Square inscribed in equilateral triangle. Perimeter = ?

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

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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Source: Beat The GMAT — Problem Solving | Sun May 04, 2014 11:43 pm

kartikc11: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Thu Feb 21, 2013 1:53 am; Thanked: 1 times

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

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

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

Hello Kartik,

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

Best Regards,
Sri

ibmagladry: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Tue May 06, 2014 3:24 pm

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

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)]

kartikc11: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Thu Feb 21, 2013 1:53 am; Thanked: 1 times

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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Rich

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Fri Jun 20, 2014 8:41 pm

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