Congruent Triangles Worksheet Answer Key – Geometry

Problem 1 Corresponding parts

The letters correspond in order: P ↔ X, Q ↔ Y, R ↔ Z. So XY = PQ = 9.5, YZ = QR = 12, m∠X = 51° and m∠Y = 77°.

m∠R = 180° − 51° − 77° = 52°, so m∠Z = 52°.

XY = 9.5, YZ = 12, ∠X = 51°, ∠Y = 77°, ∠Z = 52°

Problem 2 Name the criterion

1. SSS.
2. SAS: the angle is included.
3. ASA: the side is included.
4. AAS: the side is not between the angles.
5. HL.

(a) SSS, (b) SAS, (c) ASA, (d) AAS, (e) HL

Problem 3 Why SSA is not enough

No. The angle is not included, so this is SSA. The side of length 6 can reach the base ray in two places, so two different triangles can have these measurements. SSA is not a congruence criterion.

No: SSA does not guarantee congruence

Problem 4 Solve for x

Corresponding sides are equal: 4x + 1 = 7x − 11, so 12 = 3x and x = 4.

AB = 4(4) + 1 = 17, and DE = 7(4) − 11 = 17.

x = 4; AB = 17

Problem 5 Shared side

Both triangles contain side BD, and BD ≅ BD by the Reflexive Property. Three pairs of sides match.

BD ≅ BD; △ABD ≅ △CBD by SSS

Problem 6 Read the marks

The marked parts are AB ≅ DE, ∠A ≅ ∠D and AC ≅ DF. The angle at A lies between sides AB and AC, so it is included.

SAS

Problem 7 Angle sum in congruent triangles

∠A ≅ ∠D: 4x + 3 = 6x − 17, so 20 = 2x and x = 10. Then m∠A = 43°.

∠C = 180° − 43° − 62° = 75°, and ∠F ≅ ∠C.

x = 10; m∠F = 75°

Problem 8 Perimeter of congruent triangles

JK ≅ MN: 2x + 1 = 13, so x = 6. Check with KL ≅ NP: 3(6) − 2 = 16 ✓.

JL = 4(6) − 5 = 19. Perimeter = 13 + 16 + 19 = 48.

x = 6; JL = 19; perimeter 48

Problem 9 ASA or AAS

1. AB lies between ∠A and ∠B: ASA.
2. BC is not between ∠A and ∠B: AAS.
3. Three pairs of angles (AAA) fix only the shape, not the size: no criterion applies.

(a) ASA, (b) AAS, (c) not enough

Problem 10 Hypotenuse and leg

The hypotenuses are congruent (AC = DF = 13) and one pair of legs is congruent (AB = DE = 5), so △ABC ≅ △DEF by HL.

By the Pythagorean Theorem, BC = √(13² − 5²) = √144 = 12, and EF = 12 as a corresponding side.

HL; BC = EF = 12

Problem 11 Congruence from coordinates

Find the side lengths. △ABC: AB = 4 (horizontal), BC = 3 (vertical), AC = √(4² + 3²) = 5.

△DEF: DE = |−2 − (−6)| = 4, EF = |−3 − (−6)| = 3, DF = √(3² + 4²) = 5.

Matching sides 4, 3, 5 gives △ABC ≅ △DEF by SSS.

Congruent by SSS (sides 4, 3, 5)

Problem 12 The ambiguous SSA case

Since 6.4 < 7 < 10, the circle of radius 7 about B meets the ray from A in two points. The foot of the height is 10 cos 40° ≈ 7.66 from A, and the offset along the ray is √(7² − 6.43²) ≈ 2.77.

So AC ≈ 7.66 − 2.77 ≈ 4.89 or 7.66 + 2.77 ≈ 10.43.

Two triangles: AC ≈ 4.9 or AC ≈ 10.4

Problem 13 Prove angles congruent with SSS

∠BAD ≅ ∠BCD

Problem 14 Crossing midpoints

△AOB ≅ △COD by SAS

Problem 15 Roof truss braces

PQ = PS = 18, QR = SR = 14.5, and PR is shared. Three pairs of sides match, so △PQR ≅ △PSR by SSS.

∠PQR corresponds to ∠PSR, so by CPCTC m∠PQR = 112°.

Congruent by SSS; m∠PQR = 112°

Problem 16 Matching ramps

Both are right triangles with equal hypotenuses (10 ft) and a pair of equal legs (6 ft), so they are congruent by HL.

Height = √(10² − 6²) = √64 = 8 feet. Triangular side area = ½ · 6 · 8 = 24 square feet.

Congruent by HL; height 8 ft; area 24 ft²

Problem 17 Bisector and perpendicular

AD ≅ CD

Problem 18 Prove sides congruent with AAS

AC ≅ DF

Problem 19 Congruent triangles with algebra

Match corresponding sides: AB = XY gives 3x − 1 = 17, so x = 6. Check: BC = 2(6) + 5 = 17 ✓ and CA = 6 + 12 = 18 ✓.

Perimeter = 17 + 17 + 18 = 52. Two sides are equal, so the triangle is isosceles.

x = 6; perimeter 52; isosceles

Problem 20 Overlapping triangles

BE ≅ CD