Question : If $\frac{a^2+b^2}{c^2}=\frac{b^2+c^2}{a^2}=\frac{c^2+a^2}{b^2}=\frac{1}{k}$, $(k\neq 0)$, then $k$=?
Option 1: $2$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{2}$
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Correct Answer: $\frac{1}{2}$
Solution : Given: $\frac{a^2+b^2}{c^2}=\frac{b^2+c^2}{a^2}=\frac{c^2+a^2}{b^2}=\frac{1}{k}$ Solving the expressions, we get: $k(a^2+b^2)=c^2$ (equation 1). $k(b^2+c^2)=a^2$ (equation 2). $k(c^2+a^2)=b^2$ (equation 3). On adding all the equations, we get– $(a^2+b^2+c^2)=k(a^2+b^2+b^2+c^2+c^2+a^2)$ ⇒ $(a^2+b^2+c^2)=2k(a^2+b^2+c^2)$ ⇒ $k=\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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