Question

$ABC$ is an isosceles triangle with $AB=AC=5$ and $BC=6$. If $G$ is the centroid of $△ABC$, then $AG$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $\frac{8}{3}$

Answer 1

The correct option is D $\frac{8}{3}$

$Let\angle ABC=\theta$

Draw a median AD passing through A.

$So,$

$\frac{AG}{GD}=\frac{2}{1}-\left(1\right)$

$In△ABC$

$\cos \theta =\frac{{BC}^2+{AB}^2-{AC}^2}{2\left(BD\right)\left(AB\right)}$ (cosine formula)

$\cos \theta =\frac{6^2+5^2-5^2}{2\left(5\right)\left(6\right)}=\frac{6^2}{2\left(5\right)\left(6\right)}=\frac{6}{10}=\frac{3}{5}-\left(2\right)$

$In△ADB$

$\cos \theta =\frac{3}{5}\left(from\right(2\left)\right)$

$\frac{3}{5}=\frac{BD}{AB}(AsBD=3\&AB=5)$

$\therefore \cos \theta =\frac{BD}{AB}$

$⟹△ABD$ is a right angled triangle.

$\therefore AD=\sqrt{{\left(AB\right)}^2-{\left(BD\right)}^2}$

$⟹AD=\sqrt{5^2-3^2}=4$

$AD=AG+GD$

$\&\frac{AG}{2}=GD$

$\therefore AD=\frac{AG}{2}+AG$

$⟹4=\frac{3AG}{2}$

$⟹AG=\frac{8}{3}$