Question

a+b+c=12 and a2+b2+c2=100 find ab+bc+ca

Answer 1

$\textbf{Given:}$

$a+b+c=12\;\text{and}\;a^2+b^2+c^2=100$

$\textbf{To find:}\;ab+bc+ca$

$\text{We know that, the folwing identity}$

$\boxed{\bf\,(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)}$

$\implies(12)^2=100+2(ab+bc+ca)$

$\implies\,144=100+2(ab+bc+ca)$

$\implies\,144-100=2(ab+bc+ca)$

$\implies\,44=2(ab+bc+ca)$

$\implies\,ab+bc+ca=\frac{44}{2}$

$\implies\boxed{\bf\,ab+bc+ca=22}$

$\therefore\textbf{The value of ab+bc+ca is 22}$

Find more:

If p SQUARE + q SQUARE + r SQUARE = 20 and pq + qr + rs = 15, then the value of p + q + r is

https://brainly.in/question/4795546

Answer 2

Answer:

ab+bc+ca=22 ( ANS )

Step-by-step explanation:

Given a+b+c=12 and a2+b2+c2=100

using the formula ( a+b+c ) ^2 =a2+b2+c2+2( ab+bc+ca )

                         => ( 12 ) ^2 = 100 + 2( ab+bc+ca )

                         => 144 = 100 + 2( ab+bc+ca )

                         => 100 + 2( ab+bc+ca ) = 144

                         => 2( ab+bc+ca ) = 144 - 100

                         => ab + bc + ca = 44 / 2

                         => ab + bc + ca = 22 ( ANS )

   Hope it helps you guys.Till then take care and goodbye!