Question

If $a,b,c$ are natural numbers such that $a^2+b^2+c^2=29$ and $ab+bc+ca=26$, and $a+b+c=$_____.

Answer 1

Given: $a^2+b^2+c^2=29$ and $ab+bc+ca=26$

Using identity,

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

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

$(a+b+c{)}^2=29+2(26)$

$(a+b+c{)}^2=29+52$

$(a+b+c{)}^2=81$

$\therefore (a+b+c)=\sqrt{81}$

$(a+b+c)=\pm 9$

Since $a,b,c$ are natural numbers,

So, value of $(a+b+c)$ will be always positive.

$\therefore (a+b+c)=9$

Hence, (a) is the correct option.