Question

If $3a=4b=6c$ and $a+b+c=27\sqrt{29}$, then $\sqrt{a^2+b^2+c^2}$ is

• A. $3\sqrt{29}$
• B. $81$
• C. $87$
• D. $29$

Answer 1

The correct option is C $87$

$3a=4b=6c\Rightarrow 4b=6c\Rightarrow b=\frac{3}{2}c$ and $3a=4b\Rightarrow a=\frac{4}{3}b=\frac{4}{3}\times \frac{3}{2}c=2c$
$\therefore a+b+c=27\sqrt{29}\Rightarrow 2c+\frac{3}{2}c+c=27\sqrt{29}\Rightarrow \frac{9}{2}c=27\sqrt{29}\Rightarrow c=6\sqrt{29}$
Now $\sqrt{a^2+b^2+c^2}$
$=\sqrt{(a+b+c{)}^2-2(ab+bc+ca)}$
$=\sqrt{(27\sqrt{29}{)}^2-2(2c\times \frac{3}{2}c+\frac{3}{2}c\times c+c\times 2c)}$
$=\sqrt{729\times 29-2(3c^2+\frac{3}{2}{c}^2+2c^2)}$
$=\sqrt{729\times 29-2\times \frac{13c^2}{2}}$
$=\sqrt{729\times 29-13\times (6\sqrt{29}{)}^2}$
$=\sqrt{29(729-468)}=\sqrt{29\times 261}=\sqrt{29\times 29\times 9}$
$=29\times 3=87$