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Square inscribed in equilateral triangle. Perimeter = ?

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by gmattesttaker2 » Sun May 04, 2014 11:43 pm
Hello,

Can you please assist with this:

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

OA: [spoiler]8 sq. root (3) + 12[/spoiler]


I came up with the following diagram.


We now have diagonal of the square i.e. 4 sq. root 2

In triangle ADC, by 30 60 90 rule we have:

30 : 60 : 90
1 : sq. root(3) : 2
x : x sq. root(3) : 2x

From the diagram, side opposite to 60 degrees is 4 sq. root (2)
=> x sq. root(3) = 4 sq. root (2)
=> x = 4 sq. root(2)/sq. root(3)

Side opposite to 30 degrees i.e. CD = x = 4 sq. root(2) / sq. root(3)

Hence, base of equilateral triangle BC = 2 x ( 4 sq. root(2) / sq. root(3) )
= 8 sq. root(2) / sq. root(3)

Since all sides are equal, perimeter = 3 x ( 8 sq. root(2) / sq. root(3) )
= 8 sq. root (6)

Can you please tell me where I am going wrong?

Thanks for your help.

Regards,
Sri
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by kartikc11 » Mon May 05, 2014 12:56 am
it's pretty simple, do this:

create a triangle and square as shown, you will see that the side opposite the 30degrees in the small triangle, becomes 4 sq root 3. then the diagram should be self explanatory, if not, let me know and i can explain in more detail
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by gmattesttaker2 » Tue May 06, 2014 6:31 pm
kartikc11 wrote:it's pretty simple, do this:

create a triangle and square as shown, you will see that the side opposite the 30degrees in the small triangle, becomes 4 sq root 3. then the diagram should be self explanatory, if not, let me know and i can explain in more detail
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Hello Kartik,

Thank you very much for your excellent diagram. It explains everything clearly. Thanks again.

Best Regards,
Sri
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by ibmagladry » Wed May 07, 2014 3:39 pm
kartikc11 wrote:it's pretty simple, do this:

create a triangle and square as shown, you will see that the side opposite the 30degrees in the small triangle, becomes 4 sq root 3. then the diagram should be self explanatory, if not, let me know and i can explain in more detail
Image
Isn't it 12+[24/sqrt(3)]?

When you multiple 4+[8/sqrt(3)] by 3 it should be 4*3=12 and [8/sqrt(3)]*3=24/sqrt(3).

Alternatively, you could add and +[8/sqrt(3)]+[8/sqrt(3)]++[8/sqrt(3)]= 12+[24/sqrt(3)]
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by kartikc11 » Fri Jun 20, 2014 5:53 am
You are correct, but if you did not have 24/ sqrt 3 in the answer choices, it would be time to see how else you can calculate it.

So, keep the 8 aside, and take it as 3/sqrt3. If you work with the powers, you will get 3^1 / 3^(1/2) which then becomes 3^(1-(1/2)) = 3^(1/2) which is Sqrt 3. Bring back the 8, giving 8/sqrt3 + 12
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by [email protected] » Fri Jun 20, 2014 10:47 am
Hi Sri,

The immediate issue is that you CANNOT place a square inside an equilateral triangle in that way. Each corner of a square is 90 degrees, while each corner of an equilateral triangle is 60 degrees, so the two angles won't match. The proper way to place the square is explained by the other posts in this thread.

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by GMATinsight » Fri Jun 20, 2014 8:41 pm
The alternate approach is as mentioned below:



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