Find the perimeter of a polygon with vertices at A (0, 1), B (4, 4), C (7, 0), D (3, - 3), E (-4, - 4). Round your answer to the nearest hundredth.

Group of answer choices

40.55 units

25.00 units

32.00 units

28.47 units

It's interesting that areas from coordinates are much simpler than perimeters from coordinates. It's almost like math is telling us area is more fundamental than length.

p = AB+BC+CD+DE+EA

We use the Pythagorean theorem, square root form, on all the sides. We can skip a lot of the square roots because most of these are the hypotenuse of a 3/4/5 right triangle.

B-A = (4-0, 4-1) = (4,3) so length AB=5

C-B = (7-4,0-4) = (3,-4), so again BC=5

D-C = (3-7, - 3-0) = (-4,-3) so again CD=5

E-D = (-4-3, - 4 - - 3) = (-7, - 1). Ends our streak. DE=√ (7²+1²) = √50=5√2.

A-E = (0 - - 4,1 - - 4) = (4,5) so EA=√ (4²+5²) = √41.

Three of five were 3/4/5.

p = 5+5+5+5√2+√41 = 15+5√2+√41

I hate to ruin a nice exact answer with an approximation.

p ≈ 28.474192049298324

Answer: 28.47, last choice