If $a + b + c = 8$ $a^{2} + b^{2} + c^{2} = 40$ . Find the value of $a b + b c + c a$ .

10

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12

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14

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Solution

The correct option is B $12$
Given,
$a + b + c = 8$ and $a^{2} + b^{2} + c^{2} = 40$
$∴ {(a + b + c)}^{2} = 64$
$a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 64$
$40 + 2 (a b + b c + c a) = 64$
$a b + b c + c a = \frac{64 - 40}{2}$

$a b + b c + c a = 12$

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Similar questions

If $a^{2} + b^{2} + c^{2} = 16$ and ab + bc + ca = 10, find the value of a + b + c.

If $a + b + c = 9$ and $a b + b c + c a = 40$ . Find the value of $a^{2} + b^{2} + c^{2}$ .

a + b + c = 12

a b + b c + c a = 60

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

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

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