Finding the area of $GCD$

Question

$\DeclareMathOperator{\cm}{cm}$On a plate in the shape of an equilateral triangle $ABC$ with area $16\sqrt{3}\cm^2$, a rod $GD$ of height $8\cm$, is fixed vertically. $G$ is a point on the plate. If the areas of the triangles $AGD$ and $BGD$ are both equal to $4\sqrt{19}\cm^2$, find the area of the triangle $CGD$ (in $\cm^2$).

Answer 1

This is a sketch of steps you can take to find the answer. You will need to fill in missing details.

First, note that $AB=BC=CA=8\texttt{ cm}$. Next, we have $$AG=BG=\sqrt{19}\texttt{ cm}\,.$$ Thus, the triangle $AGB$ is isosceles, and so the line segment $GM$ is perpendicular to $AB$, where $M$ is the midpoint of $AB$. That is, $$GM^2=BG^2-BM^2=\big(\sqrt{19}\texttt{ cm}\big)^2-\left(\frac{8\texttt{ cm}}{2}\right)^2=\big(\sqrt{3}\texttt{ cm}\big)^2\,.$$ Consequently, $GM=\sqrt{3}\texttt{ cm}$. Finally, note that $CM=4\sqrt{3}\texttt{ cm}$, whence $$CG=CM-GM=3\sqrt{3}\texttt{ cm}$$ and the area of the triangle $CGD$ is $$\frac{1}{2}\,GC\cdot CG=\frac{1}{2}\,\big(8\texttt{ cm}\big)\,\big(3\sqrt{3}\texttt{ cm}\big)=12\sqrt{3}\texttt{ cm}^2\,.$$