CONGRUENCE ND SIMILRITY 1.1CONGRUENT FIGURES The figures that have the same size and the same shape, i.e. one shape fits exactly onto other is called Congruent figures. CONGRUENT TRINGLES: 1. Two triangles are congruent if they have the same size and the same shape.. If two triangles are congruent, then a) Their corresponding angles are equal and b) Their corresponding sides are equal. 3. If Δ BC Δ PQR, then P P B Q C R B PQ BC QR C PR B C Q R
TIPS FOR STUDENTS: 1. The symbol means is congruent to. The matching diagram shows how the corresponding vertices match. Δ BC Δ PQR, The corresponding vertices must be named in the correct order. e.g. Δ BC Δ QRP and Δ CB Δ PRQ TEST OF CONGRUENCY BETWEEN TWO TRINGLES: TEST DESCRIPTION SSS Two triangles are congruent if all Three Corresponding Sides are equal. If B PQ BC QR C PR Δ BC Δ PQR. P This is known as the SSS rule. ( side, side, side ) B C Q R
TEST SS DESCRIPTION Two triangles are congruent if Two Corresponding Sides and the Included angle are equal. If B PQ C PR and P then Δ BC Δ PQR. P B C Q R This is known as the SS rule. ( side, angle, side ) TEST DESCRIPTION S Two triangles are congruent if Two ngles and a Corresponding Sides are equal. If P and B Q then BC QR Δ BC Δ PQR. P B C Q R This is known as the S rule. ( angle, angle, side )
TEST DESCRIPTION S Two triangles are congruent if Two angles and the Included Sides are equal. If P and B Q then B PQ Δ BC Δ PQR. P B C Q R This is known as the S rule. ( angle, side, angle )
TEST DESCRIPTION RHS Two triangles are congruent if both triangles have a Right ngle, equal Hypotenuse and another Side which is equal. P If C R B PQ and C PR then Δ BC Δ PQR. B C Q R This is known as the RHS rule. ( right angle, hypotenuse, side ) EXMPLE1: In the Δ BE Δ CDE. BE 90 0, ED 60 0, BC 3 cm and DE 18 cm. FIND a) The length of E, b) Δ CDE. B E C 3 60 0 18 D
SOLUTION : a) Δ BE Δ CDE ( Given ) BE DE 18 cm C 60 0 Corresponding sides are equal. B 3 15 30 0 E CE 18-3 cm 15 cm E CE 15 cm 18 Corresponding sides are equal. D b) DCE BE 90 0 Corresponding sides are equal. CED 60 0 EB CED sincecorresponding sides are equal. 30 0 CED 180 0-90 0 30 0 ( sum of Δ ) EXMPLE: In the quadrilateral BCD, BE CE and E DE.
a) Prove that triangle EB is congruent to B C triangle DEC. b) Name a triangle that is congruent to triangle BD. E c) Name a triangle that is congruent to triangle BC. D SOLUTION: a) BE CE Given E DE EB DEC ( vert. opp. s ) Δ EB Δ DEC ( SS rule ) b) Δ DEC BD C BD CD ( Base s of isos. Δ EB ) D ( common sides ) Δ EB Δ DEC ( SS rule )
c) Δ DCB C DB CB DBC ( Base s of isos. Δ EBC ) BC ( common sides ) Δ BC Δ DCB ( SS rule ) 1. CONGRUENCE ND TRNSFORMTION 1. Under a transformation, an object is formed onto its image.. figure and its image are congruent under a translation, a rotation and a reflection. Translation Rotation Reflection 1.3 SIMILR FIGURES Two figures are similar if
a)the corresponding angles are equal and b)the corresponding sides are in the same ratio. Two similar figures have the same shape but not necessarily the same size. TIP FOR STUDENTS: When two figures are congruent, they are also similar. However the converse is not true. SIMILR TRINGLES: 1. Two triangles are similar if a) Their corresponding angles are equal and b) Their corresponding sides are in the same ratio.. If Δ BC is similar to Δ PQR, then P B Q and C R
P B C Q R TEST FOR SIMILRITY BETWEEN TWO TRINGLES: One of the following conditions is sufficient for two triangles to be similar. 1. Two triangles are similar if two of their corresponding angles are equal. P If P and B Q then Δ BC is similar to Δ PQR. B C Q R
. Two triangles are similar if all three corresponding sides are proportional in the same ratio. If k Where k is a constant, then Δ BC is similar to Δ PQR. P kd kf d f B e C Q ke R 3.Two triangles are similar if two of their corresponding sides are proportional and the included angle is equal. If k, and B Q, then Δ BC is similar to Δ PQR.
P kd d EXMPLE:1 B e C Q ke R In the diagram, B is parallel to DE, B 7 cm, BC 4 cm, CD 8 cm and CE 16 cm. a) Show that triangle BC is similar to triangle EDC. b) CLCULTE 7 B 4 i) The length of C, C 8 16 ii) The length of DE. D E SOLUTION: a) BC EDC (alt. s, B // DE ) BC DEC (alt. s, B // DE ) Two triangles are similar if two of their corresponding angles are equal..
Δ BC is similar to Δ EDC. b) i)since Δ BC is similar to Δ EDC. Corresponding sides of similar triangles are equal. C 4 X 16 8 8 cm ii) Corresponding sides of similar triangles are proportional. DE 8 X 7 4 14 cm EXMPLE: The diagram shows triangle CD. BE is parallel to CD, E 9 cm, DE 3 cm, BC cm and CD 6 cm.
CLCULTE a) BE, 9 b) B. B 3 E D C 6 SOLUTION: a) Δ BE is similar to Δ CD. BE CD (corr. s BE // CD ) EB DC (corr. s BE // CD ) Δ BE is similar to Δ CD Corresponding sides of similar triangles are proportional. BE X 6 4 cm
b) Corresponding sides of similar triangles are proportional. Cross - multiply 4 B 3 ( B + ) 4 B 3 B + 6 B 6 cm EXMPLE3: In the diagram triangle BC is a right angle. D lies on B and CD is perpendicular to B 5 cm, C 4 cm and BC 3cm. a) Show that triangle BC is similar C to triangle CBD. 4 3 b) Find The length of BD. D B 5
SOLUTION: a) C C 4 3 3 5 B D B b) Since Δ BC is similar to Δ CBD. Corresponding sides of similar triangles are equal. BD X 3 1.8 cm 1.4 SIMILRITY ND ENLRGEMENT 1. n enlargement is a transformation that charges the size of a figure (enlarged / reduced) without changing its shape.
. n enlargement is determined by the centre of enlargement and a scale factor. 3. The scale factor of an enlargement is the common ratio between pairs of corresponding sides of the image and the original figure. Scale Factor 4.In the diagram, Δ BC is mapped onto ' B' C' by an enlargement with centre O and scale factor. O B B C C
or 5. In the diagram, Δ BC is mapped onto B' C' by an enlargement with centre and scale factor. C C B B 6. In general, if Δ ' B' C' is an enlargement of Δ BC with centre O, then SCLE FCTOR ( or ) or ( or or ) ' O C C ' B B '
7. figure and its image under an enlargement are similar. TIPS FOR STUDENTS: 1. If the scale factor is greater than 1, then the image is enlarged.. If the scale factor is between 0 and 1, then the image is reduced. 3. If the scale factor is 1, then the image is congruent to the original figure. EXMPLE: In the diagram, Δ BC is mapped onto B' C' by an enlatgement with centre and scale factor k. B 4 cm, BB' 6 cm, BC 3.5 cm, and CC' 9 cm. FIND C ' 9 a) The value of k, C b) The length of B ' C ', 3. 5 c) The length of C. 4 B 6 B ' SOLUTION:
a)scale Factor, k. 5 b) Scale Factor..5 B'C'.5 X 3.5 8. 75 cm c) Scale Factor.5 C + 9 1.5 C 9.5 C Multiply both sides by C. C. 6 cm 1.5 RES OF SIMILR FIGURES If two figures are similar, then the ratio of their areas is equal to the square of the ratio of the lengths of any pair of corresponding sides.
If 1 and denoted the areas of the two similar figures and l1 and l denoted their corresponding lengths, then 1 l 1 l EXMPLE: In the diagram, BC is parallel to DE, C 5 cm and CE 3 cm, find the value of a), E 3 b), C 5 c). B D SOLUTION: Δ BC is similar to Δ DE BC DE (corr. s BC // DE) CB ED (corr. s BC // DE)
a) Corresponding sides of similar triangles are proportional. b) C E 5 8 1 l 1 l 5 64 c) EXMPLE: In the diagram, QR is parallel to ST and a) Find the value of, P b) Given that the area of triangle PQR is 1 cm. Find the area of the quadrilateral QRTS. Q R c) What is the special name given to S T
quadrilateral QRTS? SOLUTION: Δ PQR is similar to Δ PST PQR PST ( corr. s QR // ST ) PRQ PTS ( corr. s QR // ST ) a) PQ PS Take positive square root on both sides. 16 49 b) rea of Δ PST X 1 36.75 cm rea of Quadrilateral QRTS rea of Δ PST - rea of Δ PQR 36.75-1 4.75 cm c) Quadrilateral QRTS is a trapezium. trapezium is a quadrilateral which has a pair of parallel opposite sides.
EXMPLE: In the diagram, BCD and EFBGH are straight lines. E, GC and HD are parallel, BG 6 cm. BF cm, GC 8 cm and HD 13 cm. a) Calculate the length of GH. b) Write down the numerical value of i) ii) E F B 6 G C 8 H 1 13 D 3 SOLUTION: Δ BGC is similar to Δ BHD Corresponding sides of similar triangle are proportional.
1 13 3 8 X 6 GH + 6 10 GH 4 cm b) i) GC HD 8 1 13 3 1 l 1 l 3 5 9 5 5 64 ii) F B 10 h H D
1 1 XBFXh XHBXh rea of Δ X Base X Height X b X h h b EXMPLE: a) The surface areas of two similar figures are in the ratio 16 : 81. Given that the length of the smaller figure is 18 cm, find the length of the corresponding side of the larger figure. b) Two statues are geometrically similar and one 3 times as tall as the other. i) Given that the length of the smaller statue is 1 cm, find the length of foot of the larger statue. ii) The surface area of the larger statue is 98 cm. Calculate the surface area of the smaller statue. SOLUTION: a) LengthOfLargerFigure LengthOfSmallerFigure 1 l 1 l
81 16 Take positive square root on both sides. Length of Larger Figure X 18 40.5 cm b) i) Length of Foot of Larger Statue X 1 4 cm ii) 7 7 Surface rea of Smaller Statue 7 X 98 8 cm
1.6VOLUME OF SIMILR SOLIDES: The ratio of the volumes of two similar solids is equal to the cube of the ratio of any two corresponding length of the two solids. If V1 and V denoted the volume of the two similar solids and l1 and l denoted their corresponding lengths, then V V 1 l 1 l 3 EXMPLE: The two cones are geometrically similar. The height of the smaller cone is 9 cm while the height of the larger cone is 1 cm. a) Find the ratio of the curved surface area of the smaller cone to that of the Larger Cone. 9 1 b) Given that the volume of the smaller cone is 189 cm, find the volume of the larger cone. SOLUTION:
a) 3 9 4 1 1 l 1 l b) 1 9 3 V V 1 l 1 l 3 1 9 3 Volume of Larger Cone 1 9 3 X 189 448 cm 3 EXMPLE: The surface areas of two similar spheres are in the ratio 4:5. a) The radius of the larger sphere is 10 cm. Find the radius of the smaller sphere. b) If the mass of the smaller sphere is 1.6 kg, find the mass of the larger sphere. SOLUTION: a) ( Given ) 1 l 1 l
RadiusOFSmallerSphere RadiusOfLargerSphere 4 5 Take positive square root on both sides. Radius Of Smaller Sphere X 10 4 cm b). 5 3 V V 1 l 1 l 3 Mass of Larger Sphere 5 3 X 189 5 kg TIPS FOR STUDENTS: The mass of the sphere is proportional to its volume. EXMPLE: shop sells bottles of pasta sauce in two sizes as shown below. The two bottles are
geometrically similar. The base area of the larger bottle is 75 cm and the base area of the smaller bottle is 48 cm a) CLCULTE i) The height of the smaller bottle if the height of the larger bottle is 15 cm. ii) The ratio of the volume of pasts in the smaller bottle to the volume of pasta in the larger bottle. b) The smaller bottle costs $ 9.60. Calculate the cost of the larger bottle, given that a discount of 0 % is given for buying the larger bottle.
16 48 a) i) 5 75 HeightOfSmallerBottle HeightOfLargerBottle 1 l 1 l 16 5 Take positive square root on both sides. Height of Smaller Bottle X 15 1 cm ii) 4 5 3 V V 1 l 1 l 3 The Required Ratio is 64 : 15 b) $. Cost of Larger Bottle X $ 9.60 $ 18.75
Cost of Larger Bottle after Discount X $ 18.75 $ 15 TIPS FOR STUDENTS: The cost of the bottle of pasta is proportional to its volume. PROBLEM1: In the diagram, PTQ RSQ and QT QS. Prove that triangle PQT is Congruent to triangle RQS. Q S T U P R SOLUTION: Q S T U P R
PTQ RSQ PQT RQS (Given) ( Q is Common) QT QS (Given ) Δ PQT Δ RQS (S Rule ) PROBLEM: In the diagram, B 13.5 cm, E.5 cm and FE 9 cm. a) Write down a triangle that is congruent to triangle BE, giving reasons for your answer. b) Find the length of BC. B F D C E SOLUTION: 13.5 (13.5).5 B F 9 D C E
a) F.5-9 13.5 cm B BE FC 90 0 ( Given ) BE FC ( is Common ) Δ BE Δ FC ( S Rule ) b) C E.5-13.5 cm 9 cm PROBLEM3: re triangles BC and PQR similar? Explain your answer. a) C P R 35 0 75 0 70 0 B 75 0 Q
b) C R 3.6 4. 6.6 6.3 4.8 B P 7. Q c) P 5.6 5 0 50 0 B 3.5 C Q 7 R SOLUTION: a) YES.Δ BC is similar to Δ PQR. C P 70 0 R 35 0 75 0 70 0 B 75 0 Q
BC 80 0-35 0-75 0 ( sum of Δ) 70 0 Since B Q 75 0 and C R 70 0 Δ BC is similar and Δ PQR. b) NO. Δ BC is not similar to Δ PQR... 1.5.. 1.5.. 1.83 Since, Δ BC is not similar to Δ PQR. c) NO. Δ BC is not similar to Δ PQR. B Q PROBLEM4: Find the value of p if triangles BC and XYZ are similar. a) C X 3.6 30 0 30 0 80 0 6.8 B 70 0 Z Q P
b) Z Y 4.6 3.8 P 6.9 B 3.4 C X c) Z 13 Y 4.6 7. P B 8 C X SOLUTION: a) Since Δ BC is similar to Δ XYZ,
... Corresponding sides of similar triangles are proportional. P.. X 3.6 4.5 cm b) Since Δ BC is similar to Δ XYZ, Corresponding sides of similar triangles are proportional.... P.. X 3.8 5.7 cm c) Since Δ BC is similar to Δ XYZ, Corresponding sides of similar triangles are proportional.. P X 7. 11.7 cm
PROBLEM5: a) b) P D 4 y Q R 1.4 3 x E 4.5 8 1. B 3.5 S x T C c) d) E P 6 C S 1.8 9 4 3.6 B. D y 5 T Q R e) f) T x 5 13 R B C 7.5 4 5 D E P 6 Q x S
g) h) y B E.8 4 5 5.4 C D C 11 x 8 E x 5.5 7.5 B y D i) j) 4 F 3 B S R E x 10.8 x T 8 8 16. 16.8 4.9 P 18.6 Q C 7 D SOLUTION: a) Δ PQR is similar to Δ PST.
Corresponding sides of similar triangles are proportional. x X 3 9 cm b) Δ BE is similar to Δ CDE.... x. X 4.5 1.8 cm c) Δ BC is similar to Δ DE........ x.. X 41.8.9 cm d) Δ PQT is similar to Δ RST. x X 9 7.5 cm
x X 4 74.8 cm e) Δ BC is similar to Δ DE. Cross multiply 13 x 5 ( x + 4 ) 13 x 5 x + 0 8 x 0 x.5 cm f) Δ PQR is similar to Δ PST.. 1.5 x + 6 1.5 X 6 x + 6 9 x 3 cm g) Δ ECD is similar to Δ EB. Cross multiply
3 x ( x +.8 ) 3 x 5 x + 5.6 x 5.6 cm. y X 5.4 8.1 cm h) Δ EF is similar to Δ DEC.. Cross multiply x x + 5 x 5 cm... y + 7.5 X 7.5 y + 7.5 15 y 7.5 cm i) Δ EF is similar to Δ DEC.. x X 4.9 cm.8 cm
j) S R 10.8 x T 16. 16.8 P 18.6 Q PTQ STR ( Vert. opp. s ) QPR QSR ( s in the same segment ) QPT RST Δ PTQ is similar to Δ STR.... x.. X 18.6 cm 1.4 cm PROBLEM6: In the diagram, BC DEC, B 8 cm, BC 1 cm, CD 18 cm and CE 15 cm. a) Give reason why triangle BC and EDC are similar. b) Calculate C and DE.
E 15 8 C 1 18 B D SOLUTION: a) CB ECD ( vert. opp. s ) BC DEC ( Given) Δ BC is similar to ΔEDC. b) C X 15 Corresponding sides of similar triangles are proportional. 10 cm DE X 8 1 cm
PROBLEM7: In the diagram, Δ BC is similar to ΔPQR. Given that B 6 cm, BC 8 cm, PQ ( 1 x ) cm and QR ( 1 + x ) cm, from an equation in x and solve it. P 1 x 6 B 8 C Q 1 + x R SOLUTION: Since Δ BC is similar to Δ PQR. Cross Multiply 8 (1 - x ) 6(1 + x ) 168 8 x 16 + 6 x 4 14 x 3 x x 3 PROBLEM8: In the diagram, PWQ, PXR, WYR, XYZ and QZR are straight lines. WX is parallel to QR
and PQ is parallel to XZ. QZ 3 cm, ZR 5 cm and WQ 4 cm. a) Name two triangles, which are similar to triangle RWQ. b) Use similar triangles to calculate i) YZ, ii) PW. P W X 4 Y Q 3 Z 5 R SOLUTION: P W X 4 Q 3 Z 5 R 8
a) Δ RYZ is similar to Δ RWQ. Δ RYZ is similar to Δ WYX. RYZ RWQ ( Corr. s, YZ // WQ ) RZY RQW ( Corr. s, YZ // WQ ) WXY RZY (lt. s, WX // ZR ) WYX RYZ (Ver. opp s ) Δ RYZ and Δ WYX are similar to Δ RWQ. b) i) Δ RYZ is similar to Δ RWQ. YZ X 4.5 cm ii) WX QZ 3 cm since WQZX is a parallelogram. Δ PWX is similar to Δ PQR. Cross Multiply 8 PW 3 ( PW + 4 ) 8 PW 3 PW + 1 5 PW 1 PW.4 cm
PROBLEM9: In the diagram, PXQ, PYW, PZR, XYZ and QWR are straight lines. XZ is parallel to QR and PQ is parallel to ZW. XY 3 cm, YZ 4 cm and WZ 8 cm. a) Show that triangle XYP is similar to triangle ZYW. b) Calculate PX. c) Show that triangle RQP is similar to triangle RWZ. d) Calculate RW. P X 3 Y 4 Z 8 Q W R SOLUTION:
P (6) X 3 Y 4 Z 8 8 Q 7 W R a) XYP ZYW ( vert. opp. s ) PXY WZY (lt. s PX // ZE ) Δ XYP is similar to Δ ZYW. (Shown) b) PX X 8 6 cm c) RQP RWZ ( Corr. s, PQ // ZW ) RPQ RZW ( Corr. s, PQ // ZW ) Δ RQP is similar to Δ RWZ. ( Shown ) d) QW XZ 3 + 4 Opposite Sides of the parallelogram XQWZ are equal.
7 cm XQ ZW 8 cm PQ PX + XQ Opposite Sides of the parallelogram XQWZ are equal. 6 + 8 14 cm Since Δ RQP is similar to Δ RWZ. Cross Multiply 7 RW 4 ( RW + 7 ) 7 RW 4 RW + 8 3 RW 8 RW cm PROBLEM10: In the diagram, quadrilateral PQRS is mapped onto quadrilateral P'Q'R'S' by an enlargement with centre O and scale factor k. OQ' 6 cm, Q'Q 9 cm,qr 8 cm and P' S' 3 cm. FIND a) Find the value of k, b) The length of Q'R', c) The length of PS.
S P S' P' O Q' Q R' R SOLUTION: S P 3 S' P' O 6 Q' 9 Q R' 8 R
a)scale factor, k b) Scale factor X 8 3. cm c) Scale factor X 3 PS X 3 Take reciprocals on both sides. 7.5 cm PROBLEM11: In the diagram, pentagon PQRS is mapped onto pentagon P' Q' R' S' by an enlargement with centre O and scale factor 3.given that PQ 60 cm and O S' 450 cm.
P' FIND T' a) The length of P' Q', b) The length of OS. Q' P Q T O S R S' R' SOLUTION: a) Scale factor 3 3 X 60 180 cm b) Scale factor 3 OS X 450 Take reciprocals on both sides. 150 cm
PROBLEM1: In the diagram, figure BCDE is mapped onto figure B' C' D' E' by an enlargement with centre. DE is quadrant of a circle, centre D. BC 4.75 cm, B'C' 9.5 and D' E' 1cm. a) Find the scale factor of the enlargement. b) Take to be, find the length of the arc E. c) Given that the area of figure BCD is 110 cm, find the area of figure B' C' D' E'. E' E 1 C' C D' D 4.75 B B' SOLUTION:
E' E 1 10.5 C' C D' D 9.5 4.75 B B' a) Scale factor.. b) Scale factor DE X 1 Take reciprocals on both sides. 10.5 cm Length of arc X X X 10.5 Circumference of circle r 16.5 cm
c) TIP FOR STUDENTS: figure and its image under an enlargement are similar. Figure BCDE is similar to figure B' C' D' E'. 9.5 4.75 1 1 l 1 l rea of B'C'D'E' 4 X 110 440 cm PROBLEM13: In the diagram, BCD is quadrilateral. The diagram C and BD intersect at E. D 15 cm, E 10 cm, BC 7.5 cm and BE DCE. D is parallel to BC and triangle ED is an isosceles triangle. a) Show that triangle BE is congruent to triangle DCE.
b) Name a triangle that is similar to Δ ED. c) Show that triangle CD is isosceles triangle. d) Write down the numerical value of i), ii). 15 D 10 E B 7.5 C SOLUTION: 15 D 10 E (5) B 7.5 C
a) BE DCE ( Given ) E DE ( Sides of isos. s ) BE DCE ( Vert. opp. s ) Δ BE Δ DCE ( S rule ) b) Δ BE is similar to Δ DCE (Δ BEC is also similar to Δ ED since Δ ED is an isosceles triangle. BEC is also an isosceles triangle.) ED CEB ( Vert. opp. s ) ED ECB ( lt. s D // BC ) c). Corresponding sides of similar triangles are proportional. CE. X 10 5 cm C E + CE 10 + 5 15 cm D
Δ CD is an isosceles triangle. PROBLEM14: PSU is a triangle. QR is parallel to SU and SU and PS is parallel to RT. Give that and the area of triangle PQR is 8 cm. CLCULTE a) The value of, b) The area of trapezium QSUR, c) The area of parallelogram QSTR. P Q R X S T U SOLUTION:
P Q R parts S parts T 3 parts U a) Since Δ PQR is similar to Δ PSU, ST QR parts since QSTR is a parallelogram. b) SU QR 1 l 1 l 5
rea of Δ PSU 5 X 8 50 cm rea of trapezium QSUR rea of Δ PSU - rea of Δ PQR 50-8 4 cm c) 3 1 l 1 l rea of Δ RTU X 8 18 cm rea of parallelogram QSTR rea of trapezium QSUR - rea of Δ RTU 4-18 4 cm PROBLEM15: The diagram shows two geometrically similar solid cylinders. the height of Cylinder is 6 cm and the height of Cylinder B is 4 cm. a) If the base area of Cylinder B is 80 cm, calculate the volume of Cylinder.
b) If the mass of Cylinder is 0.18 kg, find the density of the material used to make solid, giving your answer in g/cm 3. 4 6 Cylinder Cylinder B SOLUTION: a) 6 4 1 l 1 l 1 4
1 Base rea of Cylinder X 80 4 5 cm Volume of Cylinder Base rea X Height 5 X 6 30 cm b) Density 6 g / cm 3 0.18 kg 0.18 X 1000g 180g. SUMMMRY ND KEY POINTS 1.) The figures that have the same size and the same shape, i.e. one shape fits exactly onto other is called Congruent figures..)two triangles are congruent if they have the same size and the same shape. 3.) If two triangles are congruent, then a) Their corresponding angles are equal and b) Their corresponding sides are equal. 4.) The symbol means is congruent to
5.) TEST OF CONGRUENCY BETWEEN TWO TRINGLES: a.) Two triangles are congruent if all Three Corresponding Sides are equal This is known as the SSS rule. (side, side, side) b.) Two triangles are congruent if Two Corresponding Sides and the Included angle are equal. This is known as the SS rule. (side, angle, side) c.) Two triangles are congruent if Two ngles and a Corresponding Sides are equal. This is known as the S rule. ( angle, angle, side ) d.) Two triangles are congruent if Two angles and the Included Sides are equal This is known as the S rule. (angle, side, angle) e.) Two triangles are congruent if both triangles have a Right ngle, equal hypotenuse and another Side which is equal. This is known as the RHS rule.( right angle, hypotenuse, side ) 6.) CONGRUENCE ND TRNSFORMTION: a.) Under a transformation, an object is formed onto its image. b.) figure and its image are congruent under a translation, a rotation and a reflection. Translation Rotation Reflection
7.)Two figures are similar if a)the corresponding angles are equal and b)the corresponding sides are in the same ratio. KEY POINTS: Two similar figures have the same shape but not necessarily the same size. When two figures are congruent, they are also similar. However the converse is not true. 8.) Two triangles are similar if a) Their corresponding angles are equal and b) Their corresponding sides are in the same ratio. 9.) If Δ BC is similar to Δ PQR, then P B Q and P C R B C Q R
10.) TEST FOR SIMILRITY BETWEEN TWO TRINGLES: One of the following conditions is sufficient for two triangles to be similar. a.) Two triangles are similar if two of their corresponding angles are equal. If P and B Q then Δ BC is similar to Δ PQR. P B C Q R b.) Two triangles are similar if all three corresponding sides are proportional in the same ratio. If k Where k is a constant, then Δ BC is similar to Δ PQR.
P kd kf d f B e C Q ke R c.) Two triangles are similar if two of their corresponding sides are proportional and the included angle is equal. If k, and B Q, then Δ BC is similar to Δ PQR. 11.) a.) n enlargement is a transformation that charges the size of a figure (enlarged/reduced) without changing its shape. b.) n enlargement is determined by the centre of enlargement and a scale factor. c.) The scale factor of an enlargement is the common ratio between pairs of corresponding sides of the image and the original figure. Scale Factor d.) figure and its image under an enlargement are similar.
KEY POINTS: 1. If the scale factor is greater than 1, then the image is enlarged.. If the scale factor is between 0 and 1, then the image is reduced. 3. If the scale factor is 1, then the image is congruent to the original figure. 1.) If two figures are similar, then the ratio of their areas is equal to the square of the ratio of the lengths of any pair of corresponding sides. 1 l 1 l where 1 and are the areas of the two similar figures and l1 and l are their corresponding lengths. 13.)a.) The ratio of the volumes of two similar solids is equal to the cube of the ratio of any two corresponding length of the two solids. V V 1 l 1 l 3 where V1 and V are the volumes of the two similar figures and l1 and l are their corresponding lengths. b.)the mass of the sphere is proportional to its volume.