HSG.CO.B.7: Triangle Congruence and Corresponding Parts Through Rigid Motions
In plain English: HSG.CO.B.7 is the Common Core geometry standard that asks students to use the rigid-motion definition of congruence to show that two triangles are congruent exactly when all three pairs of corresponding sides and all three pairs of corresponding angles are congruent. It is usually taught in high school Geometry.
Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent.
Common Core State Standards for Mathematics · Domain: Congruence (CO) · Cluster: Understand congruence in terms of rigid motions Also written as HSG-CO.B.7 or G-CO.7 · Official standard
In earlier units, students learned that two figures are congruent when a sequence of rigid motions (translations, rotations and reflections) carries one onto the other. This lesson connects that definition to the classic description of congruent triangles: matching sides and matching angles. Students prove both directions. If a rigid motion maps one triangle onto another, the corresponding sides and angles must be congruent, because rigid motions preserve distance and angle measure. If all six pairs of corresponding parts are congruent, students build the rigid motion themselves: translate one vertex onto its partner, rotate so a side lines up, and reflect across that side when needed.
Students work on the coordinate plane so that every image can be computed and checked with the distance formula. They also learn why the order of letters in a congruence statement matters, and why matching angles alone does not make triangles congruent. The lesson sets up HSG.CO.B.8, where students show that fewer than six pairs are enough.
Learning Objectives
By the end of this lesson, students will be able to:
Explain why a rigid motion that maps one triangle onto another makes every pair of corresponding sides and angles congruent
Construct a sequence of translations, rotations and reflections that maps a triangle onto a second triangle whose six corresponding parts are congruent
Read a congruence statement such as △JKL ≅ △MNO to name corresponding sides and angles, and use them to find missing measures
Use coordinates and the distance formula to verify that a rigid motion preserved every side length
Give a counterexample showing that congruent angles alone do not make two triangles congruent
Prior Knowledge Required
Students should already be comfortable with:
Describing a sequence of rotations, reflections and translations that shows two figures are congruent 8.G.A.2
Drawing the image of a figure under a rotation, reflection or translation on graph paper HSG.CO.A.5
Using rigid motions to decide whether two figures are congruent HSG.CO.B.6
The distance formula, from the Pythagorean Theorem in the coordinate plane 8.G.B.8
Hand out a sheet with two copies of the same scalene triangle, one of them flipped and turned. Students trace the first triangle on patty paper and try to lay the tracing exactly on the second one.
Warm-Up Prompt
"What did you have to do to the tracing to make it fit? Name every move. Then look at the tracing: which measurements could have changed while you moved it, and which could not?"
Collect answers. Many students say "slide it, turn it, and flip it over". Write the formal names next to them: translation, rotation, reflection. Ask whether the tracing ever stretched. The class should conclude that side lengths and angle sizes stayed fixed, which is the key fact for the first half of the proof.
Direct Instruction20 minutes
The claim. Write the standard as a two-way statement: △ABC ≅ △DEF (some rigid motion maps A to D, B to E and C to F) if and only if AB = DE, BC = EF, AC = DF, ∠A ≅ ∠D, ∠B ≅ ∠E and ∠C ≅ ∠F. Explain that "if and only if" means two proofs, one in each direction (Diagram 2).
Direction 1 (congruent, so parts match). Suppose a rigid motion maps A to D, B to E and C to F. Rigid motions preserve distance, so the image of segment AB is segment DE and AB = DE; the same holds for the other two sides. Rigid motions preserve angle measure, so the image of ∠A is ∠D and m∠A = m∠D; the same holds for ∠B and ∠C. This is the reason behind "corresponding parts of congruent triangles are congruent".
Direction 2 (parts match, so congruent). Suppose all six pairs are congruent. Translate △ABC so A lands on D. Rotate about D so ray AB lies along ray DE; since AB = DE, B lands exactly on E. Because ∠A ≅ ∠D, ray AC now lies along ray DF or along its mirror image across line DE. If it lies along the mirror image, reflect across line DE, which keeps D and E fixed. Now ray AC lies along ray DF and AC = DF, so C lands on F. The composition of these motions is a rigid motion that maps △ABC onto △DEF.
Correspondence matters. The statement △ABC ≅ △DEF pairs the first letters, the second letters and the third letters. It says ∠B ≅ ∠E, not ∠B ≅ ∠F. Writing the letters in the wrong order makes a different, usually false, claim.
Direction 1: a rotation preserves the parts
△ABC has A(1, 1), B(5, 1), C(1, 4). Rotate it 90° counterclockwise about the origin, using (x, y) → (-y, x). Compare the parts.
Equation: A′(-1, 1), B′(-1, 5), C′(-4, 1); AB = A′B′ = 4, AC = A′C′ = 3, BC = B′C′ = 5, and ∠A and ∠A′ are both 90°
Finding a rigid motion that shows congruence
△PQR has P(0, 0), Q(4, 0), R(1, 3). △STU has S(6, 1), T(2, 1), U(5, 4). Find a sequence that maps P to S, Q to T and R to U.
Equation: Reflect across the y-axis: (0, 0), (-4, 0), (-1, 3). Then translate by (6, 1): (6, 1), (2, 1), (5, 4). So △PQR ≅ △STU, with PQ = ST = 4, PR = SU = √10 and QR = TU = √18
Using a congruence statement
△JKL ≅ △MNO. JK = 7 cm, m∠K = 64° and m∠L = 41°. Find MN, m∠N, m∠O and m∠M.
Equation: MN = JK = 7 cm, m∠N = m∠K = 64°, m∠O = m∠L = 41°, and m∠M = m∠J = 180° - 64° - 41° = 75°
Direction 2: building the motion (Diagram 1)
△ABC has A(0, 0), B(3, 0), C(0, 4) and △DEF has D(5, 2), E(5, 5), F(9, 2). All six pairs match: AB = DE = 3, AC = DF = 4, BC = EF = 5, ∠A = ∠D = 90° and the other two angle pairs are congruent. Build the rigid motion.
Equation: Translate by (5, 2): A′(5, 2), B′(8, 2), C′(5, 6). Rotate 90° counterclockwise about D: B″(5, 5) = E, C″(1, 2). C″ is the mirror image of F across line DE (x = 5), so reflect across DE: C‴(9, 2) = F
Why angles alone are not enough
Triangle 1 has sides 3, 4, 5 and triangle 2 has sides 6, 8, 10. All three pairs of corresponding angles are congruent. Are the triangles congruent?
Equation: No. A rigid motion would keep the side of length 3 at length 3, but triangle 2 has no side of length 3. The triangles are similar, not congruent
After the fourth example, point out that the reflection was needed because the vertices A, B, C run counterclockwise while D, E, F run clockwise. Translations and rotations never change that direction; a reflection always does. This gives students a quick test for whether their sequence will need a reflection.
Guided Practice15 minutes
Pairs work on graph paper with the triangle G(-3, 1), H(-1, 1), I(-3, 5).
Reflect △GHI across the x-axis, using (x, y) → (x, -y), then translate the image by (6, 0). Record the final vertices. (G′(3, -1), H′(5, -1), I′(3, -5).)
Use the distance formula to compare all three pairs of sides. (GH = G′H′ = 2, GI = G′I′ = 4, HI = H′I′ = √20.)
Measure ∠G and ∠G′ with a protractor, or explain from the grid why both are right angles. Then state which property of rigid motions guarantees that ∠H ≅ ∠H′ and ∠I ≅ ∠I′ without measuring.
Write the congruence statement and list all six pairs of corresponding parts.
Circulate and ask each pair to say which direction of the theorem they just used (Direction 1: the motion came first, so the parts must match). Ask one pair to explain to the class why the vertex order in the congruence statement is △GHI ≅ △G′H′I′ and not △GHI ≅ △H′G′I′.
Independent Practice15 minutes
Students work alone on three problems. (1) △ABC ≅ △XYZ with AB = 14, m∠B = 52° and m∠C = 33°. Find XY, m∠Y and m∠X. (XY = 14, m∠Y = 52°, m∠X = 95°.) (2) Rotate △VWX with V(2, -1), W(6, -1), X(6, 2) by 180° about the origin, using (x, y) → (-x, -y). Give the image vertices and verify that all three side lengths are unchanged. (V′(-2, 1), W′(-6, 1), X′(-6, -2); the sides are 4, 3 and 5 in both triangles.) (3) One triangle has sides 5, 6 and 7 and another has sides 5, 6 and 8. Use the definition of congruence to explain why no rigid motion can map one onto the other. (A rigid motion preserves every length, so the image of the first triangle would still have sides 5, 6 and 7, never 8.)
Closure5-10 minutes
Exit ticket: (1) In your own words, state both directions of the theorem. (2) △DOG ≅ △CAT. Name the angle that corresponds to ∠G and the side that corresponds to OG. (∠T and AT.) (3) In Direction 2, why is a reflection sometimes needed after the translation and the rotation?
Differentiation Strategies
For Struggling Students
Let students use patty paper for every motion before they compute coordinates, so each rule matches something they have done by hand
Give a correspondence table with three rows for sides and three rows for angles, and have students fill it from the congruence statement letter by letter
Color-code matching vertices (A and D red, B and E blue, C and F green) on both triangles
For Advanced Students
Ask students to rewrite Direction 2 so it never uses a coordinate grid, naming only the rigid motions and the properties that justify each step
Ask whether the proof of Direction 2 used all six congruences, and which ones it could do without (this previews HSG.CO.B.8)
Ask students to show that the three-step map in Example 4 sends every point (x, y) to (y + 5, x + 2), and to describe that map more simply as a reflection across y = x followed by a translation by (5, 2)
Assessment Guidance
What to Look For
Strong explanations name the property that justifies each step: "rigid motions preserve distance" for sides and "rigid motions preserve angle measure" for angles. In Direction 2, look for the reason each vertex lands where it should: B lands on E because AB = DE, and C lands on F because ∠A ≅ ∠D and AC = DF, after a possible reflection. Check that students write congruence statements with matching vertex order, and that they treat "if and only if" as two separate arguments.
02
Classroom Activities
3 Activities
1
Tracing Paper Match
20 minPairs
Each pair gets a handout with 4 pairs of congruent scalene triangles drawn in different positions. Two of the pairs are mirror images of each other. Students find rigid motions with patty paper, then turn each motion into a congruence statement.
Procedure
Trace the first triangle of a pair, label its vertices on the tracing, and move the tracing onto the second triangle
Write down each move in order (for example "translate, then rotate about the matching vertex, then flip over the matching side")
Use where the labels landed to write the congruence statement, then list all six pairs of corresponding parts
Measure one pair of sides and one pair of angles with a ruler and protractor to confirm
Discussion Questions
Which pairs needed a flip? What did you notice about the order of the vertices in those pairs?
How did the motion decide which vertex matched which? Could you have written the congruence statement before moving the tracing?
Modification for Distance Learning
Students use dynamic geometry software: they build the triangles from given coordinates, apply translate, rotate and reflect tools, and share a screenshot showing the image on top of the second triangle.
2
Converse Construction Relay
20 minGroups of 3
Groups carry out Direction 2 of the proof on graph paper. Each student owns one step of the construction, and the group checks that the final image lands on the second triangle.
The 2 Triangle Pairs
Pair 1: △KLM with K(1, 1), L(4, 1), M(1, 3) and △PQR with P(7, 2), Q(7, 5), R(5, 2). (Translate by (6, 1), then rotate 90° counterclockwise about P. No reflection is needed.)
Pair 2: △STU with S(2, 0), T(5, 0), U(2, 4) and △XYZ with X(2, 6), Y(5, 6), Z(2, 2). (Translate by (0, 6); T already lands on Y, so no rotation is needed. U lands on (2, 10), so reflect across line XY, y = 6.)
Procedure
Before starting, the group checks with the distance formula that the three pairs of sides are congruent, and with the grid that the included right angles match
Student 1 translates so the first vertex lands on its partner. Student 2 rotates about that point so the second vertex lands on its partner. Student 3 decides whether a reflection is needed and, if so, reflects across the matched side
Rotate roles for the second pair. The group writes one sentence per step explaining why the vertex landed where it did
Discussion Questions
In Pair 2, why could the rotation step be skipped?
How can you tell before you start whether a reflection will be needed?
3
Claim Cards: True or Counterexample
15 minPairs
Pairs receive 6 claim cards about triangles and congruence. For each card they either justify it with the rigid-motion definition or disprove it with a specific counterexample.
The 6 Cards
"If △ABC ≅ △DEF, then BC = EF." (True: rigid motions preserve distance.)
"If △ABC ≅ △DEF, then ∠B ≅ ∠F." (False in general: ∠B corresponds to ∠E.)
"Two triangles with angles 30°, 60° and 90° are congruent." (False: one can have a 4 cm hypotenuse and the other a 10 cm hypotenuse.)
"If a rigid motion maps △PQR onto △STU, the two triangles have equal perimeters." (True: every side keeps its length.)
"If all six pairs of corresponding parts are congruent, some rigid motion maps one triangle onto the other." (True: this is Direction 2.)
"Two triangles with equal perimeters are congruent." (False: sides 3, 4, 5 and sides 4, 4, 4 both give a perimeter of 12.)
Procedure
Partners sort the cards into "always true" and "can be false" piles and must agree on each placement
For a true card, write the property of rigid motions that proves it. For a false card, sketch and label a counterexample
Pairs compare with another pair and resolve any card they placed differently
Challenge Variation
Each pair writes a new false claim about congruent triangles that sounds convincing, trades it with another pair, and asks them for a counterexample.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Building the Rigid Motion from Six Congruent Parts
Direction 2 of the theorem for △ABC with A(0, 0), B(3, 0), C(0, 4) and △DEF with D(5, 2), E(5, 5), F(9, 2), drawn to scale. A translation sends A to D, a 90° rotation about D sends B to E, and a reflection across line DE sends C to F. The dashed triangle in each panel is the position before that step.
Diagram 2: The Two Directions of the Theorem
Direction 1 starts from a rigid motion and uses the fact that rigid motions preserve distance and angle measure. Direction 2 starts from six congruent pairs and builds the rigid motion one vertex at a time.
04
Homework Assignment
~30 min
HSG.CO.B.7 Homework: Congruent Triangles and Rigid Motions
Directions: Show your coordinates and distance calculations. Whenever you claim that two parts are congruent, name the property of rigid motions or the given fact that makes it true.
Part 1: From Rigid Motions to Corresponding Parts (Problems 1-3)
△KLM has K(-2, 1), L(2, 3), M(0, -2). (a) Reflect it across the line y = x, using (x, y) → (y, x), and give K′, L′ and M′. (b) Use the distance formula to show that all three pairs of corresponding sides are congruent. (c) Explain, without measuring, why ∠L ≅ ∠L′.
△RST ≅ △XYZ. RS = 11 cm, ST = 8 cm, m∠R = 38° and m∠T = 59°. (a) Find XY, YZ, m∠X, m∠Z and m∠Y. (b) Which side of △RST corresponds to XZ? (c) Explain how you knew which parts to match.
△ABC has A(-4, 1), B(-1, 1), C(-4, 3) and △DEF has D(4, 4), E(1, 4), F(4, 2). (a) Find a sequence of rigid motions that maps A to D, B to E and C to F, and check your sequence on all three vertices. (b) Verify that AB = DE, AC = DF and BC = EF.
Part 2: From Corresponding Parts to Rigid Motions (Problems 4-6)
Write a paragraph proof: if a sequence of rigid motions maps △PQR onto △P′Q′R′ with P → P′, Q → Q′ and R → R′, then all six pairs of corresponding parts are congruent. Name the property that justifies the sides and the property that justifies the angles.
△GHI has G(0, 0), H(5, 0), I(2, 3) and △JKL has J(2, 5), K(7, 5), L(4, 2). You are told that ∠G ≅ ∠J, ∠H ≅ ∠K and ∠I ≅ ∠L. (a) Verify with the distance formula that the three pairs of sides are congruent. (b) Follow Direction 2 of the proof: translate G to J, decide whether a rotation is needed, and decide whether a reflection is needed. Give the image of each vertex after every step.
Triangle 1 has sides 5, 5 and 6; triangle 2 has sides 5, 5 and 8. (a) Show that both triangles have an area of 12 square units (hint: the altitude to the third side splits each one into two right triangles). (b) The triangles share an area and two side lengths. Use the definition of congruence to explain whether they are congruent.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Rigid Motion Images
All image coordinates correct and checked on every vertex
One coordinate error, or the sequence not checked
Images missing or incorrect
Correspondence
Sides and angles matched in the order of the congruence statement
One pair matched incorrectly
Parts matched by position or size only
Justification
Each claim cites distance or angle preservation, or a given fact
Reasons given for some claims
No reasons given
Both Directions
Clearly separates "motion gives parts" from "parts give motion"
Uses one direction correctly
Directions confused or missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Pick an answer for each multiple-choice question to check it and read the reasoning. For the short-answer questions, write your answer first, then open the model answer. Reset quiz clears all answers.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which statement is the definition of congruence used in HSG.CO.B.7?
Answer: C
Congruence is defined by rigid motions: translations, rotations, reflections and their combinations. Choice A describes similar figures as well. Choice B fails because different triangles can share a perimeter and an area. Choice D is wrong because a dilation (other than scale factor 1) changes lengths.
Question 2 of 20 · Multiple Choice
Why does a rigid motion that maps △ABC onto △DEF force AB = DE?
Answer: D
The image of segment AB is segment DE, and rigid motions preserve distance, so the lengths are equal. Choice C assumes something about side lengths that is not given. Choice B relies on appearance, which is not a mathematical reason.
Question 3 of 20 · Multiple Choice
△ABC ≅ △PQR. Which statement must be true?
Answer: B
The statement pairs letters by position: A↔P, B↔Q, C↔R. So ∠B corresponds to ∠Q. Choice A pairs the second letter with the third. Choice C pairs AB with QR instead of PQ, a common error when students match sides by position on the page.
Question 4 of 20 · Multiple Choice
△DEF has D(2, -3), E(6, -3), F(2, 2). It is rotated 180° about the origin. What is the length of E′F′?
Answer: A
Rotating by 180° gives E′(-6, 3) and F′(-2, -2), so E′F′ = √(4² + 5²) = √41. That equals EF, as it must, because rotations preserve distance. Choice B adds the horizontal and vertical distances 4 + 5 instead of using the distance formula. Choice D forgets the square root.
Question 5 of 20 · Multiple Choice
△MNP ≅ △STU. m∠M = 48° and m∠N = 71°. What is m∠U?
Answer: C
∠U corresponds to ∠P, and m∠P = 180° - 48° - 71° = 61°. Choice A matches ∠U with ∠N instead of ∠P. Choice D adds the two given angles instead of subtracting their sum from 180°.
Question 6 of 20 · Multiple Choice
A rigid motion maps △ABC onto △DEF with A → D, B → E and C → F. AB = 9, BC = 13 and AC = 10. What is DF?
Answer: D
DF is the image of AC, and rigid motions preserve length, so DF = AC = 10. Choice A uses AB, which corresponds to DE. Choice B uses BC, which corresponds to EF.
Question 7 of 20 · Multiple Choice
In Direction 1 of the proof, which fact shows that ∠C ≅ ∠F?
Answer: A
The rigid motion maps ∠C onto ∠F and does not change angle measure. Choice B is a true fact but does not match one angle to another by itself. Choice D is false: a reflection reverses orientation.
Question 8 of 20 · Multiple Choice
Triangle 1 has sides 5, 12 and 13. Triangle 2 has sides 10, 24 and 26. Their corresponding angles are congruent. Are the triangles congruent?
Answer: B
Congruence needs a rigid motion, and rigid motions preserve length, so the sides would have to match too. These triangles are similar (scale factor 2), not congruent. Choice A uses only half of the six conditions. Choice D is simply false.
Question 9 of 20 · Multiple Choice
In Direction 2 of the proof, A has been moved to D and B to E. Why must C now lie on ray DF or on the mirror image of ray DF across line DE?
Answer: C
Only two rays from D make an angle equal to m∠D with ray DE, one on each side of line DE. The length AC = DF is used in the next step, to put C at the right point of the ray. Choice D is wrong because the rotation only lined up ray AB with ray DE.
Question 10 of 20 · Multiple Choice
△ABC has A(1, 0), B(3, 0), C(1, 5), and its image has A′(0, 1), B′(0, 3), C′(5, 1). Which single rigid motion maps △ABC onto △A′B′C′?
Answer: A
Reflecting across y = x swaps the coordinates: (1, 0) → (0, 1), (3, 0) → (0, 3), (1, 5) → (5, 1). The 90° rotation, (x, y) → (-y, x), sends C to (-5, 1), not (5, 1). The translation works for A but sends B to (2, 1).
Question 11 of 20 · Multiple Choice
A rigid motion maps △KLM onto a triangle with vertices X, Y and Z. KL = 6, LM = 8 and KM = 10; XY = 8, YZ = 6 and XZ = 10. Which congruence statement is correct?
Answer: D
Lengths are preserved, so KL (6) matches YZ, LM (8) matches XY and KM (10) matches XZ. L is on the sides of length 6 and 8, so it matches Y; K matches Z and M matches X. In choice A, KL would correspond to XY = 8, which breaks the length match.
Question 12 of 20 · Multiple Choice
If △ABC ≅ △DEF, how many pairs of congruent corresponding parts are guaranteed?
Answer: B
Three pairs of sides and three pairs of angles, six in all. Choice A counts only the sides or only the angles. Choice D pairs every part with every part, which is not what correspondence means.
Question 13 of 20 · Multiple Choice
△A′B′C′ is the image of △ABC under a rigid motion. Which statement could NOT be true?
Answer: C
A rigid motion never doubles a length. Choice A can happen: a rotation about A keeps A fixed. Choice B happens under any reflection. Choice D is always true.
Question 14 of 20 · Multiple Choice
△PQR has P(-1, 2), Q(3, 2), R(3, 5). It is translated by (4, -3). What is P′R′?
Answer: A
P′(3, -1) and R′(7, 2), so P′R′ = √(4² + 3²) = 5, the same as PR. Choice B adds 4 + 3. Choice C gives the length of P′Q′ instead. Choice D measures from P′ to the original point R, mixing the original and the image.
Question 15 of 20 · Short Answer
Explain Direction 1: if a rigid motion maps △ABC onto △DEF with A → D, B → E and C → F, why are all six pairs of corresponding parts congruent?
The motion maps each side onto its partner and each angle onto its partner, and rigid motions preserve both distance and angle measure. The image of segment AB is segment DE, so AB = DE; likewise BC = EF and AC = DF. The image of ∠A is ∠D, so m∠A = m∠D; likewise for ∠B, ∠E and ∠C, ∠F.
Question 16 of 20 · Short Answer
Explain Direction 2: if AB = DE, BC = EF, AC = DF, ∠A ≅ ∠D, ∠B ≅ ∠E and ∠C ≅ ∠F, describe the rigid motion that maps △ABC onto △DEF and why each vertex lands where it should.
Translate A to D, rotate about D so B lands on E, and reflect across line DE if C is on the wrong side. The translation puts A on D. The rotation puts ray AB along ray DE, and B lands on E because AB = DE. Because ∠A ≅ ∠D, ray AC lies along ray DF or its mirror image across line DE; a reflection across DE, which fixes D and E, fixes that if needed. Then C is on ray DF with AC = DF, so C lands on F.
Question 17 of 20 · Short Answer
△RST has R(1, 1), S(4, 1), T(4, 3). △R′S′T′ has R′(-1, -1), S′(-4, -1), T′(-4, -3). Name one rigid motion that maps △RST onto △R′S′T′, then verify one pair of sides and one pair of angles.
A rotation of 180° about the origin, (x, y) → (-x, -y), sends each vertex to its image. RS = R′S′ = 3 (and ST = S′T′ = 2, RT = R′T′ = √13). ∠S and ∠S′ are both right angles, since each is formed by a horizontal and a vertical segment.
Question 18 of 20 · Short Answer
△GHJ ≅ △VWX. GH = 3x + 1 and VW = 16. Find x and explain which fact you used.
x = 5. GH corresponds to VW, and corresponding sides of congruent triangles are congruent, because the rigid motion preserves length. So 3x + 1 = 16, 3x = 15 and x = 5.
Question 19 of 20 · Short Answer
△ABC ≅ △DEF, m∠A = (2x + 10)° and m∠D = (4x - 20)°. Find x and m∠A.
x = 15 and m∠A = 40°. ∠A corresponds to ∠D, so 2x + 10 = 4x - 20, which gives 30 = 2x and x = 15. Then m∠A = 2(15) + 10 = 40°, and m∠D = 4(15) - 20 = 40° as a check.
Question 20 of 20 · Short Answer
△PQR has PQ = 5, QR = 7 and PR = 9. △STU has ST = 5, TU = 7 and SU = 8.5. Can any rigid motion map △PQR onto △STU with P → S, Q → T and R → U? Explain.
No. Such a motion would map PR onto SU and preserve its length, so SU would have to be 9. Since SU = 8.5, one pair of corresponding sides is not congruent, and by Direction 1 the triangles cannot be congruent in this correspondence (nor in any other, since 8.5 matches no side of △PQR).
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.CO.B.7 mean?
HSG.CO.B.7 means students prove that two triangles are congruent exactly when their three pairs of corresponding sides and three pairs of corresponding angles are congruent. The proof must start from the rigid-motion definition of congruence, not from a memorized rule. Students show both directions: a rigid motion forces the parts to match, and matching parts let you build a rigid motion.
Is HSG.CO.B.7 taught in Geometry or Algebra 1?
It is taught in high school Geometry, usually in the congruence unit right after rigid motions (HSG.CO.A.5 and HSG.CO.B.6). It builds on grade 8 work (8.G.A.2), where students described sequences of transformations that show two figures are congruent.
What does "if and only if" mean in this standard?
It means the statement goes both ways, so it needs two proofs. One shows that congruent triangles have congruent corresponding parts. The other shows that triangles with six congruent corresponding parts are congruent, by constructing the rigid motion.
Is HSG.CO.B.7 the same as CPCTC?
It includes CPCTC and goes further. "Corresponding parts of congruent triangles are congruent" is Direction 1 of the standard, and HSG.CO.B.7 asks students to justify it with rigid motions. Direction 2, the converse, is the part many textbooks skip.
Why does the order of letters in a congruence statement matter?
The order tells which vertex maps to which. △ABC ≅ △DEF says A goes to D, B to E and C to F, so it claims ∠B ≅ ∠E and AB = DE. Rewriting it as △ABC ≅ △EDF makes different claims, which may be false.
Do you always need a reflection to map one triangle onto another?
No, only when the triangles have opposite orientation. If the vertices of both triangles, read in corresponding order, run the same way (both clockwise or both counterclockwise), a translation and a rotation are enough. If they run opposite ways, the sequence needs one reflection.
Can two triangles with the same angles fail to be congruent?
Yes. An 8-15-17 triangle and a 16-30-34 triangle have the same angles, but a rigid motion never changes a length, so neither can be mapped onto the other. They are similar. Matching angles is only half of the six conditions.
How is HSG.CO.B.7 different from HSG.CO.B.8?
HSG.CO.B.7 uses all six pairs of parts. HSG.CO.B.8 shows that three well-chosen pairs are enough, which gives the ASA, SAS and SSS criteria. The construction in Direction 2 of this lesson is the same one students reuse to prove those criteria.
What mistakes do students make with this standard?
A common error is matching sides and angles by where they sit on the page instead of by the congruence statement. Other frequent slips are claiming two parts are congruent because they "look the same", forgetting the reflection step when orientations differ, and treating the two directions of the theorem as one argument.
How can students check their work on the coordinate plane?
They apply the transformation rules to every vertex and confirm that the final image equals the second triangle point for point. Then they compute each side with the distance formula before and after. If any length changed, a rule was applied incorrectly, since rigid motions never change length.
07
Related Standards
5 standards
These standards connect to HSG.CO.B.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.A.2Prerequisite
Congruence as a sequence of rotations, reflections and translations
