25. D 27. D 28. A 29. D. x = 7, y = 26 31. D. 120 degrees
Step-by-step explanation:
25. Since the two parallelograms are congruent, you know what ∠B and ∠Y are supplementary.
This means that 3m + 70 + 80 = 180
3m + 150 = 180
3m = 30
D. m = 10
27. Since a vertex is (2,0), you know that two other vertexes are five units away in one direct and the other vertex will be five up or down and five over.
A is 5 units over, and that's a possible other coordinate, so it can't be A.
B is 5 over, so it's a possibility and it can't be B.
C is down 5 and 5 over, so it can be the coordinate across from (2,0), so it can't be C.
D is 7 up and 2 over, which isn't within our 5 unit frame! So it must be D!
E is 5 up and 5 over, so it can still count as a possibility.
28. When you observe the coordinates, you'll notice that FG ║ HI, and FH ║ GI. This eliminates the possibility of it being a kite and ensures that it will at least be a parallelogram!
From here, you can also observe that FG ≅ HI, and FH ≅ GI. This shows you that it's not a square despite that it has right angles.
The presence of right angles shows that it's not a rhombus and instead must be a rectangle.
29. Even though the problem only identifies it as a quadrilateral, it must be a parallelogram because of the parallel sides. Opposite angles on parallelograms are congruent, so we can begin with:
6x - 8 = 4x + 6
2x = 14
x = 7
From here, you can plug "x" back into the equation to find the value of those two angles.
6(7) - 8
42 - 8
34
Now that we know that those two angles are 34 degrees, we know that they total to 68 degrees.
Since the sum of internal angles on a quadrilateral is 360 degrees, the other two anglers must be 360 - 68.
That means the total of those two angles is 292 degrees.
Since we know the angles are equal (def. of parallelogram), we know that:
146 = 5y + 16
140 = 5y
y = 26
D. x = 7, y = 26
31. The arc is 120 degrees because the angle because at the other side instead of at the center of the circle, so you double the angle.
