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Pythagoras’ Theorem CHAPTER 2 37 7. 1 cm 9 cm 12 cm 16 cm S K R P Q L ,Q WKH GLDJUDP PQRS LV D UHFWDQJOH DQG KLS LV D ULJKWDQJOHG WULDQJOH )LQG WKH OHQJWKLQFPRILQ 8. A Y X D H C 8 cm 6 cm 18 cm 3 cm E B F G 7KHGLDJUDPVKRZVDUHFWDQJXODUER[ZKLFK KDVDKRUL]RQWDOEDVHBFGCYLVDSRLQWRQ ADVXFKWKDWYD FPDQGXLVDSRLQWRQ BCDQGYHUWLFDOO\EHORZY)LQGWKHOHQJWKRI FY*LYH\RXUDQVZHU FRUUHFWWRGHFLPDO SODFHV 9. P Q S R 60° 12 cm 7KHGLDJUDPVKRZVDUHFWDQJOHPQRS)LQG D WKHOHQJWKRIPS E WKHOHQJWKRIPQ *LYH\RXUDQVZHUFRUUHFWWRGHFLPDOSODFH LIQHFHVVDU\ 10. 15 km 30 km N Ship Harbour $VKLSOHDYHVKDUERXUDQGVDLOVNP(DVW DQG WKHQ NP 1RUWK +RZ IDU LV LW IURP WKHKDUERXUWRWKHVKLS"*LYH\RXUDQVZHU FRUUHFWWRGHFLPDOSODFHV 11. 5 m 12 m $ PHWUH KLJK WHQW SROH LV VHFXUHG ZLWK ZLUHVIURPWKHWRSRIWKHSROHWRSRLQWVRQ WKH JURXQG PHWUH IURP WKH EDVH RI WKH SROH)LQGWKHOHQJWKRIHDFKZLUH 12. $K &KRQJ ZDQWHG WR FURVV D ULYHU ZKLFK LV PZLGHIURPSRLQWA to SRLQWB'XHWRWKH ULYHU FXUUHQWKHÀQDOO\ODQGHG DWSRLQWCZKLFKZDVP IURP SRLQW B DV VKRZQ LQWKHGLDJUDPDERYH:KDWZDVWKHDFWXDO GLVWDQFHWUDYHOOHGE\$K&KRQJ" 13. 15 m 5 m 8 m Rope 7KHGLDJUDPDERYHVKRZVWZRSROHVRQWKH KRUL]RQWDOJURXQG7KHGLVWDQFHEHWZHHQWKH WZRSROHVLVP*LYHQWKDWWKHKHLJKWRI WKHWZRSROHVDUHPDQGflPUHVSHFWLYHO\ ,IDURSHLVWLHGWRWKHWZRHQGVRIWKHSROHV ÀQGWKHVKRUWHVWOHQJWKRIURSHQHHGHG 14. 7KH GLDJUDP VKRZV D P ORQJ ODGGHU RQ D KRUL]RQWDO ÁRRU OHDQLQJ DJDLQVW WKH ZDOO ,I WKH WRS RI WKH ODGGHU LV P YHUWLFDOO\DERYHWKHÁRRUDQGWKHIRRWRIWKH ODGGHU LV ffi P IURP WKH ZDOO GHWHUPLQH ZKHWKHUWKHZDOOLVYHUWLFDO 15. $ VFRXW KDV WKUHH EDPERR VWHPV 7KH OHQJWKVRIWKHVWHPVDUHPPDQG P UHVSHFWLYHO\ +H IRUPV D WULDQJOH XVLQJ WKH VWHPV ,V WKH WULDQJOH D ULJKW DQJOHGDQREWXVHDQJOHGRUDQDFXWHDQJOHG WULDQJOH" A C B 1.2 m 1.5 m 0.9 m ©Praxis Publishing_Focus On Maths
3 Applications of this chapter Applications of this chapter Circles are present in real life, both in the natural world and in man-made creations. Circles are frequently used in architecture, science, construction, transportation and etc. Circles are still evident in transportation where they appear in vehicle tyres, roundabouts in roads, engine crankshafts, and road designs. Ferris wheels take the circle to vertical heights at amusement parks and carnivals. Many household items, including cups, candles, and doorknobs have circles in their designs. CIRCLES Why do you think that roundabouts are designed to be in a circular shape? 38 ©Praxis Publishing_Focus On Maths
Maths History • Circle • Centre of circle • Radius • Circumference • Chord • Diameter • Semicircle • Arc • Sector • Segment • Pi, π • Angle subtended by an arc at the centre • Angle subtended by an arc at the circumference • Identify and construct parts of a circle. • Understand and use the symmetrical properties of chords. • Find the circumference, diameter or radius, length of an arc, angle subtended at the centre, areas of circle and sector. Concept Map Key Terms Learning Outcomes Euclid of Alexandria wrote a set of books called The Elements. Book 3 of The Elements deals with circles and their properties. 39 Parts of a Circle Circumference and Area of a Circle Parts of a Circle Construction of a Circle and Parts of a Circle Symmetrical Properties of Chords Circumference of a Circle Area of a Circle Length of an Arc and Area of a Sector Circles ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 40 Look at the picture above. (a) What shapes do you see? (b) What do you know about the shapes in (a)? (c) What might be useful to know about the shapes in (a)? How can the length of a curved line be measured? Critical Thinking 2. Find the area of each diagram. (a) 12 mm 9 mm (b) 19.5 cm 15 cm 7.5 cm Flashback 1. Find the perimeter of each diagram. (a) 22.9 cm 17.3 cm (b) 28.2 cm 33 cm 1 ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 41 3.1 Parts of a Circle A Circles Objective: To explore the concept of a circle. Instruction: Do this activity in pairs. 1. 2SHQWKHÀOHConcept of a Circle using GeoGebra. 2. 'UDJWKH ¶'LVWDQFH2$· VOLGHUWR VHWWKHGLVWDQFH between point O and the red point. 3. Drag the red point to see the shape produced which are of the same distance from point O. 4. Discuss and state a conclusion that can be made. 5. 3UHVHQW\RXUÀQGLQJVLQFODVV 1 From the activity above, a circle is a set of points equidistant from D À[HG SRLQW 7KHUHIRUH D FLUFOH LV WKH ORFXV RI SRLQts moving at a constant distance frRPDÀ[HGSRLQW B Identifying parts of a circle ,IDOOSRLQWVZLWKDFRQVWDQWGLVWDQFHIURPDÀ[HGSRLQWDUHMRLQHG together, a circle will be obtained. D 7KHÀ[HGSRLQWLVNQRZQDVWKHcentre (O) of the circle. (b) The constant distance is known as the radius (r) of the circle. (c) The perimeter of the circle is known as the circumference of the circle. Look around your classroom and list down all the circular objects that you can see. INTERACTIVE ZONE Objective: To explore parts of a circle. Instruction: Do this activity in groups of four. 1. 2SHQWKHÀOHParts of a Circle using GeoGebra. 2. 6HOHFW¶&KRUGAB·IURPWKHPHQX (a) Drag point A or point B to change the length of the chord AB. (b) Explain the meaning of chord based on your observation. (c) When is the chord AB the longest? 2 Fixed point Centre Radius Circumference ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 42 1EXAMPLE 1 Find the squares of the following numbers. (a) 70 (b) 1 3 5 (c) 0.2 (d) –5 Solution: (a) 702 = 70 = 70 = 4900 (b) 8 5 2 = 8 5 = 8 5 = 64 25 (c) (0.2)2 = 0.2 = 0.2 = 0.04 (d) (–5)2 = (–5) = (–5) = 25 EXAMPLE 2 Find the square root of each of the following numbers. (a) 16 (b) 49 (c) 81 (d) 144 Solution: (a) 16 = 4 = 4 = 4 (b) 49 = 7 = 7 = 7 (c) 81 = 9 = 9 = 9 (d) 144 = 12 = 12 = 12 EXAMPLE 3 8VLQJDFDOFXODWRUÀQGWKHYDOXHVRIWKHIROORZLQJ&RUUHFW your answer to 3 decimal places where necessary. (a) 17 (b) 526 (c) 0.28 (d) 0.094 3. 6HOHFW¶0LQRUDUFAB·RU¶0DMRUDUFAB·IURPWKHPHQX&OLFNRQWKHFKHFNER[LIQHFHVVDU\ (a) Drag point A or point B to change the length of the arc AB. E ([SODLQWKHPHDQLQJRIPLQRUDUFDQGPDMRUDUFEDVHGRQ\RXUREVHUYDWLRQV 4. 6HOHFW ¶0LQRU VHFWRU OAB· RU ¶0DMRU VHFWRU OAB· IURP WKH PHQX &OLFN RQ WKH FKHFNER[ LI necessary. (a) Drag point A or point B to resize the sector. E ([SODLQWKHPHDQLQJRIPLQRUVHFWRUDQGPDMRUVHFWRUEDVHGRQ\RXUREVHUYDWLRQV 5. 6HOHFW¶0LQRUVHJPHQW·RU¶0DMRUVHJPHQW·IURPWKHPHQX&OLFNRQWKHFKHFNER[LIQHFHVVDU\ (a) Drag point A or point B to resize the segment. E ([SODLQWKHPHDQLQJRIPLQRUVHJPHQWDQGPDMRUVHJPHQWEDVHGRQ\RXUREVHUYDWLRQV 6. Discuss and state all the conclusions that can be made. 7. 3UHVHQW\RXUÀQGLQJVLQFODVV ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 43 A chordLVDVWUDLJKWOLQHVHJPHQWWKDWMRLQVDQ\WZRSRLQWVO\LQJRQWKH circumference of the circle. (a) The longest chord is the chord that passes through the centre of the circle. The longest chord is known as the diameter (d). (b) The diameter of a circle divides the circle into two congruent semicircles. An arc is the part of a circle between any two points lying on the circle. (a) The arc that subtends an angle of more than 180° at the centre is known as the major arc. (b) The arc that subtends an angle of less than 180° at the centre is known as the minor arc. A sector is the region of a circle bounded by two radii and an arc. (a) The sector that subtends an angle of more than 180° at the centre is known as the major sector. (b) The sector that subtends an angle of less than 180° at the centre is known as the minor sector. A segment is the region of a circle bounded by a chord and an arc. (a) The segment that subtends an angle of more than 180° at the centre is known as the major segment. (b) The segment that subtends an angle of less than 180° at the centre is known as the minor segment. “A circle is a polygon.” Is the statement true or false? Discuss with your friend. Critical Thinking Diameter Semicircle Chord Major arc Minor arc Minor sector Major sector Minor segment Major segment EXAMPLE 1 For each of the following, identify each stated part. (a) Point O (b) Line PQ (c) Line OP (d) Line QST (e) Line QT (f) Shaded region OQR (g) Shaded region QST Solution: (a) Centre (b) Diameter (c) Radius (d) Minor arc (e) Chord (f) Minor sector (g) Minor segment O P R Q S T ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 44 EXAMPLE 2 (a) The radius of a circle is 7 cm. Find its diameter. (b) The diameter of circle is 21 cm. Find its radius. Solution: (a) Diameter = 2=Radius = 2=7 = 14 cm (b) Radius = Diameter 2 = 21 2 = 10.5 cm Discuss with your classmates. Footballs, marbles and metal balls are in spherical shape and not in circular shape. What is the difference between a sphere and a circle? INTERACTIVE ZONE C Constructing a circle and parts of a circle EXAMPLE 3 Construct a circle with (a) a radius of 1 cm, (b) a diameter of 2.4 cm. Solution: (a) 1 cm Measure a distance of 1 cm on a pair of compasses. Fix the point of the compasses at a point as the centre of the circle. Draw with the pencil on the compasses through a complete turn. (b) Measure a distance of 1.2 cm on a pair of compasses. Fix the point of the compasses at a point as the centre of the circle. Draw the circle accordingly. • Radius = 1.2 cm Diameter 2 • Diameter = 2 = Radius ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 45 EXAMPLE 4 The diagram shows a circle with centre O. Copy and construct the diameter passing through (a) point P, (b) point Q. Solution: (a) Draw a line passing through O so that one end of the line is at P and the other end is at a point lying on the circumference. (b) Draw a line passing through O and Q so that both ends of the line are points lying on the circumference. EXAMPLE 5 The diagram shows a circle with centre O. Construct a chord of length 1.6 cm passing through point Q. Solution: Measure a distance of 1.6 cm on a pair of compasses. Fix the point of the compasses at point Q. Draw an arc that intersects the circle. 'UDZ D OLQH MRLQLQJ SRLQW Q and the intersection point drawn in step . EXAMPLE 6 Construct and shade a sector of a circle with a radius of 1 cm and the angle subtended at the centre is 110°. Solution: Construct a circle of radius 1 cm and mark the centre as O. Draw the radius OM. Use a protractor to measure an angle of 110$ from OM, that isMON = 110$. Join O to N. Shade the sector region OMN accordingly. Q O P O P O Q 1.6 cm 1.6 cm Q O 110° M 1 cm N O O Q We can draw two chords of the same length passing WKURXJK D ¿[HG SRLQW RQ WKH circumference. ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 46 “A circle can be formed from a regular polygon.” Is the statement true or false? Discuss with your friend and H[SODLQ\RXUDQVZHU D Symmetrical properties of chords Objective: To explore the symmetrical properties of chords. Instruction: Do this activity in groups of four. 1. 2SHQ WKH ÀOH Symmetrical Properties of Chords using GeoGebra. 2. 6HOHFW¶3URSHUWLHVRIGLDPHWHU·IURPWKHPHQX7KH display shows a circle with diameter AP. D 'UDJWKH¶$QLPDWLRQ·VOLGHU (b) Does the diameter divide the circle into two equal parts? 3. 6HOHFW ¶3URSHUWLHV RI FKRUG · IURP WKH PHQX The display shows the chord AB and the radius OC. (a) Drag point A, B or C and then drag the ¶$QLPDWLRQ·VOLGHU (b) Drag point C to a position so that radius OC and chord AB are perpendicular and drag the ¶$QLPDWLRQ·VOLGHU:KDWLV\RXUFRQFOXVLRQ" 4. 6HOHFW¶3URSHUWLHVRI FKRUG·IURPWKHPHQX7KH display shows the chords AB and CD as well as their respective perpendicular bisectors. (a) Drag points A, B, C or D to change the length of the chords AB and CD. (b) What can be concluded about the perpendicular bisector in relation to the centre O? 3 INTERACTIVE ZONE ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 47 5. 6HOHFW ¶3URSHUWLHV RI FKRUG ·IURPWKHPHQXThe display shows the arc AB as well as the chords AB and CD which are always the same length. (a) Drag points A or B to change the length of chord AB and the length of arc AB. E 'UDJWKH¶$QLPDWLRQ·VOLGHU (c) What can be concluded about the length of arcs AB and CD in relation to the length of chords AB and CD? 6. 6HOHFW¶3URSHUWLHVRI FKRUG·IURPWKHPHQX7KH display shows the chords AB and CD which are always the same length. (a) Drag point A or point B to change the length of chords AB and CD. E 'UDJWKH¶$QLPDWLRQ·DQG¶$QLPDWLRQ·VOLGHUV (c) What can be concluded about the distance of the chords from the centre O? 7. 3UHVHQW\RXUÀQGLQJVLQFODVV Objective: To explore the symmetrical properties of chords. Instruction: Do this activity in groups of four. Material: Scissors, paper, pencil, ruler, compasses, protractor, thread Part A 1. Draw a circle with centre O on a piece of paper. 2. Draw the diameter. 3. Cut out the circle. 4. Fold the circle along the diameter. 5. Does the diameter divide the circle into two equal parts? Part B 1. Draw a circle with centre O and draw a chord AB. 2. Using a protractor and a ruler, draw a radius which is perpendicular to the chord AB. Label M at the intersection point of the radius and the chord AB. 3. Measure the length of AM and MB. 4. Does the radius bisect the chord AB if the radius is perpendicular to the chord AB? 5. Repeat step 1. Then, use a ruler to measure the length of chord AB and mark point M on the chord AB so that AM = MB. O O A O B A O M B A M O B A M O B 4 ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 48 6. Draw a radius passing through point M. 7. Measure OMB. 8. What is your conclusion about the radius and the chord if the radius bisects the chord? Part C 1. Draw a circle with centre O on a piece of paper and cut out the circle. 2. Draw a chord PQ. 3. Fold the circle such that P and Q are aligned. 4. Unfold the circle and draw a line along the crease of the fold. This line is a perpendicular bisector of chord PQ. O O P Q O P Q P Q O P Q 5. Repeat step 2 to step 4 with chord RS. O P R S Q O R S RS O P R S Q 6. Complete the following statements. (a) Perpendicular bisector of a chord passes through the of the circle. (b) Perpendicular bisectors of two chords intersect at the of the circle. Part D 1. Draw a circle with centre O and draw two chords, AB and CD, of the same length. 2. Using a thread and a ruler, measure the lengths of arc AXB and arc CYD. What is your conclusion on the chords and the arcs? 3. Measure the distances from O to chord AB and from O to chord CD. 4. Complete the following statement. If two chords have the same length, then the chords are of the distance from the centre. 3UHVHQWDOO\RXUÀQGLQJVIURPSDUWVA, B, C and D in class. O X Y C D A B )URPWKHÀQGLQJVRI$FWLYLW\ZHFDQREVHUYHWKHV\PPHWULFSURSHUWLHVRIDFLUFOHRQFKRUGV The diameter of a circle divides a circle into two identical parts. Thus, a diameter is an axis of symmetry of a circle. If a radius is perpendicular to a chord, then the radius bisects the chord. Conversely, if a radius bisects a chord, then the radius is perpendicular to the chord. ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 49 O M P A B OP is perpendicular to AB AM = MB Perpendicular bisector of a chord passes through the centre of the circle. Thus, perpendicular bisectors of two chords intersect at the centre of the circle. Chords of the same length produce arcs with the same length. X A B D Y C If AB = CD, then, arc AXB = arc CYD If two chords have the same length, then the chords are of the same distance from the centre. Conversely, if two chords are of the same distance from the centre, then the chords have the same length. M N O A B C D AB = CD OM = ON In a circle, how are these 3 items related? • the centre of the circle, • a chord of a circle and • the perpendicular bisector of the chord. Discuss with your classmates. team work O EXAMPLE 7 The diagram on the right shows a circle. Construct the centre of the circle and label it as O. Then, determine the radius of the circle. Solution: Draw any two non-parallel chords of the circle. Construct the perpendicular bisector of each chord. Label the intersection point of the two perpendicular bisectors as O. Point O is the centre of the circle. Join point O to the circumference of the circle and measure the radius. ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 50 O 1 cm 1 2 Thus, radius = 1 cm. EXAMPLE 8 In the diagram on the right, UPQR is a semicircle with UTSR as the diameter. PQST is a rectangle. Given that PQ = 16 cm and PT FPÀQGWKHOHQJWKRIUT. Solution: Referring to the diagram below, O is the centre of the semicircle and OMN is the radius perpendicular to PMQ. Thus, PM = MQ = 16 2 = 8 cm OU = OP = 152 + 82 = 17 cm Thus, UT = OU – OT = 17 – 8 = 9 cm EXAMPLE 9 In the diagram, O is the centre of circle PQRS. OTQU is a rectangle with OT = 20 cm. PTQ is a chord of the circle with length of 30 cm. RUO and SOT are straight lines. Find the length of UR. Solution: PT = TQ = 30 2 = 15 cm OQ2 = OT 2 + TQ 2 = 202 + 152 = 400 + 225 = 625 OQ = 625 = 25 cm Therefore, OR = 25 cm and OU = 15 cm Thus, UR = OR – OU = 25 – 15 = 10 cm P Q U T S R P Q U T 15 cm S M N O R S R U TO P Q Pythagoras Theorem Radius of the circle. Opposite side of the rectangle. A semicircle is the region enclosed by a diameter and an arc of the circle. Semicircle ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 51 The diagram on the right shoZVDUDFNWR¿WDSODWH7KHZLGWKRIWKHUDFNLVcm and one part of the rack is in the shape of an arc of a circle. If the diameter of the plate is FP¿QGWKHYDOXHRIxVRWKDWWKHUDFN¿WVWKHSODWHSHUIHFWO\ 24 cm x cm Practice 3.1 Basic Intermediate Advanced Based on the diagram on the right, identify each of the following. (a) Point O (b) Line SU (c) Line OS (d) Line PS (e) Shaded region OPQ (f) Curved line QRS (g) Shaded region STU Construct a circle with a (a) radius of 2.2 cm, (b) diameter of 6.2 cm, (c) radius of 3.3 cm, (d) diameter of 4.6 cm. Each of the following diagrams shows a circle with centre O. Construct the diameter passing through (i) point P, (ii) point Q. (a) (b) O P Q O P Q (c) P O Q Construct a circle of radius 3.7 cm. Mark a point M on the circumference of the circle. Then, construct a chord that passes through point M with a length of (a) 2.6 cm, (b) 3.3 cm. (a) Construct a sector with a radius of 3.4 cm and the angle subtended at the centre is 145°. O P U T R Q S (b) Construct a sector with a radius of 2.5 cm and the angle subtended at the centre is 76°. Determine whether each of the following statements is true or false. (a) If arc PQ and arc QR of a circle are of the same length, then chord PQ and chord QR have the same length. (b) If the perpendicular bisector of chord PQ of a circle is line MN, then M or N is the centre of the circle. (c) If the perpendicular bisector of chord PQ intersects PQ at point M, then PM = MQ. (d) If O is the centre of a circle and OM = ON where M and N are two points in the circle, then the chord passing through M has the same length as the chord passing through N. 8VH HDFK RI WKH IROORZLQJ REMHFWV WR GUDZ a circle. Then, construct the centre of the circle and label with O. Hence, determine WKHUDGLXVRIWKHREMHFW (a) (b) (c) (d) The diagram on the right shows a circle PSQR with centre O. T is the midpoint of chord RS and POTQ is the diameter of the circle. Given that OP = 13 cm and RS FPÀQGWKHOHQJWKRIOT. P Q S T R O ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 52 In the diagram on the right, O is the centre of circle PQR. PSR and OSQ are perpendicular straight lines. Given that QS = 20 cm and the ratio of QS to SO LV ffl ÀQGWKH length of PR. O P Q S R The diagram on the right shows circle PQRS with centre O. Given that OQRS is a rhombus and POQ is the diameter of the circle. Explain whether arc PS = arc SR = arc RQ. Q P S R O 3.2 Circumference and Area of a Circle A Circumference of a circle Objective: To explore the relationship between the circumference and the diameter of a circle. Instruction: Do this activity in groups of four. Material: 4 circular cups of different sizes, thread, ruler, pencil, compasses 1. Take a cup and use a thread to measure the length of one complete circumference of the cup. 2. Draw the circumference of the circular cup on a piece of paper and construct the centre of the circle. Then, determine the diameter of the circle. 3. Repeat step 1 and step 2 for all the other cups and complete the following table. Cup Circumference (mm) Diameter (mm) Circumference Diameter 1 2 3 4. Observe the values of Circumference Diameter in the table. What is your conclusion? 5. 2SHQWKHÀOHRelationship Between Circumference and Diameter using GeoGebra. 6. 'UDJWKH ¶&LUFOH VL]H· VOLGHUWR FKDQJHWKH size of the circle. 7. 'UDJWKH¶5ROO·VOLGHUWRWKHULJKW 8. Record the length of circumference AB and the length of diameter AC shown in the display. 5 ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 53 9. Repeat Steps 6 to 8 for another three circle sizes. 10. Find the value of Circumference Diameter for each circle size. What is the conclusion that can be made? 11.3UHVHQW\RXUÀQGLQJVLQFODVV For all circles, the ratio Circumference Diameter LV D FRQVWDQW7KH FRQVWDQW LV NQRZQ DV ¶SL· DQG LV represented with the Greek letter /. Therefore, we have C d = / C = /d = 2/r Thus, circumference, C = 2/r or C = /d . How do you estimate the diameter of a tree? Critical Thinking d = 2r, r = radius William Jones was the earliest mathematician to use the letter / to represent the ratio of circumference to diameter in 1706. The use of / was later popularised by Leonhard Euler after 1736. EXAMPLE 10 Find the circumference of a circle with (a) radius of 7 cm, Use / § 22 7 (b) diameter of 10 cm. (Use/ § 3.142) Solution: (a) Circumference of circle = 2/r = 2 × 22 7 × 7 = 44 cm (b) Circumference of circle = /d = 3.142 × 10 = 31.42 cm EXAMPLE 11 (a) A circle has circumference of 77 cm. Find the diameter of the circle. Use / § 22 7 (b) The circumference of a circle is 204 cm. Find the radius of the circle. (Use / § 3.142) The value of /LVDSSUR[LPDWHO\ RU 22 7 . ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 54 Solution: (a) Circumference = /d 77 = 22 7 × d d = 77 × 22 7 = 24.5 cm (b) Circumference = 2/r 204 = 2 × 3.142 × r r = 204 2 × 3.142 = 32.46 cm • d = C / • r = C 2/ B Area of a circle Objective: To derive the formula for the area of a circle. Instruction: Do this activity in groups of four. 1. 2SHQ WKH ÀOH Area of a Circle using GeoGebra. 2. 'UDJWKH ¶6WUDLJKWHQ FLUFXPIHUHQFH· VOLGHUWR the right. Discuss with your group members and verify the expression of circumference AB displayed. 3. 'UDJ WKH ¶5ROO· VOLGHU WR WKH ULJKW DQG FOLFN WKH¶'LYLGHLQWRWZRSDUWV·FKHFNER['LVFXVV with your group members and verify the expression of line AM displayed. 4. 'UDJWKH¶5HDUUDQJH·VOLGHUWRWKHULJKW:KDW shape is produced when the sectors of the circle are rearranged? 5. 'UDJ WKH ¶1XPEHU RI VHFWRUV· VOLGHU WR WKH right to increase the number of sectors of the circle. What is your observation? 6. Complete each of the following to derive the formula of a circle. Area of the shape = = = Hence, area of the circle = 6 ©Praxis Publishing_Focus On Maths
55 Circles CHAPTER 3 7. &OLFNRQWKH¶5HVHW·EXWWRQLI\RXZDQWWRH[SORUHWKLVDFWLYLW\DJDLQ'UDJWKH¶1XPEHURI VHFWRUV·VOLGHULI\RXZDQWWRFKDQJHWKHVWDUWLQJQXPEHURIVHFWRUV 8. 3UHVHQW\RXUÀQGLQJVLQFODVV Area of a circle can be derived as follows: Divide the circle into sectors of the same size. A O r Straighten the circumference and arrange the sectors on the straight line. A O r B AB = 2/r Divide the sectors into two same parts. A O r M B AM = /r AB = 2/r Rearrange the sectors into nearly the shape of a rectangle. A O r M B AM = /r AB = 2/r If the circle is divided into more similar sectors, these sectors will eventually be arranged into a rectangle. Thus, Area of circle = area of rectangle = /r × r = /r 2 Thus, area of circle, A = /r 2 . ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 56 Agriculture: Crop Circles In Red Deer, Alberta, on September 17, 2001, a crop circle formation that contained 7 circles was discovered. One of the circles shown in the picture has a diameter about 10 m. This circle destroyed some wheat crop. What area of wheat crop was lost in this crop circle? Maths LINK EXAMPLE 12 Find the area of a circle with (a) radius of 14 cm, Use / § 22 7 (b) diameter of 15 cm. (Use / § 3.142) Solution: (a) Area of circle = /r 2 = 22 7 × 142 = 616 cm2 (b) Area of circle = /r 2 = 3.142 × 15 2 2 r = d 2 = 176.74 cm2 If the circumference of a circle is the same as the perimeter of a square, which shape has the greater area? EXAMPLE 13 (a) The area of a circle is 188 cm2 . Find the radius of the circle. (Use / § 3.142) (b) A circle has an area of 124 cm2 . Find the diameter of the circle. Use / § 22 7 Solution: (a) Area of circle = /r 2 188 = 3.142 × r 2 r 2 = 188 3.142 = 59.83 r = 59.83 = 7.73 cm ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 57 (b) Area of circle = /r 2 124 = 22 7 × r 2 r 2 = 124 × 22 7 = 39.45 r = 39.45 = 6.28 cm Diameter, d = 2 × 6.28 = 12.56 cm C Length of an arc and area of a sector Objective: To explore the relationship between central angle, length of arc and area of sector. Instruction: Do this activity in groups of four. Material: Manila card, pencil, compasses, protractor, ruler 1. Construct a sector with a radius of 5 cm and the angle subtended at the centre, AOB, is 120°, as shown in diagram (a). 2. Construct another circle with a radius of 5 cm and a few sectors with the angle subtended at the centre, COD, is 40°, as shown in diagram (b). 3. Cut out the sectors. Use the sectors to measure the length of arc AXB and the area of sector OAXB. 4. Complete the following. (a) COD AOB = (b) Length of arc CYD Length of arc AXB = (c) Area of sector OAXB Area of circle = 0DNHDFRQFOXVLRQRQ\RXUÀQGLQJV 5. Complete the following. (a) AOB Angle of a complete rotation = (b) Length of arc AXB Length of circumference = (c) Area of sector OAXB Area of circle = 0DNHDFRQFOXVLRQRQ\RXUÀQGLQJV 6. Repeat step 1 to step 3 with AOB = 90°, COD = 45° and AOB = 240°, COD = 60°. What is your observation? 7. 3UHVHQWDOO\RXUÀQGLQJVLQFODVV O W X B A 120° Y D C 40° O Diagram (a) Diagram (b) 7 ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 58 )URPWKHÀQGLQJVRI$FWLYLW\ZHIRXQGWKDW y x = Length of arc CYD Length of arc AXB = Area of sector OCYD Area of sector OAXB The length of arc is proportional to the angle subtended at the centre of a circle. Length of arc Circumference = Angle subtended at the centre, e 360° Length of arc = e 360° × 2/r The area of a sector is proportional to the angle subtended at the centre of a circle. Area of sector Area of circle = Angle subtended at the centre, e 360° Area of sector = e 360° × /r 2 O Y C D X B A x y Leng ht of arc θ O EXAMPLE 14 Find the value of y for each of the following. (a) (b) O 65° 5 cm y 65° Solution: (a) y = 5 cm (b) y 110° = 5 cm 10 cm y = 5 10 × 110 = 55° EXAMPLE 15 Find the arc length of a circle with radius of 14 cm and the angle subtended at the centre is 70°. (Use / § 3.142) O 5 cm 10 cm y 110° ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 59 Solution: (a) y = 5 cm (b) y 110° = 5 cm 10 cm 70° 14 cm s y = 5 10 × 110 = 55° EXAMPLE 16 In the diagram, PQ is an arc with centre O. Given that arc PQ = 5.5 cm and OP = 7 cm, ÀQGWKHYDOXHRIx. Use / § 22 7 Solution: s 2/r = e 360° 5.5 2 = 22 7 = 7 = x 360° x = 5.5 44 × 360° = 45° EXAMPLE 17 In the diagram on the right, PQR is an arc with centre O. Given that arc PQR = 45 cm, ÀQGWKHOHQJWKRIOP. (Use /ʜ Solution: s 2/r = e 360° 45 2 = 3.142 = OP = 130° 360° = 13 36 OP = 36 =45 2 = 3.142 = 13 = 19.83 cm EXAMPLE 18 In the diagram, OPQ is a sector with centre O. Given that OP FPÀQGWKH area of sector OPQ. (Use /ʜ x O P Q 68° O P Q 130° P Q O R ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 60 Solution: L /r2 = e 360° Area of OPQ 3.142 = 162 = 68° 360° Area of OPQ = 68° 360° × 3.142 × 162 = 151.9 cm2 EXAMPLE 19 The diagram shows a circle with centre O*LYHQWKDWWKH DUHD RIPDMRU VHFWRU OPQ = 385 cm2 and OP = 14 cm. Find the value of x. Use / ʜ 22 7 Solution: L /r 2 = e 360° 385 22 7 = 142 = x 360° x = 385 616 × 360° = 225° EXAMPLE 20 The diagram shows sector OPQ of a circle. Given that the area of the sector is 105 cm2 , ÀQGWKHOHQJWKRIOP. (Use /ʜ Solution: L /r 2 = e 360° 105 3.142 = OP2 = 42° 360° = 7 60 OP2 = 105 =60 3.142 = 7 = 286.44 OP = 286.44 = 16.92 cm P Q O x P O Q 42° ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 61 D Solving problems EXAMPLE 21 The diagram shows a clock face with a minute hand of length 7 cm. Calculate the distance covered by the point P at the tip of the minute hand after 2 hours. Use / ʜ 22 7 Solution: Distance covered = 2 × 2/r = 2 × 2 × 22 7 × 7 = 88 cm EXAMPLE 22 The diagram shows two semicircles of equal size in a rectangle. Calculate the area of the shaded region in cm2 . Use / ʜ 22 7 Solution: Breadth of the rectangle = Radius of the semicircle = 14 4 = 7 2 cm Area of the rectangle = 14 × 7 2 = 49 cm2 Area of the two semicircles = 22 7 × 7 2 2 = 38 1 2 cm2 Area of the shaded region = 49 – 38 1 2 = 10 1 2 cm2 EXAMPLE 23 The diagram shows a pendulum. Find the value of x. Use / ʜ 22 7 45° 11 cm x cm 12 6 11 1 10 2 9 3 8 4 7 5 7 cm P 14 cm ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 62 Solution: 11 2 = 22 7 = x = 45° 360° x = 11 × 7 44 × 360 45° = 14 cm EXAMPLE 24 7 cm E K F H G In the diagram, EFGH is a square and EFKH is a quadrant of a circle with centre E. Calculate the area, in cm2 , of the shaded region. Use / ʜ 22 7 Solution: Angle subtended by arc KH = 90° 2 = 45° Area of shaded region = 45° 360° × 22 7 × 72 = 19.25 cm2 EXAMPLE 25 The diagram shows a circular playground in a residential area. The authority wishes to cover the shaded region with grass so that there is a circular central region without grass. If the cost to cover the region with grass is $40 for every square PHWUH ÀQG WKH WRWDO FRVW WR FRYHU WKH region with grass. Use / ʜ 22 7 50 m 24 m A quadrant is one quarter of a circle. Quadrant ©Praxis Publishing_Focus On Maths
Circles CHAPTER 3 63 Solution: Let radius of bigger circle = R, radius of smaller circle = r Area of region to cover with grass = /R 2 – /r 2 = /(R 2 – r 2 ) = 22 7 (252 – 122 ) = 1511.71 m2 Thus, total cost to cover the region with grass = 1511.71 = 40 = $60 468 EXAMPLE 26 The diagram shows a sector OFGH with a radius of 10 cm. It is given that FOH = 128.3$ and the length of chord FH = 18 cm. Calculate (a) the perimeter, in cm, of the segment FGH. (b) the area, in cm2 , of the segment FGH. Use / ʜ 22 7 Solution: (a) Length of arc FGH = 128.3° 360° × 2πr = 128.3° 360° × 2 × 22 7 × 10 = 22.40 cm Perimeter of segment FGH = FH + FGH = 18 + 22.40 = 40.40 cm (b) FM = MH So, FM = 18 cm 8 2 = 9 cm 10 cm F O G H 18 cm ©Praxis Publishing_Focus On Maths
CHAPTER 3 Circles 64 OM = OF 2 ïFM 2 = 102 ïffi2 = 19 = 4.36 cm Area of segment FGH = Area of sector OFGH < Area of triangle OFH = 128.3° 360° × 2/Uɧ2 < 1 2 × FH × OM = 128.3° 360° × 22 7 × 102 < 1 2 × 18 × 4.36 = 112.01 < 39.24 = 72.77 cm2 Calculate the circumference of a circle of (a) diameter 1.4 cm, Use / ʜ 22 7 (b) diameter 35 cm, Use / ʜ 22 7 (c) radius 4 cm, (Use / ʜ 3.142) (d) radius 6.5 cm. (Use / ʜ 3.142) Using / ʜ 22 7 ÀQGWKHGLDPHWHURIDFLUFOH its circumference is (a) 22 cm, (b) 17.6 cm, (c) 154 cm, (d) 264 cm. Using / ʜ 22 7 ÀQGWKH UDGLXV RI D FLUFOH LI its circumference is (a) 220 cm, (b) 13.2 cm, (c) 176 cm, (d) 3960 cm. Calculate the area of a circle with (a) radius = 28 cm, Use / ʜ 22 7 (b) radius = 15 cm, (Use / ʜ 3.142) (c) diameter = 5.6 cm, Use / ʜ 22 7 (d) diameter = 10 cm. (Use / ʜ 3.142) Using / ʜ 22 7 ÀQGWKHUDGLXVDQGGLDPHWHU of a circle of area (a) 6.16 cm2 , (b) 13.86 cm2 . O 45° y The diagram shows a circle with centre O. State the value of y. Find the length of arc in each of the following circles, given that its angle subtended at the centre, e and radius, r. Use / ʜ 22 7 (a) e = 30°, r = 16.8 cm (b) e = 60°, r = 4.2 cm (c) e = 90°, r = 28 cm (d) e = 120°, r = 10.5 cm Calculate the area of the shaded sector in each of the following circles. Use / ʜ 22 7 (a) (b) O 5.6 cm 45° O 2.1 cm 120° (c) O 2.8 cm Practice 3.2 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
65 Circles CHAPTER 3 Using / ʜ 22 7 , calculate the circumference of a circle if its area is (a) 124.74 cm2 , (b) 38.5 cm2 . Using / ʜ 22 7 ÀQGWKHDUHDRIDFLUFOHLILWV circumference is (a) 88 cm, (b) 396 cm. Find the value of x in each of the following circles. Use / ʜ 22 7 (a) (b) 44 cm x O 31.5 cm O x 66 cm 14 cm (c) (d) x O 4.2 cm x O 6.6 cm Shaded region Shaded region = 9.24 cm2 = 53.24 cm2 Find the radius of each of the following circles. Use / ʜ 22 7 (a) (b) O 17.6 cm 40° O 44 cm (c) (d) O 66 cm O 158.4 cm 324° (e) (f) O O 216° Shaded region Shaded region = 38.5 cm2 = 207.9 cm2 In the diagram, O is the centre of the bigger circle. Using / ʜ 3.142, calculate (a) the radius of the bigger circle, (b) the area of the shaded region. Given that PQR is a semicircle, calculate the perimeter of the diagram. Use / ʜ 22 7 The diagram shows a sector of a circle. Find the perimeter of the shaded region, correct to 2 decimal places. (Use / ʜ 3.142) The wheel of a bicycle has a radius of 56 cm. Use / ʜ 22 7 (a) How far has the bicycle travelled, in m, after the wheel has made 10 complete revolutions? (b) If the bicycle travelled 440 m, how many complete revolutions has the wheel made? A string of length 396 cm makes 15 complete circles when it is ringed round a cylinder repeatedly. Find the diameter of the cylinder. Use / ʜ 22 7 The diagram shows a circle with centre O. Given that the area of the minor sector OAB is 1 7 of the area of the shaded region, calculate the area of the shaded region. Use / ʜ 22 7 P Q T R S 10 cm 6.6 cm 45° 15 cm 10 cm C D A B 15 rounds O A B 5.6 cm O 8 cm 6 cm ©Praxis Publishing_Focus On Maths
66 CHAPTER 3 Circles 7 cm 13 cm 3 cm O Calculate the area of the diagram as shown. Use / ʜ 22 7 Q P O S T 36° 3.5 cm The diagram shows two sectors, OPQ and OST, with a common centre, O. POS and QOT are straight lines. Given that PO : OS = 2 : 1, calculate the area, in cm2 , of the sector OPQ. Use / ʜ 22 7 21 The diagram shows sector OAB w i t h centre O and sector PQR with centre P. Sector OAB has a radius of 8 cm, OQ = OP = 3 cm and AOB is a right angle. Calculate (a) the perimeter of the shaded region. (b) the area of the shaded region. (Use / ʜ 3.142) O P RB A Q Summary Circumference, C C = 2/r Length of arc, s s 2/r = e 360° Area of sector, L L /r 2 = e 360° Area of circle, A A = /r 2 Parts of a circle Centre Circumference Radius Diameter Chord Minor arc Major arc Minor sector Major sector Minor segment Major segment Symmetrical properties of chords O Diameter is the D[LVRIV\PPHWU\ of the circle O M P A B OP is perpendicular to AB AM = MB O Perpendicular bisectors of two chords meet at the centre of the circle X A B D Y C If AB = CD, then arc AXB = arc CYD M N O A B C D AB = CD OM = ON Circles ©Praxis Publishing_Focus On Maths
67 Circles CHAPTER 3 Section A 1. O A B M C D N The diagram shows circle ABDC with centre O. Given that chord AB and chord CD have an equal length. Which of the following is not true? A AM = ND C OC = MB B OM = ON D Arc AB = Arc CD 2. In the diagram above, PQTU is a rectangle. RST is a semicircle with centre Q and the length of diameter RQT is the same as length of PQ. Find the length, in cm, of RQ. A 8 C 6 B 4 D 2 3. P S O Q R The diagram shows a circle with centre O. Which of the following is the minor arc of the circle? A OP C PR B PQR D PSR 4. P O 49 cm R The diagram shows two semicircles. O is the centre of the bigger semicircle. Find the perimeter of the shaded region in cm. Use / ʜ 22 7 A 280 C 348 B 320 D 444 U 8 cm T P S Q R 5. 28 cm R O PQ S In the diagram, OPQR is a square and RSP is an arc of a circle with centre O. The perimeter of the diagram, in cm, is Use / ʜ 22 7 A 132 B 146 C 176 D 188 6. The diagram shows a sector of a circle. Find its perimeter in cm. Use / ʜ 22 7 A 86 B 88 C 130 D 150 7. P Q S R T O In the diagram above, OTQS is a rectangle. PQR is an arc of a circle with centre O. OTP and OSR is a straight line. Given that TQ = 8 cm and QS = 6 cm. Calculate the length, in cm, of OTP. A 6 C 10 B 8 D 12 8. The area of the above sector is 102 2 3 cm2 . Find the value of x. Use / ʜ 22 7 A 60° B 50° C 40° D 30° 120° 21 cm O x 14 cm ©Praxis Publishing_Focus On Maths
68 CHAPTER 3 Circles 9. 60° PRO S Q The diagram shows two arcs, PQ and RS, with centre O. Given that PO FPÀQG the area of the shaded region in cm2 . Use / ʜ 22 7 A 99 B 77 C 66 D 44 10. O A B E D C The diagram above shows circle ABC with centre O. ODBE is a rectangle. Given that chord AB = 6 cm and D is the midpoint of OC. The length of BD, in cm, is A 5.2 B 5.3 C 5.8 D 5.9 Section B 1. The diagram below shows a circle with chord PQ. Draw another chord QR, hence construct the centre of the circle and label it with O. P Q 2. S P R Q T O The diagram above shows a circle with centre O. PTR is a straight line and SOTQ is the diameter of the circle. Given that TQ = 3 cm and OS FPÀQGWKHOHQJWK of PR. 3. K L M N The diagram above shows the cross section of a pipe with diameter of 40 cm. The water level was found to drop from level KL to level MN such that KL = MN = 32 cm. What is the distance of the water level that had dropped? 4. 45° P Q R S 7 cm In the diagram above, PS is an arc of a circle with centre Q. Calculate the area, in cm2 , of the whole diagram. 5. 14 cm 18 cm R T P S Q O In the diagram, OPQR is a rectangle and PTS is an arc of a circle with centre O. Find the perimeter of the whole diagram. Use / ʜ 22 7 6. The diagram above shows a square inscribed in a circle with radius of 12 cm. Find the area of the shaded region. (Use /ʜ 3.142) ©Praxis Publishing_Focus On Maths
69 Circles CHAPTER 3 7. The diagram above shows a semicircular pond. Given that the area of the pond is 962.5 m2 ÀQGWKHSHULPHWHURIWKHSRQG Use / ʜ 22 7 8. The diagram below shows a circle with centre O and radius of 14 cm. Given that the ratio of minor arc PQ WRWKHPDMRUDUFPQ is 2:7. Find the area, in cm2 , of the PDMRUVHFWRUPOQ. Use / ʜ 22 7 9. The diagram shows the cross section of a tunnel with a radius of 12 m. The segment PQ acts as a trench with the concrete road over it. The angle subtended by chord PQ at the centre is 77.36$. Calculate the area, in m2 , of the segment PQ. 10. In the diagram, KLM is an arc of a circle with centre O. The radius of the circle is 7 cm and MN = 2NT. Calculate the perimeter, in cm, of the whole diagram. Use / ʜ 22 7 11. (a) The diagram shows a circle with centre O. The length of the minor arc MN is 8.8 cm. Calculate the radius, in cm, of the circle. Use / ʜ 22 7 (b) The diagram shows a circle with centre O and radius 7 1 2 cm. O P Q T K O L M N 8 cm 8.8 cm 84° O M N O P Q 240° Calculate the length, in cm, of the minor arc PQ. Use / ʜ 22 7 12. (a) The diagram shows three semicircles with diameters SU, ST, and TU respectively.Find the area of the shaded region in cm2 . (b) The diagram shows a square PQRS, a semicircle PWS and a quadrant of a circle RUT. Find the perimeter of the shaded region. Use / ʜ 22 7 13. (a) The diagram shows two circles with centre O. Find the area of the shaded region in cm2 . Use / ʜ 22 7 (b) The diagram shows four identical circles in a rectangle. Given that the circumference of each circle is 8.8 cm, ÀQGWKHDUHDRIWKHUHFWDQJOHLQFP2 . Use / ʜ 22 7 14. The diagram shows two sectors OPQ and ORS with common centre O. Given that OP = PR, find the perimeter of t h e s h a d e d r e g i o n in m. Use / ʜ 22 7 S T 8 cm 4 cmU S UR T P Q 14 cm W 7 cm 2 cm 5 cm 252° O O P 21 m R 60° S Q 15 m P Q ©Praxis Publishing_Focus On Maths
4GEOMETRY Applications of this chapter Three-dimensional geometric solids have numerous technical applications. 'PRGHOLQJVSHFLÀFDOO\PROHFXODUPRGHOLQJQHFHVVLWDWHVXQGHUVWDQGLQJ sphere arrangements to compute properties such as volume. Solid JHRPHWU\ SOD\V PDMRU UROH LQ DUFKLWHFWXUDO GHVLJQ HQVXULQJ VDIH FRQVWUXFWLRQ DQGIRUPLQJWKH GHVLUHG VKDSH ' FRPSXWHU JUDSKLFV KDYHWUDQVIRUPHG DQLPDWLRQ YLGHR JDPHV DQG JUDSKLFV 6XUIDFH DUHD NQRZOHGJH LV FUXFLDO LQWKHIRRG DQG EHYHUDJH FRQWDLQHUPDQXIDFWXULQJ LQGXVWU\IRU FDOFXODWLQJPDQXIDFWXULQJ FRVWV +RZ GRHV VXUIDFH DUHD DIIHFW SURGXFWLRQ FRVWV LQIRRG DQG EHYHUDJH FRQWDLQHUPDQXIDFWXULQJ" 70 ©Praxis Publishing_Focus On Maths
Concept Map 71 Learning Outcomes • 8QGHUVWDQGWKHJHRPHWULFSURSHUWLHVRIFXEHVFXERLGVSULVPVS\UDPLGVF\OLQGHUV cones and spheres. • 8QGHUVWDQG WKH FRQFHSW RI VXUIDFH DUHD RI FXEHV FXERLGV SULVPV S\UDPLGV F\OLQGHUVFRQHVDQGVSKHUHV • 8QGHUVWDQG DQG XVHWKH FRQFHSW RI YROXPH RI FXEHV FXERLGV SULVPV S\UDPLGV F\OLQGHUVFRQHVDQGVSKHUHV • $SSO\WKHFRQFHSWRIDUHDDQGYROXPHWRVROYHSUREOHPVLQYROYLQJFXEHVFXERLGV SULVPVS\UDPLGVF\OLQGHUVFRQHVDQGVSKHUHV • $SSO\ WKH NQRZOHGJH RI YROXPH WR VROYH PDWKHPDWLFDOV SUREOHPV DQG UHDOOLIH SUREOHPV Key Terms Maths History Euclid of Alexandria was a great Greek mathematician and widely considered as the father of geometry. Euclid introduced Euclidean geometry around 300 BC. Euclidean geometry is set of mathematical system that was presented in his seminal work, Elements, which consists of 13 books. Properties of a Sphere Properties of a Cone Properties of a Cylinder Properties of a Pyramid Properties of a Prism Properties of a Cuboid Properties of a Cube Surface Area of a Sphere Surface Area of a Cone Surface Area of a Cylinder Surface Area of a Pyramid Surface Area of a Prism Surface Area of a Cuboid Surface Area of a Cube Volume of a Sphere Volume of a Cone Volume of a Cylinder Volume of a Pyramid Volume of a Prism Volume of a Cuboid Volume of a Cube Cylinder Pyramid Prism Cuboid Cone Sphere Application of Three-Dimensional Geometrical Shapes in Real Life Geometry Cube • Sphere • Net • Cross section • Surface area • Volume • Radius • Slanting surface • Three-dimensional • Cube • Cuboid • Prism • Pyramid • Cylinder • Cone ©Praxis Publishing_Focus On Maths
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73 Geometry CHAPTER 4 4.1 Geometrical Properties of Three-Dimensional Shapes :HKDYHVWXGLHGVKDSHVVXFKDVWULDQJOHVTXDGULODWHUDOVDQGYDULRXVSRO\JRQV7KHVHVKDSHV DUHNQRZQDVWZRGLPHQVLRQDOVKDSHVEHFDXVHWKH\KDYHRQO\OHQJWKDQGEUHDGWK$JHRPHWULF VROLGLVDWKUHHGLPHQVLRQDOVKDSHWKDWKDVOHQJWKEUHDGWKDQGKHLJKW 7KHIROORZLQJGLDJUDPVVKRZVHYHQGLIIHUHQWW\SHVRIJHRPHWULFVROLGV A Compare, contrast and classify three dimensional shapes $RQHGLPHQVLRQDOVKDSHLVDVKDSHWKDWKDVDTXDQWLW\RIOHQJWK RQO\ $WZRGLPHQVLRQDOVKDSHLVDVKDSHWKDWKDVTXDQWLWLHVRIOHQJWK DQGEUHDGWK $ WKUHHGLPHQVLRQDO VKDSH LV D VKDSH WKDW KDV TXDQWLWLHV RI OHQJWKEUHDGWKDQGKHLJKW One-dimensional (length only) Two-dimensional (length and breadth) Three-dimensional (length, breadth and height) Objective: 7RH[SORUHWKHJHRPHWULFSURSHUWLHVRIWKUHHGLPHQVLRQDOVKDSHV Instruction: 'RWKLVDFWLYLW\LQJURXSVRIIRXU Materials: Three-dimensional shapes. 1. <RXDUHJLYHQWKUHHGLPHQVLRQDOVKDSHVDVIROORZV6WXG\FDUHIXOO\DERXWWKHJHRPHWULF SURSHUWLHVRIHDFKJHRPHWULFVROLG Prism I Prism II Pyramid I Pyramid II 1 ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 74 Cylinder I Cylinder II Cone I Cone II Sphere 2. &RS\DQGFRPSOHWHWKHWDEOHEHORZ Geometric solid Number RIÁDW surfaces Number of vertices Number of edges Number of curved surfaces Have parallel surfaces (Yes / No) 3ULVP, 3ULVP,, 3\UDPLG, 3\UDPLG,, &\OLQGHU, &\OLQGHU,, &RQH, &RQH,, Sphere 3. 'LVFXVVZLWK\RXUPHPEHUVDQGVWDWHWKHQDWXUHRIWKHJLYHQJHRPHWULFVROLGV D :KLFKJHRPHWULFVROLGKDVRQO\DÁDWVXUIDFH" E :KLFKJHRPHWULFVROLGGRHVQRWKDYHDÁDWVXUIDFH" F :KLFKJHRPHWULFVROLGKDVDFRQJUXHQWVXUIDFHDQGSDUDOOHOWRWKHVROLGEDVH" G :KLFKJHRPHWULFVROLGKDVDOOIDFHVWULDQJXODULQVKDSHRWKHUWKDQWKHEDVH" H :KLFKJHRPHWULFVROLGKDVDOOIRXUVLGHGVXUIDFHVRWKHUWKDQRIWZRSDUDOOHOVXUIDFHV" I :KLFKVROLGKDVDFRPELQDWLRQRIDÁDWVXUIDFHDQGDFXUYHGVXUIDFH" 4. 3UHVHQWWKHUHVXOWVRI\RXUJURXSLQDQDSSURSULDWHPLQGPDS )RUH[DPSOH Geometric Properties of Prisms and Pyramids Prisms Pyramid ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 75 (a) Is a cuboid or a cube also a prism? (b) Is a cylinder also a prism? (c) Is a cone also a pyramid? Discuss with your friends about the properties of the solids. Then, present your answers to the class. Critical Thinking B Geometric properties of cubes, cuboids, prisms and pyramids Geometric Solid Geometric Propeties Cube $FXEHLVDWKUHHGLPHQVLRQDOVROLGZLWKHTXDOIDFHV,WDOVRKDV 12 edges and 8 vertices. Cuboid $ FXERLG LV DWKUHHGLPHQVLRQDO VROLG ZLWK UHFWDQJXODUIDFHV ,WDOVRKDVHGJHVDQGflYHUWLFHV Prism /DWHUDOIDFHVMRLQWZRRSSRVLWHSDUDOOHODQG FRQJUXHQWSRO\JRQDO EDVHV Pyramid $SRO\JRQDOEDVHZLWKWULDQJXODUIDFHVWKDWPHHWVDWYHUWH[ l l l h b l h Area of base h Area of base C Geometric properties of cylinders, cones and spheres (I) Cylinder Height Height •+DVWZRÁDWFLUFXODUIDFHVZKLFKDUHFRQJUXHQWDQGSDUDOOHO •2QHFXUYHGIDFHLVMRLQHGWRWKHEDVH • Has two curved edges and no vertices. •7KH F\OLQGHU WKDW KDV D FXUYHG IDFH SHUSHQGLFXODU WR WKH EDVHLVNQRZQDVULJKWF\OLQGHU Lateral means 'side'. Lateral surface area is the area of all the curved regions and all the ÀDW VXUIDFHV H[FOXGLQJ EDVH areas. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 76 1DPHHDFKRIWKHIROORZLQJREMHFWV+HQFH GHVFULEHEULHÁ\WKHJHRPHWULFSURSHUWLHVRI HDFKREMHFW (a) E (c) (d) (e) &RPSOHWHHDFKLIWKHIROORZLQJ D $S\UDPLGZLWKDSHQWDJRQDOEDVHKDV IDFHV vertices and edges. E $ SULVP ZLWK D TXDGULODWHUDO EDVH KDV IDFHV vertices and edges. (c) The three-dimensional shape that does QRWKDYHDQ\HGJHLV . (d) The three-dimensional shape that have RQO\ÁDWIDFHVDUH . (e) The three-dimensional shape that have SDUDOOHOIDFHVDUH . Practice 4.1 Basic Intermediate Advanced (II) Cone Height Apex •+DVDÁDWFLUFXODUEDVH •$FXUYHGIDFHZKLFKFRQYHUJHVDWDSRLQWNQRZQDVWKHDSH[ •2QHYHUWH[DQGRQHFXUYHGHGJH •7KH FRQH ZLWK LWV DSH[ YHUWLFDOO\ DERYH WKH FHQWUH RQ WKH EDVHLVNQRZQDVULJKWFRQH (III) Sphere •+DVDFXUYHGIDFH • No vertices and no edges. A three-dimensional shape ZLWKDOOÀDWIDFHVLVNQRZQDVD polyhedron. All points on the surface of a sphere are at the same distance from the centre. Is the statement true? Discuss with your classmates. Critical Thinking ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 77 4.2 Nets of Three-Dimensional Shapes A Nets of cuboids, cubes, prisms and pyramids $QHWRIDVROLGLVDSODQHÀJXUHWKDWFDQEHIROGHGXSWRIRUPWKHVROLG Objective:7RH[SORUHDQGGUDZWKHQHWVRIFXERLGVFXEHVSULVPVDQGS\UDPLGV Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU Material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n VLGHVZKDW LVWKH QXPEHU RI SRO\JRQVLQWHUPVRInLQWKHQHWVRIWKHS\UDPLGRUSULVPUHVSHFWLYHO\" 3. 3UHVHQW\RXUÀQGLQJVLQFODVV 2 &XERLG &XEH 3ULVP, 3ULVP,, 3\UDPLG, 3\UDPLG,, ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 78 7KHIROORZLQJWDEOHVKRZVWKHW\SHVRIJHRPHWULFVROLGVDQGWKHLUQHWV7KHQHWRIDVROLGLV LWVOD\RXWZKHQLWVVXUIDFHLVODLGÁDWRQDSODQH Geometric solid Net Cuboid 7KHQHWFRQVLVWVRIUHFWDQJOHV$FXERLGLVDSULVPZLWKLWVEDVHLVD rectangle. Cube 7KH QHW FRQVLVWV RI VTXDUHV$ FXEH LV D SULVP ZLWK LWV EDVH LV D square. Prism ,IWKHEDVHLVDnVLGHGSRO\JRQWKHQHWFRQVLVWVRIWZRnVLGHGSRO\JRQV RIWKHVKDSHRIWKHEDVHDQGn rectangles. )RULQVWDQFHLIn WKHQHWFRQVLVWVRIWULDQJOHVDQGUHFWDQJOHV Pyramid ,IWKHEDVHLVDnVLGHGSRO\JRQWKHQHWFRQVLVWVRIDnVLGHGSRO\JRQ RIWKHVKDSHRIWKHEDVHDQGn triangles. )RULQVWDQFHLIn WKHQHWFRQVLVWVRIDVTXDUHEDVHDQGWULDQJOHV How many types of arrangements of nets can be formed from a cube? Critical Thinking ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 79 Objective: 7RH[SORUHDQGGUDZWKHQHWVRIF\OLQGHUDQGFRQH Instruction: 'RWKLVDFWLYLW\LQJURXSVRIIRXU Materials: Three-dimensional shapes, paper, pencil 1. <RXDUHJLYHQDF\OLQGULFDOVROLGDQGDFRQLFDOVROLG5ROOWKHFXUYHGIDFHRIHDFKRIWKH VROLGVDORQJDSLHFHRISDSHUDQGWUDFHWKHSDWKZLWKDSHQFLOWRJHWDQHWRILWVFXUYHG VXUIDFH 2. :KDWLVWKHVKDSHRIWKHFXUYHGVXUIDFHRIDF\OLQGHUDQGDFRQH" 3. 'LVFXVVZLWK \RXU JURXSPHPEHUV DQG VWDWH FRQFOXVLRQVWKDW FDQ EH GUDZQ DERXWWKH QHWOD\RXWRIWKHF\OLQGHUDQGFRQH 3 B Nets of cylinders, cones and spheres Geometric solid Net Cylinder 7KHQHWFRQVLVWVRI WZR FLUFOHV RI WKH VKDSH RI WKH EDVH and a rectangle. Cone The net consists RI D FLUFOH RI WKH VKDSH RI WKH EDVH DQG D VHFWRU RI D circle. Sphere 1HWVRIDVSKHUHFDQQRWEHPDGHEHFDXVH LWGRHVQRWKDYHDQ\ÁDWVXUIDFHV Cube Î Cuboid Î Prism Î Pyramid Î ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 80 ,GHQWLI\WKHVROLGVZKLFKKDYHWKHIROORZLQJ nets. D E (c) (d) 6WDWHWKHW\SHVRI VROLGVZLWKWKHIROORZLQJ nets. (a) E (c) (d) )LOOLQWKHEODQNVLQWKHIROORZLQJWDEOH Three-dimensional shape Net (a) circle and VHFWRURID circle. E circles and rectangle. 'UDZ WKH QHW RI HDFK RI WKH IROORZLQJ geometric solids on a manila card according to the given dimensions. Hence, cut out the QHWDQGEXLOGWKHPRGHORIWKHVROLG (a) 7 cm 12 cm E 11 cm 6 cm (c) 7 cm 5 cm 6 cm 'UDZWKHQHWRIHDFKRIWKHIROORZLQJVROLGV (a) E (c) (d) 6 cm 5 cm 10 cm 13 cm 20 cm 5 cm 6 cm 4 cm 8 cm 6 cm 6 cm Practice 4.2 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 81 4.3 Surface Area of Cuboids, Cubes, Prisms and Pyramids The surface areaRIDVROLGLVWKHWRWDODUHDRIWKHH[WHUQDOVXUIDFHRIWKHVROLG7KHVXUIDFH DUHDRIDVROLGLVWKHVDPHDVWKHDUHDRILWVQHW A Deriving the formulae for the surface area of cuboids and cubes Objective:7RGHULYHWKHIRUPXODHIRUWKHVXUIDFHDUHDRIDFXERLGDQGDFXEH Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU Material: &XERLGFXEH 1. <RXDUHJLYHQDFXERLGZLWKOHQJWKRIlEUHDGWKRIb and heigKWRI h as shown in the diagram on the side. 2. 'LVFXVVZLWK\RXUJURXSPHPEHUVWRGHWHUPLQHWKHVXUIDFHDUHDRIWKHFXERLG D :KDWLVWKHDUHDRIWKHEDVHWKHDUHDRIWKHIURQWVXUIDFHDQGWKHDUHDRIWKHVXUIDFH RQWKHULJKWRIWKHFXERLG" E :KLFKVXUIDFHKDVWKHVDPHDUHDDVWKHEDVHWKHIURQWVXUIDFHDQGWKHVXUIDFHRQ WKHULJKW" F &RPSOHWHWKHIROORZLQJWRGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDFXERLGLQWHUPV RIl, b and h. $UHDRIEDVH = = $UHDRIWRSVXUIDFH = = $UHDRIODWHUDOVXUIDFH = + 2 = = 7RWDOVXUIDFHDUHD $UHDRIEDVH$UHDRIWRSVXUIDFH$UHDRIODWHUDOVXUIDFH = 3. <RXDUHJLYHQDFXEHZLWKVLGHOHQJWKRIl as shown in the diagram on the side. 4. 'LVFXVVZLWK\RXUJURXSPHPEHUVWRGHWHUPLQHWKHVXUIDFHDUHDRIWKHFXEH D 'RHVHDFKVXUIDFHRIDFXEHKDYHWKHVDPHDUHD" E +RZPDQ\HTXDOVXUIDFHVGRHVDFXEHKDYH" F &RPSOHWHWKHIROORZLQJWRGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDFXEHLQWHUPV RIl. $UHDRIRQHVXUIDFH = = 7RWDOVXUIDFHDUHD = = = 5. 3UHVHQW\RXUÀQGLQJVLQFODVV h l b 4 ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 82 7KHVXUIDFHDUHDRIDFXERLGZLWKOHQJWKlEUHDGWKb and height h is JLYHQE\ Surface area of cuboid = 2 × (base area + area of top surface + area of lateral surface) = 2(lb + lh + bh) $FXEHZLWKVLGHOHQJWKlKDVVLPLODUVXUIDFHV7KXVWKHVXUIDFHDUHDRID FXEHZLWKVLGHOHQJWKlLVJLYHQE\ Surface area of cube = 6 × (side length)2 = 6l2 B Deriving the formula for the surface area of prisms h l b l l l Objective:7RGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDSULVP Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU Material: Right prisms 1. l h l h <RXDUHJLYHQDULJKWSULVPZLWKDEDVHRIDQHTXLODWHUDOWULDQJOHDQGDULJKWSULVPZLWK DEDVHRIDUHJXODUSHQWDJRQ7KHOHQJWKRIWKHVLGHRIWKHEDVHRIHDFKSULVPLVJLYHQ E\lDQGWKHKHLJKWLVJLYHQE\hDVVKRZQLQWKHGLDJUDPDERYH 2. 'LVFXVVZLWK\RXUJURXSPHPEHUVWRGHWHUPLQHWKHVXUIDFHDUHDRIHDFKSULVP D 'RDOOWKHUHFWDQJXODUVXUIDFHVRIHDFKSULVPKDYHWKHVDPHDUHD" E :KDWLVWKHDUHDLQWHUPVRIl and hRIRQHUHFWDQJXODUVXUIDFH" F :KDWLVWKHSHULPHWHURIWKHEDVHLQWHUPVRIlRIHDFKSULVP" G &RPSOHWHWKHIROORZLQJWRGHULYHDIRUPXODIRUWKHVXUIDFHDUHDRIDULJKWSULVPZLWK nVLGHGUHJXODUSRO\JRQDVWKHEDVH 6XUIDFHDUHDRISULVP = 2 =$UHDRIFURVVVHFWLRQ =$UHDRIUHFWDQJXODUVXUIDFH = 2 =$UHDRIFURVVVHFWLRQ = =h = 2 =$UHDRIFURVVVHFWLRQ =h 3. ,VWKHVXUIDFHDUHDRIDSULVPZLWKDQLUUHJXODUSRO\JRQDVWKHEDVHDOVRXVLQJWKHVDPH IRUPXOD" 4. &RPSOHWHWKHIROORZLQJWRGHULYHDJHQHUDOIRUPXODIRUWKHVXUIDFHDUHDRIDSULVP 6XUIDFHDUHDRISULVP =$UHDRIFURVVVHFWLRQ × h 5. 3UHVHQW\RXUÀQGLQJVLQFODVV 5 ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 83 Objective:7RGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDS\UDPLG Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU Material:5LJKWS\UDPLGV 1. h l h l h l <RXDUHJLYHQWKUHHULJKWS\UDPLGVHDFKZLWKWKHEDVHRIDQHTXLODWHUDOWULDQJOHDVTXDUH DQGDUHJXODUSHQWDJRQ7KHOHQJWKRIWKHVLGHRIWKHEDVHRIHDFKS\UDPLGLVJLYHQE\ lDQGWKHVODQWKHLJKWLVJLYHQE\hDVVKRZQLQWKHGLDJUDPDERYH 2. 'LVFXVVZLWK\RXUJURXSPHPEHUVWRGHWHUPLQHWKHVXUIDFHDUHDRIHDFKS\UDPLG D 'RDOOWKHWULDQJXODUVXUIDFHsRIHDFKS\UDPLGKDYHWKHVDPHDUHD" E :KDWLVWKHDUHDLQWHUPVRIl and hRIRQHWULDQJXODUVXUIDFH" F :KDWLVWKHSHULPHWHURIWKHEDVHLQWHUPVRIlRIHDFKS\UDPLG" G &RPSOHWH WKH IROORZLQJ WR GHULYH D IRUPXOD IRU WKH VXUIDFH DUHD RI D WULDQJXODU S\UDPLG 6XUIDFHDUHDRI WULDQJXODUS\UDPLG 3. 5HSHDW6WHS D ² G WRGHULYHDIRUPXODIRUWKHVXUIDFHDUHDRIDVTXDUHS\UDPLGDQG DULJKWS\UDPLGZLWK nVLGHGUHJXODUSRO\JRQDVWKHEDVH 4. ,VWKH VXUIDFHDUHDRIDS\UDPLGZLWKDQLUUHJXODUSRO\JRQDVWKHEDVHDOVRXVLQJWKH VDPHIRUPXOD" 5. &RPSOHWHWKHIROORZLQJWRGHULYHDJHQHUDOIRUPXODIRUWKHVXUIDFHDUHDRIDS\UDPLG 6XUIDFHDUHDRIS\UDPLG $UHDRIEDVH7RWDODUHDRI 6. 3UHVHQW\RXUÀQGLQJVLQFODVV $UHDRIEDVH =$UHDRIWULDQJXODUVXUIDFH $UHDRIEDVH = 1 2 = = h $UHDRIEDVH 1 2 = = h 6 7KHVXUIDFHDUHDRIDSULVPZLWKKHLJKWhLVJLYHQE\ Surface area of prism = 2 × area of cross-section + perimeter of cross-section × h C Deriving the formula for the surface area of pyramids l h h l Area of base 7KHVXUIDFHDUHDRIDS\UDPLGLVJLYHQE\ Surface area of pyramid = area of base + total area of triangular surfaces Area of base Triangular surface ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 84 D Finding the surface areas of cuboids, cubes, prisms and pyramids EXAMPLE 1 )LQGWKHVXUIDFHDUHDRIHDFKRIWKHIROORZLQJVROLGV D E 3 cm 3 cm 3 cm 6 cm 4 cm 3 cm Solution: D 6XUIDFHDUHDRIFXEH ð VLGHOHQJWK 2 ð2 = 54 cm2 E 6XUIDFHDUHDRIFXERLG ð DUHDRIEDVHDUHDRIIURQWVXUIDFHDUHDRI ODWHUDOVXUIDFH ð ððð = 2 × (24 + 12 + 18) = 2 × 54 = 108 cm2 EXAMPLE 2 18 cm 12 cm 9 cm 7KHGLDJUDPDERYHVKRZVDSULVPZLWKWKHEDVHRIDULJKW DQJOHGWULDQJOH)LQGWKHVXUIDFHDUHDRIWKHSULVP Solution: +\SRWHQXVHRIWKHEDVH 92 + 122 = 225 = 15 cm 3HULPHWHURIWKHEDVH ffi FP 6XUIDFHDUHDRISULVP ðDUHDRIEDVHSHULPHWHURIEDVHðKHLJKW = 2 × 1 2 × 12 × 9ðfl flfl FP2 3\WKDJRUDV· theorem In the food and beverage container manufacturing industry, knowledge of surface area is essential to calculate container manufacturing costs. A rectangular prism is cut to form two triangular prisms as shown in the diagram. Is the surface area of a triangular prism half the surface area of a rectangular prism? Give your MXVWL¿FDWLRQ ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 85 If you are given an oblique prism, is the general formula for the surface area of the prism still the same as the general formula for the surface area of a right prism? Discuss with your classmates. INTERACTIVE ZONE h l EXAMPLE 3 8 cm 7 cm 7KHGLDJUDPDERYHVKRZVDS\UDPLGZLWKDVTXDUHEDVHRI VLGHFP*LYHQWKDWWKHVODQWLQJKHLJKWRIWKHODWHUDOWULDQJOH RIWKHS\UDPLGLVflFPÀQGWKHVXUIDFHDUHDRIWKHS\UDPLG Solution: 6XUIDFHDUHDRIS\UDPLG DUHDRIEDVHWRWDODUHDRIDOOWULDQJXODUVXUIDFHV DUHDRIEDVHð 1 2 ðVLGHOHQJWKRIEDVHðVODQWKHLJKW = (7 × 7) + 1 2 × (4 × 7) × 8 = 49 + 1 2 × 28 × 8 = 49 + 112 FP2 E Finding the length of side, height and slant height EXAMPLE 4 7KHWRWDOVXUIDFHDUHDRIDFXEHLVflFP2 . Find the length RIDVLGHRIWKHFXEH Solution: /HWWKHOHQJWKRIVLGHRIWKHFXEH l 7RWDOVXUIDFHDUHD l 2 fl l 2 l 2 = fl = 81 l = 81 = 9 cm ©Praxis Publishing_Focus On Maths