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3 years ago

Is EFGH below a rectangle

Mathematics

1 answer:

poizon [28]3 years ago

3 0

Answer:

D. No, because EFGH is a parallelogram but it’s diagonals are not congruent

Step-by-step explanation:

The differences between the end points of the diagonals are ...

F -H = (1, 1) -(2, -5) = (1 -2, 1 -(-5)) = (-1, 6)

G -E = (4, -2) -(-1, -2) = (4 -(-1), -2 -(-2)) = (5, 0)

The length of FH is more than 6, the length of GE is exactly 5. The diagonals are different length, so the figure cannot be a rectangle.

The midpoints of the diagonals will be in the same place if the sum of their end points is the same. (Dividing each sum by 2 gives the midpoint of that segment.)

F+H = (1, 1) +(2, -5) = (3, -4)

G+E = (4, -2) +(-1, -2) = (3, -4)

The diagonals bisect each other (have the same midpoint), so the figure is a parallelogram.

EFGH is a parallelogram, but not a rectangle: its diagonals are not congruent.

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