anonymous
  • anonymous
If K is the area of a triangle ABC and length of its two sides are 3 and 5. if c is the third side then a)\[K \le (c ^{2}+16c+64)/ 12\sqrt{3}\] b)\[K = (c ^{2}+16c+64)/ 8\] c)\[K \ge (c ^{2}+16c+64)/ 4\sqrt{3}\] d)none of these
Mathematics
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anonymous
  • anonymous
If K is the area of a triangle ABC and length of its two sides are 3 and 5. if c is the third side then a)\[K \le (c ^{2}+16c+64)/ 12\sqrt{3}\] b)\[K = (c ^{2}+16c+64)/ 8\] c)\[K \ge (c ^{2}+16c+64)/ 4\sqrt{3}\] d)none of these
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
how did you bring latex in the question?
anonymous
  • anonymous
just formed the equation here.. and pasted it at the question area=)
anonymous
  • anonymous
option a

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anonymous
  • anonymous
how??
anonymous
  • anonymous
is it correct? i sort of guessed. See when c=0, obviously K=0 and that cancels b) and c). Now generally according to my experience d)none of these is seldom correct. so a) has a good chance of being correct. ;) Well you may try cosine rule and the sine rule..But getting an inequality given in the question is appearing tough!
anonymous
  • anonymous
considering c=0 would not make this triangle a triangle right? isnt it kind of risky to thnk like that?
anonymous
  • anonymous
if c=0 u don't get a triangle at all, its the degenerated case. so area=K=0
anonymous
  • anonymous
why would there be any risk? a,b,c can be reals the question didnt mention POSITIVE reals
anonymous
  • anonymous
the solution given to the problem is this, [(s-a)(s-b)(s-c)]^1/3 <= 3s-(a+b+c) ----------- 3 (area^2 /s)^1/3<=s/3 area <= s^2/3rt3=c^2+16c+64/12rt3
anonymous
  • anonymous
but they also said that ABC is a triangle. a triangle has to have three sides
anonymous
  • anonymous
one side may be zero..and believe me it works in such questions ;) consider this not as mathematics but just "cheating" the examiner!
anonymous
  • anonymous
well okay..=) can you explain the solution to me buddy?
anonymous
  • anonymous
It was a good question. Use AM-GM inequality to (s-a),(s-b),(s-c)
anonymous
  • anonymous
i knw that for two numbers.. but what to do if there are three??
anonymous
  • anonymous
for n-numbers: \[\frac{a+b+c+...n\ terms}{n}\ge \sqrt[n]{abc...}\]
anonymous
  • anonymous
numbers need to be NON-NEGATIVE REALS
anonymous
  • anonymous
okay got it.. thanx alot!

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