 2016, 02:24
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] New post  13 Dec 2016, 09:33
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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] New post  26 Jan 2017, 12:38
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] New post  03 Aug 2017, 07:10
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] New post  09 Aug 2017, 11:44
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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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