SSC CHSL 12 Aug 2021 Shift 3 Paper

Given, Since, $△\mathit{A}\mathit{B}\mathit{C}$ is an isosceles triangle, where $\mathit{A}\mathit{C}=\mathit{B}\mathit{C}$.
Therefore, $\mathit{C}\mathit{D}$ is the median and the perpendicular to $\mathit{A}\mathit{B}$.
Since, length of centroid from the vertex $=\frac{2}{3}$ of median
$\therefore$ Median $\left(\mathit{C}\mathit{D}\right)=\frac{8\times 3}{2}=12\mathit{c}\mathit{m}$
and $\mathit{A}\mathit{D}=\frac{\mathit{A}\mathit{B}}{2}=\frac{10}{2}=5\mathit{c}\mathit{m}$
Now, in right angle triangle $\mathit{A}\mathit{D}\mathit{C}$,
$\mathit{A}{\mathit{C}}^2=\mathit{A}{\mathit{D}}^2+\mathit{C}{\mathit{D}}^2$
[By Pythagoras theorem]
$\Rightarrow \mathit{A}{\mathit{C}}^2={\left(5\right)}^2+{\left(12\right)}^2$
$=25+144=169$
$\therefore \mathit{A}\mathit{C}=\sqrt{169}=13\mathit{c}\mathit{m}$