Unit Test — Test Answers

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is the angle bisector of YEX and the perpendicular bisector of . is the angle bisector of YGZ and the perpendicular bisector of . is the angle bisector of ZFX and the perpendicular bisector of . Point A is the intersection of , , and .

Question illustration
A
Point A is the center of the circle that passes through points E, F, and G but is not the center of the circle that passes through points X, Y, and Z.
B
Point A is the center of the circle that passes through points X, Y, and Z but is not the center of the circle that passes through points E, F, and G.
C
Point A is the center of the circle that passes through points E, F, and G and the center of the circle that passes through points X, Y, and Z.
D
Point A is not necessarily the center of the circle that passes through points E, F, and G or the center of the circle that passes through points X, Y, and Z.
4

Point Z is equidistant from the vertices of ΔTUV.

Question illustration
A
Line segment T A is-congruent-to line segment T B
Option A
B
Line segment A Z is-congruent-to line segment B Z
Option B
C
BTZ BUZ
Option C
D
TZA TZB
Option D
6

Jace is making a water play table from a triangular table top. He sketched a dotted line to show where he wants to add a circular water bowl. It should go all the way to the edges of the table as shown.

Question illustration
A
He should find the midpoint of each edge of the table.
B
He should bisect each of the angles at the vertices of the triangular table top.
C
He should draw perpendicular lines through any point along the edges of the table.
D
He should draw perpendicular line segments from each corner to the opposite edge of the table.
7

Triangle A B C is shown. Lines are drawn from each point to to the opposite side and intersect at point G. Line segments A D, B E, and C F are created.

Question illustration
A
Point G cannot be the centroid because 18:6 does not equal 2:1.
B
Point G cannot be the centroid because FG should be longer than CG.
C
Point G can be the centroid because 12:6 equals 2:1.
D
Point G can be the centroid because FC is longer than FG.
8

[Figure may not be drawn to scale]

Question illustration
A
25 units
B
50 units
C
100 units
D
150 units
10

[Figure may not be drawn to scale]

Question illustration
A
12 mm
B
24 mm
C
36 mm
D
48 mm
11

Triangle J K L is shown. Lines are drawn from each point to the opposite side and intersect at point P. Line segments J O, K M, and L N are created.

Question illustration
A
LN ⊥ JK, JO ⊥ LK, and JL ⊥ MK.
B
JL = LK = KJ
C
JM = ML, LO = OK, and KN = NJ.
D
LN is a perpendicular bisector of JK, JO is a perpendicular bisector of LK, and MK is a perpendicular bisector of JL.

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