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Foundations and Reasoning: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 An endpoint and a marked point

On a number line with one unit equal to one centimeter, P has coordinate $-42$ .

M, the midpoint of nonzero segment PQ, has coordinate $-6$ .

Sketch the order of P, M, Q.

Find Q's coordinate and the length PQ, stating which answer is a location and which is a distance.
1. Hint 1
  Compare the gaps on the two sides of a midpoint.
2. Hint 2
  Find the distance from P to M, then continue that same distance beyond M.
Answer

The order is P, M, Q. Q has coordinate $30$ , a location; $PQ=72$ cm, a distance.

Full solution

The distance from P to M is
$-6-(-42)=36\text{ cm}.$

The midpoint is between the endpoints and has equal distances to them.
Continue 36 units to the right of M to obtain Q's coordinate,
$-6+36=30.$

The whole contains both equal halves, so
$PQ=36+36=72\text{ cm}.$

The coordinate 30 identifies a location on the scale, while 72 centimeters measures the endpoint gap.

Answer

The order is P, M, Q. Q has coordinate $30$ , a location; $PQ=72$ cm, a distance.

Key idea

Equal midpoint gaps locate the other endpoint, but the endpoint coordinate and whole length are different quantities.
Problem 2 Two openings on paper

On paper, draw ray VU.

With a semicircular protractor, draw ray VW so that the selected reflex opening between the rays measures $226^\circ$ .

Label the selected reflex opening and the smaller opening with their measures and types.

Explain where you place the protractor and which scale you read.
1. Hint 1
  Relate the two openings between the rays to one full turn.
2. Hint 2
  Find which opening a semicircular protractor can mark directly.
3. Hint 3
  Place the center at V and start at the zero on VU, then identify the other route between the finished rays.
Answer

Mark $134^\circ$ from VU with the protractor centered at V, then draw VW through the mark. The smaller opening is $134^\circ$ , obtuse; the other opening is $226^\circ$ , reflex.

Full solution

The two openings between the distinct ray directions fill one full turn.
The smaller opening therefore measures
$360^\circ-226^\circ=134^\circ.$

Place the semicircular protractor's center at V and align its baseline with ray VU.
Follow the scale whose zero lies on VU and mark 134 on that same scale.
Remove the tool and draw ray VW from V through the mark.

Label the opening directly marked by the protractor as 134 degrees.
It is obtuse because its measure is greater than 90 degrees and less than 180 degrees.

Select and label the other route between the rays as the requested 226-degree reflex opening.
It is greater than 180 degrees and less than 360 degrees.
The paper drawing is approximate, while the stated measures identify the intended openings.

Answer

Mark $134^\circ$ from VU with the protractor centered at V, then draw VW through the mark. The smaller opening is $134^\circ$ , obtuse; the other opening is $226^\circ$ , reflex.

Key idea

A semicircular protractor can mark the smaller opening while the selected other route records the requested reflex opening.
Problem 3 Locations on a line through a center

A circle lies in plane $\mathcal P$ with center O and radius $12.7$ cm.

A number line contained in $\mathcal P$ passes through O, whose coordinate is $-6.4$ ; one coordinate unit is one centimeter.

Find all coordinates on this line that belong to the circle.

Point T on the same line has coordinate $6.1$ .

Decide whether T is on the circle, strictly inside its disk, or outside its disk.
1. Hint 1
  On the circle, the distance from the center must equal the positive radius.
2. Hint 2
  Check a point that distance in each direction on the line, then compare OT with the radius.
Answer

The circle points on the line have coordinates $-19.1$ and $6.3$ . T is strictly inside the disk and is not on the circle.

Full solution

A point on the circle in its plane is 12.7 centimeters from O.
The two directions on the center line give
$-6.4-12.7=-19.1$
and
$-6.4+12.7=6.3.$

These exhaust the positions with that distance from O on the number line.

For T, the center distance is
$OT=6.1-(-6.4)=12.5\text{ cm}.$

Since 12.5 is less than 12.7, T is strictly inside the disk rather than on the circle or outside the disk.

Answer

The circle points on the line have coordinates $-19.1$ and $6.3$ . T is strictly inside the disk and is not on the circle.

Key idea

A planar circle condition becomes a distance test on a number line contained in its plane.
Problem 4 A required new length

On paper, draw source segments AB and CD with $AB>CD>0$ , and draw a separate ray PX in the same plane.

After drawing these givens, use only a compass that retains its opening and an unmarked straightedge to construct a point Q on the ray with $PQ=AB-CD$ .

Leave the helper circles or arcs visible.

Explain why the point you choose gives the required length.
1. Hint 1
  A compass can transfer each given length without a ruler reading.
2. Hint 2
  Place the longer length first and consider which shorter part can be removed from its far end.
Answer

A valid construction has an auxiliary point R with P, Q, R in that order, $PR=AB$ and $QR=CD$ , giving $PQ=AB-CD$ . Retain the circles centered at P and R.

Full solution

Set the compass to AB and draw a circle centered at P in the given plane.
Ray PX starts at its center and meets the circle exactly once by the accepted center-ray intersection fact.
Call that point R, so $PR=AB$ by the radius condition.

Set the compass to CD and draw a circle centered at R in the same plane.

Ray RP starts at its center and meets this circle exactly once by the same accepted fact; call that point Q.

Since $0<RQ=CD<RP$ , Q is strictly between P and R and therefore belongs to ray PX.

The second circle gives $QR=CD$ .
Betweenness gives $PQ+QR=PR$ , so
$PQ=PR-QR.$

Substituting the two constructed lengths yields
$PQ=AB-CD.$
The helper circles record the exact radius conditions; a ruler check on the paper is only approximate.

Answer

A valid construction has an auxiliary point R with P, Q, R in that order, $PR=AB$ and $QR=CD$ , giving $PQ=AB-CD$ . Retain the circles centered at P and R.

Key idea

Transferred lengths and a justified point order can construct a difference without measuring either source numerically.
Problem 5 A proposed classification

Allow selected openings with measures strictly between zero and 360 degrees.

Consider the claim, “A selected angle is obtuse if and only if its measure is greater than 90 degrees.” Decide whether the claim is valid.

Give one allowed counterexample if needed, and replace the claim with a valid biconditional using the angle's measure.
1. Hint 1
  Check each direction of the if-and-only-if claim.
2. Hint 2
  Compare the full obtuse-angle definition with the single bound stated in the claim.
Answer

False. Any measure $180^\circ\le m<360^\circ$ gives a counterexample, such as $221^\circ$ . Correct: a selected angle is obtuse if and only if $90^\circ<m<180^\circ$ , where m is its measure.

Full solution

An obtuse angle has measure greater than 90 degrees, so the forward direction holds.

A 221-degree selected opening is greater than 90 degrees but is reflex, so it satisfies the reverse direction's hypothesis and fails its conclusion.

Any allowed measure with $180^\circ\le m<360^\circ$ is straight or reflex, so it also refutes the reverse implication.

The obtuse definition requires both bounds,
$90^\circ<m<180^\circ.$

An obtuse angle satisfies these bounds, and a selected opening satisfying both bounds is obtuse by definition.
The corrected biconditional therefore holds in both directions.

Answer

False. Any measure $180^\circ\le m<360^\circ$ gives a counterexample, such as $221^\circ$ . Correct: a selected angle is obtuse if and only if $90^\circ<m<180^\circ$ , where m is its measure.

Key idea

A true one-way condition becomes a biconditional only when the reverse direction keeps every required bound.
Problem 6 Two conditions on four rays

Four distinct rays OA, OB, OC, OD occur in that order around a selected $153^\circ$ opening AOD.

Their three consecutive parts have no overlapping interiors.

Ray OB bisects angle AOC, and ray OC bisects angle BOD.

Find the measure of each of the three consecutive parts.

Give a paragraph argument that $\angle AOB\cong\angle COD$ , citing a reason for each consequential step.
1. Hint 1
  Translate each given bisector into an equality of two selected part measures.
2. Hint 2
  Compare the shared middle part with each outside part, then account for the whole opening.
Answer

Each consecutive part measures $51^\circ$ ; $\angle AOB\cong\angle COD$ .

Full solution

The first bisector makes AOB and BOC have equal measures.
The second makes BOC and COD have equal measures.
The two outside measures therefore equal each other by transitivity.

All three consecutive measures are equal and fill 153 degrees.
Each one is
$153^\circ\div3=51^\circ.$

For the requested paragraph argument: by the two bisector definitions, $m\angle AOB=m\angle BOC$ and
$m\angle BOC=m\angle COD.$
Transitivity gives
$m\angle AOB=m\angle COD.$
By the definition of congruent angles, $\angle AOB\cong\angle COD$ .

The point order selects the appropriate parts; neither equal drawn ray lengths nor a later triangle theorem is used.

The blue first ray and dashed red final ray bound the selected counterclockwise opening. Any named interior rays separate the consecutive parts. Use the stated angle measures and markings, rather than measuring the drawing.

Answer

Each consecutive part measures $51^\circ$ ; $\angle AOB\cong\angle COD$ .

Key idea

A shared part can connect two exact equalities even when the two bisectors divide different overlapping angles.
Problem 7 Lines in different planes

Line r lies in plane $\mathcal P$ and line s lies in a distinct parallel plane $\mathcal Q$ .

Identify all possible relationships between the two lines.

For each possibility, give a pair of complete edge lines of the rectangular box shown that realizes it, taking one from the bottom-face plane and one from the top-face plane.

Every face is a rectangle, opposite face planes are parallel, the bottom is ABCD, the top is EFGH, and the connecting edges are AE, BF, CG, DH.

It is also given that within each face, opposite edge lines are parallel and adjacent edge lines are perpendicular.

Justify your classifications from those givens.

Text description of this figure

The rectangular box has bottom face A B C D and top face E F G H. Its front face is A B F E and back face D C G H. Edges A E, B F, C G, D H connect corresponding bottom and top vertices. All faces are rectangles, and opposite face planes are parallel. Within each face, opposite edge lines are given parallel and adjacent edge lines perpendicular. The diagram supplies the vertex connections without selecting answer pairs.
1. Hint 1
  The parallel planes prevent any shared point, but line parallelism also requires coplanarity.
2. Hint 2
  Compare opposite edges of one side face with a bottom edge and a differently directed top edge. For a proposed common plane, check all four endpoints.
Answer

The lines can be parallel or skew. For example, bottom line AD and top line EH are parallel; bottom line AD and top line EF are skew. Other correctly justified edge pairs are acceptable.

Full solution

The two lines cannot meet because a common point would belong to both distinct parallel planes.
They cannot coincide for the same reason.

Lines AD and EH are opposite edge lines of rectangular face ADHE, so they are parallel.
Each also lies in its required bottom or top plane.

Lines AD and EF are skew.
A plane containing both would contain the noncollinear points A, D, E, so it would be the left-face plane ADHE.
But F lies in the opposite parallel right-face plane, outside that left-face plane.

Thus the givens permit both a coplanar disjoint pair and a noncoplanar pair.
Parallel or skew exhaust the possibilities.

Answer

The lines can be parallel or skew. For example, bottom line AD and top line EH are parallel; bottom line AD and top line EF are skew. Other correctly justified edge pairs are acceptable.

Key idea

Lines in parallel planes are disjoint, but a common plane is still needed before calling the lines parallel.
Problem 8 Compare the end pieces

Distinct collinear points occur in the order A, B, C, D.

In centimeters, $AC=83$ , $BD=96$ , and $BC=24$ .

Find AB and CD.

State which is longer and by how much, and explain which stated conditions justify the part equations.
1. Hint 1
  Each supplied whole contains the middle segment BC.
2. Hint 2
  Use the point order to identify the other part of each whole, then compare the two results.
Answer

$AB=59$ cm and $CD=72$ cm. CD is longer by $13$ cm. The stated point order puts B between A and C and C between B and D.

Full solution

B is between A and C, so $AB+BC=AC$ .
Subtracting the known middle length gives
$83-24=59\text{ cm}$
for AB.

C is between B and D, so $BC+CD=BD$ .
Subtracting the same middle length gives
$96-24=72\text{ cm}$
for CD.

Compare the positive part lengths:
$72-59=13\text{ cm}.$
Thus CD is longer by 13 centimeters.

The point order, rather than the appearance of overlapping strokes, makes both part equations legitimate.

The labeled points lie on one line in the stated order. Drawn spacing adds no length or equality information.

Answer

$AB=59$ cm and $CD=72$ cm. CD is longer by $13$ cm. The stated point order puts B between A and C and C between B and D.

Key idea

A common middle part can be removed from each stated whole once betweenness identifies the pieces.
Problem 9 Related statements about a pair

The allowed objects are pairs of distinct lines in ordinary three-dimensional Euclidean space.

Start with the true conditional, “If the lines are parallel, then they have no common point.” Write its converse, inverse, and contrapositive.

State whether each resulting universal claim is true or false, and justify each decision without relying on measurements.
1. Hint 1
  Keep the same allowed objects when changing the order or negating the two parts.
2. Hint 2
  For claims involving lines with no common point, check whether a common plane has also been required.
Answer

Converse: if the lines have no common point, then they are parallel, false. Inverse: if the lines are not parallel, then they have a common point, false. Contrapositive: if the lines have a common point, then they are not parallel, true.

Full solution

The converse exchanges the parts.
Skew lines are distinct and disjoint but not parallel, so they refute the converse.

The inverse negates both parts without exchanging them.
The same skew pair is not parallel yet has no common point, so it also refutes the inverse.

The contrapositive exchanges and negates the parts.
Parallel lines have no common point by definition, so lines with a common point cannot be parallel.
This is also the logical form equivalent to the true original.

Coplanarity is the condition that distinguishes a disjoint parallel pair from a disjoint skew pair.

Answer

Converse: if the lines have no common point, then they are parallel, false. Inverse: if the lines are not parallel, then they have a common point, false. Contrapositive: if the lines have a common point, then they are not parallel, true.

Key idea

Equivalent logical forms agree, while a missing geometric hypothesis can defeat both the converse and inverse.
Problem 10 A point near a shared line

Distinct planes $\mathcal P$ and $\mathcal Q$ intersect in line k.

Point T belongs to $\mathcal P$ but does not lie on k.

Choose an approach and justify your choice, then give a complete argument that T cannot belong to $\mathcal Q$ , stating the reason for each consequential step.

The blue plane P and dashed red plane Q share line k. Point T lies in P away from k. The outlines represent patches of unlimited planes.

Text description of this figure

The blue plane P and dashed red plane Q share line k. Point T lies in P away from k. The outlines represent patches of unlimited planes.
1. Hint 1
  The intersection consists of all points common to both planes.
2. Hint 2
  Consider what would follow if T belonged to the second plane as well as the first.
Answer

T does not belong to $\mathcal Q$ .

Full solution

Proof by contradiction fits this negative goal: assuming that T belongs to the second plane makes the given intersection applicable.
Keep the givens and temporarily suppose T also belongs to $\mathcal Q$ .

T would then be common to $\mathcal P$ and $\mathcal Q$ .
Their intersection is exactly line k, so T would lie on k.

This contradicts the given that T is not on k.
The temporary assumption must be rejected, and T therefore does not belong to $\mathcal Q$ .

The argument uses the full intersection statement for distinct planes, not how close T looks to the shared line in a drawing.

Answer

T does not belong to $\mathcal Q$ .

Key idea

An impossibility proof can follow by showing that the goal's negation conflicts with the stated intersection.

Multiple choice

Core practice