The vertices of a $\Delta ABC$ are $A(4, 6)$,$B(1, 5)$,and $C(7, 2)$. $A$ line is drawn to intersect sides $AB$ and $AC$ at $D$ and $E$ respectively,such that $\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4}$. Calculate the area of $\Delta ADE$ and compare it with the area of $\Delta ABC$.

(1:16) Given that,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4}$.
Since $\frac{AD}{AB} = \frac{AE}{AC}$,by the Converse of Thales Theorem (Basic Proportionality Theorem),$DE \parallel BC$.
Therefore,$\Delta ADE \sim \Delta ABC$ by $AA$ similarity criterion.
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus,$\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
Now,calculate the area of $\Delta ABC$ using the coordinates $A(4, 6)$,$B(1, 5)$,and $C(7, 2)$:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= \frac{1}{2} |4(5 - 2) + 1(2 - 6) + 7(6 - 5)|$
Area $= \frac{1}{2} |4(3) + 1(-4) + 7(1)| = \frac{1}{2} |12 - 4 + 7| = \frac{1}{2} |15| = 7.5$ square units.
Area of $\Delta ADE = \frac{1}{16} \times \text{Area}(\Delta ABC) = \frac{1}{16} \times 7.5 = \frac{7.5}{16} = \frac{15}{32}$ square units.
The ratio of the area of $\Delta ADE$ to the area of $\Delta ABC$ is $1:16$.