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In an Equilateral triangle, 3 coins of radii 1 unit each are kept in

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In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 13 Jul 2020, 08:23
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In an Equilateral triangle, 3 coins of radii 1 unit each are kept in such a way that they touch each other and also the sides of the triangle. What is the area of triangle (in sq. units)?

A. 4+5√2
B. 6+4√3
C. 4+6√3
D. 3+8√3
E. 5+3√2
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In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 15 Jul 2020, 08:27
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1
The radius of circle is one
The smaller triangle is a right angle triangle with 30, 60 and 90 degree as angle whose sides are x, √3*x, 2x
where x = 1 (on comparing)
therefore √3*x = √3

The Side of triangle = √3 + 2 + √3
=> 2√3 + 2

The area of equilateral triangle = √3/4*side^2
=> √3/4*(2√3 + 2)
=> 4√3 + 6 or 6 + 4√3
Answer is B

VERBAL1 you might be able to understand better with diagram. its pure methodological question.
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 13 Jul 2020, 09:37
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In an equilateral triangle of side x, if 3 circles of equal radii are drawn touching each other, then the relation between the side and the radius is given as x = 2r (\(\sqrt{3}\) + 1)

Here r = 1, therefore the side of the equilateral triangle = 2 * 1 (\(\sqrt{3}\) + 1) = 2(\(\sqrt{3}\) + 1)

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) * \(x^{2}\)

Therefore Area = \(\frac{\sqrt{3}}{4}\) * [\(2(\sqrt{3} + 1)] ^2\)

Area = \(\frac{\sqrt{3}}{4}\) * 4 * (3 + 2*\(\sqrt{3}\) + 1)

A = \(\sqrt{3}\) * (4 + 2 * \(\sqrt{3}\))

A = 4\(\sqrt{3}\) + 6

Option B

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 15 Jul 2020, 07:47
yashikaaggarwal , Bunuel request you'll to post a simpler solution for this one, please.
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 15 Jul 2020, 20:23
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yashikaaggarwal wrote:
The radius of circle is one
The smaller triangle is a right angle triangle with 30, 60 and 90 degree as angle whose sides are x, √3*x, 2x
where x = 1 (on comparing)
therefore √3*x = √3

The Side of triangle = √3 + 2 + √3
=> 2√3 + 2

The area of equilateral triangle = √3/4*side^2
=> √3/4*(2√3 + 2)
=> 4√3 + 6 or 6 + 4√3
Answer is B

VERBAL1 you might be able to understand better with diagram. its pure methodological question.



This is a good explanation!
I'm assuming this is the case just for an equilateral triangle. Right yashikaaggarwal?
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 15 Jul 2020, 20:37
KaramveerBakshi wrote:
yashikaaggarwal wrote:
The radius of circle is one
The smaller triangle is a right angle triangle with 30, 60 and 90 degree as angle whose sides are x, √3*x, 2x
where x = 1 (on comparing)
therefore √3*x = √3

The Side of triangle = √3 + 2 + √3
=> 2√3 + 2

The area of equilateral triangle = √3/4*side^2
=> √3/4*(2√3 + 2)
=> 4√3 + 6 or 6 + 4√3
Answer is B

VERBAL1 you might be able to understand better with diagram. its pure methodological question.



This is a good explanation!
I'm assuming this is the case just for an equilateral triangle. Right yashikaaggarwal?

Yes, because the question mention the equilateral triangle only. If it had been scalene or isosceles, it would have become complicated.

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 08:00
Where is the squre from the side?
Can anyone tell me how??

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 08:17
1
JHTIPU18 wrote:
Where is the squre from the side?
Can anyone tell me how??

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Your question is not clear. which square? the one with side 1?
I draw that. its not given. but can be inferred.
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 08:31
Side Suare from the Area.

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 08:34
The area of equilateral triangle = √3/4*side^2
=> √3/4*(2√3 + 2)

How you find the answer so easily. I can't understand.

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 08:43
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JHTIPU18 wrote:
The area of equilateral triangle = √3/4*side^2
=> √3/4*(2√3 + 2)

How you find the answer so easily. I can't understand.

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√3/4*side^2 is the area of equilateral triangle, I just find the side of triangle and put it into the formula to determine area.
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 09:40
√3/4*(2√3 + 2)
=> 4√3 + 6 or 6 + 4√3

How you can get the result? 🤔

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 09:55
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JHTIPU18 wrote:
√3/4*(2√3 + 2)^2
=> 4√3 + 6 or 6 + 4√3

How you can get the result? 🤔

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√3/4*(2√3 + 2)^2
√3/4*(12+4+8√3)
√3/4*(16+8√3)
√3/4*4(4+2√3)
√3*(4+2√3)
=> 4√3 + 6 or 6 + 4√3
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 10:00
Thank you so much! Very much appreciate!

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 10:25
CrackVerbalGMAT wrote:
In an equilateral triangle of side x, if 3 circles of equal radii are drawn touching each other, then the relation between the side and the radius is given as x = 2r (\(\sqrt{3}\) + 1)

Here r = 1, therefore the side of the equilateral triangle = 2 * 1 (\(\sqrt{3}\) + 1) = 2(\(\sqrt{3}\) + 1)

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) * \(x^{2}\)

Therefore Area = \(\frac{\sqrt{3}}{4}\) * [\(2(\sqrt{3} + 1)] ^2\)

Area = \(\frac{\sqrt{3}}{4}\) * 4 * (3 + 2*\(\sqrt{3}\) + 1)

A = \(\sqrt{3}\) * (4 + 2 * \(\sqrt{3}\))

A = 4\(\sqrt{3}\) + 6

Option B

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How is this √3 +1 is coming corresponding to 2r(√3+1)

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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 10:31
Shubham1213372 wrote:
CrackVerbalGMAT wrote:
In an equilateral triangle of side x, if 3 circles of equal radii are drawn touching each other, then the relation between the side and the radius is given as x = 2r (\(\sqrt{3}\) + 1)

Here r = 1, therefore the side of the equilateral triangle = 2 * 1 (\(\sqrt{3}\) + 1) = 2(\(\sqrt{3}\) + 1)

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) * \(x^{2}\)

Therefore Area = \(\frac{\sqrt{3}}{4}\) * [\(2(\sqrt{3} + 1)] ^2\)

Area = \(\frac{\sqrt{3}}{4}\) * 4 * (3 + 2*\(\sqrt{3}\) + 1)

A = \(\sqrt{3}\) * (4 + 2 * \(\sqrt{3}\))

A = 4\(\sqrt{3}\) + 6

Option B

Arun Kumar


How is this √3 +1 is coming corresponding to 2r(√3+1)

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its 2(√3 +1) not √3 +1
2*r*(√3 +1) = 2*1*(√3 +1)
2(√3 +1)
and rest are the stated steps
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 10:33
yashikaaggarwal wrote:
Shubham1213372 wrote:
CrackVerbalGMAT wrote:
In an equilateral triangle of side x, if 3 circles of equal radii are drawn touching each other, then the relation between the side and the radius is given as x = 2r (\(\sqrt{3}\) + 1)

Here r = 1, therefore the side of the equilateral triangle = 2 * 1 (\(\sqrt{3}\) + 1) = 2(\(\sqrt{3}\) + 1)

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) * \(x^{2}\)

Therefore Area = \(\frac{\sqrt{3}}{4}\) * [\(2(\sqrt{3} + 1)] ^2\)

Area = \(\frac{\sqrt{3}}{4}\) * 4 * (3 + 2*\(\sqrt{3}\) + 1)

A = \(\sqrt{3}\) * (4 + 2 * \(\sqrt{3}\))

A = 4\(\sqrt{3}\) + 6

Option B

Arun Kumar


How is this √3 +1 is coming corresponding to 2r(√3+1)

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its 2(√3 +1) not √3 +1
2*r*(√3 +1) = 2*1*(√3 +1)
2(√3 +1)
and rest are the stated steps


Thanks for you response.
However, I am confused on this part only --its 2(√3 +1).
How it is coming
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in  [#permalink]

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New post 16 Jul 2020, 10:38
1
yashikaaggarwal wrote:
The radius of circle is one
The smaller triangle is a right angle triangle with 30, 60 and 90 degree as angle whose sides are x, √3*x, 2x
where x = 1 (on comparing)
therefore √3*x = √3


The Side of triangle = √3 + 2 + √3
=> 2√3 + 2

The area of equilateral triangle = √3/4*side^2
=> √3/4*(2√3 + 2)
=> 4√3 + 6 or 6 + 4√3
Answer is B

VERBAL1 you might be able to understand better with diagram. its pure methodological question.

Shubham1213372 refer this
2(√3 + 1) is nothing but 2√3 + 2
where two is the rational part on the triangle side √3 is the irrational part, whose value is determined in the highlighted part above.
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Re: In an Equilateral triangle, 3 coins of radii 1 unit each are kept in   [#permalink] 16 Jul 2020, 10:38

In an Equilateral triangle, 3 coins of radii 1 unit each are kept in

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