Consider an ellipse, whose center is at the origin and its major axis is along the $$x$$-axis. If its eccentricity is $$\frac{3}{5}$$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is:

We begin by recalling the standard form of an ellipse whose centre is the origin and whose major axis lies along the $$x$$-axis:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b>0.$$

For such an ellipse the distance of each focus from the centre is denoted by $$c$$ and the eccentricity $$e$$ is defined by the relation

$$e=\frac{c}{a}.$$

According to the question the eccentricity is $$\dfrac{3}{5}$$ and the distance between the two foci is $$6$$. Since the two foci are symmetrically placed about the origin, their separation is $$2c$$, so

$$2c=6 \;\;\Longrightarrow\;\; c=3.$$

Substituting $$c=3$$ and $$e=\dfrac{3}{5}$$ into the formula $$e=\dfrac{c}{a}$$, we obtain

$$\frac{3}{5}=\frac{3}{a}\;\;\Longrightarrow\;\; a=5.$$

Next we use the fundamental relation among $$a$$, $$b$$ and $$c$$ for an ellipse:

$$c^{2}=a^{2}-b^{2}.$$

Putting $$a=5$$ and $$c=3$$, we have

$$3^{2}=5^{2}-b^{2}\;\;\Longrightarrow\;\;9=25-b^{2}\;\;\Longrightarrow\;\; b^{2}=25-9=16,$$

so

$$b=4.$$

The four vertices of the ellipse are therefore

$$(a,0)=(5,0)$$, $$(-a,0)=(-5,0)$$, $$(0,b)=(0,4)$$, $$(0,-b)=(0,-4)$$.

Joining these four points in the same order as they occur on the ellipse—$$\,(5,0)\rightarrow(0,4)\rightarrow(-5,0)\rightarrow(0,-4)\rightarrow(5,0)\,$—gives a symmetric quadrilateral whose two diagonals are the segments between opposite vertices:

Diagonal $$AC$$ from $$(5,0)$$ to $$(-5,0)$$ has length $$2a=10,$$

Diagonal $$BD$$ from $$(0,4)$$ to $$(0,-4)$$ has length $$2b=8.$$

These diagonals are perpendicular and bisect each other at the origin, so the quadrilateral is a rhombus (or kite). The area of a quadrilateral whose diagonals are perpendicular is one half the product of the lengths of the diagonals. Thus

$$ \text{Area}=\frac12\,(2a)\,(2b)=\frac12\,(10)\,(8)=\frac12\,(80)=40. $$

Hence, the correct answer is Option C.