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SAS and ASA Congruence: Core practice

10 practice problems for this lesson. 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 Name the included parts

In triangle $RST$ , name the angle included between $RS$ and $ST$ . Then name the side included between the angles at $R$ and $T$ .

Text description of this figure

Triangle RST shows the three named vertices with no chosen included part highlighted. Side and angle names can be traced directly from the vertex labels.
1. Hint 1
  Find where the two sides meet.
2. Hint 2
  For the side, join the vertices of the two named angles.
Answer

The included angle is $\angle RST$ ; the included side is $RT$ .

Full solution

Sides RS and ST meet at S.
Write S as the middle letter of the angle name: $\angle RST$ .

The angles at R and T have their vertices at R and T.
The side joining those vertices is $RT$ .

Answer

The included angle is $\angle RST$ ; the included side is $RT$ .

Key idea

The included part connects the two named parts at their common vertex or endpoints.
Problem 2 Read the matching order

Triangles KLM and NPQ have these given matching parts:
$\begin{aligned}\angle K&\cong\angle P\\\angle M&\cong\angle N\\KM&=PN\end{aligned}$
Establish congruence and write the matching triangle in the order corresponding to KLM.

Text description of this figure

Two schematic triangles show only the stated side and angle data. Corresponding supplied sides have the same tick pattern, and supplied angle pairs have the same arc pattern. The first triangle is solid blue and the second is red and dashed. Unmarked parts are not extra givens.
1. Hint 1
  Locate the side joining each pair of known angle vertices.
2. Hint 2
  Pair K with P and M with N before placing the remaining vertices.
Answer

$\triangle KLM\cong\triangle PQN$ by ASA.

Full solution

Side KM joins the known angles K and M.
Side PN joins their corresponding angles P and N.
The sides are equal, so ASA applies.

The given angle pairs identify K with P and M with N.
The remaining vertices L and Q correspond, giving:

$\triangle KLM\cong\triangle PQN.$

Answer

$\triangle KLM\cong\triangle PQN$ by ASA.

Key idea

A congruence statement records the same correspondence that made the criterion work.
Problem 3 Complete a side match

In triangles ABC and DEF, the known matching parts are:
$\begin{aligned}AC&=14\text{ cm}\\DF&=14\text{ cm}\\\angle A&\cong\angle D\end{aligned}$
The remaining corresponding side lengths are:
$\begin{aligned}AB&=5x-8\text{ cm}\\DE&=2x+19\text{ cm}\end{aligned}$
Find x so these data establish congruence in the listed order, and give the common length of AB and DE.

Text description of this figure

The schematic triangles have matching given angles at A and D, and matching sides AC and DF labeled 14 centimeters. AB carries 5x minus 8, and DE carries 2x plus 19. Those two variable sides have no equality ticks; their equality is the condition to be found.
1. Hint 1
  Check where the known angles lie relative to the two side pairs.
2. Hint 2
  Set the corresponding side lengths equal, then substitute to check positivity.
Answer

$\begin{aligned}x&=9\\AB&=37\text{ cm}\\DE&=37\text{ cm}\\\triangle ABC&\cong\triangle DEF\end{aligned}$
These results follow by SAS.

Full solution

The two side pairs meet at A and D, so the known angles are included.
The missing side match requires:

$\begin{aligned}5x-8&=2x+19.\end{aligned}$

Subtract 2x and add 8:

$\begin{aligned}3x&=27.\end{aligned}$

Divide by 3:

$\begin{aligned}x&=9.\end{aligned}$

Substitute in each expression:

$\begin{aligned}5(9)-8&=37.\\2(9)+19&=37.\end{aligned}$

Both lengths are positive.
Together with the 14 cm side pair and included angles, SAS establishes:

$\begin{aligned}\triangle ABC&\cong\triangle DEF.\end{aligned}$

Answer

$\begin{aligned}x&=9\\AB&=37\text{ cm}\\DE&=37\text{ cm}\\\triangle ABC&\cong\triangle DEF\end{aligned}$
These results follow by SAS.

Key idea

An algebraic match must give actual positive side lengths before it can support a geometric criterion.
Problem 4 Complete an angle match

Triangles UVW and XYZ have these given matching parts:
$\begin{aligned}UV&=22\text{ cm}\\XY&=22\text{ cm}\\m\angle U&=41^\circ\\m\angle X&=41^\circ\end{aligned}$
The remaining endpoint angles have these measures:
$\begin{aligned}m\angle V&=(5y+7)^\circ\\m\angle Y&=92^\circ\end{aligned}$
Find y so ASA applies in that order, and find the remaining angle in each triangle.

Text description of this figure

The schematic triangles have equal bases UV and XY labeled 22 centimeters. The given 41-degree angles at U and X have matching arcs. V carries 5y plus 7 degrees and Y carries 92 degrees, without matching arcs on that unestablished pair.
1. Hint 1
  The known side joins the given angle at its first endpoint to the angle at its second endpoint.
2. Hint 2
  Match the second endpoint angles, then use the triangle angle sum.
Answer

$\begin{aligned}y&=17\\m\angle W&=47^\circ\\m\angle Z&=47^\circ\end{aligned}.$

Full solution

ASA requires the angles at V and Y to match.
Solve:

$5y+7=92.$

Subtract 7 and divide by 5 to get:

$y=17.$

The two known angles total:

$41^\circ+92^\circ=133^\circ.$

Each remaining angle measures:

$180^\circ-133^\circ=47^\circ.$

The 22 cm sides are included between the matching 41-degree and 92-degree angles, so ASA establishes congruence.

Answer

$\begin{aligned}y&=17\\m\angle W&=47^\circ\\m\angle Z&=47^\circ\end{aligned}.$

Key idea

The side's two endpoints tell you which angle pairs are needed for ASA.
Problem 5 Construct a locked corner

You are given a 9 cm segment, a 4 cm segment, and an angle template measuring $73^\circ$ . On paper, construct triangle ABC with these measurements:
$\begin{aligned}AB&=9\text{ cm}\\AC&=4\text{ cm}\\m\angle BAC&=73^\circ\end{aligned}$
Describe your steps and justify why every successful construction is congruent to yours.
1. Hint 1
  Start with one of the prescribed sides and the vertex where both sides meet.
2. Hint 2
  Which given length starts at the vertex where you copied the angle?
Answer

Draw AB, copy the angle at A, mark C on its new ray, and join BC:
$\begin{aligned}AB&=9\text{ cm}\\AC&=4\text{ cm}\\m\angle BAC&=73^\circ\end{aligned}.$
Every successful result is congruent by SAS.

Full solution

Copy the 9 cm segment to make AB.
At A, copy the supplied 73-degree angle with ray AB as its first side.

Set the compass to 4 cm and mark C on the new ray at that distance from A.
Join B to C.

The two positive side lengths and the angle strictly between zero and 180 degrees give a nonflat triangle.
Every successful construction has these measurements:

$\begin{aligned}AB&=9\text{ cm}\\AC&=4\text{ cm}\\m\angle BAC&=73^\circ\end{aligned}$

These are exactly SAS data.
Choosing the other half-plane for C produces a reflected congruent copy.

Answer

Draw AB, copy the angle at A, mark C on its new ray, and join BC:
$\begin{aligned}AB&=9\text{ cm}\\AC&=4\text{ cm}\\m\angle BAC&=73^\circ\end{aligned}.$
Every successful result is congruent by SAS.

Key idea

An SAS construction fixes two distances on rays whose opening is prescribed.
Problem 6 Construct from two endpoint angles

Construct triangle HJK on paper from these data:
$\begin{aligned}HJ&=11\text{ cm}\\m\angle H&=34^\circ\\m\angle J&=101^\circ\end{aligned}$
You may use a ruler and protractor for the numerical drawing. Describe the construction, find angle K, and explain why an exact triangle with these measurements is fixed in shape and size.
1. Hint 1
  Decide which side must lie between the two known angles.
2. Hint 2
  Draw the new rays into the same half-plane. Check their angle sum before finding the intersection.
Answer

Draw HJ and rays into the same half-plane, meeting at K:
$\begin{aligned}HJ&=11\text{ cm}\\m\angle H&=34^\circ\\m\angle J&=101^\circ\\m\angle K&=45^\circ\end{aligned}.$
ASA fixes the shape and size.

Full solution

Draw HJ with length 11 cm.
At H, draw the 34-degree ray from ray HJ.
At J, draw the 101-degree ray from ray JH, opening into the same half-plane.

The rays' angles toward that side total:

$34^\circ+101^\circ=135^\circ.$

Since the sum is below 180 degrees, the rays meet on that side.
Call the intersection K.
A ruler and protractor give an approximate drawing of these exact data.

The remaining angle is:

$180^\circ-135^\circ=45^\circ.$

Any exact triangle with these measurements has the same two endpoint angles and included side HJ.
ASA makes all such triangles congruent.

The obtuse endpoint angle puts the third vertex beyond J, while both rays open into the same half-plane.

Answer

Draw HJ and rays into the same half-plane, meeting at K:
$\begin{aligned}HJ&=11\text{ cm}\\m\angle H&=34^\circ\\m\angle J&=101^\circ\\m\angle K&=45^\circ\end{aligned}.$
ASA fixes the shape and size.

Key idea

A positive included side and two positive endpoint angles totaling less than 180 degrees fix the triangle shape and size.
Problem 7 Locate the missing hypothesis

Use this proposed correspondence:
$\begin{aligned}A&\leftrightarrow D\\B&\leftrightarrow E\\C&\leftrightarrow F\end{aligned}$
The recorded facts are:
$\begin{aligned}AB&=DE\\AC&=DF\\\angle B&\cong\angle E\end{aligned}$
Does this record directly supply SAS? Identify the angle pair that would complete SAS with the recorded side pairs, and explain the placement.

Text description of this figure

Triangles ABC and DEF have matching ticks on AB and DE, double ticks on AC and DF, and matching angle arcs at B and E. The angles at A and D are unmarked, as in the supplied record.
1. Hint 1
  Trace both known sides back to the vertex they share.
2. Hint 2
  A given angle at just one endpoint of a side need not be between both known sides.
Answer

No. The required included-angle pair is $\angle BAC\cong\angle EDF$ .

Full solution

AB and AC meet at A.
Their included angle is $\angle BAC$ .
The corresponding sides DE and DF meet at D, with included angle $\angle EDF$ .

The recorded angle pair is at B and E, not at A and D.
It does not directly supply the included angles for these two side pairs.

Adding:

$\angle BAC\cong\angle EDF$

Completes SAS.
This identifies a gap in the direct SAS argument; it does not claim that the triangles must be noncongruent.

Answer

No. The required included-angle pair is $\angle BAC\cong\angle EDF$ .

Key idea

A criterion needs the specified arrangement of matching parts, not just a count of measurements.
Problem 8 Finish an alignment argument

Triangles ABC and DEF have these givens:
$\begin{aligned}\angle A&\cong\angle D\\AB&=DE\\\angle B&\cong\angle E\end{aligned}$
After a translation and rotation align A with D and B with E, the image C prime is on the opposite side of line DE from F. Explain how to finish a rigid-motion match, and why the final third vertices coincide.

Text description of this figure

The aligned base is shared. C prime lies above it and F below it, with the supplied matching endpoint angle pairs shown by one and two arcs. No reflection arrow or final coincident vertex is shown.
1. Hint 1
  A motion fixing every point of the base line can change which side contains the third vertex.
2. Hint 2
  After the two triangles are on the same side, compare both rays from the base endpoints.
Answer

Reflect in line DE. The equal endpoint angles then align both ray pairs, and their unique intersection puts the image of C on F.

Full solution

Reflect the aligned first triangle across DE.
This fixes D and E and puts the image of C in the same half-plane as F.

The accepted preservation of angles under rigid motions means that its ray from D coincides with DF and its ray from E coincides with EF.

The two ray lines are distinct because DEF is a triangle.
Two distinct lines have at most one common point, so the final image of C is F.

Every vertex has now been matched by rigid motions, which establishes congruence.

Answer

Reflect in line DE. The equal endpoint angles then align both ray pairs, and their unique intersection puts the image of C on F.

Key idea

The same-half-plane choice lets equal endpoint angles identify the actual rays, not merely their measures.
Problem 9 Compare rigid frame pieces

Two ideal rigid frame pieces are triangles. Each has two sides of lengths 18 cm and 27 cm meeting at an angle of $38^\circ$ . The third side joins the free endpoints. Explain whether the two pieces must fit each other exactly after rigid motions, and whether their third sides must have the same length. Treat the stated measurements as exact.

Text description of this figure

Two ideal triangular frame pieces are shown with their 18-centimeter sides marked once and their 27-centimeter sides marked twice. Both given joint angles are 38 degrees. The sides joining the free endpoints have no supplied lengths or equality marks.
1. Hint 1
  The angle is at the joint of the two known sides.
2. Hint 2
  Once all three vertices coincide, compare the endpoints of the remaining sides.
Answer

The pieces are congruent by SAS, and their third sides have equal lengths.

Full solution

The 18 cm sides correspond, the 27 cm sides correspond, and their included angles are both 38 degrees.
These are SAS data, so a rigid-motion sequence matches the triangles.

Under that match, both free endpoints of one piece land on the free endpoints of the other.
Their joining third sides therefore coincide and have equal lengths.

This conclusion uses the exact ideal measurements.
A real manufactured piece can differ slightly because its measurements are not exact.

Answer

The pieces are congruent by SAS, and their third sides have equal lengths.

Key idea

Two lengths and their locked joint angle determine the whole ideal triangular piece.
Problem 10 Compare two pairs of records

Case A. Triangles ABC and DEF have these measurements:
$\begin{aligned}AB&=12\text{ cm}\\DE&=12\text{ cm}\\AC&=19\text{ cm}\\DF&=19\text{ cm}\\m\angle A&=46^\circ\\m\angle D&=47^\circ\end{aligned}$
Case A. Can they be congruent under the listed correspondence A to D, B to E, C to F? Explain.

Case B. Triangles PQR and UVW have these measurements:
$\begin{aligned}PQ&=17\text{ cm}\\UV&=17\text{ cm}\\m\angle P&=36^\circ\\m\angle Q&=92^\circ\\m\angle U&=92^\circ\\m\angle V&=36^\circ\end{aligned}$
Case B. Decide whether they are congruent and write the second triangle in the order corresponding to PQR.

Case B: the given base lengths and endpoint angles.

Text description of this figure

This figure shows Case B only. PQ and UV each have length 17 centimeters. P and V have the supplied 36-degree angles with one arc; Q and U have the supplied 92-degree angles with two arcs. The first triangle is solid blue and the second is red and dashed.
1. Hint 1
  Case A: a rigid match preserves every corresponding angle, including the one between the known sides.
2. Hint 2
  Case B: pair equal angles before deciding the vertex order. The base may be traversed in reverse.
Answer

A: no under the listed correspondence. B: yes, $\triangle PQR\cong\triangle VUW$ by ASA.

Full solution

Case A.
A rigid match sending A to D would preserve the angle at that vertex, but 46 degrees is not 47 degrees.
Thus this listed correspondence cannot give congruence.

Case B.
The 36-degree angles pair P with V, and the 92-degree angles pair Q with U.
The included sides PQ and VU both have length 17 cm.

Case B.
ASA establishes congruence.
The remaining vertices R and W correspond, so:

$\triangle PQR\cong\triangle VUW.$

Case B.
The endpoint angles have this total:

$\begin{aligned}36^\circ+92^\circ&=128^\circ.\end{aligned}$

Case B.
This is below 180 degrees, so both sets of angles can form a triangle.

Answer

A: no under the listed correspondence. B: yes, $\triangle PQR\cong\triangle VUW$ by ASA.

Key idea

Matching records must agree in both measurement and correspondence; reversing a base can reverse the matching order.

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