Q98 / 170 · Manipal (MET) 2015 · Mathematics

In a $\triangle A B C$, if $a^2+b^2+c^2=a c+\sqrt{3} a b$, then the triangle is

  • A
    equilateral
  • B
    right angled and isosceles
  • Correct answer
    C
    right angled and not isosceles
  • D
    None of the above

Solution

We have, $a^2+b^2+c^2=a c+\sqrt{3} a b$ $\begin{aligned} & \Rightarrow \quad \frac{a^2}{4}+c^2-a c+\frac{3 a^2}{4}+b^2-\sqrt{3} a b=0 \\ & \Rightarrow \quad\left(\frac{a}{2}-c\right)^2+\left(\frac{\sqrt{3} a}{2}-b\right)^2=0\end{aligned}$ $\begin{array}{ll}\Rightarrow & \frac{a}{2}-c=0 \text { and } \frac{\sqrt{3}}{2} a-b=0 \\ \Rightarrow & a=2 c \text { and } \sqrt{3} a=2 b \\ \Rightarrow & a=\frac{2 b}{\sqrt{3}}=2 c=\lambda \\ \Rightarrow & a=\lambda, b=\frac{\sqrt{3}}{2} \lambda \text { and } c=\frac{\lambda}{2}\end{array}$ Now, $b^2+c^2=\left(\frac{\sqrt{3} \lambda}{2}\right)^2+\left(\frac{\lambda}{2}\right)^2$ $=\frac{3 \lambda^2}{a}+\frac{\lambda^2}{4}=\lambda^2$ $\therefore \quad b^2+c^2=a^2$ Hence, the triangle is right angled but not isosceles.

Hence, the correct answer is option: 'C'.

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