If (a+b+c) = 12 and a $^{2}$ +b $^{2}$ +c $^{2}$ = 108, then (ab+bc+ca) = ?

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Solution

The correct option is B
18

${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c)$

$⟹ (a b + b c + a c) = \frac{144 - 108}{2} = 18$

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Similar questions

a + b + c = 12

a^{2} + b^{2} + c^{2} = 50

a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

a + b + c = 0

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

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