Question

if a^2+b^2+c^2=200, ab+bc ca=28then the a+b+c is​

Answer 1

GiVeN :-

• $\: {a}^{2} + {b}^{2} + {c}^{2} = 200 \\ \\ ab + bc + ca = 28$

To FiNd :-

• ($a + b + c$) = ?

SoLuTiOn :-

Now,

We know that,

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

On putting the above given values in above identity , we get,

$\implies{(a + b + c)}^{2} = 200 + 2 \times (28) \\ \\ \implies \: {(a + b + c)}^{2} = 200 + 56 = 256 \\ \\ \implies \: (a + b + c) = \sqrt{256} \\ \\ \implies \: (a + b + c) = 16$

Answer 2

Answer:16

Step-by-step explanation:(a+b+c)²=a²+b²+c² +2(ab +bc + ca) (formula)

( a+b+c)²=200+2×(28)

(a+b+c)²=256

(a+b+c)=256^½

Hence, (a+b+c)=16