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P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> C H A P T E R 12 MODULE 2 PL What are the properties of parallel lines? E Geometry What are the basic properties of triangles? What are the basic properties of regular polygons? How do we use and apply similarity and Pythagoras’ theorem in two dimensions? How do we explore ratios of areas of similar figures? How do we explore ratios of volumes of similar solids? 12.1 Properties of parallel lines – a review SA M Angles 4 and 6 are called alternate angles. l3 Angles 5 and 3 are called alternate angles. Angles 2 and 6 are called corresponding angles. 1 2 l1 4 3 Angles 1 and 5 are called corresponding angles. l2 Angles 4 and 8 are called corresponding angles. 56 Angles 3 and 7 are called corresponding angles. 8 7 Angles 3 and 6 are called cointerior angles. Lines l1 and l2 are cut by a transversal l3 . Angles 4 and 5 are called cointerior angles. Angles 1 and 3 are called vertically opposite angles and are of equal magnitude. Other pairs of vertically opposite angles are 2 and 4, 5 and 7, and 6 and 8. Angles 1 and 2 are supplementary, i.e. their magnitudes add to 180◦ . l3 When lines l1 and l2 are parallel, corresponding angles are of equal magnitude, alternate angles are of equal 1 2 magnitude and cointerior angles are supplementary. 4 3 l1 Converse results also hold: If corresponding angles are equal then 5 6 l1 is parallel to l2 . 8 7 l2 If alternate angles are equal then l1 is parallel to l2 . If cointerior angles are supplementary then l1 is parallel to l2 . Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson 360 and McMenamin TI-Nspire & Casio ClassPad material in collaboration with Brown P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 361 Chapter 12 — Geometry Example 1 Parallel line properties Find the values of the pronumerals. Solution (corresponding) (alternate with a) (cointerior with d) (corresponding with b) (vertically opposite e) e° 65° d° c° b° a° E a = 65 d = 65 b = 115 e = 115 c = 115 There are lots of ways of finding these values. One sequence of reasoning has been used here. Parallel line properties PL Example 2 Find the values of the pronumerals. Solution (2x – 50)° 2x − 50 = x + 10 (alternate angles) ∴ 2x − x = 50 + 10 ∴ x = 60 Parallel line properties M Example 3 (x + 10)° Find the values of the pronumerals. Solution SA x + 100 = 2x + 80 (alternate angles) ∴ 100 − 80 = 2x − x ∴ x = 20 Also x + 100 + y + 60 = 180 (cointerior) and x = 20 ∴ y + 180 = 180 or y = 0 (x + 100)° (2x + 80)° (y + 60)° Exercise 12A Questions 1 to 5 apply to the following diagram l1 d a b c l2 h e g f l3 1 Angles d and b are: A alternate D supplementary B cointerior E vertically opposite C corresponding Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 362 Essential Further Mathematics – Module 2 Geometry and trigonometry 2 Angles d and a are: A alternate D supplementary B cointerior E vertically opposite C corresponding B cointerior E vertically opposite C corresponding 3 Angles c and h are: A alternate D supplementary A alternate D supplementary E 4 Angles b and f are: B cointerior E vertically opposite C corresponding B cointerior E vertically opposite C corresponding 5 Angles c and e are: PL A alternate D supplementary 6 Find the values of the pronumerals in each of the following: a x° z° z° 40° z° x° y° 80° (2x – 40)° 120° y° x° (x + 40)° SA y° 50° (2z)° f e d c x° y° M y° 70° b 12.2 Properties of triangles–a review a ◦ , b◦ and c◦ are the magnitudes of the interior angles of the triangle ABC. B d ◦ is the magnitude of an exterior angle at C. The sum of the magnitudes of the interior angles of b° a triangle is equal to 180◦: a ◦ + b◦ + c◦ = 180◦ . b◦ + a ◦ = d ◦ . The magnitude of an exterior angle is a° equal to the sum of the magnitudes of the two A opposite interior angles. B A triangle is said to be equilateral if all its sides are of the same length: AB = BC = CA. The angles of an equilateral triangle are all of magnitude 60◦ . c° d° C 60° 60° A 60° C Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 363 Chapter 12 — Geometry The bisector of each of the angles of an equilateral triangle meets the opposite side at right angles and passes through the midpoint of that side. B O C A A triangle is said to be isosceles if it has two sides of equal length. If a triangle is isosceles, the angles opposite each of the equal sides are equal. E B A The sum of the magnitudes of the exterior angles of a triangle is equal to 360◦: e◦ + d ◦ + f ◦ = 360◦ A triangle is said to be a right-angled triangle if it has one angle of magnitude 90◦ . B e° f° PL Example 4 C C A d° Angle sum of a triangle Find the values of the pronumerals. Solution M 20◦ + 22◦ + x◦ = 180◦ (sum angles = 180◦ ) ∴ 42◦ + x◦ = 180◦ or x = 138 138◦ + y ◦ = 180◦ (sum angles = 180◦ ) A ∴ y = 42 B 22° x° y° C 20° SA Or, to find x: Two of the angles sum to 42◦ and therefore the third angle is 138◦ . To find y ◦ . The two angles sum to 180◦ . Therefore the second is 42◦ Example 5 Angle sum of an isosceles triangle Find the values of the pronumerals. Solution 100◦ + 2x ◦ = 80◦ (sum angles = 180◦ ) ∴ 2x ◦ = 80◦ or x = 40 Or, observe the two unknown numbers are the same and must sum to 80◦ , therefore each of them has size 40◦ . B 100° A x° x° Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin C P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 364 Essential Further Mathematics – Module 2 Geometry and trigonometry Exercise 12B 1 Find the values of the pronumerals in each of the following: a b B 80° 50° x° A c Q y° x° 30° A C d B 70° y° Y 50° 30° C y° X P e g B x° D PL y° C y° b° c° A w° x° a° 75° 40° C y° E A Properties of regular polygons—a review M 12.3 Z R f B z° 40° E x° x° x° Equilateral triangle Square Regular pentagon Regular hexagon Regular octagon SA A regular polygon has all sides of equal length and all angles of equal magnitude. A polygon with n sides can be divided into n triangles. The first three polygons below are regular polygons. O O O The angle sum of the interior angles of an n-sided convex polygon is given by the formula: S = [180(n − 2)]◦ = (180n − 360)◦ The result holds for any convex polygon. Convex means that a line you draw from any vertex to another vertex lies inside the polygon. The magnitude of each of the interior angles of an n-sided polygon is given by: x= (180n − 360)◦ n Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 365 Chapter 12 — Geometry The angle bisectors of a regular polygon meet at a point O. For a regular polygon, a circle can be drawn with centre O on which all the vertices lie. O O E O Example 6 PL The sum of the angles at O of a regular polygon is 360◦ . The sum of the exterior angles of a regular polygon is 360◦ . Angle properties of an octagon The diagram opposite shows a regular octagon. a Show that x = 45. b Find the size of angle y. Solution SA M a 8x ◦ = 360◦ (sum angles at O = 360◦ ) 360◦ ∴ x◦ = = 45◦ 8 ∴ x = 45 b y ◦ + y ◦ + 45◦ = 180◦ (OBC isosceles) ∴ 2y ◦ = 135 or y = 67.5 Example 7 A H O G B x° y° C F E D Angle sum of an octagon Find the sum of the interior angles of an 8-sided convex polygon (octagon). Solution 1 Use the rule for the sum of the interior angles of an n-sided polygon: S = (180n − 360)◦ 2 In this example, n = 8. Substitute and evaluate. 3 Write down your answer. S = (180n − 360)◦ n =8 ∴ S = (180 × 8 − 360)◦ = 1080◦ The sum of the interior angles is 1080◦ . Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 366 Essential Further Mathematics – Module 2 Geometry and trigonometry Exercise 12C 1 Name each of the following regular polygons. a e d c b B C E 2 ABCD is a square. BD and AC are diagonals which meet at O. a Find the size of each of the angles at O. b What type of triangle is triangle: i BAD ii AOB D A PL 3 ABCDE is a regular pentagon. O a Find the value of: i x ii y b Find the sum of the interior angles of the regular pentagon ABCDE. A B y° x° O C D E 4 ABCDEF is a regular hexagon. Find the value of: b y E M a x 5 State the sum of the interior angles of: a a 7-sided regular polygon b a hexagon F A y° x° O B D C c an octagon SA 6 The angle sum of a regular polygon is 1260◦ . How many sides does the polygon have? 7 A circle is circumscribed about a hexagon ABCDEF. a Find the area of the circle if OA = 2 cm. b Find the area of the shaded region. 8 The diagram shows a tessellation of regular hexagons and equilateral triangles. State the values of a and b and use these to explain the existence of the tessellation. A B F O C E D b° a° 9 If the magnitude of each angle of a regular polygon is 135◦ , how many sides does the polygon have? Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 367 Chapter 12 — Geometry Pythagoras’ theorem B Pythagoras’ theorem Pythagoras’ theorem states that for a right-angled triangle ABC with side lengths a, b and c, as shown in the diagram, a 2 + b2 = c2 . c A a C b E Pythagoras’ theorem can be illustrated by the diagram shown here. The sum of the areas of the two smaller squares is equal to the area of the square on the longest side (hypotenuse). area = c2 cm2 area = b2 cm2 m b cm PL cc a cm area = a2 cm2 There are many different proofs of Pythagoras’ theorem. A proof due to the 20th President of the United States, James A. Garfield, is produced using the following diagram. Y Area of trapezium WXYZ = 12 (a + b)(a + b) Z Area of EYZ + area of EWX + area of EWZ c M b a a X c b = 12 ab + 12 c2 + 12 ab = ab + 12 c2 a2 b2 1 ∴ + ab + = ab + c2 2 2 2 2 2 2 ∴ a +b =c E W SA 12.4 Pythagorean triads A triple of natural numbers (a, b, c) is called a Pythagorean triad if c2 = a 2 + b2 . The table presents the first six such ‘primitive’ triples. The adjective ‘primitive’ indicates that the highest common factor of the three numbers is 1. Example 8 a b c 3 4 5 5 12 13 7 24 25 8 15 17 9 40 41 Pythagoras’ theorem Find the value, correct to two decimal places, of the unknown length for the triangle opposite. x cm 6.1 cm cm Evans, Lipson Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 ©5.3 Jones, TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin 11 60 61 P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 368 Essential Further Mathematics – Module 2 Geometry and trigonometry Solution Example 9 The length is 8.08 cm correct to two decimal places. PL 2 Write down your answer for the length correct to two decimal places. 2 2 x2 = 5.3 + 6.1 (Pythagoras) ∴ x = 5.32 + 6.12 = 8.080 . . . E 1 Using Pythagoras’ theorem, write down an expression for x in terms of the two other sides of the right-angled triangle. Solve for x. Pythagoras’ theorem Find the value, correct to two decimal places, of the unknown length for the triangle opposite. Solution 5.6 cm y cm M 1 Using Pythagoras’ theorem, write down an expression for y in terms of the two other sides of the right-angled triangle. Solve for y. 2 Write down your answer for the length correct to two decimal places. 5.62 + y 2 = 8.62 ∴ y 2 =8.62 − 5.62 ∴y = (Pythagoras) 8.62 − 5.62 = 6.526 . . . The length is 6.53 cm correct to two decimal places. Pythagoras’ theorem SA Example 10 8.6 cm The diagonal of a soccer ground is 130 m and the long side of the ground measures 100 m. Find the length of the short side correct to the nearest cm. Solution 1 Draw a diagram. Let x be the length of the short side. 2 Using Pythagoras’ theorem, write down an expression for x in terms of the two other sides of the right-angled triangle. Solve for x. 4 Write down your answer correct to the nearest cm. 130 m xm 100 m Let x m be the length of the shorter side. x 2 + 1002 = 1302 (Pythagoras) ∴ x 2 =1302 − 1002 = 6900 ∴x = 1302 − 1002 = 83.066 . . . Correct to the nearest centimetre, the length of the short side is 83.07 m. Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 369 Chapter 12 — Geometry Exercise 12D 1 Find the length of the ‘unknown’ side for each of the following: b a 6 cm c 10 cm 5 cm 3 cm 10 cm 11 cm C 9 cm 7 cm 33 cm B 15 cm 12 cm PL 44 cm f E e d A 2 In each of the following find the value of x correct to two decimal places. b a 4.8 cm 2.8 cm d c x cm x cm 6.2 cm 3.2 cm 9.8 cm x cm x cm 3.5 cm M 5.2 cm 4 cm 3 cm 3 Find the value of x for each of the following (x > 0). Give your answers correct to 2 decimal places. a x 2 = 62 + 42 b 52 + x 2 = 92 c 4.62 + 6.12 = x 2 4 In triangle VWX, there is a right angle at X.VX = 2.4 cm and XW = 4.6 cm. Find VW. A SA 5 Find AD, the height of the triangle. 32 cm 32 cm C D 20 cm B 6 An 18 m ladder is 7 m away from the bottom of a vertical wall. How far up the wall does it reach? 7 Find the length of the diagonal of a rectangle with dimensions 40 m × 9 m. 8 Triangle ABC is isosceles. Find the length of CB. A 8 14 C B Lipson Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 370 Essential Further Mathematics – Module 2 Geometry and trigonometry 9 In a circle of centre O, a chord AB is of length 4 cm. The radius of the circle is 14 cm. Find the distance of the chord from O. O x cm B A 10 Find the value of x. 18 cm E x cm x cm X PL 11 How high is the kite above the ground? 0m 17 Y 90 m 12 A square has an area of 169 cm2 . What is the length of the diagonal? M 13 Find the area of a square with a diagonal of length: √ a 8 2 cm b 8 cm 20 cm 14 Find the length of AB. SA 15 ABCD is a square of side length 2 cm. If CA = CE, find the length of DE. C D 12 cm 8 cm A B B A C D E 16 The midpoints of a square of side length 2 cm are joined to form a new square. Find the area of the new square. Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 371 Chapter 12 — Geometry Similar figures Similarity In this section we informally define two objects to be similar if they have the same shape but not the same size. Examples Any two circles are similar to each other. 3 cm 4 cm E C2 Any two squares are similar to each other. 3 cm S1 4 cm S2 PL C1 It is not true that any rectangle is similar to any other rectangle. For example, rectangle 1 is not similar to rectangle 2. 4 cm R1 8 cm R2 1 cm 1 cm A rectangle similar to R1 is: 8 cm 2 cm M R3 So, for to be similar, their corresponding sides must be in the same two rectangles 8 4 ratio = . 2 1 Similar triangles Two triangles are similar if one of the following conditions holds: corresponding angles in the triangles are equal SA 12.5 A 45° B' B 100° 100° 35° C A' 45° 35° C' corresponding sides are in the same ratio AB BC AC = = =k AB BC AC k is the scale factor. Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 372 Essential Further Mathematics – Module 2 Geometry and trigonometry two pairs of corresponding sides have the same ratio and the included angles are equal B B' A 45° C A' 45° C' Example 11 Similar triangles PL Find the value of length of side AC in ABC correct to two decimal places. E AB AC = AB AC If triangle ABC is similar to triangle XYZ this can be written symbolically as ABC ∼ XYZ. The triangles are named so that angles of equal magnitude hold the same position, i.e. A corresponds to X, B corresponds to Y, C corresponds to Z. 5 cm 20° A Solution SA M 1 Triangle ABC is similar to triangle AB C : two pairs of corresponding sides have the same 6.25 5 = and included angles (20◦ ). ratio 3 3.75 2 For similar triangles, the ratios of corresponding AC AB sides are equal, for example, = . AC AB Use this fact to write down an expression involving x. Solve for x. 3 Write down your answer correct to two decimal places. Example 12 B' B x cm 6.25 cm 20° 3.75 cm 3 cm C A' 3.013 cm C' Triangles similar ∴ 5 x = 3.013 6.25 5 ∴x = × 3.013 6.25 = 2.4104 The length of side AC is 2.41 cm, correct to two decimal places. Similar triangles Find the value of length of side AB in ABC. m 6c B m xc A Solution 1 Triangle ABC is similar to triangle AXY (corresponding angles are equal). X 3 cm C Y 2.5 cm Triangles similar Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 373 Chapter 12 — Geometry 3 x = x +6 5.5 ∴ 5.5x = 3(x + 6) = 3x + 18 2 For similar triangles, the ratios of corresponding AB AC sides are equal (for example, = ). AX AY Use this fact to write down an expression involving x. Solve for x. Note that, if AB = x then AX = x + 6. ∴ 2.5x = 18 or x = 3 Write down your answer. The length of side AB is 7.2 cm. ∴ E 18 = 7.2 2.5 Exercise 12E 1 Find the value of x for each of the following pairs of similar triangles. a A PL A' 4 cm 82° 5 cm 56° B C b 9 cm B' 135° 18° 18° C X d x cm 13 cm Y Z x cm 5 cm Y B M c C' 135° A 10 cm 56° C 6 cm B x cm 82° D 10 cm ° C 14 cm × 13 cm B X 8 cm 12 cm C e SA ° A 10 cm f A 6 cm ° B ° × E P 6 cm B R g C x cm 12 cm x cm 8 cm Q x cm h ° A A A 12 cm P x cm B E 2 cm ° D 4 cm i B 16 cm Q j A x cm 8 cm C 2 cm x cm P Q 1.5 P D 2 cm A 3 cm E x cm A C 2 cm 6 cm B cm B 8 cm C 10 cm C Q 2 cm Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 374 Essential Further Mathematics – Module 2 Geometry and trigonometry 2 Given that AD = 14, ED = 12, BC = 15 and EB = 4, find AC, AE and AB. D C 15 14 12 A 4 B E ° 33 m ° 0.24 m E 3 A tree casts a shadow of 33 m and at the same time a stick 30 cm long casts a shadow 24 cm long. How high is the tree? 0.3 m 4 A 20 metre high neon sign is supported by a 40 m steel cable as shown. An ant crawls along the cable starting at A. How high is the ant when it is 15 m from A? m PL 40 20 m A 5 A hill has a gradient of 1 in 20, i.e. for every 20 m horizontally there is a 1 m increase in height. If you go 300 m horizontally, how high up will you be? M 6 A man stands at A and looks at point Y across the river. He gets a friend to place a stone at X so that A, X and Y are collinear. He then measures AB, BX and XC to be 15 m, 30 m and 45 m respectively. Find CY, the distance across the river. 45 m C A 15 m X 30 m B river Y 7 Find the height, h m, of a tree that casts a shadow 32 m long at the same time that a vertical straight stick 2 m long casts a shadow 6.2 m long. 8 A plank is placed straight up stairs that are 20 cm wide and 12 cm deep. Find x, where x cm is the width of the widest rectangular box of height 8 cm that can be placed on a stair under the plank. SA nk pla 8 cm x cm 12 cm 20 cm 9 The sloping edge of a technical drawing table is 1 m from front to back. Calculate the height above the ground of a point A, which is 30 cm from the front edge. 1m 30 cm A 92 cm 80 cm Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 375 Chapter 12 — Geometry 10 Two similar rods 1.3 m long have to be hinged together to support a table 1.5 m wide. The rods have been fixed to the floor 0.8 m apart. Find the position of the hinge by finding the value of x. 1.5 m xm (1.3 – x) m E 0.8 m 11 A man whose eye is 1.7 m from the ground when standing 3.5 m in front of a wall 3 m high can just see the top of a tower that is 100 m away from the wall. Find the height of the tower. PL 12 A man is 8 m up a 10 m ladder, the top of which leans against a vertical wall and touches it at a height of 9 m above the ground. Find the height of the man above the ground. 13 A spotlight is at a height of 0.6 m above ground level. A vertical post 1.1 m high stands 3 m away, and 5 m further away there is a vertical wall. How high up the wall does the shadow reach? wall Volumes and surface areas M Volume of a prism A prism is a solid which has a constant cross-section. Examples are cubes, cylinders, rectangular prisms and triangular prisms SA 12.6 spotlight 0.6 m 1.1 m vertical post The volume of a prism can be found by using its cross-sectional area. volume = area of cross-section × height (or length) V = A×h Answers will be in cubic units, i.e. mm3 , cm3 , m3 etc. Example 13 Volume of a cylinder Find the volume of this cylinder which has radius 3 cm and height 4 cm correct to two decimal places. 3 cm 4 cm Solution 1 Find the cross-sectional area of the prism. Area of cross-section = πr 2 = π × 32 = 28.27 cm 2 Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 376 Essential Further Mathematics – Module 2 Geometry and trigonometry 2 Multiply by the height 3 Make sure that accuracy is given to the correct number of decimal places Volume = 28.27 × 4 = 113.10 cm 3 (correct to two decimal places) The formulas for determining the volumes of some ‘standard’ prisms are given here. Solid Cylinder (radius r cm, height h cm) Formula V = r 2 h E r cm h cm Cube (all edges x cm) PL V = x3 Rectangular prism (length l cm, width w cm, height h cm) V = lwh h cm w cm l cm Triangular prism h cm The triangular prism shown has a right-angled triangle base but the following formula holds for all triangular prisms l cm M b cm V = 12 bhl Volume of a pyramid The formula for finding the volume of a right pyramid can be stated as: SA Volume of pyramid = × base area × perpendicular height 1 3 h For the square pyramid shown: V = 13 x 2 h x The term right in this context means that the apex of the pyramid is directly over the centre of the base. Example 14 Volume of a pyramid Find the volume of this hexagonal pyramid with a base area of 40 cm2 and a height of 20 cm. Give the answer correct to one decimal place. Solution 20 cm V = 1 3 × A ×h = 1 3 × 40 × 20 3 Cambridge University Uncorrected • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson = Press 266.7• cm (correctSample to onepages decimal place) TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 377 Chapter 12 — Geometry Example 15 Volume of a pyramid Find the volume of this square pyramid with a square base with each edge 10 cm and a height of 27 cm. 27 cm Solution 1 V = 3x2 h 1 3 × 10 × 10 × 27 = 900 cm3 Volume of a cone 10 cm 10 cm E = Volume of cone = V = PL The formula for finding the volume of a cone can be stated as: 1 × base 3 1 r 2 h 3 area × height Volume of a sphere The formula for the volume of a sphere is: V = 43 r 3 r M where r is the radius of the sphere. Example 16 Volume of a sphere Find the volume of this sphere. Solution SA 4 Volume of sphere = πr 3 3 4 = × π × 43 3 4 cm = 268.08 cm3 (2 decimal places) Composite shapes Using the shapes above, new shapes can be made. The volumes of these can be found by summing the volumes of the component solids. Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 378 Essential Further Mathematics – Module 2 Geometry and trigonometry Example 17 Volume of a composite shape Solution 1 Use the formula V = r 2 h to find the volume of the cylinder The volume of the cylinder = π × 52 × 5 = 125π mm3 The volume of the hemisphere 2 250 = π × 53 = π mm3 3 3 PL 2 Find the volume of hemisphere using the formula that the volume of a hemisphere = 12 43 r 3 = 23 r 3 . E A hemisphere is placed on top of a cylinder to form a capsule. The radius of both the hemisphere and the cylinder is 5 mm. The height of the cylinder is also 5 mm. What is the volume of the composite solid in cubic millimeters, correct to two decimal places? Therefore the volume of the composite = 250 625 π 125π + π= = 654.498 . . . 3 3 The volume of the composite = 654.50 mm3 (correct to two decimal places) 3 Add the two together. 4 Write down your answer. M Surface area of three-dimensional shapes The surface area of a solid can be found by calculating and totalling the area of each of its surfaces. The net of the cylinder in the diagram demonstrates how this can be done. l l SA r Total area = 2r + 2rl = 2r (r + l) 2 r A = πr2 r A = 2πrl A = πr2 2πr Here are some more formulas for the surface areas of some solids. The derivation of these is left to the reader. Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 379 Chapter 12 — Geometry Solid Cylinder (radius r cm, height h cm) Formula r cm S = 2r 2 + 2r h h cm Cube (all edges x cm) E S = 6x 2 Rectangular prism (length l cm, width w cm, height h cm) S = 2(lw + lh + w h) h cm l cm PL w cm Triangular prism √ S = bh + bl + hl + l b2 + h 2 h cm b cm Surface area of a right square pyramid M Example 18 l cm Find the surface of the right square pyramid shown if the square base has each edge 10 cm in length and the isosceles triangles each have height 15 cm. SA Solution 1 Draw the net of the pyramid. 2 First determine the area of the square. 3 Determine the area of one of the isosceles triangles. 4 Find the sum of the areas of the four triangles and add to the area of the square. Area of the square = 102 = 100 cm2 The area of one triangle 1 = × 10 × 15 = 75 cm2 2 The surface area of the solid = 100 + 4 × 75 = 100 + 300 = 400 cm2 Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 380 Essential Further Mathematics – Module 2 Geometry and trigonometry Exercise 12F 1 Find the volume in cm3 of each of the following shapes, correct to two decimal places. b radius 6.3 cm and height 2.1 cm c dimensions 2.1 cm, 8.3 cm and 12.2 cm d PL E a area of cross section = 2.8 cm2 height = 6.2 cm radius 2.3 cm and length 4.8 cm 2 Find, correct to two decimal places, the surface area and volume of the solid shown given that the cross section is a right angled isosceles triangle. 4 cm 4 cm 12 cm M 3 The box shown has dimensions length: 13 cm, width: 4 cm and height 3 cm. a Find the surface area of the box in square centimetres (cm2 ). b Find the volume of the box in cubic centimetres (cm3 ). 4 Find the volumes, to two decimal places, of spheres with: a radius = 4 mm b diameter = 23 cm c radius = 3.8 m d diameter = 15 cm SA 5 Find the volume, to two decimal places, of hemispheres with: a radius = 12 cm b diameter = 32 mm c radius = 16 mm d radius = 15 cm 6 Calculate the volume of a right pyramid with a rectangular base 18 m by 15 m. The vertex of the pyramid is 20 m perpendicularly above the centre of the base. 7 Each side of the square base of one of the great Egyptian pyramids is 275 m long. Calculate the volume of the pyramid, to the nearest cubic metre, if it has a perpendicular height of 175 m. V 8 Find the surface area and volume of the right square pyramid shown. 12 cm B The length of each edge of the square base is 10 cm and the height of the pyramid is 12 cm. A C O 10 cm X D Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 381 Chapter 12 — Geometry 9 The diagram shows a capsule, which consists of two hemispheres, each of radius 2 cm, and a cylinder length 5 cm and radius 2 cm. The surface area of a sphere is given by the formula S = 4r 2 and the surface area of the curved section of a cylinder is given by the formula S = 2r h. Find the surface area and volume of the capsule. Give your answers correct to two decimal places. 5 cm E 10 Find: a the surface area b the volume of the object shown. 2 cm 3m 4m 2m 10 m PL 5m 11 The diagram opposite shows a right pyramid on a cube. Each edge of the cube is 14 cm. The height of the pyramid is 24 cm. Find: a the volume of the solid b the surface area of the solid M 24 cm 14 cm SA 12 Find: a the surface b the volume of the solid shown opposite. 7 cm 4 cm 10 cm 5 cm 20 cm 4 cm 13 The solid opposite consists of a half cylinder on a rectangular prism. Find, correct to two decimal places: a the surface area b the volume 10 cm Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 382 12.7 Essential Further Mathematics – Module 2 Geometry and trigonometry Areas, volumes and similarity Areas Some examples of similar shapes and the ratio of their areas are considered in the following. Similar circles 4 radius circle 2 = radius circle 1 3 × 42 42 4 2 Ratio of areas = = 2 = = k2 × 32 3 3 Scale factor = k = Area = × 32 3 cm Area = 3 × 2 = 6 cm2 length rectangle 2 6 = =2 length rectangle 1 3 24 Ratio of areas = = 4 = (2)2 = k 2 6 5 cm 3 cm 4 cm Area = × 4 × 3 = 6 cm2 SA 1 2 4 cm Scale factor = k = Scale factor = k = M Similar triangles Area = × 42 PL Similar rectangles 2 cm 4 cm E 3 cm Ratio of areas = height triangle 2 9 = =3 height triangle 1 3 54 = 9 = (3)2 = k 2 6 6 cm Area = 6 × 4 = 24 cm2 15 cm 9 cm 12 cm Area = 12 × 12 × 9 = 54 cm2 A similar pattern emerges for other shapes. Scaling the linear dimension of a shape by a factor of k scales the area by a factor of k 2 . Scaling areas If two shapes are similar and the scale factor is k, then the area of the similar shape = k2 × area of the original shape. Example 19 Using area scale factors with similarity The two triangles shown are similar. The base of the smaller triangle has a length of 10 cm. Its area is 40 cm2 . The base of the larger triangle has a length of 25 cm. Determine its area. 40 cm2 10 cm 25 cm Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 383 Chapter 12 — Geometry Solution 2 Write down the area of the small triangle. 3 Area of larger triangle = k 2 × area of smaller triangle. 25 = 2.5 10 Area of small triangle = 40 cm2 ∴ Area of larger triangle = 2.52 × 40 = 250 Substitute the appropriate values and evaluate. 4 Write down your answer. The area of the larger triangle is 250 cm2 . Example 20 Scale factors and area Solution 12 cm 60 cm PL The two hearts shown are similar shapes. The width of the larger heart is 60 cm. Its area is 100 cm2 . The width of the smaller heart is 12 cm. Determine its area. k= E 1 Determine the scale factor k. M 1 Determine the scale factor k. Note we are scaling down. 2 Write down the area of the larger heart. 3 Area of smaller heart = k 2 × area of larger heart. Substitute the appropriate values and evaluate. 4 Write down your answer. Area = 100 cm2 k= 12 = 0.2 60 Area of larger heart = 100 cm2 ∴ Area of smaller heart = 0.22 × 100 =4 The area of the smaller heart is 4 cm2 . SA Volumes Two solids are considered to be similar if they have the same shape and the ratio of their corresponding linear dimensions is equal. Some examples of similar volume and the ratio of their areas are considered in the following. Similar spheres 3 cm Volume = 43 × 33 = 36 cm3 Scale factor = k = radius sphere 2 4 = radius sphere 1 3 256 256 Ratio of volumes = 3 = 36 108 64 4 3 = = k3 = 27 3 4 cm Volume = 43 × 43 256 = cm3 3 Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 384 Essential Further Mathematics – Module 2 Geometry and trigonometry Similar cubes 4 cm 2 cm Scale factor = k = 2 cm 2 cm 64 =8 8 3 = (2) = k 3 4 cm 4 cm Ratio of volumes = Volume = 4 × 4 × 4 = 64 cm3 E Volume = 2 × 2 × 2 = 8 cm3 side length 2 4 = =2 side length 1 2 Similar cylinders 1 cm 3 cm radius 2 3 Scale factor = k = = =3 radius 1 1 6 cm PL 2 cm 54 Volume = × 12 × 2 Ratio of volumes = = 27 = (3)3 = k 3 2 Volume = × 32 × 6 = 2 cm3 = 54 cm3 A similar pattern emerges for other solids. Scaling the linear dimension of a solid by a factor of k scales the volume by a factor of k 3 . M Scaling volumes If two solids are similar and the scale factor is k, then the volume of the similar solid = k3 × volume of the original solid. Example 21 Similar solids SA The two cuboids shown are similar solids. The height of the larger cuboid is 6 cm. Its volume is 120 cm3 . The height of the smaller cuboid is 1.5 cm. Determine its volume. 1.5 cm 6 cm volume = 120 cm3 Solution 1 Determine the scale factor k. Note we are scaling down. 2 Write down the volume of the larger cuboid. 3 Volume smaller cuboid = k 3 × volume larger cuboid. k= 1.5 = 0.25 6 Volume larger cuboid = 120 cm3 Volume smaller cuboid = 0.253 × 120 = 1.875 Substitute the appropriate values and evaluate. 4 Write down your answer. The volume of the smaller cuboid is 1.875 cm3 . Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 385 Chapter 12 — Geometry Example 22 Similar solids V' V The two square pyramids shown are similar with a base dimensions 4 and 5 cm respectively. The height of the first pyramid is 9 cm and its volume is 48 cm3 . Find the height and volume of the second pyramid. A 9 cm C B 4 cm O A' D D' k= 5 = 1.25 4 PL Height 2 Write down the height of Pyramid 1. 3 Height Pyramid 2 = k× height Pyramid 1. Substitute the appropriate values and evaluate. 4 Write down your answer. Pyramid 2 E 1 Determine the scale factor k. Use the base measurements. 5 cm O' Pyramid 1 Solution C' B' Height 1 = 9 cm ∴ Height 2 = 1.25 × 9 = 11.25 The height of Pyramid 2 is 11.25 cm. 6 Write down your answer. The volume of Pyramid 2 is 93.75 cm3 . M Volume 5 Volume Pyramid 2 = k 3 × volume Pyramid 1 Substitute the appropriate values and evaluate. Volume 1 = 48 cm3 ∴ Volume 2 = 1.253 × 48 = 93.75 Exercise 12G SA 1 Triangle ABC is similar to triangle XYZ. The length scale factor k = 1.2. The area of triangle ABC is 6 cm2 . Find the area of triangle XYZ. 2 The two rectangles are similar. The area of rectangle ABCD is 20 cm2 . Find the area of rectangle AB C D . B A B 3 cm A Y ×1.2 C X C Z B' 5 cm D A' 3 The two shapes shown are similar. The length scale factor is 32 . The area of the shape to the right is 30 cm2 . What is the area of the shape to the left? 4 Triangle ABC is similar to triangle XYZ. YZ ZX XY = = = 2.1. AB BC CA 2 C' TheUniversity area of triangle XYZ is 20Sample cm . Find area of triangle •ABC. Cambridge Press • Uncorrected pagesthe • 978-0-521-61328-6 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin D' P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 386 Essential Further Mathematics – Module 2 Geometry and trigonometry 5 Triangles ABC and AB C are equilateral triangles. a Find the length of BF. b Find a. area of triangle AB C . c Find the ratio area of triangle ABC B' B 2 cm A 2 cm F 2 cm C a cm 2 cm A' a cm C' F' a cm E 6 The areas of two similar triangles are 16 and 25. What is the ratio of a pair of corresponding sides? 7 The areas of two similar triangles are 144 and 81. If the base of the larger triangle is 30, what is the corresponding base of the smaller triangle? PL 8 These two rectangular prisms are similar. The length scale factor is 1.8. The volume of the first solid is 20 cm3 . What is the volume of the second solid? 9 Two cones are similar. The ratio of volumes is 8 : 125. Find the ratio of the: i heights ii lengths of sloping edges iii areas of bases. 5 cm 10 cm M 10 A cone has water poured into it as shown. Find the ratio of the volume of empty space in the cone to volume of water. 11 Consider two similar cuboids that have edges where lengths are in the ratio 1 : 4. a Find the ratio of the surface area of the two cuboids. b Find the ratio of the volumes. SA 12 An inverted right circular cone of capacity 100 m3 is filled with water to half its depth. Find the volume of water. 13 The ratio of the radii of two spheres is 2 : 5. Find the ratio of: i the surface areas ii the volumes 14 Two right circular cones are as shown. Find: a the ratio of the heights of the cones b the ratio of the surface areas c the ratio of the volumes 15 cm 45 cm 10 cm 30 cm 15 The ratio of the volumes of two cubes is 1 : 27. Find: a the ratio of the surface areas of the cubes CambridgebUniversity • Uncorrected Sample the ratioPress of the edges of the cubespages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 387 Chapter 12 — Geometry Review Key ideas and chapter summary Alternate, corresponding, cointerior and vertically opposite angles Angles 4 and 6 are examples of alternate angles. Angles 2 and 6 are examples of corresponding angles. Angles 3 and 6 are examples of cointerior angles. l3 Angles 1 and 3 are examples 2 1 l1 of vertically opposite angles 4 3 l2 and are of equal magnitude. 5 6 7 When lines l1 and l2 are parallel corresponding angles are of equal magnitude, alternate angles are of equal magnitude and cointerior angles are supplementary. PL Angles associated with parallel lines crossed by a transversal line E 8 Corresponding Corresponding Alternate Alternate Converse results also hold: If corresponding angles are equal then l1 is parallel to l2 . If alternate angles are equal then l1 is parallel to l2 . If cointerior angles are supplementary then l1 is parallel to l2 . The sum of the magnitudes of the interior angles of a triangle is equal to 180◦: a ◦ + b◦ + c◦ = 180◦ . Equilateral triangle A triangle is said to be equilateral if all of its sides are of the same length. The angles of an equilateral triangle are all of magnitude 60◦ . SA M Angle sum of triangle Isosceles triangle Polygon 10 cm 10 cm 10 cm A triangle is said to be isosceles if it has two sides of equal length. 5 cm 5 cm If a triangle is isosceles the angles opposite each of the equal sides are equal. A polygon is a closed geometric shape with sides which are segments of straight lines. Examples are: 3 sides: Triangle 4 sides: Quadrilateral 5 sides: Pentagon 6 sides: Hexagon Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> Essential Further Mathematics – Module 2 Geometry and trigonometry A polygon is said to be convex if any diagonal lies inside the polygon. Regular polygon A regular polygon has all sides of equal length and all angles of equal magnitude. Sum of the interior angles The angle sum of the interior angles of an n-sided polygon is given by the formula: S = (180n − 360)◦ . Pythagoras’ theorem Pythagoras’ theorem states that for a right-angled triangle ABC with side lengths a, b and c, a 2 + b2 = c2 , where c is the longest side. E Convex polygon We informally define two objects to be similar if they have the same shape but not the same size. Conditions for similarity of triangles r Corresponding angles in the triangles are equal. r Corresponding sides are in the same ratio. PL Similar figures AB B C AC = = =k AB BC AC where k is the scale factor r Two pairs of corresponding sides have the same ratio and the included angles are equal. Volumes of solids r cm M Cylinder: V = r 2 h h cm x Cube: V = x3 SA Review 388 Surface area of solids h cm Rectangular prism: V = lw h Right-angled triangular prism V = 12 bhl w cm l cm h cm b cm l cm r cm Cylinder: S = 2r 2 + 2r h h cm Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 389 Chapter 12 — Geometry Cube: S = 6x 2 h cm Rectangular prism: S = 2 (lw + lh + w h) w cm E l cm Right-angled triangular prism √ S = bh + bl + hl + l b2 + h 2 Scaling, areas and volumes h cm b cm l cm r If two shapes are similar and the scale factor is k, PL then the area of the similar shape = k 2 × area of the original shape. k = 3 2 k2 = 9 4 r If two solids are similar and the scale factor k= 3 2 k3 = 27 8 SA M is k, then the volume of the similar solid = k 3 × volume of the original solid. Skills check Having completed this chapter you should be able to: apply the properties of parallel lines and triangles and regular polygons to find the size of an angle given suitable information find the size of each interior angle of a regular polygon with a given number of sides use the definition of objects such as triangles, quadrilaterals, squares, pentagons, hexagons, equilateral triangles, isosceles triangles to determine angles recognise when two objects are similar determine unknown lengths and angles through use of similar triangles find surface areas and volumes of solids use Pythagoras’ theorem to find unknown lengths in right-angled triangles use similarity of two- and three-dimensional shapes to determine areas and volumes Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin Review x P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> Essential Further Mathematics – Module 2 Geometry and trigonometry Multiple-choice questions Questions 1 to 3 relate to the diagram 1 Angle PRS = B 40◦ C 50◦ D 60◦ A 20◦ E 110◦ 2 Angle RPS = B 40◦ A 20◦ E 110◦ C 50◦ D 60◦ P 120° 70° E 3 Given that PS bisects angle QPR, the size of angle PQS is: B 35◦ C 40◦ D 50◦ E 60◦ A 20◦ R S Q PL Questions 4 to 6 relate to the diagram Lines m and l are parallel and cut by a transversal q. 4 The value of x is: A 65 B 125 5 The value of y is: A 65 B 125 6 The value of z is: A 65 B 125 C 62.5 D 60 C 62.5 D 60 E 55 m z° y° l q C 62.5 D 60 E 55 C M 6 cm A 8 The triangle ABC has a right angle at A. The length of side BC to the nearest cm is: A 10 B 14 C 9 D 12 E 11 C 7 cm Z A 3 cm B X. 10 YZ is parallel to Y Z and Y Y = The area of triangle XYZ is 60cm . The area of triangle X Y Z is: B 30 cm2 C 15 cm2 A 20 cm2 80 20 cm2 cm2 E D 3 3 B 9 cm 9 Triangles ABC and XYZ are similar isosceles triangles. The length of XY is: C A 4 cm B 5 cm C 4.2 cm D 8.5 cm E 7.2 cm 5 cm 5 cm B 8 cm A 125° x° E 55 7 The triangle ABC has a right angle at A. The length of side BC in cm is: A 10 B 14 C 9 D 9. 8 E 11 SA Review 390 1 Y 2 2 12 cm 12 cm X Y X Y' Z' Y Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin Z P1: FXS/ABE P2: FXS 9780521740517c12.xml CUAU031-EVANS September 6, 2008 13:34 Back to Menu >>> 391 Chapter 12 — Geometry 1.2 cm C 2.16 1.8 m xm 18 m E 12 A regular convex polygon has 12 sides. The magnitude of each of its interior angles is: B 45◦ C 60◦ D 150◦ E 120◦ A 30◦ Z C 7 cm 7 cm A B 10 cm 14 Two similar cylinders are shown. The ratio of the volume of the smaller cylinder to the larger cylinder is: 15 cm A 1:4 B 1:16 C 1:64 10 cm D 15:60 E 1:3 10 cm Y X 4 cm PL 13 Triangles ABC and XYZ are similar isosceles triangles. The length of XY correct to one decimal place is: A 4.8 cm B 5.7 cm C 4.2 cm D 8.5 cm E 8.2 cm 60 cm 40 cm 15 Each interior angle of a regular polygon measures 135◦ . The number of sides the polygon has is: A 4 B 6 C 8 D 10 E 7 D C 10 cm A B SA M 16 Each side length of a square is 10 cm. The length of the diagonal is: √ √ C 10 2 D 8 E 1.4 A 10 B 5 2 Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin Review 11 The value of x is: A 12 B 27 D 20.8 E 13.81