Question

If $\Delta$ be the area of a triangle $ABC$ and length of its two sides are $3$ and $5.$ If $c$ is the third side, then :

• A. $\Delta \le \frac{(c^2+16c+64)}{12\sqrt{3}}$
• B. $\Delta =\frac{(c^2+16c+64)}{8\sqrt{3}}$
• C. $\Delta >\frac{(c^2+16c+64)}{4\sqrt{3}}$
• D. none of these

Answer 1

The correct option is A $\Delta \le \frac{(c^2+16c+64)}{12\sqrt{3}}$

Given,

Area of the triangle $=\Delta$

Let $AB=3=a$

$BC=5=b$

Then,

$\frac{a+b+c}{2}=S$ [where s is

semiperimeter of

triangle $]$.

$\begin{array}{cc}\Rightarrow \frac{3+5+c}{2}& =S\\ \Rightarrow 3+5+C=& 25\\ & \frac{8+C}{2}=S\end{array}$

$\because$ we know, $A\cdot M⩾G\cdot M$

$\because \Delta =\sqrt{s(s-a)(s-b)(5-c)}$

Now,

$\Rightarrow \frac{(s-a)+(s-b)+(s-c)}{3}⩾\left(\right(s-a\left)\right(5-b\left)\right(s-c){)}^{1/3}$

$\Rightarrow \frac{3s-(a+b+c)}{3}⩾\left((\frac{5(5-a)(5-b)(s-c)){)}^{1/3}}{5}\right).$

$\Rightarrow \frac{35-25}{3}⩾{\left(\left(\frac{{\Delta }^2}{S}\right)\right)}^{1/3}$

$\Rightarrow {\left(\frac{5}{3}\right)}^3⩾\left(\frac{{\Delta }^2}{5}\right)$

$\Rightarrow \frac{s^4}{3^3}⩾{\Delta }^2$

$\Rightarrow \sqrt{\frac{5^4}{3^3}}⩾\Delta$

$\Rightarrow \frac{5^4}{\sqrt{3\times 3\times 3}}⩾0$

$\Rightarrow \frac{s^2}{3\sqrt{3}}⩾D[\because s=\frac{8+c}{2}]$

$={\left(\frac{8+c}{2}\right)}^2⩾\Delta$

$=(\frac{8+c{)}^2}{2}3\sqrt{3}⩾0$

$\Rightarrow \frac{c^2+64+16c}{4\cdot 3\sqrt{3}}⩾\Delta$

$\Rightarrow \left(\frac{c^2+64+16c}{12\sqrt{3}}\right)⩾\Delta$ Option A