Congruence: Constructing Regular Polygons Inscribed in a Circle (CCSS.G-CO.12)
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Common Core High School Geometry › Congruence: Constructing Regular Polygons Inscribed in a Circle (CCSS.G-CO.12)
Which sequence of steps constructs the perpendicular bisector of $AB$?
With center at $A$, draw a circle through $B$ and then draw a line tangent to the circle at $A$.
With any radius, draw an arc above $AB$ from $A$ and an arc below $AB$ from $B$; connect those two arc points with a line.
Set the compass width greater than half of $AB$. From $A$ draw arcs above and below $AB$. From $B$ with the same width draw arcs that intersect the first pair. Draw the line through the two intersection points.
Estimate the midpoint of $AB$ by eye and draw a line through it parallel to $AB$.
Explanation
To construct the perpendicular bisector, draw equal-radius arcs from both endpoints $A$ and $B$ so they intersect above and below the segment, then connect those intersections. Choice B misplaces arcs on opposite sides so they do not intersect properly and will not bisect perpendicularly.
Which sequence of steps constructs the bisector of $\angle ABC$?
From $B$, draw an arc that intersects both rays of the angle. From those two intersection points, draw arcs of equal radius that intersect inside the angle. Draw the ray from $B$ through that intersection.
From $B$, draw one large arc on one side and a smaller arc on the other, then connect those arc points with a ray.
Construct a line through $B$ perpendicular to one side of the angle.
Copy $\angle ABC$ at $B$ onto a ray at $C$, then connect $B$ to $C$.
Explanation
The angle bisector is found by first marking equal distances on both sides of the angle from the vertex, then intersecting equal-radius arcs from those points and joining the vertex to the intersection. Choice C confuses a perpendicular with a bisector.
Which sequence of steps copies segment $AB$ to start at a given point $P$?
Draw a line through $P$ parallel to $AB$; any segment on this line equals $AB$.
Draw a circle centered at $B$ through $A$; choose any diameter as the copy.
From $A$ and $B$, draw intersecting arcs and connect their intersection to $P$.
Draw a ray starting at $P$. Set the compass to the length $AB$. Mark that span on the ray to locate $Q$ so that $PQ=AB$.
Explanation
To copy a segment, draw a ray from the new starting point, set the compass to the original length, and mark off that length along the ray. Choice A assumes any segment on a parallel line matches the length, which is false.
Which construction makes a line through point $P$ on line $l$ that is perpendicular to $l$?
With center at $P$, draw any circle and connect $P$ to any point on the circle.
Mark two points on $l$ at equal distances from $P$. Using these as centers with equal radius, draw arcs above (or below) $l$ that intersect; draw the line through $P$ and that intersection.
Through $P$, draw a line parallel to $l$.
Construct the perpendicular bisector of a random segment not on $l$, then slide it to pass through $P$.
Explanation
Creating equal offsets on $l$ and intersecting arcs above or below produces a line through $P$ perpendicular to $l$. Choice C confuses parallel with perpendicular.
Which sequence of steps constructs a line through point $P$ not on line $l$ that is parallel to $l$?
Construct the perpendicular to $l$ through $P$.
Draw a circle centered at $P$; where it meets $l$ determines the parallel line.
Pick a point $Q$ on $l$ and draw the transversal $PQ$. Using the compass, copy at $P$ the angle at $Q$ formed by $l$ and $PQ$ so that the new line through $P$ makes an equal corresponding angle with $PQ$.
Construct the perpendicular bisector of $PQ$ and use it as the parallel.
Explanation
Copying the corresponding angle formed by a transversal yields a line through $P$ parallel to $l$. Choice A constructs a perpendicular, not a parallel.
Which construction creates the perpendicular bisector of segment $AB$?
With the compass centered at the midpoint of $AB$, draw a circle through $A$ and $B$; the diameter through $A$ is the bisector.
With the compass centered at $A$ and at $B$ using a radius less than half of $AB$, draw small arcs and connect their nearest intersections to $AB$.
With the compass centered at $A$ and at $B$ using the same radius greater than half of $AB$, draw arcs above and below $AB$; connect the arc intersections with a straight line.
Through $A$, draw a line perpendicular to $AB$; that line is the perpendicular bisector.
Explanation
To construct the perpendicular bisector of $AB$, draw equal-radius arcs centered at $A$ and at $B$ (radius greater than half of $AB$) so the arcs intersect above and below the segment; the line through those intersections is perpendicular to $AB$ and passes through its midpoint. Choice D makes a perpendicular through an endpoint, which need not pass through the midpoint, so it is not a bisector.
Which steps construct the bisector of $\angle ABC$?
Center the compass at $B$ to draw an arc intersecting both sides of the angle; from those two intersection points, draw equal-radius arcs that meet inside the angle; draw the ray from $B$ through that intersection.
Center the compass at $A$ and at $C$ to draw arcs that meet; connect their intersection to $B$ to bisect the angle.
Draw through $B$ a line perpendicular to $BC$; that line bisects the angle.
Copy $\angle ABC$ at $B$ onto itself by matching one side; the result is its bisector.
Explanation
The angle-bisector uses an arc from the vertex to mark equal distances on each side, then equal-radius arcs from those marks to find a point equidistant from both sides; the ray from the vertex to that point bisects the angle. Choice B incorrectly uses centers at $A$ and $C$ (the side endpoints), which does not guarantee equal distances along the sides and will not, in general, locate the bisector.
Which steps copy segment $AB$ onto ray $r$ with endpoint $P$?
Construct the perpendicular bisector of $AB$; place its midpoint on $r$ and draw a congruent segment by eye.
Set any convenient compass width; draw a circle centered at $P$; where it meets $r$ is the copy of $AB$.
With center at $A$, draw an arc through $B$ until it meets $r$; connect that point to $P$ as the copy.
Set the compass to the length $AB$; with the compass point at $P$, mark off that length along ray $r$ to locate $Q$ so that $PQ = AB$.
Explanation
To copy a segment, fix the compass to span $AB$, place the compass point at $P$, and strike an arc to meet the ray at $Q$; then $PQ = AB$. Choice B uses an arbitrary radius, so the constructed length would not match $AB$ unless by coincidence.
Which steps copy $\angle ABC$ with vertex at point $P$ on ray $p$?
Construct the perpendicular bisector of $BC$; through $P$ draw a line parallel to $BA$ to reproduce the angle.
With center at $B$, draw an arc intersecting $BA$ and $BC$; with the same radius and center at $P$, draw an arc across $p$; measure the distance between the two intersections on the first arc and mark the same distance between the intersections on the second arc; draw the ray from $P$ through that mark.
Bisect $\angle ABC$ at $B$ and then draw that bisector as the new ray through $P$.
Through $P$ draw a line parallel to $BC$; the angle between that line and $p$ equals $\angle ABC$.
Explanation
The standard angle-copy uses equal-radius arcs at the original and target vertices and transfers the chord length between arc intersections; the ray from $P$ through the transferred point matches $\angle ABC$. Choice C constructs an angle half as large, not a copy.
Which construction draws a line through point $P$ parallel to line $\ell$?
First construct a line through $P$ perpendicular to $\ell$; then construct a line through $P$ perpendicular to that new line. The second line is parallel to $\ell$.
Construct the perpendicular bisector of a segment joining $P$ to any point on $\ell$; that bisector is parallel to $\ell$.
Through $P$ draw a line that appears to make the same acute angle with $\ell$ by sight; adjust until they look equal.
Through $P$ draw a single line perpendicular to $\ell$; that line is parallel to $\ell$.
Explanation
Lines perpendicular to the same line are parallel. Build a perpendicular from $P$ to $\ell$, then a perpendicular to that at $P$; the final line is parallel to $\ell$. Choice D stops after the first perpendicular, which is not parallel to $\ell$ but perpendicular to it.