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Triangle Congruence and SSS: 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 Translate an ordered match

Given the ordered statement:
$\begin{aligned}\triangle ABC&\cong\triangle WUV.\end{aligned}$
List the three corresponding side pairs and the three corresponding angle pairs. Use the ordered statement rather than the triangles’ positions in the figure.

Use the supplied order across the changed orientation.

Text description of this figure

Congruent triangles ABC and WUV have different orientations. No matching side or angle pair is highlighted.
1. Hint 1
  Match vertices by their position in the two names.
2. Hint 2
  Use the two endpoint matches for a side, and the vertex match for an angle.
Answer

$\begin{aligned}AB&\leftrightarrow WU.\\BC&\leftrightarrow UV.\\AC&\leftrightarrow WV.\\\angle A&\leftrightarrow\angle W.\\\angle B&\leftrightarrow\angle U.\\\angle C&\leftrightarrow\angle V.\end{aligned}$

Full solution

The ordered vertex match is

$\begin{aligned}A&\leftrightarrow W.\\B&\leftrightarrow U.\\C&\leftrightarrow V.\end{aligned}$

Applying it to the endpoints gives

$\begin{aligned}AB&\leftrightarrow WU.\\BC&\leftrightarrow UV.\\AC&\leftrightarrow WV.\end{aligned}$

Applying it to each angle vertex gives

$\begin{aligned}\angle A&\leftrightarrow\angle W.\\\angle B&\leftrightarrow\angle U.\\\angle C&\leftrightarrow\angle V.\end{aligned}$

The solution establishes the side pairings before these matching ticks are added.

Answer

$\begin{aligned}AB&\leftrightarrow WU.\\BC&\leftrightarrow UV.\\AC&\leftrightarrow WV.\\\angle A&\leftrightarrow\angle W.\\\angle B&\leftrightarrow\angle U.\\\angle C&\leftrightarrow\angle V.\end{aligned}$

Key idea

One vertex correspondence determines all six corresponding parts.
Problem 2 Name the compass settings

You will construct $\triangle HJK$ with these given lengths:
$\begin{aligned}HJ&=12\text{ cm}.\\HK&=9\text{ cm}.\\JK&=16\text{ cm}.\end{aligned}$
After drawing $HJ$ , state the center and radius of each arc needed to locate $K$ , then explain why a common point has the required distances.

The base is drawn; the given distances to K still need their arcs.

Text description of this figure

Base HJ is 12 centimeters. The required distances HK 9 and JK 16 centimeters are listed separately. K, the arcs, and their centers and radii are not constructed for the student.
1. Hint 1
  Each given distance involving K describes a circle.
2. Hint 2
  Each given distance to K is measured from a point already on the page. Which point supplies each distance?
Answer

Arc centered at $H$ : radius 9 cm. Arc centered at $J$ : radius 16 cm. At a common point $K$ :
$HK=9\text{ cm}.$ $JK=16\text{ cm}.$

Full solution

Draw the first arc with center $H$ and radius $9$ cm.
Every point on it is $9$ cm from $H$ .

Draw the second with center $J$ and radius $16$ cm.
Every point on it is $16$ cm from $J$ .

At a common point both conditions hold:

$\begin{aligned}HK&=9\text{ cm}.\\JK&=16\text{ cm}.\end{aligned}$

The already drawn base is:

$\begin{aligned}HJ&=12\text{ cm}.\end{aligned}$

The three required side lengths have now been supplied.

After choosing the compass settings, their intersection has both required endpoint distances.

Answer

Arc centered at $H$ : radius 9 cm. Arc centered at $J$ : radius 16 cm. At a common point $K$ :
$HK=9\text{ cm}.$ $JK=16\text{ cm}.$

Key idea

An arc intersection satisfies two distance conditions at once.
Problem 3 Recover the correct vertex order

Two scalene triangles have these side lengths in centimeters:
$\begin{aligned}DE&=6.\\EF&=11.\\FD&=14.\end{aligned}$ $\begin{aligned}LM&=14.\\MN&=6.\\NL&=11.\end{aligned}$
Write an ordered congruence statement beginning with $\triangle DEF$ , and justify its vertex order.

Use the two given lengths meeting at each vertex; no ordered answer is printed.

Text description of this figure

The first triangle has DE 6, EF 11, and FD 14 centimeters. The second has LM 14, MN 6, and NL 11 centimeters. Only given length labels and vertex names are shown.
1. Hint 1
  At each vertex, identify the two side lengths that meet there.
2. Hint 2
  Find the vertex of LMN where the 6 and 14 cm sides meet.
Answer

The triangles are congruent by SSS:
$\begin{aligned}\triangle DEF&\cong\triangle MNL.\end{aligned}$

Full solution

At $D$ , the $6$ and $14$ cm sides meet.
They meet at $M$ in the second triangle, so $D$ matches $M$ .

At $E$ , the $6$ and $11$ cm sides meet.
They meet at $N$ , so $E$ matches $N$ and $F$ matches $L$ .

The side pairs in that order are

$\begin{aligned}DE&=MN.\\DE&=6.\\EF&=NL.\\EF&=11.\\FD&=LM.\\FD&=14.\end{aligned}$

All three side pairs match.
SSS establishes the ordered congruence:

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

Answer

The triangles are congruent by SSS:
$\begin{aligned}\triangle DEF&\cong\triangle MNL.\end{aligned}$

Key idea

When side lengths are distinct, the two lengths meeting at a vertex identify its partner.
Problem 4 Find a matching length and perimeter

The given ordered match is:
$\begin{aligned}\triangle RST&\cong\triangle UVW.\end{aligned}$
The lengths in centimeters are:
$\begin{aligned}RS&=4a-1.\\UV&=27.\\VW&=31.\\UW&=35.\end{aligned}$
Find $a$ and the perimeter of $\triangle RST$ .

The expressions and lengths are givens; the computation belongs in the solution.

Text description of this figure

Congruent triangles RST and UVW have labels RS 4a minus 1, UV 27, VW 31, and UW 35 centimeters. The value of a and requested perimeter are not shown.
1. Hint 1
  Read which length corresponds to RS before forming an equation.
2. Hint 2
  The perimeter uses the three lengths of one triangle, each equal to its partner.
Answer

$\begin{aligned}a&=7.\\\text{Perimeter: }&93\text{ cm}.\end{aligned}$

Full solution

The corresponding lengths $RS$ and $UV$ are equal:

$\begin{aligned}4a-1&=27.\end{aligned}$

Add 1 and divide by 4 to obtain

$\begin{aligned}a&=7.\end{aligned}$

The other side lengths are

$\begin{aligned}ST&=31\text{ cm}.\\RT&=35\text{ cm}.\end{aligned}$

Add all three lengths:

$\begin{aligned}27+31+35&=93.\end{aligned}$

The perimeter is $93$ cm.

Answer

$\begin{aligned}a&=7.\\\text{Perimeter: }&93\text{ cm}.\end{aligned}$

Key idea

Ordered congruence supplies the matching lengths before the arithmetic begins.
Problem 5 Separate six-part information from sufficient information

Two actual triangles have three corresponding side pairs equal in one vertex order. Must you also measure their angle pairs to establish congruence? Explain what SSS establishes and what rigid-motion preservation then tells you about the angles.
1. Hint 1
  Distinguish the information sufficient to establish a match from the consequences of that match.
2. Hint 2
  Recall what a sequence of rigid motions preserves.
Answer

No; three corresponding side matches establish congruence by SSS, and rigid motions preserve the three corresponding angle measures.

Full solution

SSS says that the three side pairs alone establish congruence.
There is no need for a separate angle measurement to complete that criterion.

Congruence means a sequence of rigid motions maps the entire first triangle onto the second.
Those motions preserve angles, so all three corresponding angle measures agree.

The complete six-part description is consistent with the shorter sufficient test; it is not a demand for six independent measurements.

Answer

No; three corresponding side matches establish congruence by SSS, and rigid motions preserve the three corresponding angle measures.

Key idea

A sufficient test can establish more matching parts than the ones it initially checks.
Problem 6 Apply SSS to two sets of measurements

Case A. One triangle has sides $15$ , $18$ , and $23$ cm; another has sides $23$ , $15$ , and $18$ cm. Decide whether the given data establish congruence.

Case B. One triangle has sides $15$ , $18$ , and $23$ cm; another has sides $15$ , $18$ , and $24$ cm. Decide whether they are congruent.

State a reason in each case.
1. Hint 1
  A different order in a side-length list does not change the triangle.
2. Hint 2
  Match equal lengths in Case A. In Case B, ask whether any rigid motion could preserve all distances.
Answer

Case A: congruent by SSS. Case B: not congruent, because the three side-length lists differ.

Full solution

Case A.
Each of the three lengths has an equal partner in the other triangle.
The pairs

$\begin{aligned}15&=15.\\18&=18.\\23&=23.\end{aligned}$

These three matches establish congruence by SSS.

Case B.
A side of length $23$ cm has no equally long side in the other triangle.
Since a rigid motion preserves every side length, it cannot map one triangle onto the other.

Answer

Case A: congruent by SSS. Case B: not congruent, because the three side-length lists differ.

Key idea

Compare all three lengths, independently of how a list or drawing orders them.
Problem 7 Compare a triangle and a four-strip frame

A square frame has four strips of length $10$ cm, joined by hinges. The same strips are moved into a slanted four-sided frame with a $60^\circ$ interior angle. Decide whether these two frames are congruent. Then explain whether three fixed strip lengths could produce two noncongruent triangles.

The same strip lengths are given; the angle data differ.

Text description of this figure

Both frames use four strips of length 10 centimeters. The blue solid frame is square. The red dashed slanted frame has a given 60-degree interior angle. The diagram makes no claim about triangle rigidity.
1. Hint 1
  An exact rigid match preserves angles as well as lengths.
2. Hint 2
  The statement about triangles has a criterion that does not apply to four-sided frames.
Answer

The frames are not congruent; one has a $60^\circ$ angle while all square angles are $90^\circ$ . Three fixed lengths cannot give two noncongruent triangles.

Full solution

Every angle of the square is $90^\circ$ .
The slanted frame has an angle of $60^\circ$ , so a rigid motion cannot match their angles.

The four side lengths stayed equal, but that was insufficient to preserve the whole shape.

For two actual triangles, three equal corresponding side lengths establish congruence by SSS.
The freedom of the four-sided frame therefore does not occur for a triangle.

Answer

The frames are not congruent; one has a $60^\circ$ angle while all square angles are $90^\circ$ . Three fixed lengths cannot give two noncongruent triangles.

Key idea

SSS belongs to triangles; matching a larger figure’s side lengths need not fix its angles.
Problem 8 Interpret three circle outcomes

Each case uses a base with distinct endpoints and two complete circles centered at those endpoints. In each case the base length and two positive radii are fixed; different cases may use different lengths.

Case A. The circles meet at two points off the base line. How many different triangle shapes, up to congruence, do the two choices produce?

Case B. The circles touch at one point on the base line. Does choosing that point produce a triangle?

Case C. The circles have no common point. Can a point on either circle alone supply a third vertex satisfying both prescribed distances?

Each panel shows its stated circle outcome; decide what it means for a third vertex.

Text description of this figure

Case A shows two off-line circle intersections. Case B shows circles touching on their common base line. Case C shows two circles with no common point. No triangle edges, shape counts, or existence verdicts are added.
1. Hint 1
  A third vertex must satisfy both distance requirements at once.
2. Hint 2
  For two off-line intersections compare their triangles by SSS; for the other cases check noncollinearity or simultaneous distances.
Answer

Case A: one shape up to congruence. Case B: no triangle. Case C: no third vertex meets both distances.

Full solution

Case A.
The circles are centered at the base endpoints, so each intersection has the same two prescribed distances from those endpoints.
The base is common.
Thus all three side pairs match, and SSS gives one shape up to congruence.

Case B.
The third point lies on the line through the two base endpoints.
A triangle requires noncollinear vertices, so this is not a triangle.

Case C.
Because the complete circles have no common point, no point has both prescribed distances.
A point on only one circle satisfies only that circle's distance requirement.

In Case A, the two triangles share the base and both prescribed endpoint distances.

Answer

Case A: one shape up to congruence. Case B: no triangle. Case C: no third vertex meets both distances.

Key idea

Existence, noncollinearity, and congruence are separate questions in a construction.
Problem 9 Use a shared diagonal

A four-sided figure has vertices $P,Q,R,S$ in that order. The given pairs and lengths are
$\begin{aligned}PQ&=RS.\\PQ&=12\text{ cm}.\\QR&=SP.\\QR&=20\text{ cm}.\end{aligned}$
Draw diagonal $PR$ . Each resulting triangle has three noncollinear vertices. Establish an ordered congruence statement for the two resulting triangles, listing all three matching side pairs.

Only the given lengths and shared diagonal are shown.

Text description of this figure

Four-sided figure PQRS is divided by diagonal PR. PQ and RS are labeled 12 centimeters; QR and SP are labeled 20 centimeters. The diagonal is common to the two triangles.
1. Hint 1
  The diagonal belongs to both triangles.
2. Hint 2
  Match the two given pairs and then write the common segment as the third pair.
Answer

$\begin{aligned}\triangle PQR&\cong\triangle RSP.\\PQ&=RS.\\QR&=SP.\\PR&=RP.\end{aligned}$
SSS establishes this match.

Full solution

The two given pairs are

$\begin{aligned}PQ&=RS.\\PQ&=12\text{ cm}.\\QR&=SP.\\QR&=20\text{ cm}.\end{aligned}$

The diagonal has the same length in either direction:

$\begin{aligned}PR&=RP.\end{aligned}$

These side pairs match $P$ with $R$ , $Q$ with $S$ , and $R$ with $P$ .
Thus, by SSS,

$\begin{aligned}\triangle PQR&\cong\triangle RSP.\end{aligned}$

After listing the given pairs, the shared diagonal supplies the third side pair.

Answer

$\begin{aligned}\triangle PQR&\cong\triangle RSP.\\PQ&=RS.\\QR&=SP.\\PR&=RP.\end{aligned}$
SSS establishes this match.

Key idea

A shared segment can supply the third side pair without a new measurement.
Problem 10 Construct and account for both placements

Using a ruler, compass, and straightedge on paper, construct $\triangle UVW$ with these lengths:
$\begin{aligned}UV&=11\text{ cm}.\\UW&=8\text{ cm}.\\VW&=10\text{ cm}.\end{aligned}$
Keep the relevant arcs.

Construct the possible third vertex on each side of the same base. Describe a rigid motion mapping one resulting triangle onto the other while keeping $U$ and $V$ fixed. Explain why there is no third, noncongruent triangle with these three side lengths.

Start with the fixed base; construct and retain both sets of arcs yourself.

Text description of this figure

Base UV is drawn with length 11 centimeters. Required distances UW 8 and VW 10 centimeters are listed separately. No arcs or third-vertex positions are supplied.
1. Hint 1
  Draw the base first, then use one distance arc from each endpoint.
2. Hint 2
  The SSS justification leaves at most one possible third vertex on either side of a fixed base.
3. Hint 3
  Which rigid motion can fix every point of a line while exchanging its two sides?
Answer

Base UV of 11 cm; arcs of radius 8 cm centered at U and radius 10 cm centered at V. The two intersections give reflected, congruent triangles. There is no third noncongruent shape.

Full solution

Draw the base:

$\begin{aligned}UV&=11\text{ cm}.\end{aligned}$

With center $U$ , draw an arc of radius $8$ cm on each side of the base.
With center $V$ , draw arcs of radius $10$ cm that cross them.

Join each crossing to $U$ and $V$ .
The circle definition guarantees

$\begin{aligned}UW&=8\text{ cm}.\\VW&=10\text{ cm}.\end{aligned}$

These lengths hold for either crossing, and the base is unchanged.

The SSS justification fixes a unique horizontal position and a unique height in size after a base is placed.
The two signs of that height are the two reflected positions.

Reflection across the line $UV$ fixes $U$ and $V$ and exchanges the two third vertices, so it maps one whole triangle onto the other.

Every actual triangle with those side lengths can have its $11$ cm side aligned to $UV$ with the vertex shared by its $11$ cm and $8$ cm sides at $U$ .
The other endpoint then matches $V$ , where the $10$ cm side joins the base.
Its third vertex must be one of the two positions satisfying those assigned distances, so no additional noncongruent shape is possible.

The two 11, 8, and 10 cm placements differ by reflection across the fixed base.

Answer

Base UV of 11 cm; arcs of radius 8 cm centered at U and radius 10 cm centered at V. The two intersections give reflected, congruent triangles. There is no third noncongruent shape.

Key idea

Three side lengths determine a triangle up to rigid motion, even when its construction has two placements.

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