Question 1143168: A square metal plate of side length 2 feet and negligible thickness rests on a table top parallel to the ground, and a point source of light is suspended 2 feet above the center of the plate. While one edge of the plate remains on the table, the opposite edge is lifted until the plate makes a 45∘
angle with the table top. What is the area, in square feet, of the shadow cast by the plate on the table top
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Since this question has been sitting here for a couple of days without any response, I will take a stab at it.
There are probably fancy mathematical ways to solve this problem; but I don't know them.
Perhaps another tutor will see my response and either confirm my solution, or show a better method for solving the problem.
So here is my solution....
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.
We can label the corners of the plate
A(1,-1,0)
B(1,1,0)
C(-1,1,0)
D(-1,-1,0)
The point light source (2 feet above the center of the plate) is at P(0,0,2).
Now keep edge AD of the plate on the table and rotate the plate up 45 degrees. Let B' and C' be the images of B and C after that rotation.
DC' and AB' 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' (-1,-1+sqrt(2),sqrt(2)).
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.
The vector PC' is [-1,-1+sqrt(2),sqrt(2)-2].
Let C'' be the projection of PC' onto the table top (z=0). The vector PC'' is a scalar multiple of the vector PC'; the multiplication factor is the ratio of the differences in the z values, which is  .
So vector PC'' is (2+sqrt(2)) times [-1,-1+sqrt(2),sqrt(2)-2] =[-2-sqrt(2),sqrt(2),-2]; and that makes C'' (-2-sqrt(2),sqrt(2),0).
By symmetry, B'' is (2+sqrt(2),sqrt(2),0).
B''C'' is a base of the isosceles trapezoid shaded area; its length is 4+2sqrt(2).
The other base of the shaded area is AD, length 2.
And the height of the pyramid is the difference in the y values of D and C'', which is 1+sqrt(2).
Finally, the area of the isosceles trapezoid shaded area is
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