# Getting Area of a region transformed by GeometricTransformation

I tried this:

Clear[b1, b2, A, p]
b1 = {-1, 1};
b2 = {1, 1};
A = {{2, 1}, {-1, 1}};
p = Parallelogram[{0, 0}, {b1, b2}];


Then drew this:

Graphics[{
{Red, p},
Blue, Thick,
Arrow[{{0, 0}, b1}],
Arrow[{{0, 0}, b2}],
{Opacity[0.6], GeometricTransformation[p, A]},
Red,
Arrow[{{0, 0}, A.b1}],
Arrow[{{0, 0}, A.b2}]
}]


Then I tried:

Area[GeometricTransformation[p, A]]


But it didn't work, giving me:

Area::reg: GeometricTransformation[p,{{2,1},{-1,1}}] is not a correctly specified region. >>

Now, I know I can get the area of the transformed region with:

Abs[Det[A.Transpose[{b1, b2}]]]


Which is 6, but I am wondering if there is a simple way (for students just beginning with Mathematica) to use the Area command in this situation.

Update: Thanks to MichaelE2, J.M., and rcollyer, I was able to transform the region:

t = TransformedRegion[p, AffineTransform[A]]


Which returned Parallelogram[{0, 0}, {{-1, 2}, {3, 0}}], which is terrific information for students.

Then I was able to draw the same image like this:

Graphics[{
{Red, p},
Blue, Thick,
Arrow[{{0, 0}, b1}],
Arrow[{{0, 0}, b2}],
{Opacity[0.6], t},
Red,
Arrow[{{0, 0}, A.b1}],
Arrow[{{0, 0}, A.b2}]
}]


Then I was able to get the area.

Area[t]


Which returned a 6.

• Hmm, Area@DiscretizeGraphics@Graphics@GeometricTransformation[p, A] fails miserably, and Normal@GeometricTransformation[p, A] fails, too. :/ Jun 26, 2016 at 3:39
• Try using TransformedRegion[] instead. Jun 26, 2016 at 3:40

You are looking for TransformedRegion. GeometricTransformation is for transforming graphics primitives, but you are looking at region functionality. This is a case where the difference is important. Simply,
t = TransformedRegion[p, AffineTransform@A]

where the AffineTransform was needed as TransformedRegion expects a function.