Lesson 4
Construction Techniques 2: Equilateral Triangles
- Let’s identify what shapes are possible within the construction of a regular hexagon.
Problem 1
This diagram is a straightedge and compass construction. \(A\) is the center of one circle, and \(B\) is the center of the other. Explain how we know triangle \(ABC\) is equilateral.
Problem 2
\(A\), \(B\), and \(C\) are the centers of the 3 circles. How many equilateral triangles are there in this diagram?
Problem 3
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.
\(AC=BC\)
\(AC=BD\)
\(CD=AB\)
\(ABCD\) is a square.
\(ABD\) is an equilateral triangle.
\(CD=AB+AB\)
Problem 4
Line segment \(CD\) is the perpendicular bisector of line segment \(AB\). Is line segment \(AB\) the perpendicular bisector of line segment \(CD\)?
Problem 5
Here are 2 points in the plane.
- Using only a straightedge, can you find points in the plane that are the same distance from points \(A\) and \(B\)? Explain your reasoning.
- Using only a compass, can you find points in the plane that are the same distance from points \(A\) and \(B\)? Explain your reasoning.
Problem 6
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. Select all statements that must be true.
\(A M = B M\)
\(C M = D M\)
\(E A = E M\)
\(E A < E B\)
\(A M < A B\)
\(A M > B M\)
Problem 7
The diagram was constructed with straightedge and compass tools. Name all segments that have the same length as segment \(AC\).
Problem 8
Starting with 2 marked points, \(A\) and \(B\), precisely describe the straightedge and compass moves required to construct the quadrilateral \(ACBD\) in this diagram.
Problem 9
In the construction, \(A\) is the center of one circle and \(B\) is the center of the other. Which segment has the same length as \(AB\)?
\(CB\)
\(CD\)
\(CE\)
\(CA\)