Question

In triangle $ABC,a=5,b=4,c=3.G$ is the centroid of triangle.If $R_1$ be the circumradius of triangle $GAB$ then the value of $90R_1^2$ must be

• A. $3250$
• B. $2530$
• C. $5320$
• D. $5230$

Answer 1

The correct option is A $3250$

$AG=\frac{2}{3}A{A}_1$
$BG=\frac{2}{3}B{B}_1$
$\Rightarrow AG=\frac{2}{3}\sqrt{2b^2+2c^2-a^2}$
$BG=\frac{2}{3}\sqrt{2a^2+2c^2-b^2}$
$\Rightarrow AG=\frac{2}{3}a,BG=\frac{2}{3}\sqrt{b^2+4c^2}$ as $a^2=b^2+c^2$
Substituting for $a=5,b=4,c=3$ we have
$\Rightarrow AG=\frac{2}{3}a=\frac{2}{3}\times 5=\frac{10}{3}$
$\Rightarrow BG=\frac{2}{3}\sqrt{b^2+4c^2}=\frac{2}{3}\sqrt{4^2+43^2}=\frac{2}{3}\times 2\sqrt{13}=\frac{4}{3}\sqrt{13}$
Also, $Ab=c=3$ and ${△}_{GAB}=\frac{1}{3}{△}_{ABC}=2$
If $R_1$ be the circumradius of $△GAB,$ then
$R_1=\frac{\left(AG\right)\left(BG\right)\left(AB\right)}{4{△}_{GAB}}$
$=\frac{10}{3}.\frac{4}{3}\sqrt{13}.3.\frac{1}{4.2}$
$=\frac{5\sqrt{13}}{3}$units.
$\therefore 90R_1^2=90\times {\left(\frac{5\sqrt{13}}{3}\right)}^2=3250$(on simplification)