Trigonometric Equations Question 378
Question: If A is the area and 2s the sum of 3 sides of triangle, then
Options:
A) $ A\le \frac{s^{2}}{3\sqrt{3}} $
B) $ A\le \frac{s^{2}}{2} $
C) $ A>\frac{s^{2}}{\sqrt{3}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- We have, $ 2s=a+b+c,A^{2}=s(s-a)(s-b)(s-c) $ $ \because \ A\text{.M}.\ge G\text{.M} $ .
Þ $ \frac{s-a+s-b+s-c}{3}\ge \sqrt[3]{(s-a)(s-b)(s-c)} $
Þ $ \frac{3s-2s}{3}\ge \frac{{{(A^{2})}^{1/3}}}{{s^{1/3}}}\Rightarrow \frac{s^{3}}{27}\ge \frac{A^{2}}{s}\Rightarrow A\le \frac{s^{2}}{3(\sqrt{3})} $ .
BETA
