Question

In a triangle ABC $a=5,b=4$ and $c=3$. 'G' is the centroid of the triangle. Circumradius of triangle GAB is equal to

• A. $2\sqrt{13}$
• B. $\frac{5}{12}\sqrt{13}$
• C. $\frac{5}{3}\sqrt{13}$
• D. $\frac{3}{2}\sqrt{13}$

Answer 1

The correct option is B $\frac{5}{12}\sqrt{13}$

$a=5$

$b=4$

$c=3$

Since, $5^2=4^2+3^2$, we can prove by Pythagoras theorem that triangle ABC is right angled.

(Image)

Now, if we take AC to be on Y axis and AB on the x-axis the co ordinates for point G will be :$(1,4/3)$

$\therefore$Length of GA(p)$=\sqrt{1^2+{\left(\frac{4}{3}\right)}^2}=\sqrt{\frac{25}{9}}=\frac{5}{3}$

Length of GB(q)$=\sqrt{{(1-3)}^2+{(\frac{4}{3}-0)}^2}=\sqrt{4+\frac{16}{9}}=\frac{2\sqrt{13}}{3}$

Length of AB(r)$=1$ unit

$\therefore$Circumradius (R)$=\frac{pqr}{\sqrt{(p+q+r)(p+q-r)(p+r-q)(r+q-p)}}$

$=\frac{\frac{5}{3}\times \frac{2\sqrt{13}}{3}\times 1}{\sqrt{\left(\frac{8+2\sqrt{13}}{3}\right)\left(\frac{2+2\sqrt{13}}{3}\right)\left(\frac{8-2\sqrt{3}}{3}\right)\left(\frac{2\sqrt{13}-2}{3}\right)}}$

$=\frac{10\sqrt{13}}{9}\times \frac{9}{12\times 2}$

$R=\frac{5}{12}\sqrt{13}$