Question

review on unit 1a constructions name: ____________ per: ____________ for 1 - 6 decide if it is true or false, if false correct the statement so it is true. 1) if $overline{qr}$ goes through the mid - point of $overline{xy}$, then $overline{qr}$ is the perpendicular bisector of $overline{xy}$. 2) if segments $overline{ab}$ and $overline{ac}$ are both radii of $odot a$, then $overline{ab}congoverline{ac}$. 3) all rectangles are regular polygons. 4) to the right in $odot a$ and $odot b$, figure acbd must be a rhombus. 5) you can create a regular pentagon with only a compass, straight edge, and pencil, by creating overlapping intersecting circles of the same radii. 6) if $overline{ef}$ is the perpendicular bisector of $overline{gh}$, then every point on $overline{ef}$ is equidistant from both points g and h.

Explanation:

Step1: Recall perpendicular - bisector definition

A perpendicular bisector of a segment must both pass through the mid - point and be perpendicular to the segment. Just passing through the mid - point is not enough. So the statement "If $\overline{QR}$ goes through the midpoint of $\overline{XY}$, then $\overline{QR}$ is the perpendicular bisector of $\overline{XY}$" is False. The corrected statement is: If $\overline{QR}$ goes through the midpoint of $\overline{XY}$ and $\overline{QR}\perp\overline{XY}$, then $\overline{QR}$ is the perpendicular bisector of $\overline{XY}$.

Step2: Recall radius property

All radii of a circle are congruent. If segments $\overline{AB}$ and $\overline{AC}$ are both radii of $\odot A$, then $\overline{AB}\cong\overline{AC}$. This statement is True.

Step3: Recall regular polygon definition

A regular polygon has all sides and all angles congruent. In a rectangle, all angles are $90^{\circ}$, but the adjacent sides are not always equal. So the statement "All rectangles are regular polygons" is False. The corrected statement is: All squares are regular polygons.

Step4: Recall rhombus definition

In $\odot A$ and $\odot B$, $AC = AD=BC = BD$ (radii of the circles). A quadrilateral with all four sides congruent is a rhombus. So the statement "To the right in $\odot A$ and $\odot B$, Figure $ACBD$ must be a rhombus" is True.

Step5: Recall regular pentagon construction

A regular pentagon can be constructed using a compass, straight - edge, and pencil, but not by creating overlapping intersecting circles of the same radii. The statement "You can create a regular pentagon with only a compass, straight edge, and pencil, by creating overlapping intersecting circles of the same radii" is False. The corrected statement is: You can create a regular pentagon with only a compass and straight - edge using a more complex construction method (such as the method based on the golden ratio).

Step6: Recall perpendicular bisector property

The perpendicular bisector of a segment is the set of all points that are equidistant from the endpoints of the segment. So the statement "If $\overline{EF}$ is the perpendicular bisector of $\overline{GH}$, then every point on $\overline{EF}$ is equidistant from both points $G$ and $H$" is True.

Answer:

1. False; If $\overline{QR}$ goes through the midpoint of $\overline{XY}$ and $\overline{QR}\perp\overline{XY}$, then $\overline{QR}$ is the perpendicular bisector of $\overline{XY}$.
2. True
3. False; All squares are regular polygons.
4. True
5. False; You can create a regular pentagon with only a compass and straight - edge using a more complex construction method (such as the method based on the golden ratio).
6. True