CHAPTER 4 Geometry 86 EXAMPLE 5 12 cm a cm 10 cm 7KH GLDJUDP VKRZV DWULDQJXODU SULVPZLWK D VXUIDFH DUHD RIflffiFP2 )LQGWKHYDOXHRIa. Solution: +HLJKWRIWULDQJOHh = 102 ²2 = 8 cm 7RWDOVXUIDFHDUHD 1 2 = 12 = 8 + (12 + 10 + 10) × a flffi ffia 32a = 800 a = 25 EXAMPLE 6 25 cm 7KH GLDJUDP VKRZV D ULJKW S\UDPLG *LYHQ WKDW WKH WRWDO VXUIDFHDUHDRIWKHS\UDPLGLVFP2 . Calculate the slant KHLJKWRIWKHS\UDPLG Solution: 7RWDOVXUIDFHDUHD DUHDRIEDVHDUHDRIVODQWLQJVXUIDFHV 1525 = (25 = 25) + 4 1 2 = 25 = h h h = ² 50 = 18 cm Geometric solids such as pyramids and prisms that have a polygonal base are named after the shape of the base. For H[DPSOH Triangular pyramid Rectangular pyramid Pentagonal prism Quadrilateral prism ©Praxis Publishing_Focus On Maths
87 Geometry CHAPTER 4 Practice 4.3 Basic Intermediate Advanced )LQG WKH VXUIDFH DUHD RI HDFK RI WKH IROORZLQJFXERLGV D E 6 cm 3 cm 9 cm 12 m 6 m 8 m )LQG WKH VXUIDFH DUHD RI HDFK RI WKH IROORZLQJFXEHV D E 5 m 16 mm )LQG WKH VXUIDFH DUHD RI HDFK WULDQJXODU prism and trapezoidal prism. D E 9 cm 4 cm 5 cm 3 cm 12 cm 18 cm 5 cm 13 cm (c) (d) 7.3 cm 6 cm 14.5 cm 9.4 cm 19 cm 4 cm 10 cm 7 cm 8 cm &DOFXODWH WKH VXUIDFH DUHD RI HDFK RI WKH IROORZLQJVROLGV D E 8 cm 12 cm 16 cm 16 cm 15 cm (c) 10 cm 12 cm )LQGWKHWRWDOVXUIDFHDUHDRIWKHVROLGVZLWK WKHIROORZLQJQHWV D E 5 cm 6 cm 20 cm 12 cm 10 cm 13 cm )RUHDFKIROORZLQJSULVP (i) )LQG WKH YDOXH RI WKH x or h in each triangular prism. LL +HQFHFDOFXODWHWKHVXUIDFHDUHD (a) 7.5 cm 18 cm x cm 25 cm E 14 cm 25 cm 72 cm h cm $FXEHKDVDWRWDOVXUIDFHDUHDRIffiFP2 . D :KDWLVWKHDUHDRIHDFKIDFH" E :KDWLVWKHOHQJWKRIHDFKVLGH" 7KHWRWDO VXUIDFH DUHD RIWKH solid as shown is 9000 cm2 . Find its slant height. 16 cm 15 cm 15 cm 38 cm 5 cm 14 cm The diagram shows a trapezoidal prism. &DOFXODWHWKHWRWDO VXUIDFHDUHDLQ FP2 RI the trapezoidal prism. 6 cm 8 cm 7KHGLDJUDPVKRZVDKH[DJRQDOSULVP)LQG WKH VXUIDFHDUHDLQ FP2 RIWKHKH[DJRQDO prism. 50 cm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 88 4.4 Surface Area of Cylinders, Cones and Spheres A Deriving the formula for the surface area of cylinders Objective: 7RGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDF\OLQGHU Instruction: 'RWKLVDFWLYLW\LQJURXSVRIIRXU Materials: &\OLQGHUVROLGDQGQHWRIF\OLQGHU 1. h r r h r <RXDUHJLYHQDULJKWF\OLQGULFDOVROLGZLWKEDVHUDGLXVRI rDQGKHLJKWRIh, as well as its QHWDVVKRZQLQWKHGLDJUDPDERYH 2. 'LVFXVVZLWK\RXWHDPPHPEHUVWRGHWHUPLQHWKHVXUIDFHDUHDRIWKHF\OLQGHU D :KDWLVWKHDUHDLQWHUPVRIrRIWKHEDVHRIWKHF\OLQGHU" E :KDWLVWKHOHQJWKRIWKHFLUFXPIHUHQFHLQWHUPVRIrRIWKHEDVHRIWKHF\OLQGHU" F :KDWLVWKHOHQJWKRIWKHUHFWDQJOHLQWKHQHWRIWKHF\OLQGHULQWHUPVRIr " G :KDWLVWKHEUHDGWKRIWKHUHFWDQJOHLQWKHQHWRIWKHF\OLQGHU" H &RS\DQGFRPSOHWHWKHIROORZLQJWRGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRID F\OLQGHU 6XUIDFHDUHDRIF\OLQGHU × $UHDRIFXUYHGIDFH = 2 × $UHDRIUHFWDQJOHLQWKHQHW = + 3. 3UHVHQW\RXUÀQGLQJVLQFODVV 7 6XUIDFHDUHDRIDF\OLQGHUFDQEHGHULYHGDVIROORZV Surface area of cylinder = 2 × base area + area of curved face = 2r 2 + area of rectangle in the net = 2r 2 + 2rh where r = the base radius and h = the height of rectangle. ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 89 B Deriving the formula for the surface area of cones Objective: 7RGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDFRQH Instruction: 'RWKLVDFWLYLW\LQJURXSVRIIRXU Materials: &RQHVROLGDQGQHWRIFRQH 1. r s r s x° <RXDUHJLYHQDULJKWFRQHVROLGZLWKEDVHUDGLXVRI rDQGWKHVODQWLQJKHLJKWRIs and its QHWDVVKRZQLQWKHGLDJUDPDERYH 2. 'LVFXVVZLWK\RXUWHDPPHPEHUVWRGHWHUPLQHWKHVXUIDFHDUHDRIWKHFRQH D :KDWLVWKHDUHDLQWHUPVRIrRIWKHEDVHRIWKHFRQH" E :KDWLVWKHFLUFXPIHUHQFHLQWHUPVRIrRIWKHEDVHRIWKHFRQH" F :KDWLVWKHOHQJWKRIDUFLQWKHQHWRIWKHFRQHLQWHUPVRIr " G &RS\DQGFRPSOHWHWKHIROORZLQJWRGHULYHWKHIRUPXODIRUWKHVXUIDFHDUHDRIDFRQH 7KHUHIRUHEDVHGRQWKHVHFWRURIFLUFOHLQWKHQHWZHJRW x = $UHDRIVHFWRU $UHDRIFLUFOHZLWKUDGLXVs x = Thus, $UHDRIVHFWRU $UHDRIFLUFOHZLWKUDGLXVs = $UHDRIVHFWRU s 2 = $UHDRIVHFWRU × s 2 = 6XUIDFHDUHDRIFRQH %DVHDUHD$UHDRIFXUYHGIDFH = $UHDRIVHFWRU = + 3. 3UHVHQW\RXUÀQGLQJVLQFODVV 8 Area of curved face of the cone Ratio of the length of arc to the circumference of circle Ratio of the area of sector to the area of circle ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 90 7KHIRUPXODIRUWKHVXUIDFHDUHDRIDFRQHFDQEHGHULYHGXVLQJWKHSURSHUWLHVRISURSRUWLRQ LQ D FLUFOH ZKHUHE\ • DUHD RI D VHFWRU LV SURSRUWLRQDOWRWKH DQJOH VXEWHQGHG E\WKH VHFWRU DWWKH FHQWUH • DUF OHQJWK LV SURSRUWLRQDO WRWKH DQJOH VXEWHQGHG E\WKH DUF DWWKH FHQWUH Surface area of cone = base area + area of curved face = r 2 + rs where r WKH EDVH UDGLXV DQG s = the slanting height. C Formula for the surface area of spheres Surface area of sphere = 4 r 2 where r LV WKH UDGLXV RIWKH VSKHUH r D Finding the surface areas of cylinders, cones and spheres EXAMPLE 7 )LQGWKHVXUIDFHDUHDVRIWKHIROORZLQJVROLGV (a) Use / ʜ 22 7 E 8VH /ʜ 30 cm 28 cm 20 cm 25 cm Solution: (a) Radius = 28 ÷ 2 = 14 cm 6XUIDFHDUHD = 2ʌr 2 + 2ʌrh = 2 × 22 7 × 142 + 2 × 22 7 × 14 × 30 = 3872 cm2 E 5DGLXV ¸ FP 6XUIDFHDUHD r 2 + ʌr s = 3.142 × 102 + 3.142 × 10 × 25 = 1099.7 cm2 ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 91 EXAMPLE 8 )LQGWKHVXUIDFHDUHDRIDVSKHUHRIUDGLXVFP Solution: 6XUIDFHDUHD ð22 7 × 3.52 = 154 cm2 EXAMPLE 9 7KHGLDJUDP VKRZVD FRQHZLWKD VXUIDFHDUHD RIFP2 )LQGWKHYDOXHRIx. (Use /ʜ Solution: 7RWDOVXUIDFHDUHD /r 2 + /r s 1131.12 = 3.142 × 102 + 3.142 × 10 × x 1131.12 = 314.2 + 31.42x 31.42x flffi x FP EXAMPLE 10 $VSKHUHKDVDVXUIDFHDUHDRIflFP2 . Find its radius and diameter. Use / ʜ 22 7 Solution: /HWLWVUDGLXVEHr cm. 6XUIDFHDUHD r 2 3850 = 4 × 22 7 × r 2 r = 3850 × 7 88 = = 17.5 cm 7KHUHIRUHGLDPHWHU ð = 35 cm EXAMPLE 11 7KH GLDJUDP VKRZV D FORVHG F\OLQGHU ZLWK D VXUIDFHDUHDRIffiffiFP2 . Find the height, in cm, RIWKHF\OLQGHUUse / ʜ 22 7 Use / ʜ 22 7 42 cm 10 cm x cm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 92 Solution: /HWWKHKHLJKWRIWKHF\OLQGHUEHh cm. 7RWDOVXUIDFHDUHD r 2 + 2rh ffiffi ð22 7 × 42 2 2 + 2 × 22 7 × 42 2 × h ffiffi h 132h = 4224 h = 4224 132 = 32 cm 20 cm 85 cm 35 cm Diagram (a) Diagram (b) h = 352 – 212 = 1225 – 441 = 784 = 28 cm E Solving problems involving surface areas EXAMPLE 12 7KHGLDJUDPVKRZVDF\OLQGHUWKDWKDVEHHQ FXWLQKDOI&DOFXODWHLWVWRWDOVXUIDFHDUHD (Use /ʜ Solution: 7RWDOVXUIDFHDUHD DUHDRIÁDWVLGH DUHDRIEDVH DUHDRIFXUYHGVXUIDFH = 2r × h + 1 1 2 r 2 + 1 2 (2rh) = 40 × 85 + 2 1 2 × 3.142 × 202 + 1 2 (2 × 3.142 × 20 × 85) = 9998.2 cm2 EXAMPLE 13 A cone with slant height 35 cm in Diagram (a) is cut into two KDOYHV 2QH RI WKH KDOYHV LV VKRZQ LQ 'LDJUDP E ,I WKH DUHDRIWKHVKDGHGEDVHLVffiFP2 ÀQGWKHKHLJKWLQFP RIWKHFRQHUse / ʜ 22 7 Solution: /HWWKHUDGLXVRIWKHFRQHEHr FPDQGWKHKHLJKWEHh cm. $UHDRIVKDGHGEDVH ffiFP2 1 2 × 22 7 × r 2 ffi r = ffi× 2 × 7 22 = 441 = 21 cm 21 cm h cm 35 cm ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 93 EXAMPLE 14 5 cm 3 cm 6 cm 6 cm $IDFWRU\KDVUHFHLYHGDQRUGHUWRPDNHffiPRGHOVZKLFK DUHPDGHIURPDOXPLQLXP VKHHWV7KH VKDSHRIWKHPRGHO LVWKHFRPELQDWLRQRIDFXERLGDQGDS\UDPLGDVVKRZQLQ WKHGLDJUDPJLYHQ)LQGWKHWRWDODUHDRIDOXPLQLXPVKHHWV required, in m2 , to make 900 models. Solution: /HWVODQWLQJKHLJKWRIS\UDPLG h h = 52 – 32 = = 4 cm 6XUIDFHDUHDRIDPRGHO EDVHDUHDWRWDODUHDRIUHFWDQJOHVWRWDODUHDRIWULDQJOHV 2 EDVHSHULPHWHUð 1 2 ðEDVHSHULPHWHUð ðð 1 2 ððð fl FP2 7RWDODUHDRIDOXPLQLXPVKHHWVWRPDNHffiPRGHOV ðffi = 140 400 cm2 = 140 400 10 000 1 m2 = 1 m × 1 m = 100 cm × 100 cm = 10 000 cm2 = 14.04 m2 Maths LINK Medicine *HRPHWU\ SOD\V DQ LPSRUWDQW UROH LQ PHGLFLQH DV LW KHOSV doctors solve various medical conditions. Three-dimensional JHRPHWU\LQSDUWLFXODULVXVHGWRPDNHGHFLVLRQVDERXWWKH GHVLJQ DQG PDQXIDFWXULQJ RI LPSODQWV DQG SURVWKHWLFV %\ considering volumes, areas, and lengths in relation to the KXPDQERG\GRFWRUVFDQXVHWKUHHGLPHQVLRQDOJHRPHWU\WR FUHDWHSURVWKHWLFVWKDWPHHWWKHVSHFLÀFUHTXLUHPHQWVRIWKHLU patients. 5 cm h 3 cm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 94 A tent has a shape which is the combination of a cylinder and a cone, as shown in the diagram above. In the structure of the tent, the height of the cylinder is 3 m and the slanting height of the cone is 3.5 m. Mutti has been provided with a canvas of area 160 m2 for the curved portions of the F\OLQGULFDO DQG FRQLFDO VKDSHV:KDW LVWKHPD[LPXP GLDPHWHU RIWKHWHQW that can be built? (Use /ʜ 3.142) 3.5 m 3 m )LQG WKH VXUIDFH DUHD RI HDFK RI WKH IROORZLQJJHRPHWULFVKDSHV (Use /ʜLIQHFHVVDU\ D E 3 cm 3 cm 3 cm 3 cm 4 cm 12 cm (c) (d) 10 cm 3 cm 11 cm 5 cm H I 3 cm 11 cm 5 cm 4.5 cm 7 cm 5 cm 6 cm (g) (h) 8 cm 7 cm 18 cm 12 cm 9 cm &DOFXODWHWKH VXUIDFH DUHDIRU HDFK RIWKH IROORZLQJVSKHUHVEDVHGRQWKHJLYHQUDGLXV (a) 12 cm (Use /ʜ E FP Use /ʜ 22 7 )LQG WKH VXUIDFH DUHD RI HDFK RI WKH IROORZLQJVROLGV*LYH\RXUDQVZHUVLQPP2 . D E 15 cm 7 cm 13 cm 6 cm 8 cm Use /ʜ 22 7 )RUHDFKRIWKHIROORZLQJÀQGWKHYDOXHRI xEDVHGRQWKHJLYHQLQIRUPDWLRQ Use /ʜ 22 7 LIQHFHVVDU\ D E 7 cm x cm 10.5 cm x cm 6XUIDFHDUHD 6XUIDFHDUHD ffiflFP2 = 775.5 cm2 (c) (d) x cm 3 cm x cm 7 cm 6XUIDFHDUHD 6XUIDFHDUHD FP2 = 202 cm2 H I x cm 6 cm 8 cm x cm 12 cm 6XUIDFHDUHD 6XUIDFHDUHD = 144 cm2 FP2 Practice 4.4 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
95 Geometry CHAPTER 4 (g) x cm 3 cm 2 cm 6XUIDFHDUHD FP2 $F\OLQGHUKDVDVXUIDFHDUHDRIflflFP2 . ,ILWVUDGLXVLVFPÀQGLWVKHLJKW (Use /ʜ 7KHVXUIDFHDUHDRIDVSKHUHLVFP2 . Find its diameter. Use /ʜ 22 7 25 cm 7 cm 80 cm &DOFXODWHWKHWRWDOVXUIDFHDUHDRIWKHVROLG as shown. Use /ʜ 22 7 The solid shows a right cone WKDW KDV EHHQ FXW LQWR KDOI &DOFXODWH LWV WRWDO VXUIDFH DUHD *LYH \RXU DQVZHU correct to 2 decimal places. (Use /ʜ 10 cm This diagram shows two identical cones MRLQHG DW WKH EDVH *LYHQ WKDW WKH VXUIDFH DUHD RI WKH VROLG LV FP2 . Find the GLDPHWHURIWKHEDVHRIWKHFRQH (Use /ʜ $ WURSK\ KDV D VKDSH ZKLFK LV WKH FRPELQDWLRQ RI D SULVP DQG D FXERLG as shown in the diagram DERYH,I DWLQ RI SDLQW LV HQRXJKWRSDLQWDQDUHDRI 4 m2 KRZ PDQ\ WURSKLHV FDQEHSDLQWHGZLWKRQHWLQ RIWKHSDLQW" O 10.5 cm 20 cm 7KHGLDJUDPDERYH VKRZVD VROLGPDGHXS RIWKHFRPELQDWLRQRIDVTXDUHEDVHFXERLG DQGDKDOIF\OLQGHUZLWKEDVHFHQWUHO. Find WKH VXUIDFH DUHD RIWKH VROLG Use /ʜ 22 7 8 cm 7KH GLDJUDP DERYH VKRZV WKUHH VSKHULFDO EDOOVÀWWLQJSHUIHFWO\LQDF\OLQGULFDOER[)LQG WKH VXUIDFH DUHD RIWKH ER[ Use /ʜ 22 7 7 cm 9 cm 6 cm 7KHGLDJUDPDERYH VKRZVD VROLGPDGHXS RIWKHFRPELQDWLRQRIDWULDQJXODUSULVPDQG D KDOI FRQH )LQG WKH VXUIDFH DUHD RI WKH solid. Use /ʜ 22 7 5 cm 15 cm 25 cm 7KH GLDJUDP DERYH VKRZV DPDLOER[ZKLFK LV PDGH XS RI WKH FRPELQDWLRQ RI D FXERLG DQG D KDOI F\OLQGHU 7KH PDLOER[ LV PDGH IURP LURQ VKHHWV(Use /ʜ D )LQG WKH DUHD RI LURQ VKHHWV UHTXLUHG WR PDNH WKHPDLOER[ E :KDWLVWKHDUHDRILURQVKHHWVUHTXLUHG in m2 WRPDNH PDLOER[HV" 9 cm 4 cm 2 cm 12 cm 40 cm 48 cm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 96 4.5 Volume of Cuboids, Cubes, Prisms and Pyramids A Volume of a cuboid and a cube 7KHYROXPHRIDFXERLGZLWKOHQJWKlEUHDGWKb and height hLVJLYHQE\ Volume of cuboid = length × breadth × height = l × b × h = lbh unit3 7KHYROXPHRIDFXEHZLWKVLGHOHQJWKlLVJLYHQE\ Volume of cube = length × breadth × height = l × l × l = l 3 unit3 B Volume of a prism Prism I Prism II Prism III &RQVLGHUWKHSULVPVDERYHZKLFKKDYHWKHVDPHEDVH7KHVHSULVPVDUHREWDLQHGE\DUUDQJLQJ FXEHVRIYROXPHXQLW3 (1 unit × 1 unit × 1 unit). Solid Prism I Prism II Prism III 7RWDOQXPEHURIXQLWFXEHV 12 24 $UHDRIEDVH XQLW2 ) 12 12 12 +HLJKWRISULVP XQLW 123 9ROXPHRISULVP XQLW3 ) 12 24 %DVHGRQWKHWDEOHDERYHWKXV Volume of prism = area of base × height = A × h = Ah unit3 h l b l l l ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 97 Objective: 7RGHULYHWKHIRUPXODIRUWKHYROXPHRIDS\UDPLG Instruction: 'RWKLVDFWLYLW\LQSDLUV 1. 2SHQWKHÀOHVolume of a Pyramid using GeoGebra. 2. 'UDJWKH VOLGHUV \HOORZPDJHQWDJUHHQEOXH UHG RQHE\RQH$OWHUQDWLYHO\ \RX FDQ FOLFNWKHEXWWRQҊ$QLPDWLRQҋ:KDWFDQ\RXREVHUYH" 3. 'LVFXVVZLWK\RXUJURXSPHPEHUVDQGH[SODLQKRZ\RXFDQDUUDQJHVL[LGHQWLFDOS\UDPLG PRGHOVWRIRUPDVTXDUHSULVP FXEH D ,VWKHWRWDOYROXPHRIWKHVL[S\UDPLGVHTXDOWRWKHYROXPHRIDVTXDUHSULVP" E ,VWKHDUHDRIWKHEDVHRIWKHS\UDPLGHTXDOWRWKHDUHDRIWKH VTXDUHSULVPIRUPHG" F :KDWLVWKHKHLJKWRIWKHVTXDUHSULVPIRUPHGLIWKHKHLJKWRIWKH S\UDPLGPRGHOLVh" 4. Taking hDVWKHKHLJKWRIWKHS\UDPLGFRPSOHWHWKHIROORZLQJWRIRUPDJHQHUDOIRUPXOD IRUWKHYROXPHRIDS\UDPLG 9ROXPHRIWKHVTXDUHSULVP $UHDRIEDVHð+HLJKW $UHDRIEDVHð +HQFHYROXPHRIS\UDPLG 9ROXPHRIVTXDUHSULVP = $UHDRIEDVe × = ð$UHDRIEDVHðh 5. 3UHVHQW\RXUÀQGLQJVLQFODVV h 9 Scan or click the above QR code to download this activity file. GeoGebra C Volume of a pyramid ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 98 7KHIRUPXODRIWKHYROXPHRIDS\UDPLGFDQEHGHULYHGDVIROORZV h 2h 2h 2h 2h 2h Diagram (a) Diagram (b) 'LDJUDP D VKRZVDULJKWS\UDPLGZLWKDVTXDUHEDVH7KHKHLJKWRIWKHS\UDPLGLVKDOIRI WKHVLGHOHQJWKRIWKHEDVH6L[RIWKHVLPLODUS\UDPLGVFDQEHDUUDQJHGLQWRDVTXDUHSULVP DFXEH DVVKRZQLQGLDJUDP E 7KHUHIRUH YROXPHRIVTXDUHSULVP DUHDRIEDVHðWKHKHLJKW DUHDRIEDVHðh 7KXVYROXPHRIS\UDPLG YROXPHRIVTXDUHSULVP YROXPHRIS\UDPLG DUHDRIEDVHðh Volume of pyramid = 1 3 × area of base × h = 1 3 × A × h = 1 3 Ah unit3 D Finding the volumes of cuboids, cubes, prisms and pyramids EXAMPLE 15 7KHGLDJUDPVKRZVFXERLGRIOHQJWKFP EUHDGWK FP DQG KHLJKW FP )LQG WKH YROXPHRIWKHFXERLG Solution: 9ROXPHRIWKHFXERLG = 4 = 3 = 72 cm3 EXAMPLE 16 7KH GLDJUDP VKRZV D FXEH RI OHQJWK fl FP )LQGWKHYROXPHRIWKHFXEH Solution: 9ROXPHRIWKHFXEH = 8 = 8 = 8 = 512 cm3 4 cm 6 cm 3 cm 8 cm 8 cm 8 cm ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 99 EXAMPLE 17 )LQGWKHYROXPHRIHDFKRIWKHIROORZLQJULJKWSULVPV D E 4.2 cm 10 cm2 13 mm 4 mm 7 mm 10 mm Solution: (a) V = Ah = 10 = 4.2 = 42 cm3 E $UHDRIEDVHA = 1 2 = (10 + 4) = 7 = 49 mm2 V = Ah = 49 × 13 PP3 EXAMPLE 18 )LQGWKHYROXPHRIHDFKRIWKHIROORZLQJULJKWS\UDPLGV D E 8 m 6 m 10 m 11 cm 8 cm 6 cm Solution: (a) V = 1 3 Ah = 1 3 = flð = 10 P3 E V = 1 3 Ah = 1 3 = 1 2 ððfl = 11 = 88 cm3 7 mm 4 mm 10 mm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 100 E Finding the height and base EXAMPLE 19 7KHGLDJUDPVKRZVDFXERLGZLWKDYROXPHRI 432 cm3 . Find the height, hRIWKHFXERLG Solution: V = lwh 432 = 9 × 4 × h h = 432 9 × 4 = 12 cm EXAMPLE 20 )LQGWKH KHLJKW RI D ULJKW SULVP ZLWK D YROXPH RI fl FP3 DQGDEDVHDUHDRIFP2 . Solution: V = Ah 480 = 32 × h h = 480 32 = 15 cm EXAMPLE 21 )LQG WKH DUHD RI EDVH RI D ULJKW SULVP ZLWK D YROXPH RI 1232 cm3 DQGDKHLJKWRIFP Solution: V = Ah 1232 = A × 14 A = 1232 14 = 88 cm2 EXAMPLE 22 5 cm 8 cm 7KHGLDJUDPVKRZVDULJKWS\UDPLGZLWKDUHFWDQJXODUEDVH *LYHQWKDWWKH YROXPH RIWKH S\UDPLG LV FP3 ÀQGWKH KHLJKWRIWKHS\UDPLG 4 cm 9 cm h ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 101 Solution: V = 1 3 Ah 120 = 1 3 = (8 = 5) = h h = 3 = 120 40 = 9 cm EXAMPLE 23 )LQG WKH DUHD RI EDVH RI WKH S\UDPLG ZLWK D YROXPH RI 378 cm3 DQGDKHLJKWRIFP Solution: V = 1 3 Ah 378 = 1 3 = A = 14 A = 3 = 378 14 = 81 cm2 F Solving problems involving volumes EXAMPLE 24 The diagram shows a rectangular DTXDULXPZKLFKLVÀOOHGZLWKFP3 RIZDWHU&DOFXODWHWKHKHLJKWRIWKHZDWHU level in the aquarium. Solution: /HWWKHKHLJKWRIZDWHUOHYHOEHh cm. 40 = 30 = h = 30 000 h = 30 000 40 = 30 = 25 7KHUHIRUHWKH KHLJKW RIWKH ZDWHU OHYHO LQWKH DTXDULXP LV 25 cm. EXAMPLE 25 The diagram shows a container in the VKDSH RI D ULJKW SULVP &DOFXODWH WKH YROXPH RI ZDWHU QHHGHGWR ÀOO XSWKH FRQWDLQHUFRPSOHWHO\ 40 cm 30 cm 9 cm 8 cm 6 cm 14 cm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 102 Solution: 9ROXPHRIZDWHUQHHGHG 1 2 flðffi ð ² = 1 2 (72) × 8 = 288 cm3 EXAMPLE 26 Given that 218.7 m# RI RUDQJH MXLFH LV SDFNHG LQWR D ULJKW S\UDPLGDOFRQWDLQHUZKLFKKDVDVTXDUHEDVH,IWKHKHLJKWRI WKHFRQWDLQHULVFPÀQGWKHOHQJWKRIHDFKVLGHRIWKHEDVH Solution: V = 1 3 Ah 218.7 = 1 3 = x2 = 10 x = 3 = 218.7 10 = 8.1 cm 7KHUHIRUH WKH OHQJWK RI HDFK VLGH RI WKH EDVH RI WKH container is 8.1 cm. x x 10 cm Practice 4.5 Basic Intermediate Advanced &DOFXODWHWKHYROXPHRIHDFKRIWKHIROORZLQJVROLGV (D E) (c) (d) H I) (g) (h) (i M 6 cm 4 cm 2 cm 9 cm 5 cm 20 cm 7 mm 8 mm 2 mm 2.6 cm 4 cm 11 cm 8 cm 6 cm 2 cm 7 cm 6 cm 5 cm 9 cm 4 cm 7 cm 45 cm2 5 mm 15 mm 3 mm 3 cm 4 cm 9 cm ©Praxis Publishing_Focus On Maths
103 Geometry CHAPTER 4 )RUHDFKRIWKHIROORZLQJÀQGWKHYDOXHRI xEDVHGRQWKHJLYHQLQIRUPDWLRQ D E 11 cm x 3 cm 8 cm 6 cm x Volume = 132 cm3 Volume = 192 cm3 (c) (d) 6 cm 12 cm 8 cm x 5 cm x 9ROXPH ffiFP3 Volume = 510 cm3 )LQGWKHKHLJKWRIHDFKRIWKHIROORZLQJULJKW prisms. (a) Volume = 210 cm3 $UHDRIEDVH FP2 E 9ROXPH flP3 $UHDRIEDVH P2 )LQG WKH DUHD RI EDVH RI HDFK RI WKH IROORZLQJULJKWSULVPV D 9ROXPH ffiFP3 Height = 15 cm E 9ROXPH PP3 Height = 44 mm )LQG WKH KHLJKW RI HDFK RI WKH IROORZLQJ S\UDPLGV (a) Volume = 24 mm3 $UHDRIEDVH PP2 E 9ROXPH flP3 $UHDRIEDVH P2 (c) 4 mm 6 mm 9ROXPH ffiPP3 (d) 8 cm Volume = 320 cm3 )LQG WKH DUHD RI EDVH RI HDFK RI WKH IROORZLQJS\UDPLGVZLWKYROXPHDQGKHLJKW DVEHORZ (a) Volume = 150 cm3 Height = 18 cm E 9ROXPH PP3 Height = 24 mm 7KHWRWDOVXUIDFHDUHDRIDFXEHLVFP2 . D )LQGWKHOHQJWKRILWVHGJH E )LQGLWVYROXPH 15 cm 18 cm 12 cm The diagram shows a rectangular container ZKLFKLVÀOOHGZLWKflFP3 RIZDWHU)LQG WKH YROXPH RI ZDWHU UHTXLUHG WR ÀOO XS WKH FRQWDLQHUFRPSOHWHO\ 1.5 m 0.9 m 0.4 m $ WXE LQ WKH VKDSH RI D ULJKW SULVP KDV GLPHQVLRQVDVVKRZQ,IWKHFDSDFLW\RIWKH WXELVOLWUHVKRZGHHSLVLWLQPHWUHV" 10 cm 18 cm 15 cm The diagram shows a container in the shape RID VTXDUH S\UDPLG7KH YROXPH RIZDWHU in the container is 400 m#. D )LQGWKHGHSWKRIWKHZDWHU E &DOFXODWHWKHYROXPHRIZDWHUQHHGHG in m#WRÀOOXSWKHFRQWDLQHUXQWLOLWLV IXOO $VWRFNFXEHKDVVLGHVRIOHQJWKFP+RZ PDQ\VWRFNFXEHVZRXOGÀOODER[PHDVXULQJ FP× 4 cm × FP" ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 104 4.6 Volume of Cylinders, Cones and Spheres A Volume of a cylinder h h r Prism Cylinder If a cylinder and a cone have the same base and also the same height, then the volume of the cylinder is three times the cone. 7KHYROXPHRIDF\OLQGHUZLWKEDVHUDGLXVr and height hFDQEHGHULYHGEDVHGRQWKHYROXPH RISULVPDVIROORZV 9ROXPHRIDF\OLQGHU V DUHDRIEDVHðKHLJKW V = r 2 × h V = r 2 h B Volume of a cone 1RWLFHWKDW D FRQH FDQ EH FRQVLGHUHG DV D S\Uamid ZLWKDFLUFXODUEDVH 7KHUHIRUH YROXPHRID FRQHZLWKEDVH UDGLXVr and height h FDQ EH GHULYHG EDVHG RQ WKH YROXPH RI S\UDPLGDVIROORZV )RUDFRQHRIEDVHUDGLXVr and height h, its volume V = 1 3 ðDUHDRIEDVHðKHLJKW V = 1 3 × r 2 × h V = 1 3 r 2h h h r Pyramid Cone $VPDOOFRQHZDVFXWRXWIURPDODUJHFRQH7KHKHLJKWRIWKHVPDOOFRQHLV 1 3 RIWKHKHLJKWRIWKHODUJHFRQH$QLWDVWDWHVWKDWWKHYROXPHRIWKHVPDOOFRQHLV 1 3 RIWKHYROXPHRIWKHODUJHFRQH,V$QLWDVVWDWHPHQWWUXHRUIDOVH" 'LVFXVVZLWK\RXUFODVVPDWHVDQGH[SODLQ\RXUDQVZHUZLWKGLDJUDPV 1 3 h h team work ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 105 C Volume of a sphere 7KHYROXPHRIDVSKHUHZLWKUDGLXVr is giveQE\ V = 4 3 r 3 . r EXAMPLE 27 )LQGWKHYROXPHRIHDFKRIWKHIROORZLQJF\OLQGHUV*LYH\RXU DQVZHUVFRUUHFWWRWKHQHDUHVWZKROHQXPEHU Use / ʜ 22 7 D E 6 cm 20 cm2 5 m 6.3 m Solution: (a) V DUHDRIEDVHðKHLJKW E V = r 2 h ð 22 7 ð 2 × 5 = 120 cm3 P3 EXAMPLE 28 )LQGWKHYROXPHRIHDFKRIWKHIROORZLQJULJKWFLUFXODUFRQHV D E 5 cm 7.2 cm 24 mm 28 mm (Use ʜ3.142) Use / ʜ 22 7 Solution: (a) V = 1 3 r 2 h = 1 3 × 22 7 × 28 2 2 × 24 = 188.52 cm3 E V = 1 3 r 2 h = 1 3 × 22 7 × 28 2 2 × 24 = 4928 mm3 Volume of a hemisphere = 1 2 × 4 3 r 3 = 2 3 r 3 r ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 106 EXAMPLE 29 D )LQGWKHYROXPHRIDVSKHUHRIUDGLXVFP E $KHPLVSKHUHKDVDUDGLXVRI 1 2 m. Calculate its volume. *LYH\RXUDQVZHUVFRUUHFWWRGHFLPDOSODFHV Solution: D 9ROXPHRIWKHVSKHUH = 4 3 r 3 = 4 3 × × 33 = 113.10 cm3 E 9ROXPHRIWKHKHPLVSKHUH = 2 3 r 3 = 2 3 × 22 7 × 1 1 2 3 = 2 3 × 22 7 × 3 2 3 = 7.07 m3 D Finding the volumes of cylinders, cones and spheres EXAMPLE 30 )LQGWKHKHLJKWRIWKHF\OLQGHUDVVKRwn ZKLFKKDVDYROXPHRIFP3 . Solution: V = r 2 h 707 = × 32 × h h = 707 × 32 = 25 cm EXAMPLE 31 $ F\OLQGHU KDV D YROXPH RI flfl FP3 DQG D KHLJKW RI FP)LQGLWVEDVHUDGLXV 3 cm ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 107 Solution: V = r 2 h 188.52 = × r 2 × 15 r = 188.52 × 15 = 4 = 2 cm EXAMPLE 32 )LQG WKH KHLJKW RI D FRQH RI YROXPH FP3 DQG D EDVH UDGLXV RI FP *LYH \RXU DQVZHU FRUUHFW WR WKH QHDUHVW ZKROHQXPEHU Solution: V = 1 3 r 2 h 1 3 × ð2 × h h = 3 × × 2 = 15 cm EXAMPLE 33 )LQG WKH EDVH UDGLXV RI D FRQH ZKLFK KDV D YROXPH RI 275 m3DQGDKHLJKWRIP Solution: V = 1 3 r 2 h 275 = 1 3 × × r 2 × 42 r = 3 × 275 × 42 = = 2.5 m EXAMPLE 34 D 7KHYROXPHRIDVSKHUHLV43 cm3 )LQGWKHUDGLXVRI the sphere. Use / ʜ 22 7 E )LQG WKH UDGLXV RI D KHPLVSKHUH RI YROXPH fl P3 . *LYH\RXUDQVZHUFRUUHFWWRGHFLPDOSODFHV (Use ʜ3.142) ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 108 Solution: (a) 4 3 r 3 = 143 4 3 × 22 7 × r 3 = 99 r = 3 99 × 3 4 × 7 22 = 3 27 = 3 4 cm E 2 3 r 3 = 80.5 2 3 × 3.142 × r 3 = 80.5 r = 3 80.5 3.142 × 3 2 Use a VFLHQWLÀF calculator. = 3.37 m 12 cm 10 cm The volume of a solid is measured in cubic units such as mm3 , cm3 and m3 . 1 cm3 = 1000 mm3 1 m3 = 1 000 000 cm3 m3 cm3 mm3 × 1 000 000 ÷ 1 000 000 × 1000 ÷ 1000 The volume of liquid is measured in litres (#) and millilitres (m#) 1 # = 1000 m# 1 m# = 1 cm3 E Solving problems involving volumes EXAMPLE 35 Find the volume, in cm3 RIRUDQJHMXLFHLQWKH F\OLQGULFDOFRQWDLQHUDVVKRZQ 8VH ʜ 3.142) Solution: 9ROXPHRIRUDQJHMXLFH ð 10 2 2 × 12 ffiFP3 EXAMPLE 36 8 cm P Q 11 cm 14 cm The diagram shows a rectangular container PÀOOHGXSZLWK ZDWHUDQGDQHPSW\F\OLQGULFDOFRQWDLQHUQ. All the water in container P is poured into container Q. Given that the radius RI FRQWDLQHUQ LV FP FDOFXODWHWKH KHLJKW LQ FP RIWKH water in container Q. Use ʜ 22 7 ©Praxis Publishing_Focus On Maths
Geometry CHAPTER 4 109 Solution: 9ROXPHRIZDWHULQFRQWDLQHUP = 11 × 8 × 14 = 1232 cm3 /HWWKHKHLJKWRIZDWHULQFRQWDLQHUQEHh cm. 22 7 × 72 × h = 1232 154h = 1232 h = 1232 154 = 8 cm EXAMPLE 37 ,QWKH GLDJUDPWKH F\OLQGULFDO FRQWDLQHU LV ÀOOHGZLWK VRPH ZDWHU DQG WKH FRQLFDO FRQWDLQHU LV IXOO\ ÀOOHG ZLWK ZDWHU ,I ERWKFRQWDLQHUVFRQWDLQHTXDOYROXPHRIZDWHUFDOFXODWHWKH KHLJKWLQFPRIWKHZDWHULQWKHF\OLQGULFDOFRQWDLQHU Use ʜ 22 7 Solution: 9ROXPHRIZDWHULQFRQLFDOFRQWDLQHU = 1 3 × 22 7 × 21 2 2 × 12 flFP3 /HWWKHKHLJKWRIZDWHULQF\OLQGULFDOFRQWDLQHUEHh cm. 22 7 × 14 2 2 × h fl 154h fl h = fl 154 = 9 cm 14 cm 12 cm 21 cm ©Praxis Publishing_Focus On Maths
CHAPTER 4 Geometry 110 EXAMPLE 38 M N 7KH GLDJUDP VKRZV VL[ LGHQWLFDO EDOOV DUUDQJHG RQ D ÁDW VXUIDFH*LYHQWKDWWKHYROXPHRIHDFKEDOOLV 2 cm3 ÀQG WKHOHQJWKRIMN in cm. Solution: V = 4 3 r 3 4 2 = 4 3 r 3 9 2 = 4 3 r 3 7KHOHQJWKRIMN = 12 × 3 2 = 18 cm. EXAMPLE 39 The diagram shows a rectangular container which contains ZDWHUWRDGHSWKRIFP:KHQDVROLGPHWDOEDOOLVSODFHG LQWR WKH FRQWDLQHU WKH GHSWK RI WKH ZDWHU LQFUHDVHV WR 12.5 cm. (Use ʜ 3.142) D )LQGWKHYROXPHRIWKHPHWDOEDOO E :KDWLVWKHUDGLXVRIWKHPHWDOEDOO"*LYH\RXUDQVZHU correct to 2 decimal places. Solution: D 9ROXPHRIWKHPHWDOEDOO OHQJWKðEUHDGWKðGLIIHUHQFHRIWKHGHSWKVRIZDWHU = 10 × 8 × (12.5 – 11) = 120 cm3 E /HWWKHUDGLXVRIPHWDOEDOOEHr cm. 4 3 r 3 = 120 4 3 × 3.142 × r 3 = 120 r = 3 120 3.142 = 3 4 FP 11 cm 10 cm 8 cm 12.5 cm r = 3 9 2 × 3 4 = 3 27 8 = 3 2 cm ©Praxis Publishing_Focus On Maths
111 Geometry CHAPTER 4 )LQG WKH YROXPH RI HDFK RI WKH IROORZLQJ solids. Use ʜ 22 7 LIQHFHVVDU\ D E 5.6 cm 13 cm 6 cm 12 cm 7 cm (c) (d) 5 cm 9 cm 6 cm 7 cm 10 cm H I 7 cm 6 cm 4 cm 5 cm 3 cm (g) (h) 10.5 cm 5 cm 5 cm 4 cm 8 cm (i) 3.5 cm 12 cm )LQG WKH YROXPH RI HDFK RI WKH IROORZLQJ solids. (a) 5 cm 7 cm 3 cm 11 cm 6 cm E) (c) 8 cm 17 cm 12 cm 5 cm 6 cm 14 cm 9 cm 150° $ SLHFH RI FDNH ZLWK D EDVH RI FLUFXODU VHFWRU LV FXW IURP D F\OLQGULFDO FDNH )LQG WKHYROXPHRIWKHSLHFHRIFDNH Use ʜ 22 7 D *LYHQWKDWWKHYROXPHRIDSULVPZLWK KHLJKWRIflFPLVflFP3 ÀQGWKHEDVH DUHDRIWKHSULVP E *LYHQ WKDW D S\UDPLG KDV D KHLJKW RI FPDQGDYROXPHRIFP3 ÀQGWKH EDVHDUHDRIWKHS\UDPLG F *LYHQ WKDW WKH KHLJKW RI D FRQH LV 19.5 mm and the volume is 325 mm3 , ÀQGWKHEDVHDUHDRIWKHFRQH D 7KHYROXPHRID ULJKWFLUFXODUF\OLQGHU RIGLDPHWHUflFPLVFP3 . Find the KHLJKWRIWKHULJKWFLUFXODUF\OLQGHU (Use ʜ E 7KH EDVH UDGLXV RI D ULJKW FLUFXODU F\OLQGHULVPDQGWKHYROXPHLVffiP3 . :KDWLVLWVKHLJKW"Use ʜ 22 7 D $ULJKWFLUFXODUF\OLQGHUKDVDYROXPHRI 1357 mm3 DQGDKHLJKWRIPP)LQG LWVEDVHUDGLXV 8VHʜ E 7KHYROXPHRIDULJKWFLUFXODUF\OLQGHULV 5 7 m3 &DOFXODWHWKHEDVHUDGLXV LILWVKHLJKWLVP Use ʜ 22 7 )LQG WKH KHLJKW RI HDFK RI WKH IROORZLQJ FRQHV ZLWK YROXPH DQG EDVH UDGLXV DV EHORZ (a) Volume = 4713 mm3 Base radius = 15 mm (Use ʜ E 9ROXPH flFP3 Base radius = 14 cm Use ʜ 22 7 (c) Volume = 754 m3 Base radius = 5 m (Use ʜ Practice 4.6 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
112 CHAPTER 4 Geometry )LQGWKHEDVHUDGLXVRIHDFKRIWKHIROORZLQJ FRQHVZLWKYROXPHDQGKHLJKWDVEHORZ (a) Volume = 4312 mm3 Height = 21 mm Use ʜ 22 7 E 9ROXPH flFP3 Height = 12 cm (Use ʜ (c) Volume = 92 m3 Height = 5 m (Use ʜ D )LQGWKHYROXPHRIHDFKRIWKHIROORZLQJ spheres, given that: (i) Radius = 1.5 cm (Use ʜ (ii) Diameter = 7 m Use ʜ 22 7 *LYH\RXUDQVZHUVFRUUHFWWRGHFLPDO places. E &DOFXODWH WKH YROXPH RI HDFK RI WKH IROORZLQJKHPLVSKHUHVJLYHQWKDWffl (i) Radius = 12 cm (Use ʜ (ii) Diameter = 3 1 2 m Use ʜ 22 7 *LYH\RXUDQVZHUVFRUUHFWWRGHFLPDO places. (a) )LQGWKHUDGLXVRIHDFKRIWKHIROORZLQJ spheres, given that: (i) Volume = 381.753 m3 (Use ʜ (ii) Volume = 53 19 81 mm3 Use ʜ 22 7 *LYH\RXUDQVZHUVFRUUHFWWRGHFLPDO places. E &DOFXODWH WKH UDGLXV RI HDFK RI WKH IROORZLQJKHPLVSKHUHVJLYHQWKDWffl (i) Volume = 13.4 cm3 (Use ʜ LL 9ROXPH 4 7 mm3 Use ʜ 22 7 *LYH\RXUDQVZHUVFRUUHFWWRGHFLPDO places. The diagram shows a F\OLQGULFDO FRQWDLQHURI UDGLXV fl FP 7KH FRQWDLQHU LV ÀOOHG with 1005 m#RIRUDQJHMXLFH (Use ʜ 11 cm D )LQGWKH KHLJKW RI RUDQJH MXLFH LQWKH container. E +RZ PXFK PRUH RUDQJH MXLFH LQ m# LV QHHGHG WR ÀOO XS WKH FRQWDLQHU FRPSOHWHO\" 5 cm 20 cm 4 cm $ F\OLQGULFDO SODVWLF WXEH LV FP ORQJ 7KH FURVVVHFWLRQ RI WKH WXEH LV DQ DQQXOXV ZLWK DQ LQWHULRU GLDPHWHU RI FPDQGDQH[WHUQDOGLDPHWHURIFP)LQG WKHYROXPHRIWKHSODVWLFPDWHULDO*LYH\RXU DQVZHUFRUUHFWWRWKHQHDUHVWZKROHQXPEHU (Use ʜ 36 cm 16 cm Q P 12 cm 10 cm A cone is divided into two parts, P and Q, as shown in the diagram. (Use ʜ D )LQGWKHKHLJKWRISDUWP. E &DOFXODWH WKH YROXPH RI SDUW Q. *LYH\RXUDQVZHUFRUUHFWWRGHFLPDO places. 12 cm 10 cm 8 cm 6 cm 7KHGLDJUDPVKRZVDS\UDPLGDQGDFRQH RIWKH VDPH YROXPH )LQGWKH EDVH UDGLXV RI WKH FRQH JLYH \RXU DQVZHU FRUUHFW WR 2 decimal places. (Use ʜ 3 cm 10 cm $OHDGF\OLQGHULVPHOWHGDQGUHFDVWWRIRUP a solid sphere as shown. Find the radius RIWKH VSKHUH JLYH \RXU DQVZHU FRUUHFWWR 2 decimal places. (Use ʜ ©Praxis Publishing_Focus On Maths
113 Geometry CHAPTER 4 Summary Summary Cuboid h b l • It has 6 rectangular faces, 12 edges and 8 vertices. • Volume = l = b = h Cube l l l • It has 6 square faces, 12 edges and 8 vertices. • Volume = l = l =l Right prism h Area of base • Surface area = 2 (Area of base) + Total area of lateral faces • Volume = Area of base = h Right pyramid h Area of base • Surface area = Area of base + Area of four slanting faces • Volume = 1 3 = Area of base = h Cylinder h r • Surface area = Area of circular face + Area of curved surface = 2r 2 + 2rh • Volume = r 2 h Cone h r • Surface area = Area of circular face + Area of curved surface = r 2 + rh • Volume = 1 3 r 2 h Sphere r • Surface area = 4r 2 • Volume = 4 3 r 3 Hemisphere r • Surface area = 2r 2 • Volume = 2 3 r 3 Geometry Section A 1. $JHRPHWULFVROLGKDVVTXDUHVXUIDFHV edges and 8 vertices. This geometric solid is a A cone B FXEH C FXERLG D S\UDPLG 2. :KLFK RI WKH IROORZLQJ WKUHHGLPHQVLRQDO VKDSH KDV DW OHDVW WKUHH ÁDW WULDQJXODU VXUIDFHV" A Prism B 3\UDPLG C &XEH D &\OLQGHU 3. V A B 7KHGLDJUDPDERYHVKRZVDULJKWFRQHVAB. :KLFKRIWKHIROORZLQJVWDWHPHQWLVQRWWUXH" A 7KH QHW RI WKH EDVH RI WKH FRQH LV D circle. B 7KHQHWRIWKH FXUYHGIDFHRIWKH FRQH is an isosceles triangle. C Points V, BDQGWKH FHQWUHRIWKHEDVH IRUPHGDULJKWDQJOHGWULDQJOH D 7KH OLQH MRLQLQJ SRLQWV and the centre RIWKHEDVHUHSUHVHQWVWKHKHLJKWRIWKH cone. ©Praxis Publishing_Focus On Maths
114 CHAPTER 4 Geometry 4. $ WKUHHGLPHQVLRQDO VKDSH KDV WZR ÁDW VXUIDFHVDQGRQHFXUYHGVXUIDFH1DPHWKH shape. A 3\UDPLG C &\OLQGHU B Cone D Sphere 5. 'LDJUDPDERYHVKRZVDULJKWSULVP:KLFK RI WKH IROORZLQJ LV WKH SRVVLEOH QHW RI WKH SULVP" A C B D 6. 5 cm 8 cm 7KH GLDJUDP VKRZV D ULJKW S\UDPLGZLWK D UHFWDQJXODU EDVH *LYHQ WKDW WKH YROXPH RIWKHS\UDPLGLVfl FP3 WKHKHLJKWRIWKH S\UDPLGLQFPLV A 4 C B 5 D 7 7. t 6 cm 4 cm 3 cm 7KH GLDJUDP VKRZV D FXERLG DQG D FXEH *LYHQWKDW ERWK FXERLG DQG FXEH KDYHWKH VDPHYROXPHÀQGWKHYDOXHRIt in cm. A 72 C 18 B D 12 8. 8 cm 7KH GLDJUDP VKRZV D ULJKW S\UDPLG ZLWK D VTXDUH EDVH RI FP2 . Calculate the VXUIDFHDUHDLQFP2 RIWKHS\UDPLG A 240 B C D 384 9. 28 cm 7KHGLDJUDPVKRZVDF\OLQGULFDOFRQWDLQHU Given that 3080 cm3 RIZDWHULVSRXUHGLQWR WKH FRQWDLQHU ÀQG WKH KHLJKW RI WKH ZDWHU level in cm. Use ʜ 22 7 A 5 B C 8 D 9 10. 7KH GLDJUDP VKRZV D F\OLQGULFDO FRQWDLQHU ZLWKDEDVH UDGLXVRIFP*LYHQWKDWWKH YROXPHRIZDWHULQWKHFRQWDLQHULVFP3 , FDOFXODWHWKHKHLJKWRIWKHZDWHUOHYHOLQFP A 7 / B 14 / C 21 / D 28 / 11. x cm 14 cm 7KHGLDJUDPVKRZVDVROLGZKLFKLVKDOIRID ULJKWFLUFXODUF\OLQGHU*LYHQWKDWWKHYROXPH RIWKHVROLGLVffi/ cm3 ÀQGWKHYDOXHRIx. A C 13 B 14 D 12 ©Praxis Publishing_Focus On Maths
115 Geometry CHAPTER 4 12. V R Q P S 7KH GLDJUDP VKRZV D ULJKW S\UDPLG RI VTXDUH EDVH ZLWK D YROXPH RI fl FP3 . *LYHQ WKDW WKH KHLJKW RI WKH S\UDPLG LV flFPÀQGWKHOHQJWKRIPQ in cm. A C 12 B 8 D 14 13. 10 cm 12 cm 7KHGLDJUDPVKRZVWKHQHWRIDULJKWFLUFXODU FRQH &DOFXODWHWKH YROXPH RIWKH FRQH LQ cm3 . A 288/ C 120/ B 144/ D ffi/ 14. 5 cm 4 cm 7KH GLDJUDP VKRZV D VROLG PDGH XS RI a right circular cone and a hemisphere. &DOFXODWHWKHYROXPHRIWKHVROLGLQFP3 . A 54/ C 39/ B 48/ D 30/ 15. 20 cm 8 cm 14 cm 7KH GLDJUDP VKRZV D VROLG PDGH XS RI a right circular cone and a right circular F\OLQGHU &DOFXODWH WKH YROXPH RI WKH VROLG in cm3 . Use ʜ 22 7 A 1848 C B 2040 D 3080 16. 7KHGLDJUDPVKRZVDULJKWFLUFXODUF\OLQGHU ZLWK D GLDPHWHU RI FP DQG D KHLJKW RI 15 cm. A right circular cone, as shown in the VKDGHGSDUWRIWKHGLDJUDPLVUHPRYHGIURP WKHF\OLQGHU,IWKHKHLJKWRIWKHULJKWFLUFXODU FRQHLV FPWKH YROXPHRIWKH UHPDLQLQJ F\OLQGHULQFP3 , is A 150 B 135 C 120 D 105 Section B 1. 7KH GLDJUDP VKRZV D ULJKW S\UDPLGZLWK D VTXDUHEDVH2QHRIWKHVXUIDFHVRIWKHQHW RIWKHS\UDPLGLVGUDZQRQWKHJULGEHORZ 7KH JULG KDV HTXDO VTXDUHV ZLWK VLGHV RI XQLW&RPSOHWHWKHQHWRIWKHS\UDPLG 8 units 5 units ©Praxis Publishing_Focus On Maths
116 CHAPTER 4 Geometry 2. )LQGWKHWRWDOVXUIDFHDUHDDQGWKHYROXPH RIWKHVROLGVZLWKWKHIROORZLQJQHWV (a) 5 cm 5 cm 5 cm E 5 cm 3 cm 13 cm 3. 2.4 m 5 m 3.7 m D )LQGWKHKHLJKWRIWKHWULDQJXODUVXUIDFH E )LQGWKHVXUIDFHDUHDRIWKHSULVP 4. (a) 66 cm 33 cm 7KH GLDJUDP DERYH VKRZV D SLHFH RI SDSHU ZKLFK LV UROOHG XS WR IRUP WKH FXUYHG IDFH RI D F\OLQGHU )LQG Use ʜ 22 7 L WKH EDVH UDGLXV RI WKH F\OLQGHU IRUPHG LL WKHYROXPHRIWKHF\OLQGHUIRUPHG E p cm 8 cm 7KH GLDJUDP DERYH VKRZV WKH QHW RI D FRQH*LYHQWKDWWKHVXUIDFHDUHDRIWKH FRQH LV fl cm2 ÀQG WKH YDOXH RI p. +HQFH ÀQGWKH YROXPH RIWKH FRQHWKDW ZLOO EH IRUPHG (Use ʜ 5. 20 cm 16 cm ,Q WKH GLDJUDP D F\OLQGULFDO KROH LV GULOOHG LQWR D ELJJHU F\OLQGULFDO VROLG RI GLDPHWHU FP ,I WKH KROH KDV D GLDPHWHU RI FPFDOFXODWHWKHWRWDOVXUIDFHDUHDRIWKH F\OLQGULFDOVROLGLQWHUPVRI. 6. (a) 7KHSULVPKDVDYROXPHRIflFP3 . What LVWKHYROXPHRIWKHS\UDPLG" E $VTXDUHS\UDPLGKDVDKHLJKWRIFP DQG D YROXPH RI FP3 . Find the side OHQJWKRIWKHEDVH 7. 12 mm T Q R P S 6 mm 8 mm 11 mm 7KH VROLG VKRZQ LV D WUDSH]RLGDO S\UDPLG )LQGWKHYROXPHRIWKHS\UDPLG 8. 12 cm 5 cm 10 cm 7KHGLDJUDP VKRZVD FRQHZLWKSDUWRILWV WRSFXWRII&DOFXODWHWKHDUHDRIWKHFXUYHG VXUIDFHRIWKHUHPDLQLQJVROLG (Use ʜ ©Praxis Publishing_Focus On Maths
117 Geometry CHAPTER 4 9. 10.5 cm 14 cm The diagram shows a cone placed on top RIDKHPLVSKHUH&DOFXODWHWKHWRWDOVXUIDFH DUHDRIWKHVROLGUse ʜ 22 7 10. D *LYHQ WKDW WKH UDGLXV RI D KHPLVSKHUH is 1 1 2 FP ILQG WKH YROXPH RI WKH hemisphere. Use ʜ 22 7 E )LQGWKHGLDPHWHURIDVSKHUHRIYROXPH 113 1 7 cm3 . Use ʜ 22 7 11. 1.4 m The diagram shows a hemisphere. Find the YROXPHRIWKHKHPLVSKHUH Use ʜ 22 7 12. )LQGWKHGLDPHWHURIDULJKWFLUFXODUF\OLQGHU RIYROXPHFP3 and height 5 cm. Use ʜ 22 7 13. ,I D ULJKW FLUFXODU FRQH KDV D YROXPH RI 825 cm3 DQGDGLDPHWHURIFPÀQGWKH KHLJKWRIWKHcone. Use ʜ 22 7 14. The diagram shows a right circular cone RIYROXPHffi cm3)LQGWKHYDOXHRIk. k cm 6 cm 15. 42 cm 33 cm 7KH GLDJUDP VKRZV D ULJKW S\UDPLG DQG D KHPLVSKHUH ,I WKH YROXPHV RI WZR VROLGV DUH HTXDO ÀQG WKH UDGLXV LQ FP RI WKH hemisphere. Use ʜ 22 7 16. 7KHGLDJUDPVKRZVWZRLGHQWLFDOEDOOVWKDW ÀWH[DFWO\LQVLGHRIDFXERLGER[,IWKHWRWDO YROXPH RI WZR EDOOV LV cm3 ÀQG WKH KHLJKWLQFPRIWKHER[ 17. 14 cm 11 cm 7 cm 28 cm 7KHGLDJUDPVKRZVDFXERLGFRQWDLQHUDQG D ULJKW FLUFXODU F\OLQGULFDO FRQWDLQHU 7KH FXERLG FRQWDLQHU LV IXOO\ ÀOOHG ZLWK ZDWHU ,I DOO WKH ZDWHU IURP WKH FXERLG FRQWDLQHU LV SRXUHG LQWR WKH F\OLQGULFDO FRQWDLQHU FDOFXODWHWKH KHLJKW LQ FP RIWKHZDWHU LQ WKHF\OLQGULFDOFRQWDLQHUUse ʜ 22 7 ©Praxis Publishing_Focus On Maths
CARTESIAN COORDINATE CARTESIAN COORDINATE 5SYSTEM Applications of this chapter 7KH PDQDJHPHQW DQG UHJXODWLRQ RI DLU WUDIÀ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©Praxis Publishing_Focus On Maths
Concept Map Learning Outcomes • 8QGHUVWDQGWKHFRQFHSWRIFRRUGLQDWHVDQG&DUWHVLDQFRRUGLQDWHV\VWHP • ,GHQWLI\WKHTXDGUDQWVRIWKH&DUWHVLDQSODQH • ,GHQWLI\WKHFRRUGLQDWHVRIDSRLQWRQWKH&DUWHVLDQSODQHDQGYLFHYHUVD • 8QGHUVWDQGWKHGLVWDQFHEHWZHHQWZRSRLQWVRQWKH&DUWHVLDQSODQH • 'HWHUPLQHWKHGLVWDQFHEHWZHHQWZRSRLQWVRQWKH&DUWHVLDQSODQH • 8QGHUVWDQGWKHGLYLVLRQRIDOLQHVHJPHQWRQWKH&DUWHVLDQSODQH • 'HWHUPLQHWKHFRRUGLQDWHVRIDSRLQWWKDWGLYLGHVDOLQHVHJPHQWLQDFHUWDLQUDWLR • 6ROYHSUREOHPVLQYROYLQJ&DUWHVLDQFRRUGLQDWHV\VWHP 119 Division of a Line Segment Distance Between Two Points Horizontal Distance and Vertical Distance Identify and Plot Points Determine Location Cartesian Coordinates Cartesian Coordinate System Rene Descartes (1596 – 1650) was a French mathematician. He proposed the Cartesian plane to determine the location of a point. • Coordinates • Cartesian plane • x-axis • y-axis • Origin • x-coordinate • y-coordinate • Quadrant • Distance • Pythagoras’ theorem • Midpoint • Division of a line segment Key Terms Maths History ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 120 Flashback 1. /LVWDOOWKHRGGQXPEHUVEHWZHHQDQG 2. /LVWDOOWKHHYHQQXPEHUVEHWZHHQDQG 3. (YDOXDWH D ð E ð F ð G ð H ð 1 Critical Thinking 4. (YDOXDWH D ¸ E ¸ F ¸ G ¸ H ¸ Do you know the use of grid lines? What are the uses of grid lines? Jason Rob Kristov Sally Height Age 8VHWKHJUDSKDERYHWRDQVZHUTXHVWLRQVEHORZ D :KRLVWKH\RXQJHVW" E :KRLVWKHWDOOHVW" F ,VWKHROGHVWSHUVRQWKHWDOOHVW" ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 121 5.1 Determining Location Coordinate LVDQ\RIDVHWRIQXPEHUVWKDWXQLTXHO\LGHQWLÀHVWKHORFDWLRQRIDSRLQWUHODWLYH WRDVHWRIÀ[HGUHIHUHQFHSRLQWV /RRNDWWKHPDSEHORZ Hospital H Bus station Bus station B t ti Park Joe’s ho Joe s house o house School A 1 2 3 4 5 6 7 B CDE F GH I J K Police station o 7KHPDSLVGLYLGHGLQWRVTXDUHVWRKHOSXVWRÀQGDORFDWLRQ(DFKVTXDUHKDVLWVRZQQDPH ZKLFKLVJLYHQE\WZRFRRUGLQDWHVDOHWWHUDQGDQXPEHU EXAMPLE 1 2 -RH SODQV WR JR WR SDUN 8VLQJ WKH PDS DERYH VWDWH WKH ORFDWLRQRI D -RH·VKRXVH E WKHSDUN Solution: (a) D5 E % EXAMPLE 2 8VLQJWKHPDSDERYHQDPHWKHEXLOGLQJORFDWHGDW* Solution: 3ROLFHVWDWLRQ :ULWH WKH KRUL]RQWDO FRRUGLQDWH ÀUVW IROORZHG E\WKH YHUWLFDO FRRUGLQDWH /RRNIRUWKHEXLOGLQJLQ FROXPQ * DQG URZ ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 122 Practice 5.1 Basic Intermediate Advanced 7KHGLDJUDPEHORZVKRZVWKHORFDWLRQVRI WKHWLFNHWFRXQWHUDQGVRPHDWWUDFWLRQVLQD WKHPHSDUN A 1 2 3 4 5 BCDEF Haunted house Ferris wheel Roller coaster Ticket counter Bumper car Carousel D (ULQ ULGHV WKH FDURXVHO DQG )HUULV ZKHHO6WDWHWKHORFDWLRQRIHDFKRIWKH DWWUDFWLRQV E 1DPHWKHDWWUDFWLRQORFDWHGDW' 7KHGLDJUDPEHORZVKRZVDWUHDVXUHPDS A 1 2 3 4 5 BCDEF D -RKQZDQWVWRÀQGWKHWUHDVXUHFKHVW :KHUHLVWKHWUHDVXUHFKHVW" E :KDWLWHPLVORFDWHGDW%" F :KDW LWHP LV LQ WKH VDPH URZ DV WKH VKDUN" 2 3 1 A B C 8VHWKHJUDSKDERYHWRDQVZHUTXHVWLRQV D DQG E D :KDWLVORFDWHGDW%" E :KDWLVWKHJULGUHIHUHQFHIRUWKHKRXVH" 2 ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 123 5.2 Cartesian Coordinates A Identifying and plotting points ,Q WKH WK FHQWXU\ D PDWKHPDWLFLDQ FDOOHG 5HQH 'HVFDUWHVSURSRVHGWKHCartesian plane to determine WKHORFDWLRQRIDSRLQW,WLVDYLVXDOPHDQVRIGHVFULELQJ ORFDWLRQVRQDSODQHE\XVLQJQXPEHUVDVFRRUGLQDWHV 7KH &DUWHVLDQ SODQH FRQVLVWV RI D KRUL]RQWDO QXPEHU OLQH DQG D YHUWLFDO QXPEHU OLQH LQWHUVHFWLQJ DW D ULJKW DQJOH 7KH KRUL]RQWDO QXPEHU OLQH LV NQRZQ DV WKH x-axis ZKHUHDV WKH YHUWLFDO QXPEHU OLQH LV NQRZQ DV the y-axis7KHLQWHUVHFWLRQSRLQWRIWKHQXPEHUOLQHV LVNQRZQDVWKHorigin $&DUWHVLDQFRRUGLQDWHLVZULWWHQDV xy ZKHUHx and yDUHDQ\QXPEHUV7KHÀUVWQXPEHU xLVFDOOHGWKHx-coordinateRIDSRLQWZKLFKVKRZVWKHKRUL]RQWDOSRVLWLRQRIWKHSRLQW7KH VHFRQGQXPEHUyLVFDOOHGWKHy-coordinateRIDSRLQWZKLFKVKRZVWKHYHUWLFDOSRVLWLRQRI WKHSRLQW)RUH[DPSOH (4, 5) The xD[LVDQGyD[LVGLYLGHWKH&DUWHVLDQSODQHLQWRIRXUquadrantsDVIROORZV x y 4 5 6 3 2 1 –1 – 2 –3 – 4 – 5 –6 –6 –5 – 4 –3 –1–2 O 21 3 456 Quadrant II Quadrant I Quadrant III Quadrant IV *LYHQWKDWDSRLQWP(xy ZKHUHxLVWKHxFRRUGLQDWHDQGyLVWKHyFRRUGLQDWHRIWKHSRLQW ,IWKHSRLQWPLVLQTXDGUDQW,x > 0 and y! ,IWKHSRLQWPLVLQTXDGUDQW,,x < 0 and y! ,IWKHSRLQWPLVLQTXDGUDQW,,,x < 0 and yfi ,IWKHSRLQWPLVLQTXDGUDQW,9x > 0 and yfi y x 4 5 6 3 2 1 2 3 –1 –2 –1–2 O 1 4 5 6 (4, 5) y-axis x-axis Origin Coordinates xFRRUGLQDWH yFRRUGLQDWH ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 124 EXAMPLE 3 5HIHUULQJ WR WKH &DUWHVLDQ SOane DV VKRZQ VWDWH WKH FRRUGLQDWHV RISRLQWV (a) K E L F M (d) N (e) P (f) Q Solution: D 7KHFRRUGLQDWHVRIK DUH ² E 7KHFRRUGLQDWHVRIL DUH F 7KHFRRUGLQDWHVRIM DUH ² G 7KHFRRUGLQDWHVRIN DUH ²² H 7KHFRRUGLQDWHVRIP DUH I 7KHFRRUGLQDWHVRIQ DUH ² y x –2 2 2 P L Q M O N K 4 6 4 –2 –4 • The letter O is usually used to label the origin. • The coordinates of the origin are (0, 0). • The y-coordinate of the point lying on the x-axis is 0. • The x-coordinate of the point lying on the y-axis is 0. )RUH[DPSOH Point Distance from the y-axis Distance from the x-axis A 3 units to the right 2 units above B 3 units to the left 2 units above C 3 units to the left 2 units below D 3 units to the right 2 units below 7KHUHIRUH WKHFRRUGLQDWHVRIADUH WKHFRRUGLQDWHVRIBDUH ² WKHFRRUGLQDWHVRICDUH ²² WKHFRRUGLQDWHVRIDDUH ² 7RSORWDSRLQWRQWKH&DUWHVLDQSODQHEHJLQDWWKHRULJLQPRYH along the xD[LVWRWKHQXPEHULQGLFDWHGE\WKHxFRRUGLQDWHDQG then along the yD[LVWRWKHQXPEHULQGLFDWHGE\WKHyFRRUGLQDWH )RUH[DPSOHWRSORWWKHSRLQWZLWKFRRUGLQDWHV PRYHXQLWV WRWKHULJKWRIWKHRULJLQDQGWKHQXQLWVXS 2 –2 –4 –2 O 2 4 x y A D B C xFRRUGLQDWH yFRRUGLQDWH x y 4 5 6 3 2 1 –1 – 2 –1–2 O 21 3 456 (3, 4) 3 units 4 units ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 125 EXAMPLE 4 3ORW WKH SRLQWV P ² DQG Q ² ffi LQWKHIROORZLQJ&DUWHVLDQSODQH,GHQWLI\ WKHTXDGUDQWVLQZKLFKWKHSRLQWVP and and QOLH Solution: Point P OLHVRQTXDGUDQW,9 Point QOLHVRQTXDGUDQW,, EXAMPLE 5 1DPH WKH VKDSH IRUPHG E\ WKH SRLQWV P Q R DQGS ²² LQD&DUWHVLDQSODQH Solution: O y 2 4 –2 –2 2 4 6 x P Q R S PQRSLVDWUDSH]LXP 1RWLFH WKDWPQ LV SDUDOOHO WRSR O y x –3 –5–10–15 3 5 10 15 6 9 –6 –9 O y P Q x –3 –5–10–15 3 5 10 15 6 9 –6 –9 B Horizontal distance and vertical distance 7KHGLVWDQFHDORQJWKHxD[LVLVhorizontal distance7KHGLVWDQFHDORQJWKHyD[LVLVvertical distance x y 4 5 3 2 1 21 3 O 4 5 N M Vertical distance Horizontal distance /RRNDWWKHGLDJUDPDERYH7KHGLVWDQFHRIHDFKJULGLVXQLW6RWKHKRUL]RQWDOGLVWDQFHRI SRLQWNIURPSRLQWMLVXQLWV7KHYHUWLFDOGLVWDQFHRISRLQWNIURPSRLQWMLVXQLWV (x, y) is also known as an ordered pair. A solution of a two variable equations can be written as an ordered pair. Cartesian coordinates is a system for the representation of a point in space in terms of its distance, measured along a set of mutually perpendicular axes, from a given origin. ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 126 EXAMPLE 6 'HWHUPLQHWKHKRUL]RQWDOGLVWDQFH DQG YHUWLFDO GLVWDQFH RI SRLQW A IURPWKHRULJLQ Solution: +RUL]RQWDOGLVWDQFHRISRLQWA IURPWKHRULJLQ XQLWV 9HUWLFDOGLVWDQFHRISRLQWAIURPWKHRULJLQ XQLWV EXAMPLE 7 )LQG WKH KRUL]RQWDO GLVWDQFH DQG YHUWLFDOGLVWDQFHEHWZHHQSRLnt C DQGSRLQWD Solution: 7KHFRRUGLQDWHVRISRLQWC LV 7KHFRRUGLQDWHVRISRLQWDLV +RUL]RQWDOGLVWDQFHEHWZHHQSRLQWCDQGSRLQWD ï XQLWV 9HUWLFDOGLVWDQFHEHWZHHQSRLQWCDQGSRLQWD ï XQLWV x y 4 2 2 O 4 6 A(5, 6) 6 x y 4 6 2 2 O 4 6 D C Note that the distance of each grid is 1 unit. Then, count the number of squares. x y 4 6 2 2 O 4 6 D 3 units 2 units C Horizontal distance = 2 units Vertical distance = 3 units C Distance between two points +RZGR\RXPHDVXUHWKHGLVWDQFHEHWZHHQWZRSRLQWVRQWKH&DUWHVLDQSODQH" y x O P Q (a) y x O P Q (b) y x O P Q (c) 'LDJUDP F VKRZVWKHFRUUHFWZD\WRPHDVXUHWKHGLVWDQFHEHWZHHQSRLQWPDQGSRLQWQ 7KHGLVWDQFHEHWZHHQWZRSRLQWVRQWKH&DUWHVLDQSODQHLVWKHGLVWDQFHPHDVXUHGIURPRQH SRLQWWRDQRWKHUSRLQWLQDVWUDLJKWOLQH (x, y) 7KH KRUL]RQWDO GLVWDQFH IURP WKH RULJLQ LV x XQLW The vertical distance from the origin is y unit. ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 127 Objective:7RGHULYHWKHIRUPXODIRUWKHGLVWDQFHEHWZHHQWZRSRLQWV Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU 1. /RRNDWWKHIROORZLQJGLDJUDPV y x O 1 1 2 3 4 P Q 234567 y x O 1 1 2 3 4 234 M x y R S 1 1 2 34 –1 –2 2 O &DOFXODWHWKHGLVWDQFHEHWZHHQWKHSDLURISRLQWVE\ D FRXQWLQJWKHQXPEHURIVTXDUHVEHWZHHQWKHSDLURISRLQWV E ÀQGLQJWKHGLIIHUHQFHEHWZHHQxFRRUGLQDWHVRUyFRRUGLQDWHV 2. 'HVFULEHWKHGLVWDQFHEHWZHHQWZRSRLQWVWKDWKDYH D DFRPPRQyFRRUGLQDWH E DFRPPRQxFRRUGLQDWH 3. 7KHGLDJUDPVKRZVWZRSRLQWVP(x y ) and R(x y 7KHGLVWDQFHEHWZHHQSRLQWP and SRLQWRFDQEHFDOFXODWHGXVLQJ3\WKDJRUDV·WKHRUHP &RPSOHWHWKHIROORZLQJ 'LVWDQFHEHWZHHQSRLQWPDQGSRLQWQ 'LVWDQFHEHWZHHQSRLQWQDQGSRLQWR 8VLQJ3\WKDJRUDV·WKHRUHP 35ɧ 34ɧ + 45ɧ ) + ( ) PR ( ) + ( ) y x P(x1 , y1 ) R(x2 , y2 ) x2 – x1 y2 – y1 Q O 1 )URP$FWLYLW\LWLVIRXQGWKDW % WKHGLVWDQFHEHWZHHQWZRSRLQWVWKDWKDYHDFRPPRQyFRRUGLQDWHLVWKHGLIIHUHQFHEHWZHHQ their xFRRUGLQDWHVWKDWLVx – x % WKHGLVWDQFHEHWZHHQWZRSRLQWVWKDWKDYHDFRPPRQxFRRUGLQDWHLVWKHGLIIHUHQFHEHWZHHQ their yFRRUGLQDWHVWKDWLVy – y % WKHGLVWDQFHEHWZHHQRULJLQSRLQW DQG xy LV [ɧ + \ɧ % WKHGLVWDQFHEHWZHHQWZRSRLQWV x y ) and (x y LV (x – x ) + (y – y ) ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 128 EXAMPLE 8 )LQGWKHGLVWDQFHEHWZHHQWKHIROORZLQJSDLUVRISRLQWV (a) P DQGQ E R ² DQGS ²² F T DQGU ffifl G V ² DQGW ²fl Solution: (a) PQ x – x ² XQLWV E RS y – y ² ² XQLWV F TU (x – x ) + (y – y ) ffi² fl² fl XQLWV (d) VW (x – x ) + (y – y ) >² ² @ ²fl² ² XQLWV GS 'LIIHUHQFH EHWZHHQ xFRRUGLQDWHV 'LIIHUHQFH EHWZHHQ yFRRUGLQDWHV EXAMPLE 9 ,QWKHGLDJUDPPQ and R are the YHUWLFHVRIDWULDQJOH&DOFXODWH D WKHSHULPHWHURIWULDQJOHPQR E WKHDUHDRIWULDQJOHPQR y x R(–3, 6) Q(–3, –2) P(3, 4) O Jackius and Lai Yee each uses the formula shown to determine the distance between two points M(x1 , y1 ) and N(x2 , y2 ). • Jackius • Lai Yee (x1 – x2 )2 + (y1 – y2 ) 2 (x1 – x2 )2 + (y2 – y1 ) 2 Their teacher stated that the formula used by Jackius was correct but the formula used by Lai Yee was wrong. Comment on the formula used by them with your classmates. INTERACTIVE ZONE 136 = ? 8VLQJVFLHQWL¿FFDOFXODWRUSUHVV 1 3 6 = 11.661903789 Pythagoras’ theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 129 Solution: (a) QR ² ² flXQLWV PQ >² ² @ >² ² @ flffiXQLWV PR >² ² @ ² ² XQLWV 7KHSHULPHWHURIWULDQJOHPQR flflffi flXQLWV E /HWSEHSRLQWWKDWOLHVRQWKH line QR VXFK WKDW OLQH PS LV SHUSHQGLFXODUWROLQHQR 7KXVPS ² ² XQLWV The area of triangle PQR ðflð XQLWV EXAMPLE 10 7KH ORFDWLRQV RI $LVKDK·V KRXVH %HQ·V KRXVH &LQG\·V KRXVH DQG WKHLU VFKRRO DUH SORWWHG LQ D &DUWHVLDQ SODQH DQGDUHORFDWHGDWA ² B ² C DQGS ² UHVSHFWLYHO\:KRVHKRXVHLVWKHQHDUHVWWRVFKRRO" Solution: AS ²² >² ² @ (–5) + 5 50 XQLWV BS ²² >² ² @ ² flXQLWV CS ² >² ² @ XQLWV &LQG\·VKRXVHLVWKHQHDUHVWWRVFKRRO R(–3, 6) Q(–3, –2) S(–3, 4) P(3, 4) Perpendicular means at or forming a right angle. ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 130 D Division of a line segment $SRLQWWKDWGLYLGHVDOLQHVHJPHQWLQDFHUWDLQUDWLRLVLOOXVWUDWHGDVIROORZV Objective:7RGHULYHWKHIRUPXODXVHGWRÀQGWKHFRRUGLQDWHVRIWKHGLYLVRURIDOLQHVHJPHQW Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU 1. 7KH GLDJUDP VKRZV D SRLQW P WKDW GLYLGHV D OLQH VHJPHQW FRQQHFWLQJ SRLQW A(x y DQG SRLQW B(x y ) in the ratio mffln 2. *LYHQWKDWWKHUDWLRmfflnIROORZVWKHIROORZLQJFRQGLWLRQffl x – x fflx – x mffln x – x x – x m n n(x – x m(x – x) nx – nx mx – mx mx + nx nx + mx (m + n)x nx + mx x 3. 5HSHDWVWHSIRUyïy ffly ïy mffln 4. :KDWLVIRUPXODXVHGWRÀQGWKHFRRUGLQDWHVRISRLQWPWKDWGLYLGHVWKHOLQHVHJPHQWAB in the ratio mffln" 5. :KDWZLOOKDSSHQWRWKHIRUPXODZKHQm n" 2 m n A B P m n : Point P is closer to point B, where AP PB. m = n m n A B P : Point P is the midpoint of the line segment AB, where AP = PB. m n A B P m n : Point P is closer to point A, where AP PB. The position of point P that divides the line segment AB in the ratio m : n The midpoint between two points in the Cartesian plane is a point that bisect a straight line joining the two points. For instance, P M Q Point M lies on the line PQ such that PM = MQ. Thus, M is the midpoint of the line PQ. y x m O n y2 – y y – y1 x2 x – x – x 1 P(x, y) B(x2 , y2 ) A(x1 , y1 ) : ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 131 )URP$FWLYLW\WKHFRRUGLQDWHVRISRLQWP(xy WKDWGLYLGHVWKHOLQHVHJPHQWMRLQLQJSRLQWV A(x y ) and B(x y ) in the ratio mffln are P(xy nx + mx m + n ny + my m + n When m nSRLQWPZLOOEHFRPHWKHPLGSRLQWRIOLQHVHJPHQWABDQGLWVFRRUGLQDWHVDUH x1 + x2 2 y1 + y2 2 EXAMPLE 11 *LYHQ WKDW A ² DQG B ffi )LQG WKH PLGSRLQW RI WKH VWUDLJKWOLQHAB Solution: 0LGSRLQWRIAB ²ffi Critical Thinking Based on the Cartesian plane, is M the midpoint between points P and Q? y P M Q x O EXAMPLE 12 7KH GLDJUDP VKRZV D VWUDLJKW OLQH MN *LYHQ WKDW WKH PLGSRLQW RI MN LV )LQGWKH YDOXHV of p and q M = x + x y + y x y N(9, q) M(p, –2) O ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 132 Solution: 0LGSRLQWRIMN x + x y + y pffi ²q pffi DQG ²q p ² q EXAMPLE 13 7KHGLDJUDPEHORZVKRZVSRLQWPZKLFKGLYLGHVCD in the UDWLRffl x y O C(–3, 8) D(11, 1) P )LQGWKHFRRUGLQDWHVRISRLQWP Solution: C(–3, 8) D(11, 1) P 4 3 P ² fl fl EXAMPLE 14 $VWUDLJKWOLQHSDVVHVWKURXJKP DQGQ ,ISRLQWTGLYLGHVWKHOLQHVHJPHQWPQZKHUHPT TQÀQG WKHFRRUGLQDWHVRISRLQWT P = nx + mx m + n ny + my m + n ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 133 Solution: )URPPT TQ PT TQ 7KHUHIRUHPTfflTQ ffl T fl The diagram shows the positions of three lamp posts, P, Q and R. Given that all the lamp posts are installed along a straight road where the distance of the lamp post Q from the lamp post P is twice the distance from the lamp post R. Express k in terms of h. x y O R(3, –3r) Q(r, h) P(k, –3) EXAMPLE 15 7KH GLDJUDP VKRZV SRLQW B ZKLFKGLYLGHVVWUDLJKWOLQHAC in the ratio ACfflBC ffl )LQGWKHFRRUGLQDWHVRISRLQWC Solution: C(x, y) B(3, –5) A(1, –8) 1 4 3 ABfflBC ffl B nx + mx m + n ny + my m + n ² x) ²fl y) ² x ²y x DQG ²y ² x ffi y 7KHUHIRUHC ffi y x C B(3, –5) A(1, –8) O Given that P is the point that divides the line segment AB, • If mAP = nPB, then AP : PB = n : m n m A B n P AP = — PB m m PB = — AB n • If mAB = nAP, then AP : PB = m : n – m m n m A B P n ©Praxis Publishing_Focus On Maths
CHAPTER 5 Cartesian Coordinate System 134 EXAMPLE 16 7KHMRXUQH\RIDFKDULW\UXQLVSORWWHGLQWKHFRRUGLQDWHVPDS x y Q(4, 8) C A(0, –3) B(10, 5) P O P and Q DUH WKH ZDWHU VWDWLRQV SURYLGHG LQ WKH PLGGOH RI HDFKVWUDLJKWOLQH)LQG D WKHFRRUGLQDWHVRIP E WKHFRRUGLQDWHVRIC Solution: (a) P LVWKHPLGSRLQWRIAB P ² E QLVWKHPLGSRLQWRIBC /HWWKHFRRUGLQDWHVRICEH xy 7KXV x DQG y + 5 fl x fl y x ² y 7KHFRRUGLQDWHVRICLV ² EXAMPLE 17 y x O (18, 8) P Q S R(12, 6) 7KH GLDJUDP DERYH VKRZV D PDS RI WKH WUDQVLW UDLOZD\ URXWHSDVVLQJWKURXJKIRXUVWDWLRQVPQR and SRLVWKH PLGSRLQWRIP and SQLVWKHPLGSRLQWRIP and R)LQG D WKHFRRUGLQDWHVRIVWDWLRQQ E WKHGLVWDQFHRIPS ©Praxis Publishing_Focus On Maths
Cartesian Coordinate System CHAPTER 5 135 Solution: D /HWWKHFRRUGLQDWHVP EH xy 7KXV xfl xfl x and yfl yfl y 7KHFRRUGLQDWHVRIPLV QLVWKHPLGSRLQWRIPR Q fl ffi E PS fl² fl² XQLWV EXAMPLE 18 y x O K(4, 6) L(4, q) J(–2, 2) 7KH GLDJUDP DERYH VKRZV D WULDQJOH JKL GUDZQ RQ D &DUWHVLDQSODQH)LQG D WKHYDOXHRIq E WKHGLVWDQFHRIKL F WKHDUHDRIWULDQJOHJKL Solution: D /HWMEHDSRLQWRQWKHOLQHKLZKHUH JMLVSHUSHQGLFXODUWRKL7KXVWKH FRRUGLQDWHVRIMLV q q q ² y x O K(4, 6) L(4, q) J(–2, 2) M(4, 2) ©Praxis Publishing_Focus On Maths