Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

In a triangle $ABC,a=5,b=4,$ and $c=3$. $G$ is the centroid of the triangle.

Then the 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 $\frac{5}{12}\sqrt{13}$

Let the coordinates of
$A=(0,0),B=(3,0).andC=(0,4)4$
Hence Centroid G=$1,\frac{4}{3}$.
Now $GA=\frac{5}{3}$ ... using distance formula.
$G{B}^2=\frac{16}{9}+4$
$=\frac{52}{9}$.
And $AB=3$
Hence
$cosA=\frac{G{A}^2+A{B}^2-G{B}^2}{2.GA.AB}$
$=\frac{\frac{25}{9}+9-\frac{52}{9}}{10}$
$=0.6$
$=\frac{3}{5}$.

Hence $sinA=\frac{4}{5}$

Hence $R=\frac{GB}{2sinA}$
$=\frac{5\sqrt{52}}{6\times 4}$
$=\frac{5\sqrt{13}}{12}$.