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by chayanika » Thu Aug 28, 2008 9:46 am
If a. b, c are the sides of a triangle, and a2 + b2 + c2 = bc + ca + ab, then the triangle is:
1)equilateral
2)isosceles
3)right angled
4)obtuse angled

Can someone help me to solve this?
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by pre-gmat » Thu Aug 28, 2008 10:49 am
It should be equilateral triangle. Try putting in the values and check it.
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by bomond » Thu Aug 28, 2008 11:03 am
Yes, there is easy way to solve this problem.

Multiply both side by 2

2a2+2b2+2c2=2bc+2ca+2ab
a2-2ab+b2+a2-2ca+c2+b2-2bc+c2=0
(a-b)^2+(a-c)^2+(b-c)^2=0
Each of them is >=0
So, a-b=0 a-c=0 b-c=0 =>a=b=c

Answer is equilateral
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by rabab » Wed Sep 10, 2008 11:12 am
bomond wrote: (a-b)^2+(a-c)^2+(b-c)^2=0
Each of them is >=0
i don't get this part. how come each of them is greater than= 0?
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by Stuart@KaplanGMAT » Wed Sep 10, 2008 12:46 pm
rabab wrote:
bomond wrote: (a-b)^2+(a-c)^2+(b-c)^2=0
Each of them is >=0
i don't get this part. how come each of them is greater than= 0?
Anything squared will be >= 0.
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