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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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15 Sep 2013, 10:08
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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) $$384$$
[Reveal] Spoiler: OA

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Collection of some good questions on Number System

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

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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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20 Nov 2013, 02:28
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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]

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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:
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]

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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:
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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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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13 Feb 2014, 10:12
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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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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 = $$9*8\sqrt[4]{3}=72\sqrt[4]{3}$$ --> $$\sqrt[4]{3}\approx{\sqrt{2}}$$.

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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13 Dec 2016, 10:33
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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 = $$9*8\sqrt[4]{3}=72\sqrt[4]{3}$$ --> $$\sqrt[4]{3}\approx{\sqrt{2}}$$.

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
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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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 = $$9*8\sqrt[4]{3}=72\sqrt[4]{3}$$ --> $$\sqrt[4]{3}\approx{\sqrt{2}}$$.

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
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
Re: In the given figure, the area of the equilateral triangle is   [#permalink] 26 Jan 2017, 13:38
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