Prac Maths. Geometry is Easy Grade 11 & 12. Seeliger ~ Mouton. Set by / Opgestel deur

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2 Prac Maths Geometry is Easy Grade & Set by / pgestel deur Seeliger ~ Mouton Geometry is Easy Grade & ~ PS JNM PULISHERS (Pty) Ltd 06 P ox WTERKLF 045 Tel: (0) Fax: (0) Fax: megaprint@icon.co.za john@systemmaths.co.za PYRIGHT RESERVE Unauthorised reproduction by photocopying or any other method STRITLY PRHIITE. KPIEREG VREHU ngemagtigde reproduksie deur fotokopiëring of enige ander metode STRENG VERE.

3 Geometry is Easy Grade & Important information Geometry is Easy This workbook has been specifically written to help grade and learners develop clear insight into the logical development of geometry. Just like the study of algebra rests upon the knowledge of sets of numbers and certain asic Laws, the study of geometry rests upon working with sets of points and certain asis Laws called xioms. xioms are obvious truths that require no proof. For example: The first two axioms are: Line axiom: Exactly one line can be drawn through any two points. Line intersection axiom: Two distinct lines intersect in exactly one point. xioms are used to prove the first theorems, on which proofs of further theorems are built. For example: Theorem states: If two lines intersect, the sum of any pair of adjacent angles is 80. Theorem states: If two lines intersect, the vertically opposite sides are equal. The most important reason for studying geometry is its great value in developing reasoning skills. The structured exercises in this book, firstly ensure that the learners establish a thorough knowledge of the definitions and terms when dealing with the size, shape and position of lines, line segments and angles. Secondly the learners apply their knowledge to solve problems involving geometry language and reasons by doing calculations. Learners then use deductive reasoning to solve problems (riders). Each step of a deliberation which cannot be directly deduced from a previous step must be answered with a reason. Examples have been provided to illustrate logical reasoning and methodical setting out of solutions. It is important that learners thoroughly examine the structure of each given figure before proceeding to answer the questions. When doing geometry involving circles one must note how many vertices of angles lie on the circle or at the centre of the circle or at any point inside the circle. Examples:. 3 is the centre of the circle tells us that we might need to work with: i) equal angles opposite equal radii ii) an angle with vertex at the centre with a partner(s) at the circle iii) a 90 angle subtended by the diameter N: o not mark all the equal angles in the diagram, but be aware that = and that = if asked to work with. System Maths

4 Geometry is Easy Grade &. 3 T 3 a) There are 4 points marked on the circle tells us that we might work with pairs of angles in the same segment. b) hords and produced intersect at T tells us that: i) = + T (ext of T) ii) = + T (ext of T) iii) = 3 (ext of cyclic quad) iv) 3 = (ext of cyclic quad) T Given that = means that,, and lie on a circle and it follows that there are 3 pairs of equal angles in the same segment. Furthermore, to work with T means T = + or T = + (ext. of ). In the pilot study, with minimum assistance from the tutor, learners learnt to focus on the diagram signals and then to use the question information in order to do calculations and solve problems. s their diagram awareness developed their reasoning skills and confidence improved. fter completing the Grade and/or Grade training questions, the respective learners tackled the difficult questions in their text books or Prac books or previous examination questions with enthusiasm and success. System Maths

5 Geometry is Easy Grade & Grade & Geometry ontents Section Topic Page Page Questions nswers Grade Revision of pre-knowledge and Theorem (line segment from centre to chord) exercises Theorems, 3 and 4 ( s on circle and at centre) exercises 6 Theorems 5 and 6 (cyclic quadrilaterals) exercises 3 4 Theorems 7 and 8 (tangents) exercises 3 7 E Mixed questions 39 8 Grade F Revision of pre-knowledge and Theorem ( ) exercises 44 0 G& Theorem (similar triangles) exercises 50 G3 Theorem 3 (mean proportion theorem) exercises 57 4 G4&5 Theorem 4 (Theorem of Pythagoras) exercises 60 5 H Mixed questions 6 6 Seeliger ~ Mouton P ox Waterkloof 045 Pretoria ongratulations, you are on your way to success in Maths! Learners say The books work!!! opyright reserved

6 Geometry is easy Grade &. Revision of pre-knowledge. Grade Geometry. Remember that each point on a circle is the same distance from a fixed point called the centre of the circle.. E omplete: If = = then:. Point is called the. Line segment is called the.3 Line segment is called the H.4 Line segment is called the.5 The dark shaded region is called a.6 The light shaded region is called a.7 Points and divide the circle into.8 hord divides the interior of the circle into.9 Straight line F is called a G F.0 Straight line GH is called a. The points EF lie on the. The distance around the circle is called.3 = t. means that t =.4 The letter π (pi) denotes the ratio 3. Lines and angles 3. 3 omplete: If = and: i) ˆ o ˆ ˆ = θ then = and = ii) ˆ o = then ˆ ˆ ˆ + = and = iii) ˆ ˆ o = = then, ˆ ˆ ˆ 3 = and + = iv) ˆ ˆ o 3 = = then, ˆ = and ˆ ˆ + = 3. omplete: i) ˆ ˆ = (ext. p of Δ) ii) Ĉ = + (ext. p of Δ) iii) ˆ + ˆ = 80 o (sum of ps of Δ) Grade & Geometry is easy

7 Geometry is easy Grade & 3.3 omplete: i) ˆ = + ( ) o ii) ˆ o ˆ = x and = x ( ) means ˆ = and = ( ) 3.4 The Theorem of Pythagoras states: In any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. In Δ: b a c b = a + c or a = b c or c = b a (Pyth) 3.5 alculate the value of x. i) G: = x cm and = 9 cm. In Δ: ii) G: = x + cm, = 5x cm and = 5x cm Grade & Geometry is easy

8 Geometry is easy Grade & 3 ircle Theorems. Theorem (a) In À, M = M means Theorem (b) M M z (À centre and mid-pt of chord) In À, T z means T = T (perp. from centre to chord) hence: T P If PT z and = then T M M is the centre of ÀTP (perp. bisector of chord) Example: Question: In, KM z PT, PT = 30 cm and M = 7 cm. alculate the length of KM. nswer: raw T Then T = M = 7 cm. KM z PT PK = KT = 5 cm (perp. from centre bisects chord) In ΔKT: K = T KT (Pyth) = cm P K M T = 64 cm K = 8 cm KM = 7 8 cm = 9 cm Grade & Geometry is easy

9 Geometry is easy Grade & 4 Exercise. If P = T = and = 3 calculate the length of. P = T (given) ( ) In ΔP : = = = P T = =. If T = 8, = 3 and z T T calculate the length of. In ΔT : T = ( ) T = = = = 3. alculate the value of x in each circle with centre. = 3. G: = 80 mm, = 30 mm, z and = x mm. In Δ: = = ( ) Grade & Geometry is easy

10 Geometry is easy Grade & 5 3. G: = =, P = 8 and = x. P i) Remember that the distance from a point to a line is given by the length of the perpendicular line segment from the X point to the line. ii) hords that are equal in length are equidistant from the centre of the circle. Thus, if = in circle then X = Y. iv) onversely, chords that are equidistant from the centre of a circle are equal in length. Y Thus, if X = Y in circle then =. 4. P M T G: In M, = P = 6 cm, MT z, T = 8 cm and MT = 6 cm. omplete without giving reasons. i) ˆ MP = ii) = iii) hord = chord iv) MT = 6 cm means PM = Grade & Geometry is easy

11 Geometry is easy Grade & 6. Study: In Δ: subtends ˆ at. subtends ˆ at and subtends ˆ at. Note: PQR ˆ is subtended by PR at Q. and XYZ ˆ is subtended by XZ at Y. 5. Given that is the centre of circle, means: i) The vertex of ˆ is at the centre of the circle. We say that arc subtends ˆ at the centre. ii) The vertex of ˆ is at point on the circle. We say that subtends ˆ at the circle. Exercise. R omplete if is the centre of circle PTR.. Radius T subtends at P.. Radius T subtends at R. 3.3 Radius P subtends at T..4 Radius P subtends at R..5 Radius R subtends at T. P T.6 Radius R subtends at P..7 hord PT subtends at and at R..8 hord RT subtends at and at P..9 hord PR subtends at and at T.. omplete if is the centre of circle. 3. rc subtends at and at.. hord subtends at and at..3 hord subtends at and at..4 Radius subtends at and at..5 Radius subtends at and at. Grade & Geometry is easy

12 Gr & Geometry Gr & Meetkunde NSWERS NTWRE Set by / pgestel deur Seeliger~Mouton Tel (0) PS opyright Reserved Kopiereg Voorbehou

13 Gr & Geometry / Meetkunde nswers/ntwoorde Grade / Graad 3. i). T = 4 (perp from centre to chord) Revision of pre-knowledge ii) = ₁ + (ext. of ) (loodlyn van midpt na koord) Hersiening van vorige kennis (buite van ) T² = (Pyth). centre / middelpunt iii) 80 (sum of s of ) = 5. radius / radius (straal) (som van e van ) = T = 5.3 diameter / middellyn 3.3 i) + ₂ (ext. of ) =5 3=.4 chord / koord (buite van ).5 segment / segment ii) ₂ = (ext. of ) 3. = = 40mm (perp from centre to chord).6 sector / sektor (buite van ) (loodlyn van midpt na koord).7 arcs / boë iii) = (sides opp equal s of ) ² = mm².8 segments / segmente (sye teenoor gelyke e van ) = 500mm².9 secant / snylyn 3.5 i) = 50mm.0 tangent / raaklyn ² + ² = ² (Pyth) In : = mm. circle / sirkel ² + ² = 6 ² ² = 80² + 40² mm² (Pyth) ² = 6 ². circumference / omtrek = mm².3.4 t= ² = 8 ² =9 = 8000 mm² circumference : diameter or = 0 80 mm omtrek : middellyn of ii) ² + ² = ² (Pyth) =40 5mm ( +) + (5 ) = 5 ² ² ² 0 + = 5 ² 3. In : ( centre & mid-pt of chord) 3. ² = 0 (midpt & midpt van koord) i) = θ & = 80 θ ( )( 5) =0 = P = + 8 (radii) = =5 ² = ² ² (Pyth) ii) ₁+ = 80 α & = 90 α Exercise / efening. PT ( centre & mid-pt of chord) ² = ( +8) 44 iii), ₃ = (midpt & midpt van koord) ² = ² & ₁+ = 80 α ² = P² P² (Pyth) 0 = 6 80 = = 5 iv), = ₂ & ₁+ ₂= 80 β = 5 5= = 5 = 3 5 = 8

14 Gr & Geometry / Meetkunde nswers/ntwoorde 4.. ₁ = + (ext of ). ₁ =4 =4 (given / gegee) i) MP = 90 =55 (buite van ) ₁ = ( at centre =. at ) ii) = 6 M ₁ = 0 = y ( at centre =. at ) (midpt =. op ) iii) (midpt =. op ) +T = ₁ (ext of T) iv) PM = 6cm + T = (buite van T) 3. M ₁ = 00 ( at centre =. at ) T = = Exercise / efening (midpt =. op ) = T (sides opp. eq. s of ). P ₁. R ₁.3 T ₁ M ₂ = 60 (revolution / omwenteling) (sye teenoor gelyke e van ).4 R ₂.5 T ₂.6 P ₂ ₁+ ₁ = 0.7 ₁& R T.8 ₂ & P T.9 ₃ & T R ₁ = ₁ = 0 (radii M & ) 3. ₁ = ₁ ( at centre =. at ) ₃ = ₃ (midpt =. op ). ₁ & ₁. ₂ & ₂.3 ₃ & ₁ 4. M ₁ = ₁ ( atcentre=. at ) but/maar.4 ₂ & ₁.5 ₂ & ₁ = 60 (midpt =. op ) ₁ = ₃ (vertically opp. s) M ₃ = ₁ ( atcentre=. at ) ₁ = ₃ (regoorst e) Exercise 3 / efening 3 = 70 (midpt =. op ) M ₂ = (M str ) 4. Let / laat P = = 50 (M gestrekte ) ₁ = ( at centre =. at ) Example / Voorbeeld (midpt =. op ) ₂+ ₂ = 80 (sum of s of ) ₁ = (som van e van ). P ₁ = but/maar i) + ₂ ii) ₁ = ₂ T ₁ = + ₁ Q ₂ =R ₂ ( s opp equal radii) & = ₁ iii) ₂ = ₁ + ₂ iv) ₂ = 90 ( e teenoor gelyke radii) T ₁+ = v) ₁ vi) ₂ T ₁ + = ₁ ₂ + = = 90 i) P ₁ + P ₂ ii) R ₁ + R ₂ iii) R ₂ Exercise 5 / efening 5 3 Exerise 6 / efening 6 iv) R ₁ v) P ₂ vi) PT R. QT = QP (given / gegee) Exercise 4 / efening 4 P =T ( soppequal sidesof ). = 50 ( e teenoor gelyke sye v ). = 36. R = 35 ( at centre=. at ) Q ₁ =P +T (ext of PQT) 3. = (midpt =. op ) (buite van PQT) = 87 P =R = 35 (alt s & ) Q ₁ =P =T (verw e en MP RQ) ₁ =Q ₁ ( atcentre=. at ) =4T (midpt =. op )

PracMaths. Trigonometry is Easy Grades 10 & 11. Seeliger ~ Mouton. Set by / Opgestel deur

PracMaths. Trigonometry is Easy Grades 10 & 11. Seeliger ~ Mouton. Set by / Opgestel deur PracMaths Trigonometry is Easy Grades 0 & Set by / Ogestel deur Seeliger ~ Mouton Trigonometry is Easy Grades 0 & ~ PS JNM PULISHERS (Pty) Ltd 07 PO ox 955 WTERKLOOF 045 Tel: (0) 460 907 Fax: (0) 460 0848