2 Pythagoras’ Theorem magenta yellow 95 100 50 75 Pythagoras’ Theorem The converse of Pythagoras’ Theorem Problem solving using Pythagoras’ Theorem Three-dimensional problems More difficult problems (Extension) 25 0 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 cyan 5 A B C D E Contents: black Y:\HAESE\SA_10-6ed\SA10-6_02\033SA10-6_02.CDR Monday, 4 September 2006 2:17:55 PM PETERDELL SA_10-6 34 PYTHAGORAS’ THEOREM (Chapter 2) Right angles (90o angles) are used in the construction of buildings and in the division of areas of land into rectangular regions. The ancient Egyptians used a rope with 12 equally spaced knots to form a triangle with sides in the ratio 3 : 4 : 5: This triangle has a right angle between the sides of length 3 and 4 units, and is, in fact, the simplest right angled triangle with sides of integer length. corner take hold of knots at arrows make rope taut line of one side of building OPENING PROBLEM Karrie is playing golf in the US Open but hits a wayward tee shot on the opening hole. Her caddy paces out some distances and finds that Karrie has hit the ball 250 m, but is 70 m from the line of sight from the tee to the hole. A marker which is 150 m from the pin is further up the fairway as shown: 0 Tee Caddy 150 m marker Hole 70 m 250 m Ball Consider the following questions: 1 How far is the caddy away from the tee? cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 2 From where the caddy stands on the fairway, what distance is left to the 150 m marker if he knows the hole is 430 m long? 3 How far does Karrie need to hit her ball with her second shot to reach the hole? black Y:\HAESE\SA_10-6ed\SA10-6_02\034SA10-6_02.CDR Wednesday, 26 July 2006 11:46:26 AM PETERDELL SA_10-6 35 PYTHAGORAS’ THEOREM (Chapter 2) INTRODUCTION A right angled triangle is a triangle which has a right angle as one of its angles. se tenu o hyp The side opposite the right angle is called the hypotenuse and is the longest side of the triangle. legs The other two sides are called the legs of the triangle. Around 500 BC, the Greek mathematician Pythagoras formulated a rule which connects the lengths of the sides of all right angled triangles. It is thought that he discovered the rule while studying tessellations of tiles on bathroom floors. Such patterns, like the one illustrated, were common on the walls and floors of bathrooms in ancient Greece. A PYTHAGORAS’ THEOREM c In a right angled triangle, with hypotenuse c and legs a and b, a c2 = a2 + b2 . b In geometric form, the Theorem of Pythagoras is: GEOMETRY PACKAGE In any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. c2 c Can you see how Pythagoras may have discovered the rule by looking at the tile pattern above? a c a b magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 b2 cyan a2 black Y:\HAESE\SA_10-6ed\SA10-6_02\035SA10-6_02.CDR Wednesday, 26 July 2006 11:47:01 AM PETERDELL b SA_10-6 36 PYTHAGORAS’ THEOREM (Chapter 2) Over 400 different proofs of the Pythagorean Theorem exist. Here is one of them: Proof: a On a square we draw 4 identical (congruent) right angled triangles, as illustrated. A smaller square in the centre is formed. b b c Suppose the legs are of length a and b and the hypotenuse has length c: c Since total area of large square = 4 £ area of one triangle + area of smaller square, c a c (a + b)2 = 4( 12 ab) + c2 2 2 a b 2 ) a + 2ab + b = 2ab + c ) a2 + b2 = c2 a b When using Pythagoras’ Theorem we often see surds, which are square root p numbers like 7: Note: Example 1 Self Tutor Find the length of the hypotenuse in: x cm 2 cm 3 cm The hypotenuse is opposite the right angle and has length x cm. If x2 =pk, then x = § k` , p but we reject ¡ k` as lengths must be positive! x2 = 32 + 22 x2 = 9 + 4 x2 = 13 p fas x > 0g i.e., x = 13 p ) the hypotenuse is 13 cm. ) ) ) EXERCISE 2A 1 Find the length of the hypotenuse in the following triangles, leaving your answer in surd (square root) form if appropriate: a b c 4 cm x km x cm x cm 7 cm 8 km 5 cm cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 13 km black Y:\HAESE\SA_10-6ed\SA10-6_02\036SA10-6_02.CDR Thursday, 3 August 2006 10:07:08 AM PETERDELL SA_10-6 PYTHAGORAS’ THEOREM (Chapter 2) Example 2 37 Self Tutor The hypotenuse has length 6 cm. ) x2 + 52 = 62 fPythagorasg ) x2 + 25 = 36 ) x2 = 11 p ) x = 11 fas x > 0g p ) third side is 11 cm long. Find the length of the third side of: 6 cm x cm 5 cm 2 Find the length of the third side of the following right angled triangles. Where appropriate leave your answer in surd (square root) form. a b c x km 11 cm 6 cm x cm 1.9 km 2.8 km x cm 9.5 cm Example 3 Self Tutor The hypotenuse has length x cm. p ) x2 = 22 + ( 10)2 fPythagorasg ) x2 = 4 + 10 ) x2 = 14 p ) x = § 14 p But x > 0, ) x = 14. Find x in the following: ~`1`0 cm 2 cm x cm 3 Find x in the following: a b c 3 cm ~`7 cm ~`2 cm x cm x cm x cm ~`1`0 cm ~`5 cm Example 4 2 x + Solve for x: Qw_ cm ) yellow 95 100 50 75 25 0 5 95 100 50 ) 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 x cm 1 4 Self Tutor 2 =1 black Y:\HAESE\SA_10-6ed\SA10-6_02\037SA10-6_02.CDR Wednesday, 26 July 2006 11:48:03 AM PETERDELL fPythagorasg =1 x2 = ) 1 cm magenta 2 x2 + ) cyan ¡ 1 ¢2 3 4 q x = § 34 q x = 34 fas x > 0g SA_10-6 38 PYTHAGORAS’ THEOREM (Chapter 2) 4 Solve for x: a b c 1 cm m x cm Qw_ cm Qw_ cm Ö̀3 2 x cm 1m xm Ew_ cm Example 5 Self Tutor (2x)2 ) 4x2 ) 3x2 ) x2 = x2 + 62 fPythagorasg = x2 + 36 = 36 = 12 p ) x = § 12 p But x > 0, ) x = 12. Find the value of x: 2x m xm 6m 5 Find the value of x: a b 9 cm c 26 cm 2x cm x cm 2x cm 2x m 3x m ~`2`0 m 3x cm Example 6 Self Tutor 5 cm A Find the value of any unknowns: x cm y cm D 6 cm B 1 cm C In triangle ABC, the hypotenuse is x cm. fPythagorasg ) x2 = 52 + 12 2 ) x = 26 p ) x = § 26 p ) x = 26 fx > 0g cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 In ¢ACD, the hypotenuse is 6 cm. p fPythagorasg ) y 2 + ( 26)2 = 62 2 ) y + 26 = 36 ) y 2 = 10 p ) y = § 10 p fy > 0g ) y = 10 black Y:\HAESE\SA_10-6ed\SA10-6_02\038SA10-6_02.CDR Wednesday, 30 August 2006 11:13:17 AM DAVID3 SA_10-6 39 PYTHAGORAS’ THEOREM (Chapter 2) 6 Find the value of any unknowns: a b c 1 cm 2 cm x cm 3 cm y cm 7 cm 3 cm 4 cm x cm y cm y cm 2 cm 2 cm x cm 7 Find x: a b 3 cm (x-2)¡cm 4 cm 5 cm 13 cm x cm 8 Find the length of AC in: A 9m 5m B D C 9 Find the distance AB, in the following figures. (Hint: It is necessary to draw an additional line or two on the figure in each case.) a b c C D 1m M 3 cm 4 cm B 4m N 5m 7m 6m 3m B A B A A B THE CONVERSE OF PYTHAGORAS’ THEOREM If we have a triangle whose three sides have known lengths, we can use the converse of Pythagoras’ Theorem to test whether (or not) it is right angled. THE CONVERSE OF PYTHAGORAS’ THEOREM 2 2 2 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 If a triangle has sides of length a, b and c units and a + b = c , then the triangle is right angled. black Y:\HAESE\SA_10-6ed\SA10-6_02\039SA10-6_02.CDR Wednesday, 26 July 2006 11:50:46 AM PETERDELL GEOMETRY PACKAGE SA_10-6 40 PYTHAGORAS’ THEOREM (Chapter 2) Example 7 Self Tutor Is the triangle with sides 6 cm, 8 cm and 5 cm right angled? 52 + 62 = 25 + 36 = 61 The two shorter sides have lengths 5 cm and 6 cm, and But 82 = 64 ) 52 + 62 6= 82 and hence the triangle is not right angled. EXERCISE 2B 1 The following figures are not drawn accurately. Which of the triangles are right angled? a b c 7 cm 9 cm 9 cm 12 cm 5 cm 5 cm 4 cm 8 cm 15 cm d e f 3 cm ~`2`7 m ~`7 cm ~`4`8 m 8m 15 m 17 m ~`1`2 cm ~`7`5 m 2 If any of the following triangles (not drawn accurately) is right angled, find the right angle: A C a b c A 2 cm 8m 1 cm B ~`2`4 km B C ~`5 cm 5 km ~`2`0`8 m B C 12 m 7 km A PYTHAGOREAN TRIPLES The simplest right angled triangle with sides of integer length is the 3-4-5 triangle. The numbers 3, 4, and 5 satisfy the rule 32 + 42 = 52 . 5 3 4 The set of integers fa, b, cg is a Pythagorean triple if it obeys the rule a2 + b2 = c2 : magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 50 75 25 0 5 95 100 50 75 25 0 5 cyan 100 f5, 12, 13g, f7, 24, 25g, f8, 15, 17g. Other examples are: black Y:\HAESE\SA_10-6ed\SA10-6_02\040SA10-6_02.CDR Wednesday, 26 July 2006 11:52:34 AM PETERDELL SA_10-6 PYTHAGORAS’ THEOREM (Chapter 2) Example 8 41 Self Tutor Show that f5, 12, 13g is a Pythagorean triple. We find the square of the largest number first. 132 = 169 and 5 + 122 = 25 + 144 = 169 2 52 + 122 = 132 ) i.e., f5, 12, 13g is a Pythagorean triple. 3 Determine if the following are Pythagorean triples: a f8, 15, 17g b f6, 8, 10g d f14, 48, 50g e f1, 2, 3g c f Example 9 f5, 6, 7g f20, 48, 52g Self Tutor Find k if f9, k, 15g is a Pythagorean triple. Let 92 + k2 = 152 ) 81 + k2 = 225 ) k2 = 144 p ) k = § 144 ) k = 12 fPythagorasg fk > 0g 4 Find k if the following are Pythagorean triples: a f8, 15, kg b fk, 24, 26g d f15, 20, kg e fk, 45, 51g c f f14, k, 50g f11, k, 61g 5 For what values of n does fn, n + 1, n + 2g form a Pythagorean triple? 6 Show that fn, n + 1, n + 3g cannot form a Pythagorean triple. INVESTIGATION PYTHAGOREAN TRIPLES SPREADSHEET Well known Pythagorean triples are f3, 4, 5g, f5, 12, 13g, f7, 24, 25g and f8, 15, 17g. Formulae can be used to generate Pythagorean triples. SPREADSHEET cyan magenta yellow Y:\HAESE\SA_10-6ed\SA10-6_02\041SA10-6_02.CDR Friday, 28 July 2006 3:29:17 PM PETERDELL 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 An example is 2n + 1, 2n2 + 2n, 2n2 + 2n + 1 where n is a positive integer. A spreadsheet will quickly generate sets of Pythagorean triples using such formulae. black SA_10-6 42 PYTHAGORAS’ THEOREM (Chapter 2) What to do: 1 Open a new spreadsheet and enter the following: fill down 2 Highlight the formulae in B2, C2 and D2 and fill down to Row 3. You should now have generated two sets of triples: 3 Highlight the formulae in Row 3 and fill down to Row 11 to generate 10 sets of triples. 4 Check that each set of numbers is actually a triple by adding two more columns to your spreadsheet. In E1 enter the heading ‘aˆ2+bˆ2’ and in F1 enter the heading ‘cˆ2’. In E2 enter the formula =B2ˆ2+C2ˆ2 and in F2 enter the formula =D2ˆ2. 5 Highlight the formulae in E2 and F2 and fill down to Row 11. Is each set of numbers a Pythagorean triple? [Hint: Does a2 + b2 = c2 ?] 6 Your task is to prove that the formulae f2n + 1, 2n2 + 2n, 2n2 + 2n + 1g will produce sets of Pythagorean triples for positive integer values of n. We let a = 2n + 1, b = 2n2 + 2n and c = 2n2 + 2n + 1. Simplify c2 ¡b2 = (2n2 +2n+1)2 ¡(2n2 +2n)2 using the difference of two squares factorisation, and hence show that it equals (2n + 1)2 = a2 . C PROBLEM SOLVING USING PYTHAGORAS’ THEOREM Right angled triangles occur frequently in problem solving and often the presence of right angled triangles indicates that Pythagoras’ Theorem is likely to be used. SPECIAL GEOMETRICAL FIGURES All of these figures contain right angled triangles where Pythagoras’ Theorem applies: magenta yellow 95 100 50 0 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 cyan 75 rectangle Construct a diagonal to form a right angled triangle. d 25 In a rectangle, a right angle exists between adjacent sides. l 5 na o iag black Y:\HAESE\SA_10-6ed\SA10-6_02\042SA10-6_02.CDR Wednesday, 26 July 2006 11:53:48 AM PETERDELL SA_10-6 PYTHAGORAS’ THEOREM (Chapter 2) 43 In a square and a rhombus, the diagonals bisect each other at right angles. rhombus square In an isosceles triangle and an equilateral triangle, the altitude bisects the base at right angles. isosceles triangle equilateral triangle Things to remember ² ² ² ² ² ² Draw a neat, clear diagram of the situation. Mark on known lengths and right angles. Use a symbol, such as x, to represent the unknown length. Write down Pythagoras’ Theorem for the given information. Solve the equation. Write your answer in sentence form (where necessary). Example 10 Self Tutor A rectangular gate is 3 m wide and has a 3:5 m diagonal. How high is the gate? Let x m be the height of the gate. fPythagorasg Now (3:5)2 = x2 + 32 ) 12:25 = x2 + 9 ) 3:25 = x2 p ) x = 3:25 fas x > 0g ) x + 1:80 Thus the gate is approximately 1:80 m high. 3m xm 3.5 m EXERCISE 2C.1 1 A rectangle has sides of length 8 cm and 3 cm. Find the length of its diagonals. 2 The longer side of a rectangle is three times the length of the shorter side. If the length of the diagonal is 10 cm, find the dimensions of the rectangle. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 3 A rectangle with diagonals of length 20 cm has sides in the ratio 2 : 1. Find the a perimeter b area of the rectangle. black Y:\HAESE\SA_10-6ed\SA10-6_02\043SA10-6_02.CDR Wednesday, 26 July 2006 11:54:14 AM PETERDELL SA_10-6 44 PYTHAGORAS’ THEOREM (Chapter 2) Example 11 Self Tutor A rhombus has diagonals of length 6 cm and 8 cm. Find the length of its sides. The diagonals of a rhombus bisect at right angles. Let a side be x cm. fPythagorasg ) x2 = 32 + 42 2 ) x = 9 + 16 ) x2 = 25 p ) x = § 25 ) x=5 fx > 0g x cm 3 cm 4 cm i.e., the sides are 5 cm in length. 4 A rhombus has sides of length 6 cm. One of its diagonals is 10 cm long. Find the length of the other diagonal. 5 A square has diagonals of length 10 cm. Find the length of its sides. 6 A rhombus has diagonals of length 8 cm and 10 cm. Find its perimeter. Example 12 Self Tutor A man travels due east by bicycle at 16 kmph. His son travels due south on his bicycle at 20 kmph. How far apart are they after 4 hours, if they both leave point A at the same time? 64 km A 80 km After 4 hours the man has travelled 4 £ 16 = 64 km and his son has travelled 4 £ 20 = 80 km. Thus x2 = 642 + 802 fPythagorasg i.e., x2 = 4096 + 6400 ) x2 = 10 496 p ) x = 10 496 fas x > 0g ) x + 102 ) they are 102 km apart after 4 hours. man x km N W son E S 7 A yacht sails 5 km due west and then 8 km due south. How far is it from its starting point? cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 8 Town A is 50 km south of town B and town C is 120 km east of town B. Is it quicker to travel directly from A to C by car at 90 kmph or from A to C via B in a train travelling at 120 kmph? black Y:\HAESE\SA_10-6ed\SA10-6_02\044SA10-6_02.CDR Wednesday, 26 July 2006 11:54:38 AM PETERDELL SA_10-6 PYTHAGORAS’ THEOREM (Chapter 2) 45 9 Two runners set off from town A at the same time. If one runs due east to town B and the other runs due south to town C at twice the speed, they arrive at B and C respectively two hours later. If B and C are 50 km apart, find the speed at which each runner travelled. Example 13 Self Tutor An equilateral triangle has sides of length 6 cm. Find its area. The altitude bisects the base at right angles. ) a2 + 32 = 62 ) a2 + 9 = 36 ) a2 = 27 p ) a = § 27 p ) a = 27 6 cm a cm fPythagorasg fa > 0g 1 2 base £ height p = 12 £ 6 £ 27 p = 3 27 cm2 + 15:6 cm2 f1 d.p.g Now, area = 3 cm So, area is 15:6 cm2 : 10 Find any unknowns in the following: a b c 45° x cm 2 cm h cm 7 cm x cm 60° 1 cm y cm 30° x cm y° 12 cm 11 An equilateral triangle has sides of length 12 cm. Find the length of one of its altitudes. 12 An isosceles triangle has equal sides of length 8 cm and a base of length 6 cm. Find the area of the triangle. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 13 When an extension ladder rests against a wall it reaches 4 m up the wall. The ladder is extended a further 0:8 m without moving the foot of the ladder and it now rests against the wall 1 m further up. How long is the extended ladder? black Y:\HAESE\SA_10-6ed\SA10-6_02\045SA10-6_02.CDR Wednesday, 26 July 2006 11:55:01 AM PETERDELL 1m 4m SA_10-6 46 PYTHAGORAS’ THEOREM (Chapter 2) p 14 An equilateral triangle has area 16 3 cm2 . Find the length of its sides. 15 Revisit the Opening Problem on page 34 and answer the questions posed. TRUE BEARINGS When using true bearings we measure the direction of travel by comparing it with the true north direction. Measurements are always taken in the clockwise direction. Imagine you are standing at point A, facing north. You turn clockwise through an angle until you face B. The bearing of B from A is the angle through which you have turned. That is, the bearing of B from A is the measure of the angle between AB and the ‘north’ line through A. north 72° B north A In the diagram at right, the bearing of B from A is 72o from true north. We write this as 72o T or 072o . B If we want to find the true bearing of A from B, we place ourselves at point B and face north and then measure the clockwise angle through which we have to turn so that we face A. The true bearing of A from B is 252o . 252° A Example 14 Self Tutor A helicopter travels from base station S on a true bearing of 074o for 112 km to outpost A. It then travels 134 km on a true bearing of 164o to outpost B. How far is outpost B from base station S? Let SB be x km. From the diagram alongside, in triangle SAB ]SAB = 90o . x2 = 1122 + 1342 ) x2 = 12 544 + 17 956 ) x2 = 30 500 p ) x = 30 500 ) x + 175 In bearings problems, notice the use of the properties of parallel lines for finding angles. N fPythagorasg N 112 km 74° A 164° 74° 16° S 134 km fas x > 0g x km i.e., outpost B is 175 km from base station S. B EXERCISE 2C.2 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 1 Two bushwalkers set off from base camp at the same time. If one walks on a true bearing of 049o at an average speed of 5 kmph and the other walks on a true bearing of 319o at an average speed of 4 kmph, find their distance apart after 3 hours. black Y:\HAESE\SA_10-6ed\SA10-6_02\046SA10-6_02.CDR Thursday, 3 August 2006 10:08:10 AM PETERDELL SA_10-6 PYTHAGORAS’ THEOREM (Chapter 2) 47 2 James is about to tackle an orienteering course. He has been given these instructions: ² the course is triangular and starts and finishes at S ² the first checkpoint A is in a direction 056o from S ² the second checkpoint B is in a direction 146o from A ² the distance from A to B is twice the distance from S to A ² the distance from B to S is 2:6 km. Find the length of the orienteering course. 3 A fighter plane and a helicopter set off from airbase A at the same time. If the helicopter travels on a bearing of 152o and the fighter plane travels on a bearing of 242o at three times the speed, they arrive at bases B and C respectively 2 hours later. If B and C are 1200 km apart, find the average speed of the helicopter. D THREE-DIMENSIONAL PROBLEMS Pythagoras’ Theorem is often used when finding lengths in three-dimensional solids. Example 15 Self Tutor A 50 m rope is attached inside an empty cylindrical wheat silo of diameter 12 m as shown. How high is the wheat silo? 12 m 50 m 12 m Let the height be h m. ) ) hm 50 m h2 + 122 = 502 h2 + 144 = 2500 ) h2 = 2356 p ) h = 2356 ) h + 48:5 i.e., the wheat silo is fPythagorasg fas h > 0g fto 1 dec. placeg 48:5 m high. EXERCISE 2D 1 A cone has a slant height of 17 cm and a base radius of 8 cm. How high is the cone? 2 Find the length of the longest nail that could be put entirely within a cylindrical can of radius 3 cm and height 8 cm. 3 A 20 cm nail just fits inside a cylindrical can. Three identical spherical balls need to fit entirely within the can. What is the maximum radius of each ball? cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 In three-dimensional problem solving questions we often need the theorem of Pythagoras twice. We look for right angled triangles which have two sides of known length. black Y:\HAESE\SA_10-6ed\SA10-6_02\047SA10-6_02.CDR Wednesday, 26 July 2006 11:57:04 AM PETERDELL SA_10-6 48 PYTHAGORAS’ THEOREM (Chapter 2) Example 16 Self Tutor A room is 6 m by 4 m at floor level and the floor to ceiling height is 3 m. Find the distance from a floor corner point to the opposite corner point on the ceiling. The required distance is AD. We join BD. fPythagorasg In ¢BCD, x2 = 42 + 62 2 2 2 In ¢ABD, y = x + 3 fPythagorasg ym 2 2 2 2 ) y = 4 +6 +3 ) y2 = 61 p D ) y = § 61 p ) the required distance is 61 + 7:81 m. A 3m B 4m xm 6m C But y > 0 4 A cube has sides of length 3 cm. Find the length of a diagonal of the cube. diagonal 5 A room is 7 m by 5 m and has a height of 3 m. Find the distance from a corner point on the floor to the opposite corner of the ceiling. 6 A rectangular box is 2 cm by 3 cm by 2 cm (internally). Find the length of the longest toothpick that can be placed within the box. 7 Determine the length of the longest piece of timber which could be stored in a rectangular shed 6 m by 5 m by 2 m high. Example 17 Self Tutor A pyramid of height 40 m has a square base with edges 50 m. Determine the length of the slant edges. Let a slant edge have length s m. Let half a diagonal have length x m. x2 + x2 = 502 ) 2x2 = 2500 ) x2 = 1250 Using xm 40 m sm xm 50 m fPythagorasg s2 = x2 + 402 fPythagorasg ) s2 = 1250 + 1600 2 40 m ) s = 2850 p fas s > 0g ) s = 2850 ) s + 53:4 fto 1 dec. placeg Using xm sm 50 m xm cyan magenta yellow Y:\HAESE\SA_10-6ed\SA10-6_02\048SA10-6_02.CDR 17 August 2006 09:02:48 DAVID2 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 i.e., each slant edge is 53:4 m long. black SA_10-6 PYTHAGORAS’ THEOREM (Chapter 2) 49 E 8 ABCDE is a square-based pyramid. E, the apex of the pyramid is vertically above M, the point of intersection of AC and BD. If an Egyptian Pharoah wished to build a square-based pyramid with all edges 100 m, how high (to the nearest metre) would the pyramid reach above the desert sands? C D M A B 9 A symmetrical square-based pyramid has height 10 cm and slant edges of 15 cm. Find the dimensions of its square base. B 10 A cube has sides of length 2 m. B is at the centre of one face, and A is an opposite vertex. Find the direct distance from A to B. A E MORE DIFFICULT PROBLEMS (EXTENSION) PRINTABLE EXERCISE Click on the icon to obtain a printable exercise on more difficult problems requiring a solution using Pythagoras’ Theorem. REVIEW SET 2A 1 Find the lengths of the unknown sides in the following triangles: 2 cm 4 cm a b c 5 cm x cm x cm 9 cm 7 cm 2x cm x cm A 2 Is the following triangle right angled? Give evidence. 1 4 C Ö̀1̀7 B 3 Show that f5, 11, 13g is not a Pythagorean triple. 4 A rectangle has diagonal 15 cm and one side 8 cm. Find the perimeter of the rectangle. 5 An isosceles triangle has equal sides of length 12 cm and a base of length 8 cm. Find the area of the triangle. 6 A boat leaves X and travels due east for 10 km. It then sails 10 km south to Y. Find the distance and bearing of X from Y. cyan magenta yellow Y:\HAESE\SA_10-6ed\SA10-6_02\049SA10-6_02.CDR 17 August 2006 09:08:20 DAVID2 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 7 What is the length of the longest toothpick which can be placed inside a rectangular box that is 3 cm £ 5 cm £ 8 cm? black SA_10-6 50 PYTHAGORAS’ THEOREM (Chapter 2) 8 Two rally car drivers set off from town C at the same time. A travels in a direction 63oT at 120 kmph and B travels in a direction 333oT at 135 kmph. How far apart are they after one hour? REVIEW SET 2B 1 Find the value of x in the following: a b x cm xm Ö̀7 cm 5 cm 2x c Ö̀4̀2 5m 5x 6m B 2 Show that the following triangle is right angled and state which vertex is the right angle: 5 2 C A ~`2`9 3 A rectangular gate is twice as wide as it is high. If a diagonal strut is 3:2 m long, find the height of the gate to the nearest millimetre. 4 If a softball diamond has sides of length 30 m, determine the distance a fielder must throw the ball from second base to reach home base. 5 Town B is 27 km in a direction 134o T from town A, and town C is 21 km in a direction 224o T from town B. Find the distance between A and C. 6 If a 15 m ladder reaches twice as far up a vertical wall as the base is out from the wall, determine the distance up the wall to the top of the ladder. 7 Can an 11 m long piece of timber be placed in a rectangular shed of dimensions 8 m by 7 m by 3 m? Give evidence. cyan magenta 52 km A yellow Y:\HAESE\SA_10-6ed\SA10-6_02\050SA10-6_02.CDR 17 August 2006 09:14:45 DAVID2 95 100 50 X 75 25 0 B 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 8 Straight roads from towns A and B intersect at right angles at X. A and B are 52 km apart. Mika leaves A at the same time as Toshi leaves B. Mika cycles at 24 km/h and Toshi jogs at 10 km/h, towards X, and they arrive at X at the same time. For how long were they travelling? black SA_10-6

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