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7.14 Congruent Triangles

EDEXCEL 1MA1 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State one similarity and one difference between a pair of congruent shapes and a pair of similar shapes.

(Total for Question 1 is 2 marks)

Write down the four standard conditions that can be used to prove that two triangles are congruent, stating briefly what each one means.

(Total for Question 2 is 4 marks)

For each part below, decide whether the given information proves the two triangles are congruent, and if so, state the condition used.

(a)Triangle ABC and triangle PQR are drawn such that AB = PQ = 6 cm, BC = QR = 8 cm, and angle ABC = angle PQR = 90 degrees (the angle between the two given sides in each triangle). State the condition that proves triangle ABC is congruent to triangle PQR.(1)

(b)Triangle DEF and triangle GHI are drawn such that angle D = angle G, angle E = angle H and angle F = angle I (all three pairs of angles are equal), but no side lengths are marked as equal. State whether triangle DEF must be congruent to triangle GHI, giving a reason.(1)

(c)Triangle JKL and triangle STU are right-angled at K and T respectively. The hypotenuses JL = SU = 10 cm, and one further pair of sides JK = ST = 6 cm. State the condition that proves triangle JKL is congruent to triangle STU.(1)

(d)Two triangles have two pairs of corresponding sides equal in length and one pair of corresponding angles equal, but the equal angle is not the angle included between the two given sides (the 'SSA' case). Which of the following statements is correct?(1)

A) The triangles must always be congruent
B) The triangles can never be congruent
C) The triangles may or may not be congruent - SSA is not a valid congruence condition
D) The triangles are congruent only if the equal angle is obtuse

(Total for Question 3 is 4 marks)

Triangle JKL has JK = 5.2 cm, KL = 7.8 cm and angle JKL = 62 degrees. Triangle MNO has MN = 7.8 cm, NO = 5.2 cm and angle MNO = 62 degrees.

(a)State whether triangle JKL is congruent to triangle MNO.(1)

(b)Give a full geometric justification for your answer, stating the congruence condition used and the correct correspondence of vertices.(2)

(Total for Question 4 is 3 marks)

Triangle ABC is isosceles, with AB = AC. M is the midpoint of BC, and the line AM is drawn. Prove that triangle ABM is congruent to triangle ACM, and hence show that angle ABC = angle ACB.

(Total for Question 5 is 4 marks)

ABCD is a kite, with AB = AD and CB = CD. The diagonal AC is drawn. Prove that triangle ABC is congruent to triangle ADC, and hence show that AC bisects both angle BAD and angle BCD.

(Total for Question 6 is 4 marks)

Triangle ABC is congruent to triangle DEF, with vertices corresponding as A to D, B to E and C to F. AB = (2x + 5) cm and DE = (4x - 7) cm are corresponding sides. Form and solve an equation to find x, and hence find the length of AB.

(Total for Question 7 is 4 marks)

Triangle ABC has vertices A(1, 1), B(5, 1) and C(5, 4). Triangle DEF has vertices D(-2, -2), E(-2, -6) and F(-5, -6), all coordinates in cm.

Diagram NOT accurately drawn

(a)Calculate the lengths of all three sides of each triangle.(3)

(b)State the condition that proves triangle ABC is congruent to triangle DEF, and write down the correspondence of vertices.(2)

(Total for Question 8 is 5 marks)

ABCD is a parallelogram, so AB is parallel to DC and AD is parallel to BC. The diagonal AC is drawn.

(a)Prove that triangle ABC is congruent to triangle CDA.(4)

(b)Hence state two facts about the sides or angles of parallelogram ABCD that follow from this congruence.(2)

(Total for Question 9 is 6 marks)

Two ladders are placed against the same vertical wall, standing on horizontal ground. Ladder 1 has its foot at point P and its top at point Q on the wall. Ladder 2 has its foot at point R and its top at point S on the wall (on the same side as Ladder 1). Let W be the point where the wall meets the ground. The two ladders are the same length (PQ = RS) and reach the same height up the wall (QW = SW).

Diagram NOT accurately drawn

(a)Prove that triangle PWQ is congruent to triangle RWS.(3)

(b)The ladders are each 4.5 m long, and each reaches a height of 4.2 m up the wall. Calculate the distance of each ladder's foot from the wall, giving your answer to 1 decimal place.(2)

(Total for Question 10 is 5 marks)

Triangle ABC is isosceles, with AB = AC. D is a point on AB and E is a point on AC such that AD = AE. The lines DC and BE are drawn. Prove that triangle ADC is congruent to triangle AEB, and hence show that DC = EB.

(Total for Question 11 is 4 marks)

A hiker wants to find the width of a straight river, from a fixed point A on the near bank to a post B on the far bank, without crossing the river. AB is perpendicular to the near bank. She uses the following method: she walks along the near bank from A to a point C, then continues the same distance again in the same direction to a point D, so that C is the midpoint of AD. At D she turns through 90 degrees away from the river and walks until she reaches a point E such that B, C and E lie exactly on a straight line. She then measures the distance DE.

Diagram NOT accurately drawn

(a)Prove that triangle ABC is congruent to triangle DEC.(3)

(b)Given that DE = 23.6 m, state the width of the river, AB.(1)

(Total for Question 12 is 4 marks)

L is a point that is the same distance from two rescue boats, P and Q (that is, LP = LQ). M is the midpoint of PQ, and LM is drawn. Prove that triangle LPM is congruent to triangle LQM, and hence prove that LM is perpendicular to PQ.

(Total for Question 13 is 5 marks)

O is the centre of a circle. AB and CD are chords of the circle with AB = CD. M and N are the feet of the perpendiculars from O to AB and CD respectively (OM perpendicular to AB, ON perpendicular to CD). You may assume that a perpendicular from the centre of a circle to a chord bisects that chord. Prove that OM = ON.

(Total for Question 14 is 5 marks)

Triangle ABC is congruent to triangle DEF (correspondence A to D, B to E, C to F), proved using the condition SAS: AB corresponds to DE, angle A corresponds to angle D (the included angle), and AC corresponds to DF. AB = (5x - 2) cm, DE = (2x + 7) cm, angle A = (3y + 10) degrees and angle D = (y + 50) degrees. Find the values of x and y, and state the resulting length AB and the size of angle A.

(Total for Question 15 is 5 marks)

ABCD is a trapezium with AB parallel to DC, and AD = BC (an isosceles trapezium). E and F are the points on DC directly below A and B respectively, so that AE is perpendicular to DC and BF is perpendicular to DC. Prove that triangle ADE is congruent to triangle BCF, and hence show that angle ADC = angle BCD.

(Total for Question 16 is 6 marks)

In triangle ABC, D is a point on BC. E is the foot of the perpendicular from D to AB, and F is the foot of the perpendicular from D to AC, with DE = DF. Prove that AD bisects angle BAC.

(Total for Question 17 is 5 marks)

Quadrilateral ABCD has diagonals AC and BD which intersect at M, such that AM = MC, BM = MD, and AC is perpendicular to BD. Prove that triangle AMB is congruent to triangle CMB, to triangle CMD, and to triangle AMD, and hence prove that ABCD is a rhombus (all four sides equal in length).

(Total for Question 18 is 6 marks)

Mark scheme · 7.14 Congruent Triangles

Question 1

B1 similarity: both have equal corresponding angles / both are the same shape oe
B1 difference: congruent shapes are also the same size (all corresponding sides equal, scale factor 1), whereas similar shapes can be different sizes (scale factor not equal to 1) oe
Answer: Similar: same shape (equal angles). Different: congruent shapes must also be the same size; similar shapes need not be.

Question 2

B1 SSS - all three pairs of corresponding sides are equal
B1 SAS - two pairs of corresponding sides and the included angle between them are equal
B1 ASA (or AAS) - two pairs of corresponding angles and a corresponding side are equal
B1 RHS - both triangles are right-angled, with equal hypotenuses and one further pair of equal sides
Answer: SSS, SAS, ASA (or AAS), RHS

Question 3

(a) B1 SAS oe - two sides and the included angle equal
(a) Answer: SAS
(b) B1 no oe - AAA only proves the triangles are similar, not congruent, since no side length is fixed
(b) Answer: No - not necessarily congruent (only similar)
(c) B1 RHS oe - right angle, hypotenuse and one other side equal
(c) Answer: RHS
(d) B1 C oe
(d) Answer: C

Question 4

(a) B1 yes cao
(a) Answer: Yes
(b) B1 SAS identified with correctly matched pairs: JK = NO (5.2 cm), KL = MN (7.8 cm), angle K = angle N = 62 degrees (the included angle)
(b) B1 correct correspondence stated, e.g. triangle JKL = triangle ONM (oe, any correct equivalent vertex order)
(b) Answer: Triangle JKL = Triangle ONM (SAS)

Question 5

B1 AB = AC (given)
B1 BM = CM (M is the midpoint of BC)
B1 AM = AM (common side); hence triangle ABM = triangle ACM (SSS)
B1 angle ABM = angle ACM (corresponding angles in congruent triangles), so angle ABC = angle ACB (cso)
Answer: Triangle ABM = Triangle ACM (SSS); therefore angle ABC = angle ACB

Question 6

B1 AB = AD (given)
B1 CB = CD (given)
B1 AC = AC (common side); hence triangle ABC = triangle ADC (SSS)
B1 angle BAC = angle DAC and angle BCA = angle DCA (corresponding angles in congruent triangles), so AC bisects angle BAD and angle BCD (cso)
Answer: Triangle ABC = Triangle ADC (SSS); AC bisects angle BAD and angle BCD

Question 7

M1 forms the equation 2x + 5 = 4x - 7 (or equivalent), using that corresponding sides of congruent triangles are equal
A1 x = 6 cao
M1 substitutes x = 6 back into the expression for AB (or DE)
A1 AB = 17 cm cao
Answer: x = 6, AB = 17 cm

Question 8

(a) M1 valid method, e.g. counting squares or distance = √(x2-x1)² + (y2-y1)², used for at least one side of each triangle
(a) A1 AB = DE = 4 cm and BC = EF = 3 cm
(a) A1 AC = DF = 5 cm (oe using Pythagoras: √4² + 3²)
(a) Answer: AB = 4 cm, BC = 3 cm, AC = 5 cm; DE = 4 cm, EF = 3 cm, DF = 5 cm
(b) B1 RHS oe - right angle (at B and at E), hypotenuse (AC = DF = 5 cm) and one other side (AB = DE = 4 cm, or BC = EF = 3 cm) all equal
(b) B1 correct correspondence, e.g. triangle ABC = triangle DEF (A to D, B to E, C to F)
(b) Answer: RHS; triangle ABC = triangle DEF

Question 9

(a) B1 angle BAC = angle DCA (alternate angles, AB parallel to DC)
(a) B1 angle BCA = angle DAC (alternate angles, AD parallel to BC)
(a) B1 AC = CA (common side)
(a) B1 hence ASA, triangle ABC = triangle CDA (cso)
(a) Answer: Triangle ABC = Triangle CDA (ASA)
(b) B1 AB = CD (or BC = AD) - opposite sides of a parallelogram are equal (corresponding sides in congruent triangles)
(b) B1 angle ABC = angle ADC (or angle BAD = angle BCD) - opposite angles of a parallelogram are equal (corresponding angles in congruent triangles)
(b) Answer: AB = CD, BC = AD; angle ABC = angle ADC

Question 10

(a) B1 angle PWQ = angle RWS = 90 degrees (wall is vertical, ground is horizontal)
(a) B1 PQ = RS (given, hypotenuse) and QW = SW (given, one other side)
(a) B1 hence RHS, triangle PWQ = triangle RWS (cso)
(a) Answer: Triangle PWQ = Triangle RWS (RHS)
(b) M1 PW = √4.5² - 4.2² oe (correct use of Pythagoras)
(b) A1 PW = RW = 1.6 m awrt
(b) Answer: 1.6 m

Question 11

B1 AD = AE (given)
B1 AC = AB (given, triangle ABC is isosceles)
B1 angle DAC = angle EAB (the same angle, angle BAC, since D lies on AB and E lies on AC); hence SAS, triangle ADC = triangle AEB
B1 hence DC = EB (corresponding sides in congruent triangles, cso)
Answer: Triangle ADC = Triangle AEB (SAS); DC = EB

Question 12

(a) B1 AC = CD (by construction, C is the midpoint of AD)
(a) B1 angle BAC = angle EDC = 90 degrees (both perpendicular to the bank, by construction)
(a) B1 angle ACB = angle DCE (vertically opposite angles, since A, C, D and B, C, E are each straight lines); hence ASA, triangle ABC = triangle DEC (cso)
(a) Answer: Triangle ABC = Triangle DEC (ASA)
(b) B1 AB = 23.6 m (corresponding sides of congruent triangles, cao)
(b) Answer: 23.6 m

Question 13

B1 LP = LQ (given)
B1 PM = QM (M is the midpoint of PQ)
B1 LM = LM (common side); hence SSS, triangle LPM = triangle LQM
B1 angle LMP = angle LMQ (corresponding angles in congruent triangles), and angle LMP + angle LMQ = 180 degrees (angles on a straight line PQ)
B1 hence angle LMP = angle LMQ = 90 degrees, so LM is perpendicular to PQ (cso)
Answer: Triangle LPM = Triangle LQM (SSS); LM is perpendicular to PQ

Question 14

B1 AM = (1/2) AB and CN = (1/2) CD (perpendicular from centre bisects the chord); since AB = CD, AM = CN
B1 OA = OC (both radii of the circle)
B1 angle OMA = angle ONC = 90 degrees (given)
B1 hence RHS, triangle OAM = triangle OCN
B1 hence OM = ON (corresponding sides in congruent triangles, cso)
Answer: Triangle OAM = Triangle OCN (RHS); OM = ON

Question 15

M1 forms the equation 5x - 2 = 2x + 7 (using AB = DE)
A1 x = 3 cao
M1 forms the equation 3y + 10 = y + 50 (using angle A = angle D)
A1 y = 20 cao
B1 AB = 13 cm and angle A = 70 degrees (both correct, ft their x and y)
Answer: x = 3, y = 20, AB = 13 cm, angle A = 70 degrees

Question 16

B1 AD = BC (given, hypotenuse of each right-angled triangle)
B1 angle AED = angle BFC = 90 degrees (given, by construction)
B1 AE = BF (A and B are the same perpendicular distance from line DC)
B1 reason: since AB is parallel to DC, the perpendicular distance between the two parallel lines is constant
B1 hence RHS, triangle ADE = triangle BCF
B1 hence angle ADE = angle BCF, i.e. angle ADC = angle BCD (corresponding angles in congruent triangles, cso)
Answer: Triangle ADE = Triangle BCF (RHS); angle ADC = angle BCD

Question 17

B1 angle AED = angle AFD = 90 degrees (given)
B1 DE = DF (given)
B1 AD = AD (common side, the hypotenuse of each triangle); hence RHS, triangle ADE = triangle ADF
B1 hence angle DAE = angle DAF (corresponding angles in congruent triangles)
B1 therefore AD bisects angle BAC (cso)
Answer: Triangle ADE = Triangle ADF (RHS); AD bisects angle BAC

Question 18

B1 AM = CM and BM = DM (given, diagonals bisect each other), and all four angles at M are 90 degrees (diagonals perpendicular)
B1 triangle AMB = triangle CMB (SAS: AM = CM, angle AMB = angle CMB = 90, MB common); hence AB = CB
B1 triangle CMB = triangle CMD (SAS: CM common, angle CMB = angle CMD = 90, MB = MD); hence CB = CD
B1 triangle CMD = triangle AMD (SAS: CM = AM, angle CMD = angle AMD = 90, MD common); hence CD = AD
B1 combines results: AB = CB = CD = AD
B1 therefore all four sides of ABCD are equal, so ABCD is a rhombus (cso)
Answer: AB = BC = CD = DA; ABCD is a rhombus

Congruent Triangles - Worksheets, Questions and Revision

7.14 Congruent Triangles

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18