Free Math Worksheets

- How large must the compass radius be for the arcs to intersect twice? - Where does a perpendicular bisector cross the original segment? - Compare the distances from any point on the perpendicular bisector to the endpoints.

1. Draw \(\overline{AB}=7.6\,\text{cm}\). 2. Open the compass to a radius greater than \(3.8\,\text{cm}\). With center \(A\), draw arcs above and below the segment. Without changing the compass width, repeat with center \(B\). 3. Draw the line through the two arc intersection points. This line is the perpendicular bisector of \(\overline{AB}\). 4. The midpoint divides the segment into two parts of length \(7.6\div2=3.8\,\text{cm}\). 5. Every point on the perpendicular bisector is equidistant from \(A\) and \(B\).

a) Draw equal-radius arcs from \(A\) and \(B\) with radius greater than \(3.8\,\text{cm}\), then draw the line through their two intersections. b) Each part is \(3.8\,\text{cm}\). c) Every point on the perpendicular bisector is the same distance from \(A\) as from \(B\).

- What happens to an angle measure when the angle is bisected? - Repeat the same construction on one of the smaller angles. - Track the angle created after each construction.

1. To construct \(w_1\), draw an arc centered at the vertex that intersects both sides of the angle. From those two intersection points, draw equal-radius arcs that intersect inside the angle. Draw a ray from the vertex through that intersection. 2. The first bisection creates two angles of measure \(74^\circ\div2=37^\circ\). 3. Apply the same construction to one \(37^\circ\) angle to create \(w_2\). 4. The smallest angle measures \(37^\circ\div2=18.5^\circ\).

a) Construct \(w_1\) using equal-radius arcs from points on the two sides of the angle. b) Repeat the angle-bisector construction on one \(37^\circ\) angle to construct \(w_2\). c) The smallest angle measures \(18.5^\circ\).

- Make a table of the number of bisections and the resulting angle measure. - Each bisection divides the current angle measure by \(2\). - Continue until the result is below \(10^\circ\).

1. The first bisection gives \(90^\circ\div2=45^\circ\). 2. The second bisection gives \(45^\circ\div2=22.5^\circ\). 3. The third bisection gives \(22.5^\circ\div2=11.25^\circ\), which is still greater than \(10^\circ\). 4. The fourth bisection gives \(11.25^\circ\div2=5.625^\circ\), which is less than \(10^\circ\). 5. Therefore, four bisections are required.

a) \(45^\circ\). b) \(22.5^\circ\). c) Four bisections; the resulting angle is \(5.625^\circ\).