Answer
$A(S)=\sqrt{1+a^2+b^2} A(D) $
Work Step by Step
The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt{1+(a)^{2}+(b)^{2}} d A \\ =\iint_{D} \sqrt{1+a^2+b^2} dA \\=\sqrt{1+a^2+b^2} \iint_{D} dA $
We are given that $A(D)$ is the area of that projection.
This means that $\iint_{D} dA=A(D)$
Therefore, we have:
$A(S)=\sqrt{1+a^2+b^2} A(D) $
The result has been proved.
