SAMPLE MODULE 2. Geometry

Transcription

1 H P T E R 12 MODULE 2 Geometry What are the properties of parallel lines? 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 ngles 4 and 6 are called alternate angles. ngles 5 and 3 are called alternate angles. l 3 ngles 2 and 6 are called corresponding angles. 1 2 l 1 ngles 1 and 5 are called corresponding angles. 4 3 ngles 4 and 8 are called corresponding angles. l 2 5 ngles 3 and 7 are called corresponding angles ngles 3 and 6 are called cointerior angles. ngles 4 and 5 are called cointerior angles. Lines l 1 and l 2 are cut by a transversal l 3. ngles 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. ngles 1 and 2 are supplementary, i.e. their magnitudes add to 180. When lines l 1 and l 2 are parallel, corresponding angles l 3 are of equal magnitude, alternate angles are of equal magnitude and cointerior angles are supplementary l 1 onverse results also hold: If corresponding angles are equal then 5 6 l 1 is parallel to l l 2 If alternate angles are equal then l 1 is parallel to l 2. If cointerior angles are supplementary then l 1 is parallel to l 2. TI-Nspire & asio lasspad material in collaboration with rown 360 and McMenamin

2 hapter 12 Geometry 361 Example 1 Parallel line properties Find the values of the pronumerals. e 65 a = 65 (corresponding) d c d = 65 (alternate with a) b = 115 (cointerior with d) b a e = 115 (corresponding with b) c = 115 (vertically opposite e) There are lots of ways of finding these values. One sequence of reasoning has been used here. Example 2 Parallel line properties Find the values of the pronumerals. Example 3 2x 50 = x x x = x = 60 (alternate angles) Parallel line properties Find the values of the pronumerals. x = 2x + 80 (alternate angles) = 2x x x = 20 lso x y + 60 = 180 (cointerior) and x = 20 y = 180 or y = 0 Exercise 12 Questions 1 to 5 apply to the following diagram d a l 1 c b h e l 2 g f l 1 ngles d and b are: 3 alternate cointerior corresponding D supplementary E vertically opposite (2x 50) (x + 10) (x + 100) (2x + 80) (y + 60)

3 362 Essential Further Mathematics Module 2 Geometry and trigonometry 2 ngles d and a are: alternate cointerior corresponding D supplementary E vertically opposite 3 ngles c and h are: alternate cointerior corresponding D supplementary E vertically opposite 4 ngles b and f are: alternate cointerior corresponding D supplementary E vertically opposite 5 ngles c and e are: alternate cointerior corresponding D supplementary E vertically opposite 6 Find the values of the pronumerals in each of the following: a b c d z y 70 x z x y 50 e 40 y x z x y Properties of triangles a review f y 80 (2z) (2x 40) (x + 40) a, b and c are the magnitudes of the interior angles of the triangle. d is the magnitude of an exterior angle at. 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 c d equal to the sum of the magnitudes of the two opposite interior angles. triangle is said to be equilateral if all its sides are of the same length: = =. The angles of an equilateral triangle are all of 60 magnitude

4 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. 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. The sum of the magnitudes of the exterior angles of a triangle is equal to 360 : e + d + f = 360 triangle is said to be a right-angled triangle if it has one angle of magnitude 90. Example 4 ngle sum of a triangle Find the values of the pronumerals. hapter 12 Geometry 363 f e x = 180 (sum angles = 180 ) 42 + x = 180 or x = y = 180 (sum angles = 180 ) 20 x y y = 42 Or, to find x: Two oftheangles sum to 42 and therefore the third angle is 138.Tofindy.Thetwoangles sum to 180.Therefore the second is 42 Example 5 ngle sum of an isosceles triangle Find the values of the pronumerals x = 80 (sum angles = 180 ) 2x = 80 or x = 40 Or, observe the two unknown numbers are the same and must sumto80,therefore each of them has size O x x d

5 364 Essential Further Mathematics Module 2 Geometry and trigonometry Exercise 12 1 Find the values of the pronumerals in each of the following: a e z x y y w x b P x f Q 30 x b c a 75 R c x y 40 g y d E X y y x 12.3 Properties of regular polygons a review Equilateral triangle Square Regular pentagon Regular hexagon Regular octagon 50 Y x Z regular polygon has all sides of equal length and all angles of equal magnitude. 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. onvex 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: D x = (180n 360) n

6 hapter 12 Geometry 365 The angle bisectors of a regular polygon meet at a point O. Foraregular polygon, a circle can be drawn with centre O on which all the vertices lie. O 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. Example 6 O ngle properties of an octagon The diagram opposite shows a regular octagon. a Show that x = 45. b Find the size of angle y. a 8x = 360 (sum angles at O = 360 ) x = 360 = 45 8 x = 45 b y + y + 45 = 180 ( O isosceles) 2y = 135 or y = 67.5 Example 7 ngle sum of an octagon Find the sum of the interior angles of an 8-sided convex polygon (octagon). 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. G F H E O O x y S = (180n 360) n = 8 S = ( ) = 1080 Thesum of the interior angles is D

7 366 Essential Further Mathematics Module 2 Geometry and trigonometry Exercise 12 1 Name each of the following regular polygons. a b c d e 2 D is a square. D and 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 D ii O 3 DE is a regular pentagon. a Find the value of: y i x ii y x O b Find the sum of the interior angles of the regular pentagon DE. E D 4 DEF is a regular hexagon. F Find the value of: y a x b y E x O 5 State the sum of the interior angles of: a a 7-sided regular polygon b ahexagon c an octagon 6 The angle sum of a regular polygon is 1260.How many sides does the polygon have? 7 circle is circumscribed about a hexagon DEF. a Find the area of the circle if O = 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. E F D b a O D O D 9 If the magnitude of each angle of a regular polygon is 135,how many sides does the polygon have?

8 12.4 Pythagoras theorem Pythagoras theorem Pythagoras theorem states that for a right-angled triangle with side lengths a, b and c, asshown in the diagram, a 2 + b 2 = c 2. 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). There are many different proofs of Pythagoras theorem. proof due to the 20th President of the United States, James. Garfield, is produced using the following diagram. Y b E a a c c Z X b W Pythagorean triads hapter 12 Geometry 367 c b area = c 2 cm 2 rea of trapezium WXYZ = 1 (a + b)(a + b) 2 rea of EYZ + area of EWX + area of EWZ = 1 2 ab c ab = ab c2 a2 b2 + ab = ab c2 a 2 + b 2 = c 2 triple of natural numbers (a, b, c)iscalled a Pythagorean triad if c 2 = a 2 + b 2. 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 Pythagoras theorem Find the value, correct to two decimal places, of the unknown length for the triangle opposite. c cm b cm a cm area = a 2 cm 2 a area = b 2 cm 2 a b c cm ambridge University Press Uncorrected Sample pages Jones, cm Evans, Lipson

9 368 Essential Further Mathematics Module 2 Geometry and trigonometry 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. 2 Write down your answer for the length correct to two decimal places. Example 9 Pythagoras theorem Find the value, correct to two decimal places, of the unknown length for the triangle opposite. 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. Example 10 Pythagoras theorem x 2 = (Pythagoras) x = = The length is 8.08 cm correct to two decimal places. 8.6 cm y cm 5.6 cm y 2 = (Pythagoras) y 2 = y = = The length is 6.53 cm correct to two decimal places. 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. 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 100 m x m Let x m be the length of the shorter side. x = (Pythagoras) x 2 = = 6900 x = = orrect to the nearest centimetre, the length of the short side is m.

10 hapter 12 Geometry 369 Exercise 12D 1 Find the length of the unknown side for each of the following: a d 10 cm 9 cm 7 cm 6 cm b e 5 cm 4 11 cm 33 cm c f 10 cm 3 cm 15 cm 2 In each of the following find the value of x correct to two decimal places. a 4.8 cm 3.2 cm b 6.2 cm 2.8 cm c 5.2 cm 9.8 cm d 3.5 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 = b 52 + x 2 = 9 2 c = x 2 12 cm 4 In triangle VWX, there is a right angle at X.VX = 2.andXW = 4.6 cm. Find VW. 5 Find D, the height of the triangle. 32 cm 32 cm 20 cm 6 n 18 m ladder is7mawayfrom 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 9m. 8 Triangle is isosceles. Find the length of. D 3 cm 8 14 ambridge University Press Uncorrected Sample pages Jones, Evans, Lipson

11 370 Essential Further Mathematics Module 2 Geometry and trigonometry 9 In a circle of centre O,achord is of length. The radius of the circle is 1. Find the distance of the chord from O. 10 Find the value of x. 11 How high is the kite above the ground? 18 cm 12 square has an area of 169 cm 2. What is the length of the diagonal? 13 Find the area of a square with a diagonal of length: a 8 2cm b 8cm 14 Find the length of. 15 D is a square of side length 2 cm. If = E, find the length of DE. 8 cm D O 170 m 90 m 20 cm X Y 12 cm 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. D E

12 12.5 Similar figures hapter 12 Geometry 371 Similarity In this section we informally define two objects to be similar if they have the same shape but not the same size. Examples ny two circles are similar to each other. 1 2 ny two squares are similar to each other. 3 cm 3 cm S 1 S 2 It is not true that any rectangle is similar to any other rectangle. For example, rectangle 1 is not similar to rectangle 2. R 1 rectangle similar to R 1 is: 1 cm 8 cm R 3 8 cm R 2 2 cm 1 cm So, for ( two rectangles to be similar, their corresponding sides must be in the same 8 ratio 2 = 4 ). 1 Similar triangles Two triangles are similar if one of the following conditions holds: corresponding angles in the triangles are equal corresponding sides are in the same ratio = = = k k is the scale factor. ' ' '

13 372 Essential Further Mathematics Module 2 Geometry and trigonometry two pairs of corresponding sides have the same ratio and the included angles are equal ' ' = If triangle is similar to triangle XYZ this can be written symbolically as XYZ. The triangles are named so that angles of equal magnitude hold the same position, i.e. corresponds to X, corresponds to Y, corresponds to Z. Example 11 Similar triangles Find the value of length of side in correct to two decimal places. 1 Triangle is similar to triangle :two pairs ( of corresponding sides have the same 5 ratio 3 = 6.25 ) and included angles (20 ) For similar triangles, the ratios of corresponding sides are equal, for example, =. 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 Similar triangles Find the value of length of side in. 1 Triangle is similar to triangle XY (corresponding angles are equal). ' 5 cm 20 3 cm Triangles similar 6.25 cm ' ' cm x = x = = The length of side is 2.41 cm, correct to two decimal places. 3 cm 2.5 cm Y Triangles similar 6 cm X ' cm

14 2 For similar triangles, the ratios of corresponding sides are equal (for example, X = Y ). Use this fact to write down an expression involving x. Solve for x. Note that, if = x then X = x Write down your answer. Exercise 12E hapter 12 Geometry 373 x x + 6 = x = 3(x + 6) = 3x x = 18 or x = = 7.2 The lengthofsideis7.2cm. 1 Find the value of x for each of the following pairs of similar triangles. a ' 82 5 cm 9 cm ' ' b Z cm 10 cm X 5 cm Y c e f h D 2 cm Y X 13 cm 8 cm 12 cm 10 cm E 2 cm 6 cm D E 3 cm 2 cm d i 13 cm Q 6 cm g 1 D 10 cm 8 cm R 12 cm P 12 cm 16 cm P E P Q 8 cm 2 cm Q 6 cm 8 cm j 10 cm P 1.5 cm Q 2 cm

15 374 Essential Further Mathematics Module 2 Geometry and trigonometry 2 Given that D = 14, ED = 12, = 15 and E = 4, find, E and. 3 tree casts a shadow of 33 m and at the same time a stick 30 cm long casts a shadow 2 long. How high is the tree? 33 m 0.24 m 0.3 m 4 20metre high neon sign is supported by a 40 m steel cable as shown. n ant crawls along the cable starting at. How high is the ant when it is 15 m from? 5 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? 6 man stands at and looks at point Y across the river. He gets a friend to place a stone at X so that, X and Y are collinear. He then measures, X and X to be 15 m, 30 m and 45 m respectively. Find Y, the distance across the river. Y 45 m 14 D m X E 30 m river m 15 m 7 Find the height, h m, of a tree that casts a shadow 32 m long at the same time that a vertical straight stick2mlong casts a shadow 6.2 m long. 8 plank is placed straight up stairs that are 20 cm wide and 12 cm deep. Find x,where is the width of the widest rectangular box of height 8 cm that can be placed on a stair under the plank. 9 The sloping edge of a technical drawing table is 1 m from front to back. alculate the height above the ground of a point,which is 30 cm from the front edge. 80 cm plank 8 cm 20 cm 30 cm 1 m 12 cm 92 cm

16 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. hapter 12 Geometry m x m 0.8 m (1.3 x) m 11 man whose eye is 1.7 m from the ground when standing 3.5 m in front of a wall3mhigh can just see the top of a tower that is 100 m away from the wall. Find the height of the tower. 12 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 spotlight is at a height of 0.6 m above ground level. vertical post 1.1 m high stands3maway,and5m further away there is a vertical wall. How high up the wall does the shadow reach? 12.6 Volumes and surface areas Volume of a prism spotlight 0.6 m 1.1 m vertical post prism is a solid which has a constant cross-section. Examples are cubes, cylinders, rectangular prisms and triangular prisms The volume of a prism can be found by using its cross-sectional area. volume = area of cross-section height (or length) V = h nswers will be in cubic units, i.e. mm 3,cm 3,m 3 etc. Example 13 Volume of a cylinder Find the volume of this cylinder which has radius 3 cm and height correct to two decimal places. 3 cm 1 Find the cross-sectional area of the prism. rea of cross-section = πr 2 = π 3 2 = cm 2 wall

17 376 Essential Further Mathematics Module 2 Geometry and trigonometry 2 Multiply by the height Volume = Make sure that accuracy is given to the = cm 3 (correct to two correct number of decimal places decimal places) The formulas for determining the volumes of some standard prisms are given here. Solid ylinder (radius r cm, height h cm) h cm r cm ube (all edges ) Rectangular prism (length l cm, width w cm, height h cm) l cm h cm w cm Triangular prism h cm b cm l cm Volume of a pyramid Formula V = r 2 h V = x 3 V = lwh The triangular prism shown has a right-angled triangle base but the following formula holds for all triangular prisms V = 1 2 bhl The formula for finding the volume of a right pyramid can be stated as: Volume of pyramid = 1 base area 3 perpendicular height h For the square pyramid shown: V = 1 3 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 cm 2 and a height of 20 cm. Give the answer correct to one decimal place. V = 1 3 h = = cm 3 (correct to one decimal place) 20 cm

18 hapter 12 Geometry 377 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. V = 1 3 x 2 h = = 900 cm 3 Volume of a cone The formula for finding the volume of a cone can be stated as: Volume of cone = 1 base area height 3 V = 1 3 r 2 h Volume of a sphere The formula for the volume of a sphere is: V = 4 3 r 3 where r is the radius of the sphere. Example 16 Find the volume of this sphere. Volume of sphere = 4 3 πr 3 Volume of a sphere = 4 3 π 43 omposite shapes = cm 3 (2 decimal places) 10 cm 27 cm r 10 cm Using the shapes above, new shapes can be made. The volumes of these can be found by summing the volumes of the component solids.

19 378 Essential Further Mathematics Module 2 Geometry and trigonometry Example 17 Volume of a composite shape 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? 1 Use the formula V = r 2 h to find the Thevolume of the cylinder volume of the cylinder = π = 125π mm 3 2 Find the volume of hemisphere using Thevolume of the hemisphere the formula that the volume of a ( hemisphere = r 3) = 2 = 2 3 r 3 3 π 53 = π mm3. 3 dd the two together. Therefore the volume of the composite = 125π π = 625 π = Thevolume of the composite = mm 3 4 Write down your answer. (correct to two decimal places) 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. r l Total area = 2 r rl = 2 r(r + l) r = πr 2 2πr l = 2πrl r = πr 2 Here are some more formulas for the surface areas of some solids. The derivation of these is left to the reader.

20 hapter 12 Geometry 379 Solid ylinder (radius r cm, height h cm) r cm h cm ube (all edges ) Rectangular prism (length l cm, width w cm, height h cm) h cm w cm l cm Triangular prism h cm Example 18 b cm l cm Formula S = 2 r rh S = 6x 2 S = 2(lw + lh + wh) S = bh + bl + hl + l b 2 + h 2 Surface area of a right square pyramid 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. 1 Draw the net of the pyramid. 2 First determine the area of the square. rea of the square = 10 2 = 100 cm 2 3 Determine the area of one of the The areaofonetriangle isosceles triangles. = = 75 cm2 2 4 Find the sum of the areas of the four Thesurface area of the solid triangles and add to the area of the = = square. = 400 cm 2

21 380 Essential Further Mathematics Module 2 Geometry and trigonometry Exercise 12F 1 Find the volume in cm 3 of each of the following shapes, correct to two decimal places. a radius 6.3 cm and height 2.1 cm c area of cross section = 2.8cm 2 height = 6.2cm b dimensions 2.1 cm, 8.3 cm and 12.2 cm d 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. 3 The box shown has dimensions length: 13 cm, width: and height 3 cm. a Find the surface area of the box in square centimetres (cm 2 ). b Find the volume of the box in cubic centimetres (cm 3 ). 12 cm 4 Find the volumes, to two decimal places, of spheres with: a radius = 4mm b diameter = 23 cm c radius = 3.8 m d diameter = 15 cm 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 alculate 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. alculate 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 The length of each edge of the square base is 10 cm and the height of the pyramid is 12 cm. O 10 cm D X

22 hapter 12 Geometry 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 = 4 r 2 and the surface area of the curved section of a cylinder is given by the formula S = 2 rh. Find the surface area and volume of the capsule. Give your answers correct to two decimal places. 10 Find: a the surface area b the volume of the object shown. 11 The diagram opposite shows a right pyramid on a cube. Each edge of the cube is 1. The height of the pyramid is 2. Find: a the volume of the solid b the surface area of the solid 12 Find: a the surface b the volume of the solid shown opposite. 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 7 cm 3 m 5 m 4 m 2 m 2 2 cm 10 m 1 10 cm 5 cm 5 cm 20 cm 10 cm

23 382 Essential Further Mathematics Module 2 Geometry and trigonometry 12.7 reas, volumes and similarity reas Some examples of similar shapes and the ratio of their areas are considered in the following. Similar circles 3 cm rea = 3 2 Similar rectangles 3 cm 2 cm rea = 3 2 = 6cm 2 Similar triangles 5 cm rea = = 6cm 2 radius circle 2 Scale factor = k = radius circle 1 = 4 3 ( ) 42 Ratio of areas = 3 = = = k rea = 4 2 length rectangle 2 Scale factor = k = length rectangle 1 = 6 3 = 2 6 cm Ratio of areas = 24 6 = 4 = (2)2 = k 2 rea = 6 4 = 2 2 height triangle 2 Scale factor = k = height triangle 1 = 9 3 = 3 3 cm 15 cm Ratio of areas = 54 9 cm 6 = 9 = (3)2 = k 2 12 cm rea = = 5 2 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 = k 2 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 cm 2. The base of the larger triangle has a length of 25 cm. Determine its area. 40 cm2 10 cm 25 cm

24 hapter 12 Geometry Determine the scale factor k. k = = Write down the area of the small triangle. rea of small triangle = 40 cm 2 3 rea of larger triangle = k 2 area of rea of larger triangle = smaller triangle. = 250 Substitute the appropriate values and evaluate. 4 Write down your answer. The areaofthelarger triangle is 250 cm 2. Example 20 Scale factors and area The two hearts shown are similar shapes. The width of the larger heart is 60 cm. Its area is 100 cm 2. The width of the smaller heart is 12 cm. Determine its area. 12 cm 60 cm rea = 100 cm 2 1 Determine the scale factor k. Note we are k = = 0.2 scaling down. 2 Write down the area of the larger heart. rea of larger heart = 100 cm 2 3 rea of smaller heart = k 2 area of rea of smaller heart = larger heart. = 4 Substitute the appropriate values and evaluate. 4 Write down your answer. The areaofthesmaller heart is 2. 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 = = 36 cm 3 Scale factor = k = radius sphere 2 radius sphere 1 = Ratio of volumes = 3 36 = 256 Volume = = 64 ( ) 4 3 = = k 3 = 256 cm

25 384 Essential Further Mathematics Module 2 Geometry and trigonometry Similar cubes 2 cm 2 cm 2 cm Volume = = 8cm 3 Similar cylinders 1 cm 2 cm Volume = = 2 cm 3 side length 2 Scale factor = k = side length 1 = 4 2 = 2 Ratio of volumes = 64 8 = 8 = (2) 3 = k 3 Volume = = 6 3 Scale factor = k = radius 2 radius 1 = 3 1 = 3 3 cm 6 cm Ratio of volumes = 54 2 = 27 = (3)3 = k 3 Volume = = 54 cm 3 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. Scaling volumes If two solids are similar and the scale factor is k, then the volume of the similar solid = k 3 volume of the original solid. Example 21 Similar solids The two cuboids shown are similar solids. The height of the larger cuboid is 6 cm. Its volume is 120 cm 3. The height of the smaller cuboid is 1.5 cm. Determine its volume. 1.5 cm volume = 120 cm3 1 Determine the scale factor k. Note we are k = 1.5 scaling down. 6 = Write down the volume of the larger cuboid. Volume larger cuboid = 120 cm 3 3 Volume smaller cuboid = k 3 volume larger cuboid. Volume smaller cuboid = = Substitute the appropriate values and evaluate. 4 Write down your answer. Thevolume of the smaller cuboid is cm 3. 6 cm

26 hapter 12 Geometry 385 Example 22 Similar solids 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 cm 3.Find the height and volume of the second pyramid. 1 Determine the scale factor k. Use the base measurements. V 9 cm O D ' ' O' Pyramid 1 Pyramid 2 k = 5 4 = 1.25 Height 2 Write down the height of Pyramid 1. Height 1 = 9cm 3 Height Pyramid 2 = k height Pyramid 1. Height 2 = = Substitute the appropriate values and evaluate. 4 Write down your answer. V' ' 5 cm D' Volume Theheight of Pyramid 2 is cm. 5 Volume Pyramid 2 = k 3 volume Pyramid 1 Substitute the appropriate values and evaluate. Volume 1 = 48 cm 3 Volume 2 = = Write down your answer. The volume of Pyramid 2 is cm 3. Exercise 12G 1 Triangle is similar to triangle XYZ. The length scale factor k = 1.2. The area of triangle is6cm 2.Find the area of triangle XYZ. 2 The two rectangles are similar. The area of rectangle D is 20 cm 2.Find the area of rectangle D. 3 The two shapes shown are similar. The length scale factor is 3. The area of the shape to the 2 right is 30 cm 2. What is the area of the shape to the left? 1.2 X ' 3 cm 5 cm D ' 4 Triangle is similar to triangle XYZ. XY = YZ = ZX = 2.1. The area of triangle XYZ is 20 cm 2.Find the area of triangle. Y Z ' D'

27 386 Essential Further Mathematics Module 2 Geometry and trigonometry 5 Triangles and are equilateral triangles. a Find the length of F. b Find a. 2 cm 2 cm c Find the ratio area of triangle area of triangle. F 2 cm ' a cm 6 The areas of two similar triangles are 16 and 25. What is the ratio of a pair of corresponding sides? ' F' a cm 2 cm a cm 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? 8 These two rectangular prisms are similar. The length scale factor is 1.8. The volume of the first solid is 20 cm 3. 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. 10 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 onsider 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. 12 n inverted right circular cone of capacity 100 m 3 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.Findtheratio 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 10 cm 15 cm 5 cm 10 cm 45 cm ' 15 The ratio of the volumes of two cubes is 1:27. Find: a the ratio of the surface areas of the cubes b the ratio of the edges of the cubes 30 cm

28 hapter 12 Geometry 387 Key ideas and chapter summary lternate, corresponding, cointerior and vertically opposite angles ngles associated with parallel lines crossed by a transversal line ngle sum of triangle Equilateral triangle Isosceles triangle Polygon ngles 4 and 6 are examples of alternate angles. ngles 2 and 6 are examples of corresponding angles. ngles 3 and 6 are examples of cointerior angles. l ngles 1 and 3 are examples 3 l 1 2 of vertically opposite angles l and are of equal magnitude When lines l 1 and l 2 are parallel corresponding angles are of equal magnitude, alternate angles are of equal magnitude and cointerior angles are supplementary. orresponding orresponding lternate lternate onverse results also hold: If corresponding angles are equal then l 1 is parallel to l 2. If alternate angles are equal then l 1 is parallel to l 2. If cointerior angles are supplementary then l 1 is parallel to l 2. The sum of the magnitudes of the interior angles of a triangle is equal to 180 : a + b + c = 180. 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 cm 10 cm 10 cm triangle is said to be isosceles if it has two sides of equal length. If a triangle is isosceles the angles 5 cm 5 cm opposite each of the equal sides are equal. polygon is a closed geometric shape with sides which are segments of straight lines. Examples are: 3 sides: Triangle 4 sides: Quadrilateral Review 5 sides: Pentagon 6 sides: Hexagon

29 388 Essential Further Mathematics Module 2 Geometry and trigonometry Review onvex polygon Regular polygon Sum of the interior angles Pythagoras theorem Similar figures polygon is said to be convex if any diagonal lies inside the polygon. regular polygon has all sides of equal length and all angles of equal magnitude. The angle sum of the interior angles of an n-sided polygon is given by the formula: S = (180n 360). Pythagoras theorem states that for a right-angled triangle with side lengths a, b and c, a 2 + b 2 = c 2,where c is the longest side. We informally define two objects to be similar if they have the same shape but not the same size. onditions for similarity of triangles orresponding angles in the triangles are equal. orresponding sides are in the same ratio. = = = k Volumes of solids where k is the scale factor Two pairs of corresponding sides have the same ratio and the included angles are equal. ylinder: V = r 2 h ube: V = x 3 Rectangular prism: V = lwh Surface area of solids x l cm Right-angled triangular prism h V = 1 2 cmb bhl cm ylinder: S = 2 r rh h cm h cm r cm l cm r cm h cm w cm

30 hapter 12 Geometry 389 ube: S = 6x 2 Rectangular prism: S = 2(lw + lh + wh) x l cm Right-angled triangular prism S = bh + bl + hl + l h b 2 + h cmb 2 cm Scaling, areas and volumes If two shapes are similar and the scale factor is k, then the area of the similar shape = k 2 area of the original shape. Skills check k = 3 2 k 2 = 9 4 If two solids are similar and the scale factor is k, then the volume of the similar solid = k 3 volume of the original solid. k = 3 2 k 3 = 27 8 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 l cm h cm w cm Review

31 390 Essential Further Mathematics Module 2 Geometry and trigonometry Review Multiple-choice questions Questions 1 to 3 relate to the diagram 1 ngle PRS = D 60 E ngle RPS = D 60 E Given that PS bisects angle QPR, the size of angle PQS is: D 50 E 60 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: D 60 E 55 5 The value of y is: D 60 E 55 6 The value of z is: D 60 E 55 7 The triangle has a right angle at. The length of side in cm is: D 9. 8 E 11 8 The triangle has a right angle at. The length of side to the nearest cm is: D 12 E 11 9 Triangles and XYZ are similar isosceles triangles. The length of XY is: 4cm 5cm 4.2 cm D 8.5 cm E 7.2 cm 5 cm 6 cm 5 cm P R S 7 cm q 125 z y x 8 cm 9 cm Z 12 cm m l Q 12 cm 3 cm X Y 10 YZ is parallel to Y Z and Y Y = 1 2 YX. The area of triangle XYZ is 60cm 2. The area of triangle XY Z is: X 20 cm 2 30 cm 2 15 cm 2 Z' D 20 3 cm2 E 80 Y' 3 cm2 Z Y

32 hapter 12 Geometry The value of x is: 1.2 cm D 20.8 E m x m 18 m 12 regular convex polygon has 12 sides. The magnitude of each of its interior angles is: D 150 E Triangles and XYZ are similar isosceles triangles. Z The length of XY correct to one decimal place is: 4.8 cm 5.7 cm 4.2 cm 7 cm 7 cm 10 cm 10 cm D 8.5 cm E 8.2 cm X Y 14 Two similar cylinders are shown. The ratio of the volume of the smaller cylinder to the larger cylinder is: 15 cm 60 cm 1:4 1:16 1:64 D 15:60 E 1:3 10 cm 40 cm 15 Each interior angle of a regular polygon measures 135. The number of sides the polygon has is: D 10 E 7 16 Each side length of a square is 10 cm. The length of the diagonal is: D 8 E 1.4 D 10 cm Review