Question

If a, b, c are sides of a triangle and $P=\frac{(a+b+c{)}^2}{ab+bc+ca}$, then

• A. P $\in$ [1, 2]
• B. P $\in$ [3, 4]
• C. P $\in$ [2, 4]
• D. None of these.

Answer 1

Answer: (A)

P $\in$ [1, 2]

P=$\frac{(a+b+c{)}^2}{ab+bc+ca}$

P=$\frac{{(\frac{\sin A}{k}+\frac{\sin B}{k}+\frac{\sin C}{k})}^2}{\frac{\sin A}{k}\times \frac{\sin B}{k}+\frac{\sin B}{k}\times \frac{\sin C}{k}+\frac{\sin C}{k}\times \frac{\sin A}{k}}$

P=$\frac{{(\sin A+\sin B+\sin C)}^2}{\sin A\sin B+\sin B\sin C+\sin C\sin A}$

Let $A=30°,B=60°andC=90°$

P=$\frac{{(\frac{1}{2}+\frac{\sqrt{3}}{2}+1)}^2}{(\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times 1+1\times \frac{1}{2})}$

P=$\frac{{(3+\sqrt{3})}^2}{(2+3\sqrt{3})}\Rightarrow \frac{{\left(4.73\right)}^2}{7.19}$ By taking $\sqrt{3}$=1.73

clearly, $Pϵ[1,2]$

Hence, $Pϵ[1,2]$ is correct.