<PRE><B>Question 1143168</B>
&lt;br&gt;
Since this question has been sitting here for a couple of days without any response, I will take a stab at it.&lt;br&gt;
There are probably fancy mathematical ways to solve this problem; but I don&#39;t know them.&lt;br&gt;
Perhaps another tutor will see my response and either confirm my solution, or show a better method for solving the problem.&lt;br&gt;
So here is my solution....&lt;br&gt;
Picture the metal plate in the xy-plane in a 3D coordinate system, with the center of the plate at the origin.  Have the coordinate system in the usual configuration -- positive z axis vertical up; right-hand rule.&lt;br&gt;
We can label the corners of the plate
A(1,-1,0)
B(1,1,0)
C(-1,1,0)
D(-1,-1,0)&lt;br&gt;
The point light source (2 feet above the center of the plate) is at P(0,0,2).&lt;br&gt;
Now keep edge AD of the plate on the table and rotate the plate up 45 degrees.  Let B&#39; and C&#39; be the images of B and C after that rotation.&lt;br&gt;
DC&#39; and AB&#39; form 45-45-90 right triangles with the table top; the hypotenuse is 2 and each leg is sqrt(2).  That makes the coordinates of C&#39; (-1,-1+sqrt(2),sqrt(2)).&lt;br&gt;
The edges of the metal plate are straight lines, so the boundaries of the shadow will be straight lines.  So the shaded area will be in the shape of a trapezoid.  And since the light source was centered above the plate, the shape will be an isosceles trapezoid.&lt;br&gt;
The vector PC&#39; is [-1,-1+sqrt(2),sqrt(2)-2].&lt;br&gt;
Let C&#39;&#39; be the projection of PC&#39; onto the table top (z=0).  The vector PC&#39;&#39; is a scalar multiple of the vector PC&#39;; the multiplication factor is the ratio of the differences in the z values, which is {{{2/(2-sqrt(2)) = 2+sqrt(2)}}}.&lt;br&gt;
So vector PC&#39;&#39; is (2+sqrt(2)) times [-1,-1+sqrt(2),sqrt(2)-2] =[-2-sqrt(2),sqrt(2),-2]; and that makes C&#39;&#39; (-2-sqrt(2),sqrt(2),0).&lt;br&gt;
By symmetry, B&#39;&#39; is (2+sqrt(2),sqrt(2),0).&lt;br&gt;
B&#39;&#39;C&#39;&#39; is a base of the isosceles trapezoid shaded area; its length is 4+2sqrt(2).&lt;br&gt;
The other base of the shaded area is AD, length 2.&lt;br&gt;
And the height of the pyramid is the difference in the y values of D and C&#39;&#39;, which is 1+sqrt(2).&lt;br&gt;
Finally, the area of the isosceles trapezoid shaded area is&lt;br&gt;
{{{A = ((b1+b2)/2)h = ((2+(4+2sqrt(2)))/2)(1+sqrt(2)) = (3+sqrt(2))(1+sqrt(2)) = 5+4sqrt(2)}}}</PRE>