anonymous 5 years ago 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

1. anonymous

how did you bring latex in the question?

2. anonymous

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

3. anonymous

option a

4. anonymous

how??

5. 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!

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. 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

10. anonymous

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

11. anonymous

one side may be zero..and believe me it works in such questions ;) consider this not as mathematics but just "cheating" the examiner!

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

numbers need to be NON-NEGATIVE REALS

17. anonymous

okay got it.. thanx alot!