document.write( "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∘
\n" ); document.write( " angle with the table top. What is the area, in square feet, of the shadow cast by the plate on the table top
\n" ); document.write( "

Algebra.Com's Answer #764017 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "Since this question has been sitting here for a couple of days without any response, I will take a stab at it.

\n" ); document.write( "There are probably fancy mathematical ways to solve this problem; but I don't know them.

\n" ); document.write( "Perhaps another tutor will see my response and either confirm my solution, or show a better method for solving the problem.

\n" ); document.write( "So here is my solution....

\n" ); document.write( "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.

\n" ); document.write( "We can label the corners of the plate
\n" ); document.write( "A(1,-1,0)
\n" ); document.write( "B(1,1,0)
\n" ); document.write( "C(-1,1,0)
\n" ); document.write( "D(-1,-1,0)

\n" ); document.write( "The point light source (2 feet above the center of the plate) is at P(0,0,2).

\n" ); document.write( "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.

\n" ); document.write( "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)).

\n" ); document.write( "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.

\n" ); document.write( "The vector PC' is [-1,-1+sqrt(2),sqrt(2)-2].

\n" ); document.write( "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 \"2%2F%282-sqrt%282%29%29+=+2%2Bsqrt%282%29\".

\n" ); document.write( "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).

\n" ); document.write( "By symmetry, B'' is (2+sqrt(2),sqrt(2),0).

\n" ); document.write( "B''C'' is a base of the isosceles trapezoid shaded area; its length is 4+2sqrt(2).

\n" ); document.write( "The other base of the shaded area is AD, length 2.

\n" ); document.write( "And the height of the pyramid is the difference in the y values of D and C'', which is 1+sqrt(2).

\n" ); document.write( "Finally, the area of the isosceles trapezoid shaded area is

\n" ); document.write( "
\n" ); document.write( "
\n" );