Chapter Test · nothing is marked until you submit

Angles and Parallel Lines: 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)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 Partners of two equal angles

Two distinct lines intersect at O. Angles U and V are opposite sectors of that crossing. Angle E supplements U, and angle F supplements V. Given $m\angle E=133.5^\circ$ , find the measures of U, V, F and justify the relationship between E and F.
1. Hint 1
  The two supplementary pairs have the same total, while U and V remove equal amounts from that total.
2. Hint 2
  Find U from its stated pair first. Transfer its measure to V, then determine V's supplementary partner.
Answer

$m\angle U=m\angle V=46.5^\circ$ ; $m\angle F=133.5^\circ$ ; $\angle E\cong\angle F$ .

Full solution

Since E supplements U, their measures total 180 degrees.
Thus
$m\angle U=180^\circ-133.5^\circ.$
The result is $46.5^\circ$ .

The opposite sectors U and V are vertical angles of two distinct intersecting lines, so they are congruent.
Thus
$m\angle V=46.5^\circ.$
Since F supplements V,
$m\angle F=180^\circ-46.5^\circ.$
This gives $133.5^\circ$ .

Both E and F have the same measure, so they are congruent.
Equivalently, they are supplements of the congruent angles U and V.
No common vertex is needed for that reasoning.

Answer

$m\angle U=m\angle V=46.5^\circ$ ; $m\angle F=133.5^\circ$ ; $\angle E\cong\angle F$ .

Key idea

Supplements of congruent angles are congruent because they remove equal measures from the same total.
Problem 2 The two crossings

All lines lie in one plane. Distinct lines $\ell,m$ are crossed by transversal $t$ at distinct points. The upper northwest, northeast, southeast, southwest sectors are $1,2,3,4$ ; the lower sectors in the same order are $5,6,7,8$ .

The exact givens are $m\angle1=134.2^\circ$ and $m\angle8=45.8^\circ$ . No parallelism is initially given. Determine the relationship between $\ell$ and $m$ , and decide whether $t$ is perpendicular to either. Justify both decisions.

Text description of this figure

The upper and lower crossings use the stated eight-sector convention. No parallel arrows or right-angle markings are given. The exact measures are supplied in the statement.
1. Hint 1
  Separate the positions of angles 1 and 8 from the numerical relationship between them.
2. Hint 2
  Check whether this pair meets a proved converse. Then examine the individual sector at each crossing before declaring perpendicularity.
Answer

$\ell\parallel m$ ; $t$ is perpendicular to neither $\ell$ nor $m$ .

Full solution

Angles 1 and 8 are same-side exterior angles at the distinct crossings.
Their exact measures satisfy
$134.2^\circ+45.8^\circ=180^\circ.$

The same-side exterior converse therefore establishes $\ell\parallel m$ .
The common plane and distinct crossings are part of the givens.

At the upper crossing, angle 1 measures $134.2^\circ$ , which is not right.
At the lower crossing, angle 8 measures $45.8^\circ$ , which is not right.
Two perpendicular full lines make all four sectors right, so t is perpendicular to neither crossed line.

Answer

$\ell\parallel m$ ; $t$ is perpendicular to neither $\ell$ nor $m$ .

Key idea

A supplementary same-side pair can prove parallelism while its individual measures rule out perpendicularity.
Problem 3 Three proportional openings

A nondegenerate triangle has interior angle measures in the ratio $2:5:8$ . Find the three measures, classify the triangle by angles, and explain why it cannot also have a right interior angle.
1. Hint 1
  All three positive measures share one ratio unit and one triangle total.
2. Hint 2
  Represent the measures as 2k, 5k, and 8k degrees, and determine the common unit.
3. Hint 3
  After classifying the largest angle, compare it with a hypothetical right angle while remembering that the third interior angle must be positive.
Answer

The measures are $24^\circ,60^\circ,96^\circ$ . The triangle is obtuse and has no right interior angle.

Full solution

Let the measures in degrees be 2k, 5k, and 8k with $k>0$ .
The triangle sum gives
$15k=180.$
Dividing by 15 yields $k=12$ .

The first measure is
$2(12)^\circ=24^\circ.$
The second is
$5(12)^\circ=60^\circ.$
The third is
$8(12)^\circ=96^\circ.$
These positive measures total 180 degrees.

The 96-degree interior angle is obtuse, so the triangle is obtuse.
If another interior angle were right, those two alone would total
$96^\circ+90^\circ=186^\circ.$
The third interior angle is positive, so the total could not be 180 degrees.
Thus a right interior angle is impossible here.

Answer

The measures are $24^\circ,60^\circ,96^\circ$ . The triangle is obtuse and has no right interior angle.

Key idea

A triangle ratio is constrained by the 180-degree total, and an obtuse angle leaves no room for a right angle.
Problem 4 Two crossing routes

Distinct lines $\ell,m$ and distinct transversals $t,u$ lie in one plane. Each transversal crosses both lines at distinct points. In the schematic, $\ell$ is above $m$ . Along t, A is the upper northeast sector and B the lower northeast sector. Along u, C is the upper southwest sector and D the lower northwest sector.

The exact measures of A and B are both $72.4^\circ$ . The degree measures of C and D are $4k+2.5$ and $7k+12.5$ , respectively.

Part A. Determine the relationship between $\ell$ and $m$ from the stated information and justify it.

Part B. Find k and the measures of C and D, explaining how your conclusion from Part A supports the equation.

Text description of this figure

Two distinct lines l and m are each crossed by t and u. Along t, A and B occupy northeast sectors at the two crossings. Along u, C is upper southwest and D lower northwest. No parallel arrows or derived measures are shown.
1. Hint 1
  A relationship between sectors on one transversal can establish a line fact that the other transversal may then use.
2. Hint 2
  Part A: identify the positions of A and B and compare their exact measures.
3. Hint 3
  Part B: after establishing parallelism, identify the family of C and D and write the relationship for that family.
Answer

Part A: $\ell\parallel m$ . Part B: $k=15$ , $m\angle C=62.5^\circ$ , and $m\angle D=117.5^\circ$ .

Full solution

Part A.
A and B are corresponding angles for t.
Their exact equal measures establish congruence, so the corresponding-angle converse gives $\ell\parallel m$ .

Part B.
For u, C and D are same-side interior angles of these now-established parallel lines.
Their measures sum to 180 degrees.
Adding the expressions gives $4k+7k$ and a constant total of $2.5+12.5=15$ , so
$11k+15=180.$

Part B.
Subtracting 15 gives $11k=165$ , so $k=15$ .

Part B.
Substitution into C gives $4(15)+2.5=62.5$ , so C measures $62.5^\circ$ .
Substitution into D gives $7(15)+12.5=117.5$ , so D measures $117.5^\circ$ .

Part B.
The measures are positive and total 180 degrees, as required.
The measures along t were not equated to those along u; only the proved line relationship passed between the two routes.

Answer

Part A: $\ell\parallel m$ . Part B: $k=15$ , $m\angle C=62.5^\circ$ , and $m\angle D=117.5^\circ$ .

Key idea

A converse along one transversal can establish the parallelism needed for a forward theorem along another.
Problem 5 A segment from the top corner

In nondegenerate triangle $ABC$ , point D lies strictly between B and C, and AD bisects the interior angle BAC. Point E lies on the continuation of CB beyond B, so C, B, E are collinear in that order.

The givens are $m\angle ABE=147.5^\circ$ and $m\angle ACB=58.5^\circ$ . Find $m\angle BAC$ and $m\angle ADB$ , with a reason for each intermediate measure.

Text description of this figure

A is off base BC, D lies between B and C, and segment AD is drawn. CB continues past B to E, so C, B, E occur in that order. AD is given to bisect angle BAC. No solved angles are drawn.
1. Hint 1
  The smaller triangle containing ADB needs an angle at A and an angle at B.
2. Hint 2
  The supplied opening at B is outside the triangle; first identify the interior opening next to it.
3. Hint 3
  Find the whole angle at A in ABC before using the bisector to supply the part needed in ADB.
Answer

$m\angle BAC=89^\circ$ and $m\angle ADB=103^\circ$ .

Full solution

The order C, B, E makes BC and BE opposite rays.
The interior angle ABC and exterior angle ABE form a linear pair.
Thus
$m\angle ABC=180^\circ-147.5^\circ,$
giving $32.5^\circ$ .

The whole triangle gives the remaining angle measure in degrees:
$180-32.5-58.5=89.$
Hence
$m\angle BAC=89^\circ.$

Since AD bisects angle BAC, its part BAD measures
$89^\circ\div2=44.5^\circ.$

Because D lies between B and C, rays BD and BC agree.
Thus triangle ADB has the same $32.5^\circ$ angle at B as the whole triangle.
Its remaining angle measure in degrees is
$180-32.5-44.5=103.$
Thus
$m\angle ADB=103^\circ.$

Both triangles have positive angle measures.
The smaller triangle checks with $32.5+44.5+103=180$ degrees.
The side continuation, bisector, and point order justify the successive transfers.

Answer

$m\angle BAC=89^\circ$ and $m\angle ADB=103^\circ$ .

Key idea

Identify the interior neighbor of an exterior angle before passing through a bisector into a smaller triangle.
Problem 6 Readings along a route

A robot leaves A, follows the boundary of nondegenerate triangle ABC in the order A to B to C to A, and takes one non-reflex left exterior turn at each corner. It finishes the turn at A facing its original direction along AB. A running counter starts at zero when the robot first leaves A and adds every left turn without resetting.

The counter reads $126.7^\circ$ immediately after the turn at B, $258.9^\circ$ immediately after the turn at C, and $360^\circ$ immediately after the final turn at A.

Part A. Find the individual exterior turn measure at each of A, B, C.

Part B. Jules says the three displayed readings are the triangle's three exterior angle measures. Explain the error.

Part C. Find the interior angle at C. Check that the three recovered turns and their adjacent interior angles are compatible with a nondegenerate triangle.

Text description of this figure

A triangular route is followed from A to B to C and back to A. Arrows identify the travel order. All cumulative counter readings appear separately in the statement; no angle values appear on the sketch.
1. Hint 1
  A running total at a corner includes earlier turns. The local turn is the increase since the preceding reading.
2. Hint 2
  For the turn at C, subtract the reading after B from the reading after C. Treat the final return to A the same way.
3. Hint 3
  Each recovered exterior turn must lie between zero and 180 degrees. Its adjacent interior angle is its supplement.
Answer

Part A: A $101.1^\circ$ , B $126.7^\circ$ , C $132.2^\circ$ . Part B: cumulative totals. Part C: C $47.8^\circ$ ; interiors A $78.9^\circ$ , B $53.3^\circ$ ; turn total $360^\circ$ , interior total $180^\circ$ .

Full solution

Part A.
Before B the running total is zero.
Therefore the first individual turn, at B, measures $126.7^\circ$ .

Part A.
The reading after C includes the turns at B and C.
Subtract the previous total:
$e_C=258.9^\circ-126.7^\circ.$
Thus the turn at C measures $132.2^\circ$ .

Part A.
The last increase is the turn at A.
It measures
$e_A=360^\circ-258.9^\circ,$
giving $101.1^\circ$ .

Part B.
Jules has confused a running total with an individual change.
The reading $258.9^\circ$ contains two turns, and $360^\circ$ contains all three.
Neither is a single non-reflex exterior angle.
Only the first reading also equals an individual turn, because no earlier turn has been counted.

Part C.
The exterior turn at C and its interior neighbor form a linear pair.
Therefore
$m\angle ACB=180^\circ-132.2^\circ,$
so that interior angle measures $47.8^\circ$ .

Part C.
The turns $101.1^\circ,126.7^\circ,132.2^\circ$ are each strictly between zero and 180 degrees and total $360^\circ$ .
Their supplements at A, B, C are $78.9^\circ,53.3^\circ,47.8^\circ$ .
These are positive and total $180^\circ$ , so the records pass the triangle-angle checks.
All turns are left turns in the stated boundary order, so they also describe one consistent triangular traversal.

Answer

Part A: A $101.1^\circ$ , B $126.7^\circ$ , C $132.2^\circ$ . Part B: cumulative totals. Part C: C $47.8^\circ$ ; interiors A $78.9^\circ$ , B $53.3^\circ$ ; turn total $360^\circ$ , interior total $180^\circ$ .

Key idea

An exterior turn is an individual direction change; recover it from consecutive cumulative readings before checking the triangle totals.
Problem 7 A line through the marked point

All lines lie in one Euclidean plane. Distinct lines a and b are perpendicular to t at distinct points P and Q. Point R lies on a and differs from P.

Line c passes through R and differs from a. Decide whether c can avoid b, and give a complete justification from the stated givens. The starting diagram shows a, b, t, and R; c has not been chosen for the sketch.

Text description of this figure

Lines a and b are given perpendicular to t at distinct P and Q. R is a different point on a. The arbitrary line c through R is left for the argument.
1. Hint 1
  First determine the relationship forced between a and b by their two right-angle crossings.
2. Hint 2
  That relationship locates R relative to b. Consider what would follow if c had no common point with b.
3. Hint 3
  Count the distinct lines through R that would then be parallel to b, and compare that count with the accepted uniqueness foundation.
Answer

Line c cannot avoid b; it meets b.

Full solution

The distinct coplanar perpendiculars a and b make congruent corresponding right angles with t.
The corresponding-angle converse gives $a\parallel b$ .

Since R lies on a, it is outside b.
Otherwise a and b would meet at R, contrary to their parallelism.

Suppose c avoided b.
The two lines would be distinct, because c contains R while b does not.
As disjoint coplanar lines, they would satisfy $c\parallel b$ .

Both a and c would then be parallels to b through the external point R.
They are given different, contradicting the accepted uniqueness of a parallel through an external point in the plane.

Reject the temporary assumption that c avoids b.
Therefore c meets b.
The proof uses the line relationships and uniqueness, rather than the reach of a finite drawn segment.

Answer

Line c cannot avoid b; it meets b.

Key idea

Two perpendiculars can supply a parallel, after which uniqueness forces every different coplanar line through the external point to cross.
Problem 8 Two remaining corners

Nondegenerate Euclidean triangles $ABC$ and $DEF$ satisfy $\angle CAB\cong\angle FDE$ and $\angle ABC\cong\angle DEF$ .

Give a complete argument that $\angle ACB\cong\angle DFE$ . Establish any triangle-angle total your argument needs from accepted foundations and previously proved line-angle facts, instead of assuming a triangle-angle total or a third-angle result. Include a labeled sketch for any line you add.
1. Hint 1
  The two unmatched angles will be equal remainders if each triangle has the same justified total.
2. Hint 2
  For one arbitrary triangle, consider a line through one vertex parallel to the opposite side. Track the other two angles along their separate transversals.
3. Hint 3
  Use straight-angle addition at the auxiliary line, then substitute the matching triangle measures. The same established total applies to the other triangle.
4. Hint 4
  Remove the two equal given measures from those equal totals and interpret the remaining equality as congruence.
Answer

$\angle ACB\cong\angle DFE$ .

Full solution

Begin with arbitrary triangle ABC.
Since its vertices are noncollinear, C is outside line AB.
The accepted parallel foundation supplies a line through C parallel to AB.
Choose X and Y on its opposite rays so the order below this line is CX, CA, CB, CY.

For transversal AC, alternate interior angles give $\angle XCA\cong\angle CAB$ .
For transversal BC, they give $\angle BCY\cong\angle ABC$ .
These are forward results because the new line was chosen parallel to AB.

Write p, q, r for the consecutive measures of XCA, ACB, BCY.
They fill the straight angle XCY, so
$p+q+r=180^\circ.$

Let a, b, c be the interior measures at A, B, C.
The two congruences give $p=a$ and $r=b$ , while $q=c$ .
Substitution yields
$a+b+c=180^\circ.$
Because ABC was arbitrary, this reasoning establishes the same total for every nondegenerate Euclidean triangle, including DEF.

Let d, e, f be the interior measures at D, E, F.
The same established total gives
$d+e+f=180^\circ.$
The original givens give $a=d$ and $b=e$ .

The two remaining measures are therefore
$c=180^\circ-a-b$
and
$f=180^\circ-d-e.$
Replacing d by a and e by b makes the right sides equal.
Thus $c=f$ , which is exactly $\angle ACB\cong\angle DFE$ .

Answer

$\angle ACB\cong\angle DFE$ .

Key idea

A common total must be justified before equal given parts can force equal remaining triangle angles.
Problem 9 A line through the remaining vertex

On paper, draw a nondegenerate triangle $PQR$ . Using a compass that retains its opening and an unmarked straightedge, construct a full line through Q parallel to line PR.

Retain the compass arcs, label any added points, and name the exact angle pair that justifies the line relationship. Explain why the angle positions and the distinct crossings meet the chosen test's conditions.

Text description of this figure

Triangle PQR has three noncollinear vertices. The starting picture supplies only its sides; the requested construction is left for the student.
1. Hint 1
  One existing triangle side can connect the external vertex to the opposite side's line.
2. Hint 2
  Extend RQ beyond Q. Consider copying the opening at R onto the continuation ray from Q, with the copied ray on P's side of RQ.
3. Hint 3
  Retain the equal-radius arcs and chord-transfer arc. Identify the copied angles at the two crossings before citing a converse.
Answer

A constructed line $\ell$ through Q with $\ell\parallel PR$ , retained copy arcs, and a valid converse pair; for example, $\angle SQX$ and $\angle QRP$ .

Full solution

Extend RQ beyond Q to an added point S, so R, Q, S occur in that order.
Noncollinearity places Q outside line PR, and RQ is the prospective transversal.

At R, draw a compass arc meeting rays RQ and RP.
Keep the same radius and draw an arc centered at Q from starting ray QS.
Transfer the chord between the first arc's two crossings onto the new arc, choosing the intersection on P's side of line RQ.
Call it X and draw ray QX.

The taught angle-copy procedure supplies $\angle SQX\cong\angle QRP$ .
Its correctness proof is deferred to lesson 4.5; the copying fact is accepted here.
The equal-radius arcs and chord-transfer arc record that procedure.

Extend QX to full line $\ell$ .
Since QS continues the direction RQ and X lies on P's side of RQ, SQX and QRP occupy corresponding positions.

The crossings at Q and R are distinct, and all constructed objects lie in the triangle's plane.
The corresponding-angle converse gives $\ell\parallel PR$ .
The lines are distinct because Q is outside PR.

A different correctly performed construction is acceptable if its retained arcs establish an exact angle condition and its stated pair satisfies a proved converse.
Approximate visual agreement alone is not the justification.

Answer

A constructed line $\ell$ through Q with $\ell\parallel PR$ , retained copy arcs, and a valid converse pair; for example, $\angle SQX$ and $\angle QRP$ .

Key idea

A parallel construction needs an exact copied-angle record and the positions that make a converse apply.
Problem 10 Distances on a site plan

A site plan has distinct parallel boundary lines a and b in one plane. Point P lies on a, point H lies on b, and segment PH is given perpendicular to both. Point X lies strictly between P and H. The actual lengths are $PH=28.4$ meters and $PX=11.7$ meters; the diagram is schematic.

Part A. Name the perpendicular foot from X on each boundary line and find the distance from X to each line.

Part B. Find the distance from P to line a.

Part C. Give the perpendicular separation of a and b measured along the specified segment PH. On your sketch, mark the segments that record the two distances in Part A.

Text description of this figure

Parallel lines a and b have a specified joining segment PH perpendicular to both. P is on a and H on b. X is between them. The given lengths are PH 28.4 meters and PX 11.7 meters; drawn spacing adds no length fact.
1. Hint 1
  A distance to a line uses a perpendicular segment ending on that line. Each part of PH stays on its perpendicular line.
2. Hint 2
  Part A: identify which endpoint is on each boundary. Use the stated point order to find the unlisted part of PH.
3. Hint 3
  Part B concerns a point already on its line. Part C uses the whole specified perpendicular joining segment.
Answer

Part A: feet P and H; distances to a and b are $11.7$ m and $16.7$ m; mark XP and XH. Part B: $0$ m. Part C: $28.4$ m.

Full solution

Part A.
Segment XP lies on PH, is perpendicular to a, and ends at P on a.
Thus P is the foot from X on a, and the distance is $XP=11.7$ meters.
Mark XP.

Part A.
Segment XH is perpendicular to b and ends at H on b, so H is that foot.
Since X lies between P and H, segment addition gives $PX+XH=PH$ .
Thus
$XH=28.4-11.7$
and the distance from X to b is $16.7$ meters.
Mark XH.

Part B.
Point P belongs to line a, so its distance to a is zero meters.

Part C.
PH joins the two parallel lines and is given perpendicular to both.
Its $28.4$ -meter length is therefore the specified perpendicular separation.
This uses the distance definition, not a later proof about shortest lengths or equal perpendicular widths.

Answer

Part A: feet P and H; distances to a and b are $11.7$ m and $16.7$ m; mark XP and XH. Part B: $0$ m. Part C: $28.4$ m.

Key idea

The perpendicular feet and stated point order identify the full distance and its parts, while a point already on a line has distance zero.

Multiple choice

Core practice