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 Transformed square (Posted on 2012-10-25)
Consider the transformation T on the coordinate plane where T:(x,y)→(ax+by,cx+dy).

Let S be a square with vertices (1,1),(-1,1), (-1,-1) and (1,-1)

Find the area of the image of S under T in terms of a,b,c, & d.

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Well, in problem "Transformed Circle", we dealt with a special case of this transformation where a = d = 2 and b = c = 1.  We established that it was a linear transformation, that line segments with matching slope and length have a different but still matching slope and length under T, and that relative distances on the line are preserved.  All those conclusions apply also to this general case, using the same logic, which I will not repeat.

This means that a square gets transformed under T into a parallelogram.  Its vertices are (a+b,c+d), (-a+b,-c+d), (a-b,c-d), and (-a-b,-c-d).

It is not too hard to calculate this area.  Draw the square whose opposite corners are (a+b,c+d), and (-a-b,-c-d).  This has area 4(a+b)(c+d).  Draw a vertical and a horizontal line from the other vertices of the transformed quadrilateral to the near side of the square.  This forms two squares and 4 triangles that must be subtracted to calculate the area of the quadrilateral.  The squares each have areas of 4bc, two triangles each have areas of 2bd, and the other two triangles each have areas of 2ac.  So the area of the quadrilateral = 4(a+b)(c+d) - 8bc - 4bd - 2ac = 4(ad-bc).

Wait a second!  What if ad - bc < 0?  Well, then two of the vertices are reversed and the calculation is the same except that the two squares that get subtracted out each have an area of 4ad.  The area is 4(bc-ad).

Extra credit:  Since the original square had an area of 4, the area under T increased by a multiple of |ad-bc|.  By an argument I made previously in "Transformed Circle" , T will multiply the area of any closed shape on the x-y plane by |ad-bc|.

 Posted by Steve Herman on 2012-10-25 19:00:53

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