Question

of a2+b2+c2:-200 and ab+BC+CA:-28. find a+b+c​

Answer 1

Step-by-step explanation:

${a}^{2} + {b}^{2} + {c}^{2} = 200 \: and \: ab + bc + ca$

$= 28 = > 2(ab + bc + ca) = 56 = >$

$a^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 256$

$= > {(a + b + c)}^{2} = 256 = >$

$(a + b + c) = \sqrt{256} = 16$

Hope it helps you.

Answer 2

Answer:

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