a2+b2c2a2+b2c2\frac{{{a^2} + {b^2}}}{{{c^2}}} = b2+c2a2" role="presentation">b2+c2a2b2+c2a2\frac{{{b^2} + {c^2}}}{{{a^2}}} = c2+a2b2" role="presentation">c2+a2b2c2+a2b2\frac{{{c^2} + {a^2}}}{{{b^2}}} = 1k," role="presentation">1k,1k,\frac{1}{k}{\text{,}} (k≠0)" role="presentation">(k≠0)(k≠0)\left( {k \ne 0} \right) then k = ?" /> a2+b2c2a2+b2c2\frac{{{a^2} + {b^2}}}{{{c^2}}} = b2+c2a2" role="presentation">b2+c2a2b2+c2a2\frac{{{b^2} + {c^2}}}{{{a^2}}} = c2+a2b2" role="presentation">c2+a2b2c2+a2b2\frac{{{c^2} + {a^2}}}{{{b^2}}} = 1k," role="presentation">1k,1k,\frac{1}{k}{\text{,}} (k≠0)" role="presentation">(k≠0)(k≠0)\left( {k \ne 0} \right) then k = ?" /> a2+b2c2a2+b2c2\frac{{{a^2} + {b^2}}}{{{c^2}}} = b2+c2a2" role="presentation">b2+c2a2b2+c2a2\frac{{{b^2} + {c^2}}}{{{a^2}}} = c2+a2b2" role="presentation">c2+a2b2c2+a2b2\frac{{{c^2} + {a^2}}}{{{b^2}}} = 1k," role="presentation">1k,1k,\frac{1}{k}{\text{,}} (k≠0)" role="presentation">(k≠0)(k≠0)\left( {k \ne 0} \right) then k = ?" />

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If $\frac{a^{2} + b^{2}}{c^{2}}$ $\frac{{{a^2} + {b^2}}}{{{c^2}}}$ = $\frac{b^{2} + c^{2}}{a^{2}}$ $\frac{{{b^2} + {c^2}}}{{{a^2}}}$ = $\frac{c^{2} + a^{2}}{b^{2}}$ $\frac{{{c^2} + {a^2}}}{{{b^2}}}$ = $\frac{1}{k},$ $\frac{1}{k}{\text{,}}$ $(k \neq 0)$ $\left( {k \ne 0} \right)$ then k = ?

12
21
30
4 $\frac{1}{2}$ $\frac{1}{2}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a^{2} + b^{2}}{c^{2}} = \frac{b^{2} + c^{2}}{a^{2}} = \frac{c^{2} + a^{2}}{b^{2}} = \frac{1}{k} \\ Put a = b = c = 1 \\ \Rightarrow \frac{1 + 1}{1} + \frac{1 + 1}{1} + \frac{1 + 1}{1} = \frac{1}{k} \\ \Rightarrow 2 = 2 = 2 = \frac{1}{k} \\ \Rightarrow k = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{{{a^2} + {b^2}}}{{{c^2}}} = \frac{{{b^2} + {c^2}}}{{{a^2}}} = \frac{{{c^2} + {a^2}}}{{{b^2}}} = \frac{1}{k} \cr & {\text{Put }}a = b = c = 1 \cr & \Rightarrow \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} = \frac{1}{k} \cr & \Rightarrow 2 = 2 = 2 = \frac{1}{k} \cr & \Rightarrow k = \frac{1}{2} \cr}$

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=

\frac{b^{2} + c^{2}}{a^{2}}

$\frac{{{b^2} + {c^2}}}{{{a^2}}}$ =

\frac{c^{2} + a^{2}}{b^{2}}

$\frac{{{c^2} + {a^2}}}{{{b^2}}}$ =

\frac{1}{k},

$\frac{1}{k}{\text{,}}$

(k \neq 0)

$\left( {k \ne 0} \right)$ then k = ?", "text": "If

\frac{a^{2} + b^{2}}{c^{2}}

$\frac{{{a^2} + {b^2}}}{{{c^2}}}$ =

\frac{b^{2} + c^{2}}{a^{2}}

$\frac{{{b^2} + {c^2}}}{{{a^2}}}$ =

\frac{c^{2} + a^{2}}{b^{2}}

$\frac{{{c^2} + {a^2}}}{{{b^2}}}$ =

\frac{1}{k},

$\frac{1}{k}{\text{,}}$

(k \neq 0)

$\left( {k \ne 0} \right)$ then k = ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{a^{2} + b^{2}}{c^{2}} = \frac{b^{2} + c^{2}}{a^{2}} = \frac{c^{2} + a^{2}}{b^{2}} = \frac{1}{k} \\ Put a = b = c = 1 \\ \Rightarrow \frac{1 + 1}{1} + \frac{1 + 1}{1} + \frac{1 + 1}{1} = \frac{1}{k} \\ \Rightarrow 2 = 2 = 2 = \frac{1}{k} \\ \Rightarrow k = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{{{a^2} + {b^2}}}{{{c^2}}} = \frac{{{b^2} + {c^2}}}{{{a^2}}} = \frac{{{c^2} + {a^2}}}{{{b^2}}} = \frac{1}{k} \cr & {\text{Put }}a = b = c = 1 \cr & \Rightarrow \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} = \frac{1}{k} \cr & \Rightarrow 2 = 2 = 2 = \frac{1}{k} \cr & \Rightarrow k = \frac{1}{2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{a^{2} + b^{2}}{c^{2}} = \frac{b^{2} + c^{2}}{a^{2}} = \frac{c^{2} + a^{2}}{b^{2}} = \frac{1}{k} \\ Put a = b = c = 1 \\ \Rightarrow \frac{1 + 1}{1} + \frac{1 + 1}{1} + \frac{1 + 1}{1} = \frac{1}{k} \\ \Rightarrow 2 = 2 = 2 = \frac{1}{k} \\ \Rightarrow k = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{{{a^2} + {b^2}}}{{{c^2}}} = \frac{{{b^2} + {c^2}}}{{{a^2}}} = \frac{{{c^2} + {a^2}}}{{{b^2}}} = \frac{1}{k} \cr & {\text{Put }}a = b = c = 1 \cr & \Rightarrow \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} = \frac{1}{k} \cr & \Rightarrow 2 = 2 = 2 = \frac{1}{k} \cr & \Rightarrow k = \frac{1}{2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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