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Hi All, Please help.

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by saurabhkamal1981 » Sat Dec 04, 2010 1:37 am
Hi Instructors,

Please help in the following question, which i encountered while practicing BTG practice questions. I have attached the image and the answer options are as follows:-

What is the perimeter of ABCDE ?

A). 36 + 6 sqrt 3
B). 48
C). 36 + 8 sqrt 3
D). 36 + 12 sqrt 3
E). 60

Answer is (E), but i am confused because in the explanation a line is drawn between A and D thus making them a 90 degree angle, and then taking triangle ADE where angles A and D is 60 degree, which concludes that angle E is also 60 degree thus making the triangle equilateral.

My question is:- If two angles and sides are equal they should be an isosceles triangle, then how we can conclude angle E as 60 degree; thus making the triangle equilateral.
or
To get the value of AE and DE we have to define the triangle as equilateral


Please help.

Waiting for your reply

Thanks & regards
Saurabh
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by zorya » Sat Dec 04, 2010 2:07 am
I'm not very sure of where your confusion lies. I did not deduce the value of angle E as per what you said.

Instead what I did was draw a line through E parallel to AB and CD and using the alternate angle theorem, I got angle E = 30+30=60 degrees.

Does this solution make things clearer for you? Let me know.
My GMAT score: 740 [Q:48, V:42]
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by Night reader » Sat Dec 04, 2010 2:38 am
saurabhkamal1981 wrote:Hi Instructors,

Please help in the following question, which i encountered while practicing BTG practice questions. I have attached the image and the answer options are as follows:-

What is the perimeter of ABCDE ?

A). 36 + 6 sqrt 3
B). 48
C). 36 + 8 sqrt 3
D). 36 + 12 sqrt 3
E). 60

Answer is (E), but i am confused because in the explanation a line is drawn between A and D thus making them a 90 degree angle, and then taking triangle ADE where angles A and D is 60 degree, which concludes that angle E is also 60 degree thus making the triangle equilateral.

My question is:- If two angles and sides are equal they should be an isosceles triangle, then how we can conclude angle E as 60 degree; thus making the triangle equilateral.
or
To get the value of AE and DE we have to define the triangle as equilateral


Please help.

Waiting for your reply

Thanks & regards
Saurabh
Dear saurabhkamal1981, thanks for bringing this problem up, I haven't tested myself with BTG problems yet.

this problem might be rated 500-level, as it tests two simple notes of the Geo concept about the right triangles.

two simple but main notes are the angle relationship in a right triangle 45:45:90 which gives you 1 : 1 : 1*Sqrt(2)-hypotenuse ratio between three sides of the triangle, and

30:60:90 - this gives you 1 : 1*Sqrt(3) : 2 relationship-hypotenuse

So if you concentrate on the image you can recognize there two right triangles with the common line/base divided by 2 - that is AD, hypotenuses AD and AE. They are equal 1*2 or just coefficient of the side at 30 degrees in a right triangle. The latter is 12/2 or 6.

So basically 2*6 (hypotenuse) * 2 (two hypotenuses) + 12*3 (three sides in rectangle)

this would be 60

Answer E.
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by saurabhkamal1981 » Sat Dec 04, 2010 3:33 am
zorya wrote:I'm not very sure of where your confusion lies. I did not deduce the value of angle E as per what you said.

Instead what I did was draw a line through E parallel to AB and CD and using the alternate angle theorem, I got angle E = 30+30=60 degrees.

Does this solution make things clearer for you? Let me know.
Hi All,

Thanks for your reply. I just researched and got the value of E.

So, here's my take:-

If a line is drawn between A and D, this will make <A and <D as a right angle (90 degree). Now take triangle AED and notice that <A and <D are 60 degree angles(90=60+30(given). We need a value of <E. So, sum of interior angles of a triangle is 180 degree.

<A + <D + <E = 180
60 + 60 + <E = 180
<E = 180 - 60 - 60
<E= 60

Therefore triangle AED is an equilateral triangle, and side AE and ED are 12 since AD is 12
Answer is E.

Please let me know your thoughts on this.

Thanks & Regards
Saurabh
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