Find the area of the shaded region in the triangle GDT.

Question

In the graph, $T$ is the point of tangency, the area of the equilateral region $ABT$ is $4\sqrt3$ and $GA=VB$. Calculate the area of the shaded region.(S:$4\sqrt3$)

$S_{\triangle ABT}=\frac{l^2\sqrt3}{4}\implies 4\sqrt3=\frac{l^2\sqrt3}{4}\\ \therefore l = 4 =AB=BT=AT$

$AT^2=AD \cdot AV=AD \cdot (4+VB)\implies 16=AD \cdot (4+VB)$

I can't finish.

Answer 1

1. We know that $4=TA$ and $VB=AG$, hence $4+VB=TG$ and $AD.TG=16$.
2. We know that $\angle DAT=60^\circ$. Therefore the altitute of $\triangle DGT$ on the side $TG$ is $h=AD\sin60^\circ=\frac{\sqrt3}{2}AD$. Hence, $\color{red}{h.TG=8\sqrt3}.$

Now, use the area formula $A=\frac12 h.TG.$

Answer 2

Hints towards a solution.

• Show that $GT = VA$.
• Show that $[GDT] = \frac{1}{2} AD \times AV \times ?? = \frac{1}{2} AT^2 \times ?? = 4 \sqrt{3}$.

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Hacky solution if all that you care about is the answer, and assume it's a constant:

Take the case where $AB$ is tangential to the circle, then $ABT = GDT$ so the shaded area is $ 4 \sqrt{3}$.

Side note: From the diagram, I found it hard to believe that they had the same area. (Yes I know it's need not be drawn to scale.)

Answer 3

You've made a good start. Extend $GAT$ to $H$ so $\lvert TH\rvert = \lvert GA\rvert$, plus add the lines $HD$ and $HV$, as indicated below.

Since $\lvert GT\rvert = \lvert AH\rvert$, using the one-half base times height formula for triangles, we get

$$S_{\triangle GDT} = S_{\triangle ADH} \tag{1}\label{eq1A}$$

Also, since $\measuredangle VAH = 60^{\circ}$ and $\lvert AB\rvert + \lvert VB\rvert = \lvert AT\rvert + \lvert HT\rvert$, this means $\triangle AHV$ is equilateral, so its height from any of its sides is, using $\lvert AB\rvert = 4$ you've already determined,

$$\frac{\sqrt{3}}{2}(4 + \lvert VB\rvert) \tag{2}\label{eq2A}$$

Using $AD$ as the base, we get using the one-half base times height formula for triangles again, plus \eqref{eq2A} and your result of $16 = \lvert AD\rvert\cdot(4 + \lvert VB\rvert)$, that

$$\begin{equation}\begin{aligned} S_{\triangle ADH} & = \frac{1}{2}\cdot\lvert AD\rvert\cdot \frac{\sqrt{3}}{2}(4 + \lvert VB\rvert) \\ & = \frac{\sqrt{3}}{4}\cdot\lvert AD\rvert\cdot(4 + \lvert VB\rvert) \\ & = \frac{\sqrt{3}}{4}\cdot 16 \\ & = 4\sqrt{3} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Thus, from \eqref{eq1A}, we have

$$S_{\triangle GDT} = 4\sqrt{3} \tag{4}\label{eq4A}$$