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# A square with area 16 is perfectly inscribed inside an equ

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A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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12 Mar 2013, 22:31
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71% (02:31) correct 29% (02:57) wrong based on 197 sessions

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A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?

A. (83√3)3+4
B. 4√3+4
C. 8√3+12
D. 24+12
E. (32√3)3+12

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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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12 Mar 2013, 23:25
1
emmak wrote:
A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?
a) (83√3)3+4
b) 4√3+4
c) 8√3+12
d) 24+12
e) (32√3)3+12

As Bunuel has mentioned that GMAT doesn't test Trigonometry, I will give a geometric approach for this problem. Let's take the base of the triangle. The middle portion is of 4 units. The small right angle triangle formed to the left of the square has 4 units opposite 60 degrees. Thus, the side opposite 30 degrees is $$4/\sqrt{3}$$. This will be same for the small right angle triangle on the right of the square too. Thus the total length of the base = 2*$$4/\sqrt{3}$$+4. Thus the perimeter of the triangle is $$3*(8/\sqrt{3} +4) = 8*\sqrt{3}+12.$$

C.
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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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21 Nov 2013, 00:00
mau5 wrote:
emmak wrote:
A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?
a) (83√3)3+4
b) 4√3+4
c) 8√3+12
d) 24+12
e) (32√3)3+12

As Bunuel has mentioned that GMAT doesn't test Trigonometry, I will give a geometric approach for this problem. Let's take the base of the triangle. The middle portion is of 4 units. The small right angle triangle formed to the left of the square has 4 units opposite 60 degrees. Thus, the side opposite 30 degrees is $$4/\sqrt{3}$$. This will be same for the small right angle triangle on the right of the square too. Thus the total length of the base = 2*$$4/\sqrt{3}$$+4. Thus the perimeter of the triangle is $$3*(8/\sqrt{3} +4) = 8*\sqrt{3}+12.$$

C.

Hey mau5, could you explain how you knew that the opposite side of 30 degrees was $$4/\sqrt{3}$$? I follow your approach, but I don't get how you know that.

Thanks!
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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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21 Nov 2013, 00:19
2
unceldolan wrote:
mau5 wrote:
emmak wrote:
A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?
a) (83√3)3+4
b) 4√3+4
c) 8√3+12
d) 24+12
e) (32√3)3+12

As Bunuel has mentioned that GMAT doesn't test Trigonometry, I will give a geometric approach for this problem. Let's take the base of the triangle. The middle portion is of 4 units. The small right angle triangle formed to the left of the square has 4 units opposite 60 degrees. Thus, the side opposite 30 degrees is $$4/\sqrt{3}$$. This will be same for the small right angle triangle on the right of the square too. Thus the total length of the base = 2*$$4/\sqrt{3}$$+4. Thus the perimeter of the triangle is $$3*(8/\sqrt{3} +4) = 8*\sqrt{3}+12.$$

C.

Hey mau5, could you explain how you knew that the opposite side of 30 degrees was $$4/\sqrt{3}$$? I follow your approach, but I don't get how you know that.Thanks!

A right angle triangle where the angles are 30:60:90 , the respective opposite sides are always in the ratio $$1:\sqrt{3}:2.$$

As the side opposite 60 degrees is 4, hence, the side opposite 30 degrees would be$$\frac{4}{\sqrt{3}}$$

For getting a thorough understanding, go through this : math-triangles-87197.html
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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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26 Jun 2015, 00:31
mau5 wrote:
emmak wrote:
A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?
a) (83√3)3+4
b) 4√3+4
c) 8√3+12
d) 24+12
e) (32√3)3+12

As Bunuel has mentioned that GMAT doesn't test Trigonometry, I will give a geometric approach for this problem. Let's take the base of the triangle. The middle portion is of 4 units. The small right angle triangle formed to the left of the square has 4 units opposite 60 degrees. Thus, the side opposite 30 degrees is $$4/\sqrt{3}$$. This will be same for the small right angle triangle on the right of the square too. Thus the total length of the base = 2*$$4/\sqrt{3}$$+4. Thus the perimeter of the triangle is $$3*(8/\sqrt{3} +4) = 8*\sqrt{3}+12.$$

C.

Isn't the side equals to 4 under root 3 instead of 4/ under root 3..
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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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26 Jun 2015, 04:55
Can someone draw this?
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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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18 Sep 2016, 10:18
3
Following $$1:\sqrt{3}:2$$ rule in $$30:60:90$$ angled triangle.

Attachment:

photo_3.png [ 1.38 MiB | Viewed 1197 times ]

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Re: A square with area 16 is perfectly inscribed inside an equ  [#permalink]

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19 Nov 2017, 06:56
emmak wrote:
A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?

A. (83√3)3+4
B. 4√3+4
C. 8√3+12
D. 24+12
E. (32√3)3+12

No need to use trigonometry. Here's what I did.
Attachments

Triangle.jpg [ 3.37 MiB | Viewed 571 times ]

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Re: A square with area 16 is perfectly inscribed inside an equ &nbs [#permalink] 19 Nov 2017, 06:56
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