SOLUTIONS TO TOPIC 6: MENSURATION

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Oliver Anderson
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1 ! i PR =! P +! +! R = 1 a + 1 a = ) j! PRj = jj = jaj fd rhomus ) jaj = jjg! ii SQ =! SD +! D +! Q = 1 + a 1 = a ) j! SQj = jaj From a we have! PS =! QR an! PQ =! SR ) we eue PQRS is a arallelogram. lso, from, the iagonals are equal in length. ) PQRS is a retangle. a T 5 mas triangle 0 to triangle 5 T 7 mas triangle 5 to triangle ) T 5 then T 7 is equivalent to T. T 1 mas triangle 0 to triangle 1 T mas triangle 1 to triangle 5 ) T 1 then T is equivalent to T 5. T 6 mas triangle 0 to triangle 6 T 6 mas triangle 6 to triangle ) T 6 then T 6 is equivalent to T. T mas triangle 0 to triangle T mas triangle to triangle 5 ) T then T is equivalent to T 5. SOLUTIONS TO TOPI 6: MENSURTION 1 a 7 mm = (7 10) m =7:m 9:75 km =(9: ) m = 9750 m m = ( ) m = ( ) km = km Distane etween light oles = a Perimeter = =m Perimeter = 1:5+:5++ =1m a mm = ( 100) m =0: m 0:059 ha =(0: ) m : km =(0: ) m = m 5:8 m =(5:8 100) m =(5: ) mm = 50 mm = 00 m =0m Perimeter = :5+ =11m m 15. m 5. m m 1:85 ha = (1:85 100) km =0:185 km 0: km =(0: ) m =(0: ) mm = mm e 60 m = ( ) m = m f m = ( ) ha =:95 ha 5 The retangle has erimeter = :+ : =11:m ) the erimeter of the square is also 11: m, an hene the length of its sies is 6 1:6 m = m 11: m =:8 m. ) the numer of oes on the allet = a P =z P = a + P = +5 + q =8 +q 8 a ¼r =6: ) r = 6: ¼ r ) r = 6: ¼ fr >0g ) r ¼ :09 The raius of the irle is :0 m. =¼r ¼ ¼ :09 ¼ 1: The irumferene of the irle is 1: m. 9 a rea =9 8 5 =57m rea = ase height =9 6 =5m 9 m = a The ue has 6 iential faes, eah with area =16 16 = 56 m : ) the surfae area of the ue =6 56 = 156 m : q 8 m 5 m m amrige IGSE International Mathematis (0607) Etene 108 Eam Prearation & Pratie Guie

2 The net of the rism is: The figure has: 1 mm ² 1 square ase mm 6 mm 1 ² triangular faes 1 1 mm 1 =6 8 = 178 mm =6 1 = 756 mm =8 1 = 1008 mm fottom an tog fsiesg ffront an akg ) total surfae area = = = 698 mm 11 a :71 litres =(:71 100) l = 71 l 1 m y m m ml = ( ) litres = 58:15 litres 1 m 09. m h +1 =5 fpythag.g ) h + 1 = 15 h mm ) h = 78 ) h =8 fh >0g ) total surfae area = = 116 mm 1 mm 5 mm 15 a Distane aroun semi-irle = 1 ¼ 10 ¼ 15:7 mm ) total erimeter ¼ 15:7+ 10 ¼ 5:7 mm rea = ¼ 5 ¼ 19 mm Perimeter =:+:+length of ar =:+ 60 ¼ : ¼ 7:7 m rea = 60 ¼ : ¼ :8 m 1 m 16. m 6. m = :6 =1: ) y =1: +0:9 ) y =:5 ) y =1:5 fy >0g ) guar rail length =1:+ 1:5+ 0:9+ 1+ =1m 0 m 70 m area of kite = (area of ) = = 100 m 1 a The figure has: ² retangular faes ² triangular faes h +55 = m 0 m fpythag.g ) h + 05 = h m ) h = 9075 ) h ¼ 95: fh >0g 110 m 55 m ) total surfae area ¼ (0 110) : ¼ m 110 m 16 a 7:5 m =(7: ) m = m mm = ( ) m = ( ) m =:9 m 500 m = ( ) mm = mm 17 of milk use eah week = 5 75 ml = ml =17:65 l 18 m h m.5 m 5. m 175. m h +1:75 =:5 ) h +:065 = 6:5 ) h =:1875 ) h ¼ 1:785 fh >0g ) area of en ¼ :5+ 1 :5 1:785 ¼ 10:1 m ) total surfae area ¼ 10:1 + (6 ) + (6 :5) ¼ 7: m So, 7: m of sheet metal is require. Eam Prearation & Pratie Guie 109 amrige IGSE International Mathematis (0607) Etene

³ µ 19 rea = ¼r 60 0 a = 50 60 ¼ ¼ :9 m D is a arallelogram ) D =60mm, D =8mm.

3 ³ µ 19 rea = ¼r 60 0 a = ¼ ¼ :9 m D is a arallelogram ) D =60mm, D =8mm. ) in D, D +8 =60 ) D = 60 8 =6mm fd > 0g Now D = D =90 ± falternate anglesg ) area of arallelogram = area D + area D = = 178 mm area of arallelogram = 178 mm ) 60 h = 178 h =8:8 1 a The shere has raius 0 m. ) surfae area = ¼ 0 a ¼ m surfae area =¼r +¼rh = ¼ 0 + ¼ 0 f:8 m = mg ¼ m 50 m 60 mm D h mm 5 Perimeter of figure = (erimeter of large semi-irle) + (erimeter of small semi-irle) = 1 ¼ ¼ 1 ¼ 11 m rea of figure =(area of large semi-irle) (area of 1 small semi-irle) = 1 ¼ 18 1 ¼ 6 ¼ 5 m 6 The eah alls have raius 18 m. ) surfae area of all = ¼ 18 = 196 ¼ m ) surfae area of 00 alls = ¼ ¼ m So, 81: m of ruer is neee. 7 a = length with eth =:5 : :5 ¼ 6:8 m 1 m = ¼r h = ¼ 86 ¼ mm = ¼r = ¼ :5 ¼ 65: m 6 m ¼ 81: m e = area of en height =:6 5 =90m = 1 ¼r h = 1 ¼ :5 8 ¼ 5: m 110 m h m 0 m m h +0 = ) h = 0 fh>0g ) h ¼ 9:9 ³ ) area of to ¼ 9:9 ¼ 50 m area of four sies = (110 55) + ( 55) + (50 55) = 1 0 m 8 a Surfae area =¼rh + ¼r 9 = ¼ 6 8+¼ 6 ¼ 15 m l =1 +8 ) l = 1 +8 ) l ¼ 1: 85. m fl >0g ) surfae area = ¼rl ¼ ¼ 1 1: ¼ 1 mm 1 m 8 mm 1 mm l mm 1000 m = mm ) numer of resistors = 0 = m 500 ml 0:5 l area of ie = ¼ 8:5 =7:5 ¼ m area of late =1 1 = 1 m Now 7:5¼ ¼ 0:515 ¼ 51:5% 1 So, the ie overs 51:5% of the late. amrige IGSE International Mathematis (0607) Etene 110 Eam Prearation & Pratie Guie

4 0 of flask = ¼r h = ¼ : 16: ¼ 600 m ) aaity of flask ¼ 600 ml = setor area + area of = ¼q + 1 q q = q ¼ q 1 of ookies = ¼r h = ¼ :5 1 =6:5 ¼ m of ough = = 00 m 00 ¼ 1, so 1 ookies an e mae from the lok. 6:5¼ = area of traezium area of semi-irle ³ a + = 1 ¼a (a + ) ¼a = 60 The attern an e ivie into 5 m y 5 m squares as shown: 10 m 10 g The square has area 5 5 = 65 m. The small tiles are 5 m y 5 m, an there are 5 of them in the square. So, the small tiles ouy a total of 5 (5 5) = 15 m. ) the roortion of the area overe y the smallest tile is 15 = 0%. 65 a Total surfae area = (10 10) + (10 0) 10 m " front an ak = 1000 m For the front an ak: " to, ottom an sies ainte area = (10 1) + (8 1) = 6 m For the to, ottom an sies: 1 m 5 m 5 m 1 m ainte area = (0 1) + (8 1) = 56 m ) total ainte area = = 96 m Unainte area = = 70 m a = area of 5 semi-irles =5 1 ¼r = 5 ¼r 0 m 5 m 8 m 10 m 8 m 1 m 1 m 10 m rea of eah square = g rea of eah irle setor = ¼g The inner heagon is mae u of 6 equilateral triangles of length g. ) area of heagon =6 1 g g sin 60± =g = g ) =6 g ³ 60 = 6+¼ + g 5 Surfae area = ¼rl + ¼r ¼g + g = ¼ 7:5 + ¼ 7:5 ¼ 978 mm 6 a Total mass =1:08 tonnes = 10 kg ) mass of eah ost = 10 kg =18kg 60 = ¼r h = ¼ 0:08 1:8 f8 m =0:08 mg ¼ 0:06 m 7 a = area of hollow yliner + area of shere ³ ³ =¼ l +¼ = ¼l + ¼ = ¼(l + ) In the triangular faes, h + = y ) h = y fh>0g ) area of eah triangular fae = 1 y = y 60 ) = y + ( )+( ) triangular faes sies ase = y +8 + ³ = y + h y Eam Prearation & Pratie Guie 111 amrige IGSE International Mathematis (0607) Etene

5 8 90 m 18 m E D 5 m 55 m of eah hanle = ¼ =6¼ m of shaft = ¼ 1:5 1 =7¼ m ) total volume of oor hanle = 6¼ 7¼ =99¼ ¼ 11 m E = D forresoning anglesg E = D =90 ± ) s E an D are similar, an E D = fsame ratiog ) E 18 = ) E =11 ) volume of uket = volume of large one volume of small one = 1 ¼(18) 90 1 ¼(11) 55 ¼ 567 m ) the uket has aaity 567 ml. In hours or 1 minutes, the uket loses 1 1: = 16 ml of water. ) the amount of water remaining ¼ ( ) ml ¼ 51 ml ¼ : litres a 55 litres = ml ) the water has volume m. Suose the water rises to a height of h m. ) ¼ 0 h = ) h = 00¼ ) h ¼ :77 ) the water is 50 :77 = 6: m from the to. of sae remaining in aquarium ¼ ¼ 0 6: ¼ 781:85 m ¼ mm of eah marle = ¼ 6 ¼ 90:8 mm So, ¼ 8660 marles an e ae efore the 90:8 aquarium overflows. SOLUTIONS TO TOPI 7: OORDINTE GEOMETRY 9 a V = 1 area of ase height = 1 ah V = volume of hemishere = 1 ¼ = ¼ V = area of traezium length ³ a +5a = h =ah 1 a (-, -) y (1, 1) O D(, 0) (5, ) 0 Let the raius of the wege e r m. Now volume = 60 m ) 1 ¼r 6:1 = 60 ) r = 180 6:1¼ r 180 ) r = fr >0g 6:1¼ ) r ¼ 9: ) the raius of the wege is 9: m.! =, an = + =5units! =, an = + =5units! D =, an D = ( ) +( ) =5units! D =, an D = ( ) +( ) =5units ³ a 1 rea of en = 1 ¼ +a a +a a = ¼a 8 +a +6a ³ +8 ¼ = a 8 ³ +8 ) V = a ¼ l 8 ³ +8 = a ¼ l 8 a a a a a a ll four sies are equal in length, so the oints form a rhomus. 5 graient = 5 = 5 - graient = = graient = 0 whih is unefine amrige IGSE International Mathematis (0607) Etene 11 Eam Prearation & Pratie Guie

Chapter 15 VECTORS EXERCISE 15A. Scale: 1 cm 10 km h 1 1 cm 10 ms 1 N. 35 ms km h-1 W S. Scale: N cm 10 m. 60 ms -1.

Chapter 15 VECTORS EXERCISE 15A. Scale: 1 cm 10 km h 1 1 cm 10 ms 1 N. 35 ms km h-1 W S. Scale: N cm 10 m. 60 ms -1. Chate VECTORS EXERCISE a Sale: Sale: m 0 km h m 0 ms W E ms - 0 km h- S W S E W 0 0 E Sale: m 0 m 60 ms - m unway S Sale: m 0 ms a Sale: m newtons Sale: m newtons 7 newtons W S E 60 newtons W S E a Sale: