A B C D E. Pythagoras Theorem

Transcription

1 2 Pythagoras Theorem ontents: D E Pythagoras Theorem The onverse of Pythagoras Theorem Problem solving using Pythagoras Theorem Three-dimensional problems More diffiult problems (Extension)

2 34 PYTHGORS THEOREM (hapter 2) Right angles (90 o angles) are used in the onstrution of buildings and in the division of areas of land into retangular regions. The anient Egyptians used a rope with 12 equally spaed 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 fat, the simplest right angled triangle with sides of integer length. orner take hold of knots at arrows make rope taut line of one side of building OPENING PROLEM Karrie is playing golf in the US Open but hits a wayward tee shot on the opening hole. Her addy paes out some distanes 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. marker whih is 150 m from the pin is further up the fairway as shown: Tee addy 150 m marker Hole 250 m 70 m all onsider the following questions: 1 How far is the addy away from the tee? 2 From where the addy stands on the fairway, what distane 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 seond shot to reah the hole?

3 PYTHGORS THEOREM (hapter 2) 35 INTRODUTION right angled triangle is a triangle whih has a right angle as one of its angles. The side opposite the right angle is alled the hypotenuse and is the longest side of the triangle. The other two sides are alled the legs of the triangle. round 500, the Greek mathematiian Pythagoras formulated a rule whih onnets the lengths of the sides of all right angled triangles. It is thought that he disovered the rule while studying tessellations of tiles on bathroom floors. Suh patterns, like the one illustrated, were ommon on the walls and floors of bathrooms in anient Greee. hypotenuse legs PYTHGORS THEOREM b a In a right angled triangle, with hypotenuse and legs a and b, 2 = a 2 + b 2. In geometri form, the Theorem of Pythagoras is: 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. GEOMETRY PKGE 2 a an you see how Pythagoras may have disovered the rule by looking at the tile pattern above? b a a 2 b 2 b

4 ` ` 36 PYTHGORS THEOREM (hapter 2) Over 400 different proofs of the Pythagorean Theorem exist. Here is one of them: Proof: On a square we draw 4 idential (ongruent) right angled triangles, as illustrated. smaller square in the entre is formed. Suppose the legs are of length a and b and the hypotenuse has length : Sine total area of large square =4 area of one triangle + area of smaller square, (a + b) 2 =4( 1 2 ab)+2 ) a 2 +2ab + b 2 =2ab + 2 ) a 2 + b 2 = 2 Note: When using Pythagoras Theorem we often see p numbers like 7: b a b a surds, whih are square root a b a b Example 1 Find the length of the hypotenuse in: 2m 3m The hypotenuse is opposite the right angle and has length. ) x 2 = ) x 2 =9+4 ) x 2 =13 i.e., x = p 13 fas x>0g ) the hypotenuse is p 13 m. 2 = If x k, then x = p k, but we rejet p k as lengths must be positive! EXERISE 2 1 Find the length of the hypotenuse in the following triangles, leaving your answer in surd (square root) form if appropriate: a b 4m 7m 5m x km 8km 13 km

5 Example 2 Find the length of the third side of: 6m 5m PYTHGORS THEOREM (hapter 2) 37 The hypotenuse has length 6 m. ) x =6 2 fpythagorasg ) x 2 +25=36 ) x 2 =11 ) x = p 11 fas x>0g ) third side is p 11 m long. 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 x km 6m 11 m 2.8 km 1.9 km 9.5 m Example 3 Find x in the following: ~`1`0 m 2m The hypotenuse has length. ) x 2 =2 2 +( p 10) 2 fpythagorasg ) x 2 =4+10 ) x 2 =14 ) x = p 14 ut x>0, ) x = p Find x in the following: a b 3m ~`7 m ~`2 m ~`5 m ~`1`0 m Example 4 Solve for x: x =1 2 fpythagorasg ) x =1 1m Qw_ m ) x 2 = 3 4 q ) x = ) x = 3 4 q 3 4 fas x>0g

6 38 PYTHGORS THEOREM (hapter 2) 4 Solve for x: a b 1m Qw_ m Qw_ m Ew_ m 2 1m m x m Example 5 Find the value of x: (2x) 2 = x fpythagorasg ) 4x 2 = x x m 6m 2x m ) 3x 2 =36 ) x 2 =12 ) x = p 12 ut x>0, ) x = p Find the value of x: a b 9m 2x m 26 m 3x m 2x m 2x m 3x m ~`2`0 m Example 6 Find the value of any unknowns: 5m y m 1m In triangle, the hypotenuse is. ) x 2 = fpythagorasg ) x 2 =26 ) x = p 26 ) x = p 26 fx >0g In D, the hypotenuse is 6 m. ) y 2 +( p 26) 2 =6 2 fpythagorasg ) y 2 +26=36 ) y 2 =10 ) y = p 10 ) y = p 10 fy >0g D 6m

7 PYTHGORS THEOREM (hapter 2) 39 6 Find the value of any unknowns: a b 2m 1m 7 Find x: a 2m y m 3m b 4m y m 7m ( x ) m 3m 2m y m 3m 4m 5m 13 m 8 Find the length of in: 5m 9m D 9 Find the distane, in the following figures. (Hint: It is neessary to draw an additional line or two on the figure in eah ase.) a b D 4m 3m 6m 4m 7m M 3m 5m 1m N THE ONVERSE OF PYTHGORS THEOREM If we have a triangle whose three sides have known lengths, we an use the onverse of Pythagoras Theorem to test whether (or not) it is right angled. THE ONVERSE OF PYTHGORS THEOREM If a triangle has sides of length a, b and units and a 2 + b 2 = 2, then the triangle is right angled. GEOMETRY PKGE

8 40 PYTHGORS THEOREM (hapter 2) Example 7 Is the triangle with sides 6 m, 8 m and 5 m right angled? The two shorter sides have lengths 5 m and 6 m, and =25+36 =61 ut 8 2 =64 ) =8 2 and hene the triangle is not right angled. EXERISE 2 1 The following figures are not drawn aurately. Whih of the triangles are right angled? a b 7m 9m 12 m 9m 5m 5m 4m 15 m 8m d e f 3m ~`1`2 m ~`7 m ~` `7 m ~`7`5 m ~`4`8 m 15 m 8m 17 m 2 If any of the following triangles (not drawn aurately) is right angled, find the right angle: a b 2 m 1m 8m ~`2`0`8 m ~`2`4 km 5km ~`5 m 12 m 7km PYTHGOREN TRIPLES The simplest right angled triangle with sides of integer length is the triangle. The numbers 3, 4, and 5 satisfy the rule = The set of integers fa, b, g is a Pythagorean triple if it obeys the rule a 2 + b 2 = 2 : Other examples are: f5, 12, 13g, f7, 24, 25g, f8, 15, 17g.

9 PYTHGORS THEOREM (hapter 2) 41 Example 8 Show that f5, 12, 13g is a Pythagorean triple. We find the square of the largest number first = 169 and = = 169 ) =13 2 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 f f5, 6, 7g f20, 48, 52g Example 9 Find k if f9, k, 15g is a Pythagorean triple. Let k 2 =15 2 fpythagorasg ) 81 + k 2 = 225 ) k 2 = 144 ) k = p 144 ) k =12 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 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 annot form a Pythagorean triple. INVESTIGTION PYTHGOREN TRIPLES SPREDSHEET Well known Pythagorean triples are f3, 4, 5g, f5, 12, 13g, f7, 24, 25g and f8, 15, 17g. Formulae an be used to generate Pythagorean triples. SPREDSHEET n example is 2n +1, 2n 2 +2n, 2n 2 +2n +1 where n is a positive integer. spreadsheet will quikly generate sets of Pythagorean triples using suh formulae.

10 42 PYTHGORS THEOREM (hapter 2) What to do: 1 Open a new spreadsheet and enter the following: fill down 2 Highlight the formulae in 2, 2 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 hek that eah set of numbers is atually a triple by adding two more olumns to your spreadsheet. In E1 enter the heading aˆ2+bˆ2 and in F1 enter the heading ˆ2. In E2 enter the formula =2ˆ2+2ˆ2 and in F2 enter the formula =D2ˆ2. 5 Highlight the formulae in E2 and F2 and fill down to Row 11. Is eah set of numbers a Pythagorean triple? [Hint: Does a 2 + b 2 = 2?] 6 Your task is to prove that the formulae f2n +1, 2n 2 +2n, 2n 2 +2n +1g will produe sets of Pythagorean triples for positive integer values of n. We let a =2n +1, b =2n 2 +2n and =2n 2 +2n +1. Simplify 2 b 2 =(2n 2 +2n+1) 2 (2n 2 +2n) 2 using the differene of two squares fatorisation, and hene show that it equals (2n +1) 2 = a 2. PROLEM SOLVING USING PYTHGORS THEOREM Right angled triangles our frequently in problem solving and often the presene of right angled triangles indiates that Pythagoras Theorem is likely to be used. SPEIL GEOMETRIL FIGURES ll of these figures ontain right angled triangles where Pythagoras Theorem applies: diagonal retangle In a retangle, a right angle exists between adjaent sides. onstrut a diagonal to form a right angled triangle.

11 PYTHGORS THEOREM (hapter 2) 43 In a square and a rhombus, the diagonals biset eah other at right angles. square rhombus In an isoseles triangle and an equilateral triangle,thealtitude bisets the base at right angles. isoseles triangle equilateral triangle Things to remember ² Draw a neat, lear diagram of the situation. ² Mark on known lengths and right angles. ² Use a symbol, suh as x, to represent the unknown length. ² Write down Pythagoras Theorem for the given information. ² Solve the equation. ² Write your answer in sentene form (where neessary). Example 10 retangular gate is 3 m wide and has a 3:5 m diagonal. How high is the gate? 3m 3.5 m x m Let x m be the height of the gate. Now (3:5) 2 = x fpythagorasg ) 12:25 = x 2 +9 ) 3:25 = x 2 ) x = p 3:25 fas x>0g ) x + 1:80 Thus the gate is approximately 180 : m high. EXERISE retangle has sides of length 8 m and 3 m. Find the length of its diagonals. 2 The longer side of a retangle is three times the length of the shorter side. If the length of the diagonal is 10 m, find the dimensions of the retangle. 3 retangle with diagonals of length 20 m has sides in the ratio 2:1. Find the a perimeter b area of the retangle.

12 44 PYTHGORS THEOREM (hapter 2) Example 11 rhombus has diagonals of length 6 m and 8 m. Find the length of its sides. 3m 4m The diagonals of a rhombus biset at right angles. Let a side be. ) x 2 = fpythagorasg ) x 2 =9+16 ) x 2 =25 ) x = p 25 ) x =5 fx >0g i.e., the sides are 5 m in length. 4 rhombus has sides of length 6 m. One of its diagonals is 10 m long. Find the length of the other diagonal. 5 square has diagonals of length 10 m. Find the length of its sides. 6 rhombus has diagonals of length 8 m and 10 m. Find its perimeter. Example 12 man travels due east by biyle at 16 kmph. His son travels due south on his biyle at 20 kmph. How far apart are they after 4 hours, if they both leave point at the same time? 64 km man fter 4 hours the man has travelled 4 16 = 64 km and his son has travelled 4 20 = 80 km. Thus x 2 = fpythagorasg 80 km son x km W N E i.e., x 2 = ) x 2 = ) x = p fas x>0g ) x S ) they are 102 km apart after 4 hours. 7 yaht sails 5 km due west and then 8 km due south. How far is it from its starting point? 8 Town is 50 km south of town and town is 120 km east of town. Is it quiker to travel diretly from to by ar at 90 kmph or from to via in a train travelling at 120 kmph?

13 9 Two runners set off from town at the same time. If one runs due east to town and the other runs due south to town at twie the speed, they arrive at and respetively two hours later. If and are 50 km apart, find the speed at whih eah runner travelled. PYTHGORS THEOREM (hapter 2) 45 Example 13 n equilateral triangle has sides of length 6 m. Find its area. a m The altitude bisets the base at right angles. ) a =6 2 fpythagorasg ) a 2 +9=36 ) a 2 =27 ) a = p 27 6m ) a = p 27 fa >0g Now, area = 1 2 base height 3m = p 27 =3 p 27 m :6 m 2 f1 d.p.g So, area is 15:6 m 2 : 10 Find any unknowns in the following: a b m h m 2m y 12 m y m 1m n equilateral triangle has sides of length 12 m. Find the length of one of its altitudes. 12 n isoseles triangle has equal sides of length 8 m and a base of length 6 m. Find the area of the triangle. 13 When an extension ladder rests against a wall it reahes 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? 1m 4m

14 46 PYTHGORS THEOREM (hapter 2) 14 n equilateral triangle has area 16 p 3 m 2. Find the length of its sides. 15 Revisit the Opening Problem on page 34 and answer the questions posed. TRUE ERINGS When using true bearings we measure the diretion of travel by omparing it with the true north diretion. Measurements are always taken in the lokwise diretion. Imagine you are standing at point, faing north. You turn lokwise through an angle until you fae. The bearing of from is the angle through whih you have turned. That is, the bearing of from is the measure of the angle between and the north line through. In the diagram at right, the bearing of from is 72 o from true north. We write this as 72 o Tor072 o. If we want to find the true bearing of from, we plae ourselves at point and fae north and then measure the lokwise angle through whih we have to turn so that we fae. The true bearing of from is 252 o. north 72 north 252 Example 14 heliopter travels from base station S on a true bearing of o 074 for 112 km to outpost. It then travels 134 km on a true o bearing of 164 to outpost. How far is outpost from base station S? Let S be x km. From the diagram alongside, in triangle S ]S =90 o. x 2 = fpythagorasg ) x 2 = ) x 2 = ) x = p fas x>0g ) x i.e., outpost is 175 km from base station S. S N km x km In bearings problems, notie the use of the properties of parallel lines for finding angles. N km EXERISE Two bushwalkers set off from base amp at the same time. If one walks on a true bearing of 049 o at an average speed of 5 kmph and the other walks on a true bearing of 319 o at an average speed of 4 kmph, find their distane apart after 3 hours.

15 PYTHGORS THEOREM (hapter 2) 47 2 James is about to takle an orienteering ourse. He has been given these instrutions: ² the ourse is triangular and starts and finishes at S ² the first hekpoint is in a diretion 056 o from S ² the seond hekpoint is in a diretion 146 o from ² the distane from to is twie the distane from S to ² the distane from to S is 2:6 km. Find the length of the orienteering ourse. 3 fighter plane and a heliopter set off from airbase at the same time. If the heliopter travels on a bearing of 152 o and the fighter plane travels on a bearing of 242 o at three times the speed, they arrive at bases and respetively 2 hours later. If and are 1200 km apart, find the average speed of the heliopter. D THREE-DIMENSIONL PROLEMS Pythagoras Theorem is often used when finding lengths in three-dimensional solids. Example m rope is attahed inside an empty ylindrial wheat silo of diameter 12 m as shown. How high is the wheat silo? 12 m 50 m 50 m 12 m Let the height be h m. ) h =50 2 fpythagorasg ) h = 2500 h m ) h 2 = 2356 ) h = p 2356 fas h>0g ) h + 48:5 fto 1 de. plaeg i.e., the wheat silo is 48:5 m high. EXERISE 2D 1 one has a slant height of 17 m and a base radius of 8 m. How high is the one? 2 Find the length of the longest nail that ould be put entirely within a ylindrial an of radius 3 m and height 8 m m nail just fits inside a ylindrial an. Three idential spherial balls need to fit entirely within the an. What is the maximum radius of eah ball? In three-dimensional problem solving questions we often need the theorem of Pythagoras twie. We look for right angled triangles whih have two sides of known length.

16 48 PYTHGORS THEOREM (hapter 2) Example 16 room is 6 mby4 m at floor level and the floor to eiling height is 3 m. Find the distane from a floor orner point to the opposite orner point on the eiling. 3m 4m x m 6m y m D The required distane is D. We join D. In D, x 2 = fpythagorasg In D, y 2 = x fpythagorasg ) y 2 = ) y 2 =61 ) y = p 61 ut y > 0 ) the required distane is p :81 m. 4 ube has sides of length 3 m. Find the length of a diagonal of the ube. diagonal 5 room is 7 mby5 m and has a height of 3 m. Find the distane from a orner point on the floor to the opposite orner of the eiling. 6 retangular box is 2 m by 3 m by 2 m (internally). Find the length of the longest toothpik that an be plaed within the box. 7 Determine the length of the longest piee of timber whih ould be stored in a retangular shed 6 mby5 mby2 m high. Example 17 pyramid of height 40 m has a square base with edges 50 m. Determine the length of the slant edges. s m x m 50 m 40 m Let a slant edge have length s m. Let half a diagonal have length x m. Using x 2 + x 2 =50 2 fpythagorasg x m x m ) 2x 2 = 2500 ) x 2 = m Using s 2 = x fpythagorasg ) s 2 = s m 40 m ) s 2 = 2850 ) s = p 2850 fas s > 0g x m ) s + 53:4 fto 1 de. plaeg i.e., eah slant edge is 53:4 m long.

17 PYTHGORS THEOREM (hapter 2) DE is a square-based pyramid. E, the apex of the pyramid is vertially above M, the point of intersetion of and D. 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 reah above the desert sands? symmetrial square-based pyramid has height 10 m and slant edges of 15 m. Find the dimensions of its square base. 10 ube has sides of length 2 m. is at the entre of one fae, and is an opposite vertex. Find the diret distane from to. D E M E MORE DIFFIULT PROLEMS (EXTENSION) lik on the ion to obtain a printable exerise on more diffiult problems requiring a solution using Pythagoras Theorem. PRINTLE EXERISE REVIEW SET 2 1 Find the lengths of the unknown sides in the following triangles: a 2m b 4m 5m 7m 9m 2x m 2 Is the following triangle right angled? Give evidene Show that f5, 11, 13g is not a Pythagorean triple. 4 retangle has diagonal 15 m and one side 8 m. Find the perimeter of the retangle. 5 n isoseles triangle has equal sides of length 12 m and a base of length 8 m. Find the area of the triangle. 6 boat leaves X and travels due east for 10 km. It then sails 10 km south to Y. Find the distane and bearing of X from Y. 7 What is the length of the longest toothpik whih an be plaed inside a retangular box that is 3 m 5 m 8 m?

18 50 PYTHGORS THEOREM (hapter 2) 8 Two rally ar drivers set off from town at the same time. travels in a diretion o 63 T at 120 kmph and travels in a o diretion 333 T at 135 kmph. How far apart are they after one hour? REVIEW SET 2 1 Find the value of x in the following: a b 2x 5m 7 m x m 5m 5x 6m 2 Show that the following triangle is right angled and state whih vertex is the right angle: 2 ~`2`9 5 3 retangular gate is twie 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 distane a fielder must throw the ball from seond base to reah home base. 5 Town is 27 km in a diretion 134 o T from town, and town is 21 km in a diretion 224 o T from town. Find the distane between and. 6 If a 15 m ladder reahes twie as far up a vertial wall as the base is out from the wall, determine the distane up the wall to the top of the ladder. 7 an an 11 m long piee of timber be plaed in a retangular shed of dimensions 8 mby 7 m by 3 m? Give evidene. 8 Straight roads from towns and interset at right angles at X. and are 52 km apart. Mika leaves at the same time as Toshi leaves. Mika yles 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? 52 km X