12 multiple-choice questions, progressively harder.
What defines the perpendicular bisector of a nonzero segment?
Solution
Correct answer: B
Both conditions are required: the line passes through the midpoint and meets the segment at a right angle.
M is the midpoint of AB, and PM is perpendicular to AB. Which statement follows?
Correct answer: A
P lies on the perpendicular bisector of AB, so the forward theorem gives equal distances to A and B.
Q is equally far from distinct points A and B. Where must Q lie?
Correct answer: D
The converse places every point with equal endpoint distances on the perpendicular bisector. Q can be the midpoint or a point off AB.
A segment is 18 cm long. How far is its midpoint from each endpoint?
Correct answer: C
A midpoint divides the segment into two equal lengths.
P lies on the perpendicular bisector of AB. PA is 16 cm. What is PB?
The two endpoint distances are equal by the forward theorem. Therefore PB is 16 cm.
Equal-radius circles centered at A and B are used to construct the perpendicular bisector of a 10 cm segment. Which radius gives two distinct crossings?
The radius must be greater than half the endpoint separation.
Only 6 cm is greater than 5 cm.
What does a locus describe?
A locus contains every point satisfying the condition and excludes every point that fails it.
Which triangle lines meet at the circumcenter?
The concurrency theorem proved in this lesson identifies the common point of the three side perpendicular bisectors.
Which distances are guaranteed equal at a triangle circumcenter O?
AB=BCAB=BCAB=BC
BC=CABC=CABC=CA
OA=OBOA=OBOA=OB
OB=OCOB=OCOB=OC
The circumcenter is equally far from all three vertices. Equal distances to side lines describe a different condition.
Where is the circumcenter of a right triangle?
The right-case midpoint-parallel argument puts the center at the hypotenuse midpoint.
What is the circumcircle of a nondegenerate triangle?
A circumcircle has the circumcenter as its center and one common vertex distance as its radius.
For one given rotation, A maps to a different point A prime. Why does the center lie on the perpendicular bisector of their chord?
Rotation preserves distance from its center. The converse turns those equal distances into membership of the chord bisector.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.