WILL MARK THE BRAINILEST Find the perimeter of a quadrilateral with vertices at C (-2, 4), D (-3, 1), E (1,0), and F(-1,2). Round your answer to the nearest hundredth when
necessary
10 units
10.77 units
11.52 units
12.35 units

When we have a polygon with vertices on the lattice points (integer coordinates) the area will always an integer or half an integer.  So the answer must be

10 units

But let's work it out; I'll show you a trick how called The Shoelace Formula.

The Shoelace Formula says the signed area of any 2d polygon is half the sum of the cross products of the sides.  The cross product of (a,b) and (c,d) is ad-bc.  A counterclockwise ordering gives a positive area, that's CDEF

C (-2, 4)    (-2)(1)-4(-3) = -2 - -12 = 10

D (-3, 1)     (-3)(0) - (1)(1) =               -1

E (1,0)        1(2) - (0)(-1) =                 2

F(-1,2)        (-1)(4) - 2(-2) =                0

C (-2, 4)

Total                                             11

Area                                              11/2

I repeated C to get all four sides; FC is a side.

Anyway, that area of 11/2 is none of the above.  There's probably a typo in the coordinates; I'd still go with

10 units

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

step-by-step explanation: