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

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- anonymous

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- anonymous

how did you bring latex in the question?

- anonymous

just formed the equation here.. and pasted it at the question area=)

- anonymous

option a

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## More answers

- anonymous

how??

- 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

considering c=0 would not make this triangle a triangle right? isnt it kind of risky to thnk like that?

- anonymous

if c=0 u don't get a triangle at all, its the degenerated case. so area=K=0

- anonymous

why would there be any risk? a,b,c can be reals the question didnt mention POSITIVE reals

- 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

but they also said that ABC is a triangle. a triangle has to have three sides

- 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

well okay..=)
can you explain the solution to me buddy?

- anonymous

It was a good question.
Use AM-GM inequality to (s-a),(s-b),(s-c)

- anonymous

i knw that for two numbers.. but what to do if there are three??

- anonymous

for n-numbers:
\[\frac{a+b+c+...n\ terms}{n}\ge \sqrt[n]{abc...}\]

- anonymous

numbers need to be NON-NEGATIVE REALS

- anonymous

okay got it.. thanx alot!

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