If $a + b + c = 12$ and $a^{2} + b^{2} + c^{2} = 50$ , then ab + bc + ca = .

84

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88

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47

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94

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Solution

The correct option is C $47$
It is given that $a + b + c = 12$

Using identity ,
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + a c + b c)$ , we have

$12^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + a c + b c)$

Substituting the value of $a^{2} + b^{2} + c^{2}$ which is 50,we get
$144 = 50 + 2 (a b + a c + b c)$
$⟹$ $2 (a b + a c + b c) = 144 - 50$
$⟹$ $2 (a b + a c + b c) = 94$
$⟹ a b + a c + b c = 47$

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Similar questions

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

a + b + c = 0

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

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

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