If $a + b + c = 12$ and $a^{2} + b^{2} + c^{2} = 50$ ; find $a b + b c + c a$

22

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53

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36

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47

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Solution

The correct option is D $47$
We know,
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
Given,
$a + b + c = 12$ and
$a^{2} + b^{2} + c^{2} = 50$
Therefore,
$=> (12)^{2} = 50 + 2 (a b + b c + c a)$
$=> 144 = 50 + 2 (a b + b c + c a)$
$=> - 2 (a b + b c + c a) = 50 - 144$
$=> - 2 (a b + b c + c a) = - 94$
$=> (a b + b c + c a) = \frac{94}{2}$
$=> (a b + b c + c a) = 47$

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Similar questions

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

(a c + b c + c a) = 47

(a + b + c)

If $a + b + c = 10$ and $a^{2} + b^{2} + c^{2} = 36$ , then $a b + b c + c a =$ .......... .

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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