Find the value of y required for
the polygon below to be a
parallelogram.
Plz help!!!!
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This is algebra disguised as geometry. I object. There are much better ways to combine algebra and geometry.
Anyway, in a parallelogram the diagonals meet at their midpoints, so bisect each other. So we have two pairs of congruent segments; geometry over.
3x + 1 = x+27
79 - y = 2y + 22
We can ignore the one about x.
79 - 22 = 2y + y
57 = 3y
y = 19
Answer: 19