
Geometry | Unit 1 | Lesson 3: Construction Techniques 1: Perpendicular Bisectors | Practice Problems
Authored by Illustrative Mathematics
Mathematics
6th Grade
CCSS covered
Used 3+ times

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5 questions
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1.
MULTIPLE SELECT QUESTION
30 sec • 1 pt
This diagram is a straightedge and compass construction. \(A\) is the center of one circle, and \(B\) is the center of the other. Select all the true statements.
Line \(CD\) is perpendicular to segment \(AB\)
Point \(M\) is the midpoint of segment \(AB\)
The length \(AB\) is the equal to the length \(CD\).
Segment \(AM\) is perpendicular to segment \(BM\)
\(CB+BD > CD\)
Tags
CCSS.HSG.CO.C.9
2.
OPEN ENDED QUESTION
3 mins • 1 pt
In this diagram, line segment \(CD\) is the perpendicular bisector of line segment \(AB\). Assume the conjecture that the set of points equidistant from \(A\) and \(B\) is the perpendicular bisector of \(AB\) is true. Is point \(E\) closer to point \(A\), closer to point \(B\), or the same distance between the points? Explain how you know.
Evaluate responses using AI:
OFF
Tags
CCSS.HSG.CO.C.9
3.
OPEN ENDED QUESTION
3 mins • 1 pt
Starting with 2 marked points, \(A\) and \(B\), precisely describe the straightedge and compass moves required to construct the triangle \(ABC\) in this diagram.
Evaluate responses using AI:
OFF
Tags
CCSS.HSG.CO.D.12
4.
MULTIPLE SELECT QUESTION
30 sec • 1 pt
This diagram was created by starting with points \(C\) and \(D\) and using only straightedge and compass to construct the rest. All steps of the construction are visible. Select all the steps needed to produce this diagram.
Construct a circle centered at \(A\).
Construct a circle centered at \(C\).
Construct a circle centered at \(D\).
Label the intersection points of the circles \(A\) and \(B\).
Draw the line through points \(C\) and \(D\).
Tags
CCSS.HSG.CO.C.9
5.
MULTIPLE SELECT QUESTION
30 sec • 1 pt
This diagram was constructed with straightedge and compass tools. \(A\) is the center of one circle, and \(C\) is the center of the other. Select all true statements.
\(AB=BC\)
\(AB=BD\)
\(AD=2AC\)
\(BC=CD\)
\(BD=CD\)
Tags
CCSS.HSG.CO.B.7
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