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Gr 9 Maths: Content Area 3 & 4 Geometry & Measurement (2D) QUESTIONS Mostly past ANA exam content • Geometry of Straight Lines • Triangles: Basic facts • Congruent Δ • Similar Δ s s All questions have been graded to facilitate concept development. GOOD LUCK! • Quadrilaterals • Polygons Theorem of Pythagoras Area and Perimeter of 2D shapes Compiled by Anne Eadie & Gretel Lampe THE ANSWER SERIES tel: (021) 671 0837 fax: (021) 671 2546 faxtoemail: 088 021 671 2546 www.theanswer.co.za Questions: Geometry of Straight lines 2. GEOMETRY OF STRAIGHT LINES ( Solutions on page A1) Calculate the size of the largest angle. 5. 1 x – 6° Show all your steps with reasons. x – 9° x + 15° In the figure, Bˆ 3 = 35º and BE || CF. ˆ Determine the size of Bˆ and BCF. (4) A Refer to page Q13 for details on parallel lines & angles. 3. 1. Complete the following: . . . . . . . angles. C 43° Statement Reason Bˆ 1 = (1) ˆ = BCF B (3) D b F E (1) 3.2 Angles around a point add up to . . . . . . . 1.1 a 1 2 3.1 Angles which add up to 90º are called Calculate the sizes of the angles marked a to d. Give reasons for your answers. A C B 3 (3) 4. 6. Complete each of the following statements: A T 4.1 D̂ and F̂ are complementary angles if ____________________________________ . 1.2 c 4.2 The sum of the interior angles of a triangle is 58° equal to _____________________________ . 12° (2) P Q d (1) (1) 2 1 3 2 C 1 D Calculate the size of  , Bˆ 1 and Bˆ 2 . Statement 4.4 A trapezium is a quadrilateral with one pair of ___________________ sides. B In the figure above, AB || TC, Cˆ 1 = 65º and Cˆ 2 = 43º. 4.3 The sum of the exterior angles of any polygon is equal to ___________________________ . 1.3 (1) Reason (1) 4.5 The diagonals of a rectangle are _________ S Copyright © The Answer 112° R T (3) in length. (1) (4) Q1 Questions: Geometry of Straight lines 7. 10. Give reasons for each of your statements in the questions below. Find the size of angles a to g (in that order), giving reasons. g In the figure PQ || RS, Qˆ 1 , Qˆ 2 and Qˆ 3 are equal to 2 x, 3 x and 4 x respectively. R̂ = y and Ŝ = z. 3 An acute angle is one that lies between 0º and 90º. 35° a b 1 2 Important Vocabulary 60° T P STRAIGHT LINE GEOMETRY c Q d An obtuse angle is one that lies between 90º and 180º. e A reflex angle is one that lies between 180º and 360º. f (7) A straight angle = 180º 11. In the sketch, AB is a straight line. A revolution = 360º When the sum of 2 angles = 90º, we say the angles are complementary. Determine the value of x + y. R 8. y z When the sum of 2 angles = 180º, we say the angles are supplementary. S 7.1 Calculate the value of x. (3) 7.2 Calculate the value of y. (3) 7.3 Calculate the value of z. (3) x+y x + 30° B 120° P State, giving reasons, whether PQ || RS. D U Q 2 4 3 Adjacent angles have a common vertex and a common ˆ 2ˆ and 3, ˆ 3ˆ and 4ˆ or 1ˆ and 4. ˆ arm, e.g. 1̂ and 2, A Vertically opposite angles lie opposite each other, x C R 76° 4 angles are formed : 1 ˆ 2, ˆ 3, ˆ 4 ˆ 1, ˆ e.g. 1̂ and 3ˆ or 2ˆ and 4. (4) The FACTS When 2 lines intersect: Hint: Draw a third line, through B, parallel to the given parallel lines. T (4) 110° B (4) 9. When 2 lines intersect, y A B 3x – 10° C x 12. Calculate, with reasons, the value of x. Calculate, with reasons, the value of x. A A right angle = 90º V 104° S W (4) For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.32 Q2 adjacent angles are supplementary vertically opposite angles are equal. See the end of the questions for more on straight lines. Copyright © The Answer Questions: Triangles 4. TRIANGLES: BASIC FACTS 7. In ΔEDF, DF is produced to C. The size of Ê is . . . ? ( Solutions on page A3) Using the figure below, calculate the size of the angles a, b and c (in this order). AD = BD = BC; E A a ˆ = 72º ADB Reasons must be provided for all Geometry statements. 1. In the figure below, ΔANT is an equilateral triangle. Calculate the size of Tˆ and Tˆ . 1 3x 2 D A 2 1 P 2. T B 60º C 140º D 20º x H W 3.1 The size of Tˆ ˆ 3.2 The size of M 2 Copyright © The Answer 44° (3) A 106° x (2) (3) P B 110° C 50° D 8.2 a ˆ = 110º. In the figure above, B̂ = 50º and ACD M 1 The size of  is . . . . . . 2 R 1 Determine the values of x, a, b and c in the figures below. 8.1 6. N In ΔPRT alongside, M is the midpoint of PR and MR = MT. If P̂ = 25º, calculate with reasons: C Determine the size of  in terms of x. 1 2 E (6) In ΔABC, AB = AC and Ĉ = x. S 70° 1 2 c C 8. A A (1) [10] D B ˆ = 70º; In the figure below, CS || HN, EAW ˆ = x. AE = AW and CAE C C 72° b (4) Determine the value of x. 3. A 40º 5. N B 4x 5x F 2 1 T (1) A 50º B 60º C 110º D 160º (1) Q3 c 28° b 44° (6) Questions: Triangles 9. Calculate the values of x and y if TRIANGLES: Study the following very carefully Bˆ 2 = x, Dˆ 2 = y, Dˆ 1 = 44º, Cˆ 1 = 75º and AD || BC. A 2 y 1 1 44° 2 D CLASSIFICATION OF TRIANGLES . . . B Triangles are classified according to their sides or their angles (or both). x 2 1 75° • Sides C E (3) equilateral Δ isosceles Δ INTERIOR AND EXTERIOR ANGLES . . . 1 2 3 1̂ , 2̂ and 3̂ are interior angles of the triangle y x x̂ is an exterior ø ŷ is not an exterior ø scalene Δ An exterior angle is formed between one side of a triangle and the produced (extension) of another. A 10. C 1 3 sides equal B 95° 2 sides equal no sides equal • Angles acute-angled Δ right-angled Δ obtuse-angled Δ 30° E D 3 acute angles ˆ = 95º In the above figure AB || ED, ACD and D̂ = 30º. Determine the size of Ê and  . For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.24 1 right angle (90°) 1 obtuse angle • Sides and Angles (3) This is an isosceles, right-angled triangle This is an isosceles, acute-angled triangle This is a scalene, obtuse-angled triangle. 4 BASIC FACTS • FACT 1 The sum of the interior angles of a triangle = 180° A Aˆ + Bˆ + Cˆ = 180° B C • FACT 2 : 2 The exterior angle of a triangle equals 3 the sum of the interior opposite angles. • FACT 3 In an isosceles triangle, the base angles are equal. B 1̂ = 2ˆ + 3ˆ 1 A 1 If AB = AC, then 1̂ = 2̂ 2 If 1̂ = 2̂, then AB = AC The converse states: If 2 angles of a triangle are equal, then the sides opposite them are equal. • FACT 4 The angles of an equilateral triangle all equal 60°. Q4 Converse: C 60° 60° 60° Copyright © The Answer Questions: Congruent Triangles 3. CONGRUENT ΔS 6. Why is ΔABC ≡ ΔDCB? A ( Solutions on page A4) D In the figure below ΔKNQ and ΔMPQ have a common vertex Q. P is a point on KQ and N is a point on MQ. KQ = MQ and PQ = QN. B See the notes on Congruency and Similarity on page A5 P 1. K A S, S, S B 90º, Hyp, S (RHS) C S, ø , S D ø, ø, S E F A C 1 4. In the figure below Dˆ 1 = Bˆ 2 = 90º and AD = BC. 2 P 1 2 B Q R B 2 D C C 1 Q 2 A Reason (2) State which triangle is congruent to ΔABC. C (4) 1 D Prove that ΔABD ≡ ΔCDB. 2. N Prove with reasons that ΔKNQ ≡ ΔMPQ. Which triangle is congruent to ΔPQR? Statement M 7. ΔABC, D and E are points on BC such that BD = EC and AD = AE. A 5. In the figure below, AB = AC and BD = CD. A P 1 2 1 2 A B Q B S T Copyright © The Answer D R B V (2) C 5.1 Prove that ΔABD ≡ ΔACD. (4) ˆ 5.2 Prove that DA bisects BAC (2) Q5 D E C 7.1 Why is BE = CD? (1) 7.2 Which triangle is congruent to ΔABE? (1) Questions: Congruent Triangles 8. In the given figure, P and T are points on a circle with centre M. P N T 10. 9.2 Prove that ΔABC ≡ ΔDEF. Statement 2 1 Reason AB = AC 2 1 N is a point on a chord PT such that MN ⊥ PT. 1 A and M B In the figure alongside 1 2 2 1 BD = CD D 1 C Prove that PN = NT. Statement Reason (5) 9.3 Why is Bˆ = Eˆ ? Statement Reason Bˆ = Eˆ 11. (1) (8) 9. A 1 C Prove that ΔABD ≡ ΔACD. (4) 10.2 Prove that ΔABE ≡ ΔACE. (4) 10.3 Prove that Eˆ 1 = Eˆ 2 = 90º. (3) 10.4 Hence, state the relationship between AE and BC. (1) In the figure below, PS || QR. Which ONE of the following statements is true for this figure? P S T Note: It has (already) been given that AB = ED. Statement 1 F 10.1 9.4 Use your answer in Question 9.3 to derive a further relationship between AB and ED. E D Q Reason R A ΔPTS ≡ ΔPQT B B ΔPTS ≡ ΔRTQ In the above diagram, AC = DF, AB = DE and BF = CE. (2) 9.1 Prove that BC = EF. Statement 2 1 E C ΔPTS ||| ΔSRT D ΔPTS ||| ΔRTQ Reason (1) For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.28 (2) Q6 Copyright © The Answer Questions: Similar Triangles 3. SIMILAR Δ S ( Solutions on page A6) In ΔPQR and ΔSTR in the figure alongside, PQ || ST, PR = 10 cm, ST = 3 cm and SR = 6 cm. 3.1 See the notes on Congruency and Similarity on page A5 3.2 B 5. P 10 E R Q T A 3 6 Prove that ΔPQR ||| ΔSTR 1 S Calculate the length of PQ. F 1 2 D (4) (3) 2 C In the figure, B̂ = Ĉ , AD = 9 cm, AE = 7 cm and CE = 21 cm. 1. Examine ΔDEF and ΔKLM. 4. D In ΔNML below, P and Q are points on the sides MN and LN respectively such that QP || LM. MN = 16 cm, QP = 3 cm and LM = 8 cm. K 14 cm 7 cm E 20 cm L F 10 cm M Complete the following calculations if ΔDEF ||| ΔKLM. x = ________ Copyright © The Answer D 2. Pˆ = ……………. 3. F Qˆ = ……………. 1 ∴ ΔQPN ||| Δ …. 4 cm 10 cm Calculate the length of BD. Statement Reason 1. N̂ …………………………… E C 5.2 Complete the following (give reasons for the statements): 1 6 cm P In ΔQPN and ΔLMN A 15 cm 1 Prove with reasons that ΔQPN ||| ΔLMN. (3) Calculate the length of AB if ΔABC ||| ΔEDF: B 2 1 N 4.1 14 x = 7 2 (6) DE EF DF = = (proportional sides of similar triangles) KL LM 2. Reason 12 cm Q Prove that ΔABD ||| ΔACE. Statement M L x cm 5.1 (4) 4.2 …………………………… ………………………….. Hence, calculate the length of PN. Q7 (5) …………………………… (4) (3) For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.28 Questions: Quadrilaterals 3. QUADRILATERALS ( Solutions on page A7) D 2 1 1 1. ABCD is a parallelogram. Calculate the size of B̂ . A x + 50° 5. In the figure below, ABCD is a square and ATB is an equilateral triangle. 1 4 A 2 D C 3 T 1 B 1 2 C 2 B Look at parallelogram ABCD above and complete the table. 1 A 2x – 20° C 2. D (4) In the figure below, DEFG is a rhombus and Ê = 156°. D 1 2 156° E 4. 2 B Statement 3.1 Name two isosceles triangles. (2) 3.2 Calculate the size of Dˆ 2 . (3) 3.3 Calculate the size of Tˆ 4 . (2) In ΔADB and ΔCBD PRTW is a square. ΔPQR and ΔRTS are equilateral. ˆ Calculate x (RQS) Q G 2 Reason Dˆ 1 = ______ Alternate ø's and AD || BC Bˆ 1 = ______ Alternate ø's and AB || DC BD = BD Common side 1 F â ΔADB ≡ Δ______ ____________ x Calculate the size of: P ˆ 2.1 EFG â AD = ______ and Corresponding sides of s AB = ______ congruent Δ R 2.2 Fˆ2 S (4) 2.3 Ĝ W Statement Reason 2.1 ˆ = EFG (2) 2.2 Fˆ2 = (2) Ĝ = (2) 2.3 T NB: Study 'Quadrilaterals' on page Q12 very carefully. Q8 (7) 6. A parallelogram with at least one angle equal to 90° is called a __________ A kite. B rhombus. C trapezium. D rectangle. (1) Copyright © The Answer Questions: Quadrilaterals 7. A B 2 T 3 1 9. 2 P 90° 1 B 120° C C 100° D 108° The bisectors of B̂ and Ĉ of parallelogram ABCD intersect at T. Points B, T and D do not lie on a straight line. P is a point on DC such that ˆ = 90º. TPD 10. 7.1 Prove that Tˆ 2 = 90º. (5) 7.2 Which triangle is similar to ΔBCT? (2) 7.3 If BC = 2TC and TP = 4 cm, calculate the length of BT. 8. (1) What is the size of each angle in a regular hexagon? A 90° B 120° C 100° D 108° (1) (3) For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.26 In the given quadrilateral AE = ED and BE = EC, therefore: A NOTES What is the size of each angle in a regular pentagon? A 1 2 D POLYGONS D E B A ΔAEB ||| ΔCED B ΔAED ||| ΔBEC C ΔAEB ≡ ΔDEC D ΔAED ≡ ΔBEC Copyright © The Answer C (2) Q9 Questions: Theorem of Pythagoras 5. THEOREM OF PYTHAGORAS Calculate the length of AD. ( Solutions on page A9) 1. 8 cm Triangle B 12 cm C D B (2) C Show that B̂ = 90°. base b ℓ A A right-angled triangle has 6 cm C y D 2.1 Calculate x. (3) 2.2 Calculate y. (3) Give reasons. 3. The area of 2 ΔTUW = 30 cm and UW = 12 cm. ℓ: length b: breadth T 2 The perimeter of a rectangle = ℓ + b + ℓ + b = 2ℓ + 2b = 2(ℓ + b) The area of a rectangle = ℓ % b = ℓb Square B The perimeter of a square = 4 % s = 4s C s The area of a square = s % s = s2 This theorem states: In a right-angled triangle . . . the square of the hypotenuse equals the sum of the squares on the other two sides. See the Quadrilaterals on page Q12 for the areas of all other quadrilaterals. i.e. In ΔABC, B̂ = 90° U 3.2 the perimeter of ΔTUW 12 cm W (2) (3) 2 2 Circle 2 So: AC = AB + BC The converse theorem states the reverse: A If in any ΔABC, 4. hypotenuse The Theorem of Pythagoras Calculate: 3.1 TU one angle of 90º. Here, B̂ = 90°. The side opposite the right angle (90°) is called the hypotenuse. Here, AC is the hypotenuse. base × height base Rectangle A Right-angled triangle B The area of height a triangle = base 17 cm x C a height height In ΔABC: AB = 9 cm, BC = 12 cm and AC = 15 cm. For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.33 A b 10 cm 5 cm 6. The perimeter of this triangle = (a + b + c) units. A c C 12,8 cm B 8 cm PERIMETER AND AREA FORMULAE B 6 cm D 14 cm Determine the length of AC if AB = 5 cm and BC = 12 cm. (4) 2.1 A A 2 cm A In ΔABC, AB ⊥ BC. In rectangle ABCD, AB = 8 cm and diagonal AC = 10 cm. A ladder is standing against the wall. If the ladder reaches a height of 12 m up the wall and has its foot 5 m away from it, calculate the length of the ladder. (3) 2 2 B Q10 centre radius (r) 2 AC = AB + BC , then B̂ = 90°. circumference C diameter The circumference of a circle: = πd = π(2r) = 2 π r The area of a circle: = π r2 Copyright © The Answer Questions: Measurement: 2D 4. MEASUREMENT: 2D In the figure below, AP = 5 m, AS = SB = 2 m and PS ⊥ AB. ( Solutions on page A10) 1. 2. AB, the diameter of the given circle, is 12 cm. Use π = 3,14 to answer the following questions, correct to two decimal places. 3. A 5m C A (3) If the length of the side of a square is 0,12 cm then the area = 8. 4.1 Calculate the length of PS correct to 2 decimal places. (3) 4.2 Calculate the length of PT if PT = 3 % AB. (1) 4.3 What kind of quadrilateral is APBT ? (2) (2) A 5. (1) The circumference of a circle is 52 cm. Calculate the area of the circle correct to 2 decimal places. (4) 12 cm Two circles have the same centre. The smaller circle has a radius of 20 cm. The larger circle has a radius of 30 cm. (2) B 3 cm (2) 8.2 The area of the shaded section. (3) D R T C r In parallelogram ABCD, AB = 5 cm, AD = 12 cm, BT = 3 cm and AT ⊥ BC. 60 m 3.1 How many times must he run around the field in order to run a distance of at least 4 km? 30 8.1 the circumference of the smaller circle. 5 cm 100 m 20 Calculate: 9. Peter runs around the field with the following dimensions: Copyright © The Answer 7. 4.4 Calculate the area of the figure correct to 2 decimal places. 2 3.2 Calculate the area of this field, correct to two decimal places. Write down the value of k if the area of the enlarged rectangle = k % the area of the original rectangle. B 1.2 Calculate the perimeter of the semi–circle ACB. Use π = 3,14. T S 2m B (4) 0,24 cm 2 0,144 cm 2 1,44 cm 2 0,0144 cm The length of a rectangle is doubled. 2m P 1.1 Calculate the area of the circle. A B C D 6. 5.1 Calculate the length of AT. (3) 5.2 Determine the area of the parallelogram. (3) 9.1 Show that the area of the shaded ring is equal to 2 2 π(R – r ). (2) 9.2 Determine the area of the shaded ring in terms of π if R = 14 cm and r = 8 cm. (2) 5.3 Calculate (4) (4) 5.3.1 the perimeter of trapezium ADCT. (1) 5.3.2 the area of trapezium ADCT. (3) Q11 For further practice in this topic – see The Answer Series Gr 9 Mathematics 2 in 1 on p. 1.26 QUADRILATERALS The arrows indicate various ‘ROUTES’ from ‘any’ quadrilateral to the square, the ‘ultimate quadrilateral’. Pathways of definitions and properties A Rectangle A Parallelogram A Trapezium Definition of a rectangle A parallelogram with one right angle Definition of a parallelogram A quadrilateral with 2 pairs of opposite sides parallel Definition of a trapezium a f c e parallel • 2 pairs of opposite sides The Angles equal The Sides • 1 pair of opposite sides parallel • all 4 angles equal 90º The Angles Definition of a square The Diagonals . . . • 2 pairs of opposite angles equal d The Square sides parallel • 2 pairs of opposite sides equal • 2 pairs of opposite sides Properties of a trapezium b • 2 pairs of opposite The Sides A quadrilateral with 1 pair of opposite sides parallel Properties of a rectangle The Sides Properties of a parallelogram 'Any' Quadrilateral Quadrilaterals play a prominent role right through to Grade 12! • bisect each other equally (the diagonals are equal to each other!) The Diagonals . . . s Sum of the ø of any quadrilateral = 360° A rectangle with one pair of adjacent sides equal OR • bisect each other A rhombus with one angle of 90º Properties of a square A square contains ALL the accumulated A Kite properties of sides, angles and diagonals!!! Properties of a kite A Rhombus The Sides • 2 pairs of adjacent sides equal The Sides • all 4 sides equal The Angles • the following pair of angles will be equal because of isosceles triangles as a result of adjacent sides equal Definition of a kite A quadrilateral with 2 pairs of adjacent sides equal Properties of a rhombus Definition of a rhombus A parallelogram with one pair of adjacent sides equal The Diagonals . . . • cut perpendicularly • the LONG DIAGONAL bisects the short diagonal and the opposite angles OR A kite with 2 pairs of opposite sides parallel Q12 QUADRILATERALS See how the properties accumulate as you move from left to right. i.e. the first quad has no special properties and each successive quadrilateral has all preceding properties. The Angles • 2 pairs of opposite angles equal The Diagonals . . . • cut perpendicularly • bisect each other • bisect the opposite angles Copyright © The Answer Geometry of Straight lines MORE STRAIGHT LINE GEOMETRY Angles that 'alternate' are on opposite sides of the transversal. Important Vocabulary When 2 lines are cut by another line (a transversal), two families of angles are formed: 1 2 4 3 1 4 8 5 6 interior 'alternate' angles. These are These are interior exterior angles. angles. 3 5 They are NOT necessarily equal. the transversal 5 6 These are pairs of 5 6 8 7 ˆ 3, ˆ 4ˆ and 5, ˆ 7, ˆ 2, ˆ 6, ˆ 8ˆ 1, 4 3 4 1 1 2 7 8 7 exterior 'alternate' angles. 3 Not usually used. 8 These pairs of angles correspond. 1 Each of these groups are 'co-' angles 2 These are pairs of 2 2 5 6 6 4 3 7 8 Note: They are NOT necessarily equal. 7 The FACTS i.e. they are on the same side of the transversal 4 5 These are pairs of co-interior angles. When 2 PARALLEL lines are cut by a transversal, then the corresponding angles are equal, the (interior) alternate angles are equal, and the co-interior angles are supplementary. 3 6 & conversely: They are NOT necessarily supplementary. 1 8 Copyright © The Answer 7 5 6 8 7 If the corresponding angles are equal, or if the (interior) alternate angles are equal, or if the co-interior angles are supplementary, then the lines are parallel. 2 These are pairs of co-exterior angles. 1 2 4 3 Recognise these angles in unfamiliar situations. Not usually used. Q13 Gr 9 Maths: Content Area 3 & 4 Geometry & Measurement (2D) ANSWERS • Geometry of Straight Lines • Triangles: Basic facts • Congruent Δ • Similar Δ s s • Quadrilaterals • Polygons Theorem of Pythagoras Area and Perimeter of 2D shapes Compiled by Anne Eadie & Gretel Lampe THE ANSWER SERIES tel: (021) 671 0837 fax: (021) 671 2546 faxtoemail: 088 021 671 2546 www.theanswer.co.za Solutions: Geometry of Straight lines GEOMETRY OF STRAIGHT LINES 5. s Bˆ 1 (= Bˆ 3 ) = 35º . . . vertically opposite ø s ˆ = Bˆ BCF 1 . . . corresponding ø ; BE || CF . . . vertically opposite ø b̂ = â . . . corresponding ø ; AB || CD ĉ = 180º – (12º + 58º) = 110º 1.3 2. s B 3 A 35° . . . alternate ø ; PQ || SRT s â d̂ = 180º – 112º . . . co-interior ø supplementary; PS || QR = 68º 6. 7.1 F E s A 7.2 T 3.2 360º D̂ + F̂ = 90º 4.2 180º 4.2 360º 4.4 parallel 4.5 equal Copyright © The Answer S s 2x + 3x + 4x = 180º . . . ø on a straight line s y (= Qˆ 2 ) = 3x . . . alternate ø ; PQ || RS . . . x = 20º in Question 7.1 s 7.3 z (= Qˆ 1 ) = 2x = 40º 8. (3x – 10º) + (x + 30º) = 180º . . . corresponding ø ; PQ || RS 43° B  = Cˆ 2 2 1 2 3 C 1 65° D s . . . alternate ø ; AB || TC Bˆ 1 = Cˆ 1 â 4x + 20º = 180º s . . . corresponding ø ; AB || TC Subtract 20º : â 4x = 160º Divide by 4: â x = 40º s . . . co-interior ø ; AB || CD = 65º Bˆ 2 = 180º – Bˆ 1 4.1 4x z = 60º = 43º complementary Q â 9x = 180º â x = 20º s x – 9º + x – 6º + x + 15º = 360º . . . ø about a point add up to 360º â 3x = 360º â x = 120º 3 y R . . . adjacent ø on a straight line add up to 180º ˆ = 112º PQR 2 C 1 2 s â The largest angle = x + 15° = 135° 3. 3x s â = 43º = 43º 1.2 2x 1 P = 35º 1.1 T 7. s . . . ø on a straight line 9. = 115º ˆ = 180º – 76º PUV s . . . ø on a straight line = 104º Be sure to study 'Straight Line Geometry' (page Q2) - vocabulary and facts and 'More Straight Line Geometry' (page Q13) - vocabulary and facts A1 ˆ = PUV ˆ â RVW â PQ || RS s . . . corresponding ø equal Solutions: Geometry of Straight lines 10. â = 60º . . . vertically opposite angles b̂ = 35º . . . alternate ø ; || lines ĉ = 35º . . . base ø of isosceles Δ NOTES s s d̂ = 180º – ( â + ĉ ) s . . . sum of the ø of a Δ = 180º – (60º + 35º) = 85º exterior ø of a Δ = the sum of s ê = â – 35º ... the interior opposite ø = 25º f̂ = ( b̂ + ĉ ) s . . . corresponding ø ; || lines = 70º or f̂ = ĉ + 35º = 70º . . . ext ø of Δ = sum of int. opp. ø s s ĝ = ê . . . alternate ø ; || lines = 25º s . . . ø on a straight line x + ( x + y) + y = 180º 11. â 2x + 2y = 180º i.e. â 2(x + y) = 180º Divide by 2: 12. â x + y = 90º 120º + 110º + x = 2 % 180º s . . . 2 pairs of co-interior ø ; parallel lines â 230º + x = 360º Subtract 230º: â x = 130º 120° A 110° B x C A2 Copyright © The Answer Solutions: Triangles 5. TRIANGLES: BASIC FACTS s . . . ø opposite equal sides in an isosceles Δ B̂ (= Ĉ ) = x 2. s 6. . . . ø on a straight line are supplementary  = 110º – 50º ... = 60º 8.2 exterior ø of Δ = sum s of interior opposite ø NB : . . . AE = AW; ø opposite equal sides in an isosceles Δ ˆ ) = 55º â Eˆ 2 (= W 1 â x (= Eˆ 2 ) = 55º 7. s . . . alternate ø ; CS || HN ĉ = b̂ + 90º s . . . the base ø of an isosceles Δ are equal ˆ In ΔABD: â = ABD 1 (180º – 72º) 2 1 = (108º) 2 â â = Often, in geometry riders, there are several possible methods. = 16º + 90º = 106º s . . . sum of the ø of a Δ 9. = 54º 3.1 Tˆ1 = 25º 3.2 ˆ = P̂ + Tˆ M 2 1 s . . . ø opposite equal sides MT and MP in an isosceles triangle b̂ = 72º + â = 126º . . . exterior ø of Δ MPT ˆ ĉ = BDC = 2(25º) = 50º 4. 4x + 5x = 180º â 9x = 180º â x = 20º ˆ = Ê + D̂ OR: Ê = 5x – 3x EFC = 2x â 5x = Ê + 3x â Ê = 2x s the interior ø = 40º â Answer: A Copyright © The Answer 10. Ê = 95º – 30º = 65º = 115º exterior ø of Δ . . . equals the sum of A3 exterior ø of Δ = sum s of interior opposite ø . . . exterior ø of ΔBDC =s sum of interior opposite ø s = 31º  = 180º – Ê = 27º ... . . . alternate ø ; AD || BC s . . . the base ø of an isosceles Δ are equal 1 (180º – b̂ ) 2 1 = (54º) 2 s x̂ = 75º – 44º = 31º ŷ = x̂ ext ø of ΔABD = the sum of ... s the interior opposite ø ĉ = . . . ø on a straight line s . . . ø opposite equal sides in an isosceles Δ â b̂ = 44º – 28º = 16º An interior angle = the exterior ø – the other interior ø s s . . . sum of the ø of a Δ = 180º b̂ + 28º = 44º . . . sum of the ø of a Δ = 110º â + 44º = 90º â â = 90º – 44º = 46º â Answer: B s ˆ = 180º – 70º Eˆ 2 + W 1 ˆ But Eˆ 2 = W 1 NB : See comment in Question 6. . . . ø of an equilateral Δ all = 60º â Tˆ2 = 120º . . . exterior ø of Δ = sums of interior opposite ø . . . sum of the ø of Δ = 180º s Tˆ1 = 60º x = 106º – 44º = 62º s â  = 180º – 2x 1.1 8.1 . . . exterior ø of ΔDECs = sum of interior opposite ø s . . . co-interior ø are supplementary because AB || ED Solutions: Congruent Triangles CONGRUENT TRIANGLES (Symbol: ≡) 1.1 ΔDEF 6. In ΔKNQ and ΔMPQ (1) NQ = PQ . . . given (2) KQ = MQ . . . given (3) Nˆ 2 = Nˆ 1 = 90º ΔSTV . . . SøS 3.1 SøS â Answer: C â ΔMPN ≡ ΔMTN 1 M 7.1 But: BD = EC . . . given (3) BD is common . . . 90º, Hyp, S (RHS) Note: Observe the layout of a congruency proof. 5.1 . . . given (2) BD = CD . . . given N 7.2 5.2 â Aˆ 1 = Aˆ 2 . . . given . . . proved in Question 7.1 s ø opposite equal sides ˆ = ADC ˆ (2) AEB ... in isosceles ΔADE (3) AE = AD . . . given . . . SøS Study the proof carefully! ˆ i.e. DA bisects BAC . . . given â BC = EF In ΔABE and ΔACD (1) BE = CD BC = BF + FC But: BF = CE 9.2 â Answer: ΔACD In ΔABC and ΔDEF (1) AB = DE . . . given (2) AC = DF . . . given (3) BC = EF . . . proved in Question 9.1 â ΔABC ≡ ΔDEF 9.3 9.4 . . . SSS B̂ = Ê because they are corresponding angles of the congruent triangles in Question 9.2 B̂ and Ê are alternate angles & B̂ = Ê in Question 9.3 Order is important in congruency layout: . . . SSS ... 9.1 NB: Always give reasons! (3) AD is common â ΔABD h ΔACD Q & EF = CE + FC â ΔABE h ΔACD In ΔABD and ΔACD (1) AB = AC . . . RHS . . . corresponding sides of congruent triangles â PN = NT â BE = CD . . . given â ΔABD h ΔCDB 2 P BE = BD + DE & CD = EC + DE In ΔABD and ΔCDB (2) AD = CB . . . given that MN ⊥ PT 1 2 2. (1) Bˆ 2 = Dˆ 1 = 90º . . . radii of the circle (2) MN is common â ΔKNQ ≡ ΔMPQ . . . SøS . . . SøS We need to prove congruent triangles! In ΔMPN and ΔMTN (1) MP = MT K (3) Q̂ is common NB: The letters must be in the correct order so that equal sides and angles correspond. 4. 8.1 NB: Always give reasons! s *corresponding ø of s congruent Δ in Question 5.1 * Nothing to do with s corresponding ø on || lines • The letters must be in the same order in both triangles, corresponding to the equal sides and angles of the triangles; • In the facts (1), (2) and (3), the sides and angles of the first triangle must come first. A4 â AB || ED . . . converse fact See the notes on Congruency and Similarity on page A5 Copyright © The Answer Solutions: Congruent Triangles 10.1 In ΔABD and ΔACD (1) AB = AC . . . given (2) BD = CD . . . given ≡ means 'is congruent to' 11. . . . given P 1 P B C Q R  = P̂ , B̂ = Q̂ and Ĉ = R̂ and AB = PQ, AC = PR and BC = QR S Note the order of the lettering Two triangles are congruent if they have 1 1 R • 3 sides the same length . . . SSS • 2 sides & an included angle equal . . . SøS • a right angle, hypotenuse & a side equal . . . RHS s . . . corresponding ø of congruent triangles in Question 10.2 But Eˆ1 + Eˆ 2 = 180º i.e. ΔABC ≡ ΔPQR means that T . . . SøS Q 10.3 Eˆ1 = Eˆ 2 1 A have the same shape and size. All 3 angles and all 3 sides are equal. The sketch with all equal angles filled in. (There are no equal sides) (2) Aˆ 1 = Aˆ 2 . . . corresponding angles in congruent triangles in Question 10.1 (3) AE is common â ΔABE ≡ ΔACE Congruent Triangles . . . ||| means 'is similar to' (i.e. same SHAPE but not necessarily same SIZE . . . SSS 10.2 In ΔABE and ΔACE (1) AB = AC of triangles whereas: (3) AD is common â ΔABD ≡ ΔACD Congruency (≡) and Similarity (|||) (i.e. same SHAPE and SIZE) • 2 angles and a side equal Answer: D . . . øøS . . . angles on a straight line â Eˆ1 = Eˆ 2 = 90º Study the (easy) logic very carefully! 10.4 AE ⊥ BC [i.e. AE is perpendicular to BC ] Proof: In ΔPTS and ΔRTQ s Similar Triangles . . . s have the same shape, but not necessarily the same size. (1) Pˆ1 = Rˆ 1 . . . alternate ø ; PS æ QR (2) Sˆ 1 = Qˆ 1 . . . alternate ø ; PS æ QR ˆ = RTQ ˆ & (3) PTS . . . vertically opposite ø â ΔPTS ||| ΔRTQ s . . . øøø A All 3 angles are equal. i.e. ΔABC ||| ΔPQR means that  = P̂ , B̂ = Q̂ and Ĉ = R̂ B P C Q R The sides are not necessarily equal, but are proportional: AB AC BC = = PQ PR QR Note the order of the lettering Copyright © The Answer A5 Solutions: Similar Triangles 3.1 SIMILAR Δ (Symbol: |||) 'S In ΔPQR and ΔSTR (1) P̂ = Ŝ . . . alternate ø ; PQ || ST (2) Q̂ = T̂ . . . alternate ø ; PQ || ST s . . . vertically opposite ø â ΔPQR ||| ΔSTR 3.2 Multiply by 16 : . . . øøø â PN = ( ) PQ PR QR = = . . . proportional sides of similar triangles ST SR TR PQ 10 â = Choose the sides for which 3 6 â ( ) PN QP QN = = . . . proportional sides of MN LM LN similar triangles PN 3 = â Choose the sides for which 16 8 â the lengths have been given. s ˆ = SRT ˆ & (3) PRQ Be sure to read the notes on Congruency and Similarity on pg. A5 4.2 s 2 3 × 16 8 â PN = 6 cm the lengths have been given. 1.1 Multiply by 3: DF DE EF then = = KL LM KM 10 × 3 â PQ = 6 (1)  is common â PQ = 5 cm (2) ABˆ D = ACˆ E â 14 7 â x = = x 12 24 . . . proportional sides of similar triangles ( ) . . . 14 = 2 = ? 7 1 12 5.1 (3) 3 L: Note the ORDER of the letters in similar Δ : ΔDEF ||| ΔKLM ¥ D̂ = K̂ , Ê = L̂ and F̂ = M̂ and this determines the proportional sides 4.1 (1) N̂ is common s (2) Pˆ1 = M̂ . . . corresponding ø ; QP || LM & (3) Qˆ 1 = L̂ . . . corresponding ø ; QP || LM s If ΔABC ||| ΔEDF, â ED DF AB 15 = 6 10 ( EF ) Multiply by 6: â AB = . . . øøø . . . proportional sides of similar triangles Choose the sides for which you have the lengths! 8 cm Q 1 A Dˆ 1 = Eˆ1 ( ) â BD AD AB = = CE AE AC â BD 9 = 21 7 2 F 21 cm 1 2 D C s . . . sum of the ø of a Δ . . . øøø . . . proportional sides of similar triangles Choose the sides with known lengths. â BD = 9 × 21 7 3 â BD = 27 cm 3 cm 2 1 Multiply by 21 : M L 5.2 7 cm E 9 cm â ΔABD ||| ΔACE In ΔQPN and ΔLMN â ΔQPN ||| ΔLMN then AB = BC = AC In ΔABD and ΔACE rd s 2. B If ΔDEF ||| ΔKLM, i.e. ΔDEF is similar to ΔKLM, 2 1 P 16 cm 15 × 6 10 â AB = 9 cm N A6 Copyright © The Answer Solutions: Quadrilaterals 3.2 QUADRILATERALS 1. s â 3x + 30º = 180º Subtract 30º: Divide by 3: But Dˆ 2 = Tˆ1 â 3x = 150º â x = 50º â Dˆ 2 = 75º 3.3 â  = 50º + 50º = 100º â B̂ = 180º – 100º s . . . co-interior ø ; AC || BD Ĉ = 2(50º) – 20º = 80º s â B̂ = 80º . . . opposite ø of a || 2.1 = 12º 2.3 3.1 Ĝ = 156º the diagonals of a rhombus ... s bisect the ø of the rhombus s = AD Cˆ 1 = 15º s . . . sum of the ø of Δ DTC 1 2 1 4 T A 2 Copyright © The Answer 60° s Dˆ 1 = Bˆ 2 . . . alternate ø ; AD || BC in parallelogram Bˆ 1 = Dˆ 2 . . . alternate ø ; AB || DC in parallelogram BD = BD . . . common side s 1 C P 60° Note: We have just proved, using congruency, that both pairs of opposite sides of a parallelogram are equal in length. 60° R 60° 3 60° S 60° W 60° . . . øøS corresponding sides s . . . of congruent Δ & AB = DC 2 30° In ΔADB and ΔCBD: â ΔADB ≡ ΔCBD x & Similarly: BT = AB = BC 1 5. 60° . . . sides of square ø of isosceles Δ; sum s of the ø of a Δ = 180º ... = 15º Q . . . sides of equilateral Δ s . . . angles opposite equal sides in ΔQRS â AD = BC 4. D . . . ø of square = 1 (180º – 150º) 2 = 1 (30º) 2 s = 180º – 30º = 150º opposite ø of a rhombus m (or || ) are equal s . . . sides of equilateral ΔPQR . . . sides of square . . . sides of equilateral ΔRTS ˆ â x = RSQ ˆ = 90º . . . Dˆ 2 = 75º and ADC â Tˆ4 = 180º – 2(15º) ΔATD and ΔBTC . . . AT = AB QR = PR = RT = RS . . . half of 150º = 15º & Similarly: ... s Dˆ 1 = 90º – 75º s . . . ø of equilateral Δ of ø about ˆ = 360º – (60º + 90º + 60º) . . sum . â QRS a point = 360º = 150º . . . ø opposite equal sides in Δ ATD Similarly: Tˆ3 = 75º and Tˆ2 = 60º . . . ø of equilateral Δ ATB OR: s = 24º 2.2 are equal . . . co-interior ø ; DE || GF in rhombus ˆ = 180º – 156º EFG Fˆ2 = 1 (24º) 2 m ˆ = 90º & PRT sum of ø about â Tˆ4 = 360º – (75º + 60º + 75º) . . . a point = 360º = 360º – 210º = 150º = 80º OR: ˆ = TRS ˆ = 60º QRP . . . ø of equilateral Δ ˆ = 90º in square â Aˆ 1 = 30º . . . DAB s â Dˆ 2 + Tˆ1 = 180º – 30º . . . sum of the ø of a Δ = 150º . . . co-interior ø ; AB || CD (x + 50º) + (2x – 20º) = 180º Aˆ 2 = 60º 60° T B A7 6. D rectangle If one angle equals 90º, then, because of co-interior angles and parallel lines, so do the others equal 90º. â We only need 'at least one angle equal to 90º.' Solutions: Quadrilaterals 7.1 Something new to experience! ( Bˆ 1 + Bˆ 2 ) + ( Cˆ 1 + Cˆ 2 ) = 180º A T 2 s co-interior ø ; . . . AB || DC in parallelogram 3 P D Let Bˆ 1 = Bˆ 2 = x and Cˆ 1 = Cˆ 2 = y . . . ... 1 given that the angles were bisected x 2 1 x B 8. Answer: C C A very useful technique in geometry! 1) AE = BE . . . given 2) EB = EC . . . given 3) ˆ = DEC ˆ AEB 9. Divide by 2: â x + y = 90º ab 7.2 g = 540º . . . angles on a straight line â Each ø = In ΔBCT and ΔTCB: 540° = 108º 5 â Answer: D 1) Cˆ 1 = Cˆ 2 (= y) 2) ˆ (= 90º) Tˆ2 = TPC 3) â Bˆ 1 = Tˆ1 rd . . . 3 ø of Δ â ΔBCT ||| ΔTCP . . . øøø BT BC ⎛ CT ⎞ = ⎜= ⎟ TP TC ⎝ CP ⎠ BT 2TC = â 4 TC ... Multiply by 4: â BT = 2 % 4 = 8 cm d (a + i + h) + (b + g + f) + (c + d + e) = 3 % 180º ΔTCP â . . SøS e f h s . . . sum of the ø of a Δ ˆ = 180º – 90º = 90º TPC 7.3 c i Bˆ 1 + Cˆ 1 = 90º â Tˆ2 = 90º s . . . vertically opposite ø â ΔAEB ≡ ΔDEC i.e. 2(x + y) = 180º â In ΔBTC: NOTES In ΔAEB and ΔDEC: y y2 1 â 2x + 2y = 180º ΔAEB ≡ ΔDEC proportional sides of similar triangles Choose the sides whose lengths you have been given. 10. s The sum of the ø of the hexagon = 4 % 180º = 720º â Each ø = 720° = 120º 6 â Answer: B A8 Copyright © The Answer Solutions: Theorem of Pythagoras THEOREM OF PYTHAGORAS 1. AC = 13 cm . . . 5 : 12 : 13 Pythagoras 'trip' 2 OR: In ΔABC: 2 2 AC = AB + BC 2 2 = 5 + 12 = 25 + 144 = 169 ... 3.1 * Thm of Pythag.; B̂ = 90º So: . . . OR: TU × 12 = 30 2 â 6 % TU = 30 6 Divide by 6: 3.2 TW = 13 cm â TU = 5 cm . . . Pythag 'trip' 5 : 12 : 13 â The perimeter of ΔTUW = 5 cm + 12 cm + 13 cm = 30 cm 2 The length of the ladder = 13 m OR: (length of the ladder) 2 2 = 5 + 12 , etc. 2 ; 5 : 12 : 13 ; 8 : 15 : 17 12 m Answer: B 6 cm In ΔADC: DC = 8 cm â AD = 6 cm and even multiples: . . . opposite sides of rectangle . . . Pythag 'trip': 3 : 4 : 5 = 6 : 8 : 10 6 : 8 : 10 2.1 In ΔABC: x = 10 cm ... Pythag 'trip': 3 : 4 : 5 = 6 : 8 : 10 2.2 In ΔABD: BD = 15 cm . . . Pythag 'trip': 8 : 15 : 17 â y = 15 cm – 6 cm = 9 cm 2 2 2 OR: 2.1 x = 8 + 6 , etc. 2 2 5 : 12 : 13 5m 5. the TRIP(LET)S : . . . Pythag 'trip': 2 3 +4 = 9 + 16 = 25 = 5 2 2 2 5 + 1 2 = 25 + 144 = 169 = 1 3 2 2 2 8 + 1 5 = 64 + 225 = 289 = 1 7 3:4:5 h×b = area of a Δ 2 â TU = 5 cm When applying the Theorem of Pythagoras, there are some well-known 'trips' which are useful to know and use instead of long calculations. 2 Multiply by 2: Divide by 12: 4.1 e.g. ... â TU % 12 = 60 â AC = 13 cm * Note: TU × 12 = 30 2 2 2.2 BD = 17 – 8 , etc. . . . Theorem of Pythag 6. This sum requires us to apply the converse of the Theorem of Pythagoras, i.e. If the square on one side of a triangle equals the sum of the squares on the other two sides, then the angle opposite the first side is a right angle. 2 2 AC = 15 = 225 2 2 2 2 & AB + BC = 9 + 12 = 81 + 144 = 225 2 2 2 â AC = AB + BC â B̂ = 90° . . . the converse of the Theorem of Pythagoras Copyright © The Answer A9 NOTES Solutions: Measurement: 2D 3.2 MEASUREMENT: 2D 1.1 = 6 000 m 2 5.1 area of 2 semi-circles 2 + + .900 m = (100 % 60)m 1 % diameter = 6 cm Radius, r = 2 2 The area of the field = area of rectangle + 2 2 .30 m . . . Area of circle = r 2 2 2 . . . Theorem of Pythagoras = 16 â AT = 4 cm 2 2 2 2 In ΔAPS: PS = 5 – 2 . . . Theorem of Pythagoras = 25 – 4 = 21 The circumference of the (full) circle = 2r â The 'circumference' of the semi-circle = r = 3,14 % 6 = 18,84 cm â PS = 21 l 4,58 m â The perimeter of the shape ACB = 18,84 cm + 12 cm = 30,84 cm . . . diameter, AB = 12 cm 2 2 . . . Pythag 'trip': 3 : 4 : 5 = 25 – 9 2 l 8 827,43 m 4.1 2. AT = 4 cm OR: AT = 5 – 3 2 â Area = r = 3,14 % 6 2 = 113,04 cm 1.2 In ΔABT: 2 The area of a square = s = (0,12) = 0,0144 cm 2 â Answer: D Observe carefully: 2 12 ⎞ 12 12 144 2 (0,12) = ⎛⎜ = 0,0144 × = ⎟ = 100 100 10 000 ⎝ 100 ⎠ 5.2 The area of parallelogram ABCT = base % height = BC % AT = AD % 4 cm See the shifting of ΔABT = 12 cm % 4 cm as shown: 2 = 48 cm 4.2 PT = 3 % AB = 3 % 4 m = 12 m 4.3 A kite 4.4 Method 1: Using the formula 1 The area = the product of the diagonals 2 1 = (PT % AB) 2 1 = (12 % 4) 2 2 = 24 m A 12 cm D . . . 2 pairs of adjacent sides equal 5 cm B 3 cm T C S Parallelogram ABCD = rectangle ATSD & Area of rectangle ATSD = length % breadth = 12 cm % 4 cm 2 = 48 cm Method 2: Without the formula The area = ΔPAT + ΔPBT 3.1 s The perimeter of the field = 2 % 100 m + 2 % semi-circles = 200 m + 2 % 3,14 % 30 m = 388,4 m . . . Circumference of circle = 2r = 2 % ΔPAT . . . because the 2 Δ are congruent ⎛1 ⎞ = 2 × ⎜ PT . AS⎟ 2 ⎝ ⎠ = 12 m % 2 m 2 = 24 m 4 000 m = 10,298 . . . 388,4 m . . . for at least 4 km! â Number of laps = â 11 laps A10 5.3.1 DC = AB = 5 cm s m . . . opposite ø of || TC = BC – BT = 12 cm – 3 cm = 9 cm â The perimeter of trapezium ADCT = AD + DC + TC + AT = 12 cm + 5 cm + 9 cm + 4 cm = 30 cm Copyright © The Answer Solutions: Measurement: 2D 7. 5.3.2 Method 1: Using the formula The circumference of a circle, 2r = 52 cm 52 26 = 2π π â r = The area of trapezium ADCT 1 (sum of the || sides) % the distance between them = 2 1 = (AD + TC) % AT 2 1 = (12 + 9) % 4 2 2 = 42 cm â The area of the circle 2 26 2 = r = % ⎛⎜ ⎞⎟ ⎝ π⎠ = % = 262 π 262 π2 2 l 215,18 cm Method 2: Without the formula . . . correct to 2 decimal places The area of trapezium ADCT = Area of ΔATC + Area of ΔADC 1 1 = (9 % 4) + (12 % 4) * 2 2 = 18 + 24 * 1 . ( 9 + 12) . 4, like the formula above! 2 2 = 42 cm A 8.1 The circumference of a circle = 2r â The circumference of the smaller circle = 2 % % 20 l 125,66 cm 8.2 The area of the shaded region D 12 cm = the area of the full circle – the area of the inner circle 2 2 = 30 – 20 2 l 1 570,80 cm 4 cm T 9 cm C 9.1 6. = the area of the full circle – the area of the inner ring 2 2 = R – r 2 2 = (R – r ) Let the area of the original rectangle = ℓ % b If the length is doubled, then the area of the enlarged rectangle = 2ℓ % b = 2(ℓb) â k = 2 b ℓ Copyright © The Answer The area of the shaded ring 9.2 2 2 Area of the shaded ring = (14 – 8 ) 2 = 132 cm Note: Give the answer 'in terms of ' A11 NOTES