Class VIII Maths Mensuration Worksheet - CBSE

Class VIII Maths Chapter 11 Mensuration CBSE Worksheets Maths Class 8 Chapter 11 Mensuration 1. Find the area of the square with the side: a. 8 cm b. 5 m c. 7 km 2. Find...

Class VIII Maths Chapter 11 Mensuration CBSE Worksheets Maths Class 8 Chapter 11 Mensuration 1. Find the area of the square with the side: a. 8 cm b. 5 m c. 7 km 2. Find the area of the rectangle with a. length= 8 cm and breadth= 5 cm b. length= 5 cm and breadth= 2 cm c. length= 9 km and breadth= 1 cm 3. Find the area of the given trapeziums: a. Length of parallel sides= 4 cm, 3 cm and height= 7 cm b. Length of parallel sides= 3 cm, 7 cm and height= 9 cm c. Length of parallel sides= 8 cm, 6 cm and height= 10 cm 4. Find the area of the semicircle with a radius of 5 cm. Frank Solutions www.vedantu.com 1 Class VIII Maths Chapter 11 Mensuration 5. Find the area of the given figure: 6. What is the cost of painting a wall with a length of 15 m and a breadth of 20 m if the cost of painting is Rs. 10 per m²? 7. A farmer wants to plough her land. The radius of the land is 50 metres. Find the cost of labour if ploughing per metre costs Rs. 10. 8. Find the area of the below trapezium: Frank Solutions www.vedantu.com 2 Class VIII Maths Chapter 11 Mensuration 9. Find the area of a rhombus whose diagonals are 15 cm and 12 cm long. 10. Find the area of the given circle: 11. Find the area of the cube whose side measures 6 cm. 12. Find the area of the shaded region where g is 8 cm and f is 10 cm. Frank Solutions www.vedantu.com 3 Class VIII Maths Chapter 11 Mensuration 13. A tank is in the shape of a cuboid whose length is 7 m, breadth is 5 m and height is 20 m. Find the quantity of water that can be stored in the tank. 14. Find the area of the given sphere: 15. A tile is in the shape of a rectangle whose base is 20 cm and the corresponding length is 10 cm. How many such tiles are required to cover a rectangular floor of an area of 1000 square metres? 16. Find the area of the given cuboid: Frank Solutions www.vedantu.com 4 Class VIII Maths Chapter 11 Mensuration 17. The area of a rhombus is 120 square centimetres and one of the diagonals is 12 cm. Find the other diagonal. 18. Find the area of the given square: 19. Find the area of the triangle whose base is 5 cm and height is 12 cm. 20. Find the area of the given figure: Frank Solutions www.vedantu.com 5 Class VIII Maths Chapter 11 Mensuration 21. Find the volume of the cuboid with dimensions 8 cm, 5cm, and 7 cm. 22. Find the area of the given figure: 23. The diagonals of a rhombus are 7 cm and 10 cm. Find its area. 24. Find the area of the given figure: Frank Solutions www.vedantu.com 6 Class VIII Maths Chapter 11 Mensuration 25. In a monument, there are 12 cylindrical pillars. The radius of each post is 2 m and the height is 20 m. Find the total cost of painting the curved surface area of all pillars at the rate of painting per metre square is Rs. 50. Answers: 1. a. Area of square= side × side= 8×8= 64 cm² b. Area of square= side × side= 5×5=25 cm² c. Area of square= side × side= 7×7=49 cm² 2. a. Area of rectangle= l × b= 8×5= 40 cm² b. Area of rectangle= l × b= 5×2= 10 cm² c. Area of rectangle= l × b= 9×1= 9 cm² 3. a. Area of trapezium= h(sum of parallel sides) / 2= 7(4+3)/2= 49/2= 24.5 cm² b. Area of trapezium= h(sum of parallel sides) / 2= 9(7+3)/2= 90/2= 45 cm² c. Area of trapezium= h(sum of parallel sides) / 2= 10(8+6)/2= 140/2= 70 cm² 4. Area of semicircle= πr²/2 Frank Solutions www.vedantu.com 7 Class VIII Maths Chapter 11 Mensuration = 3.14×5×5/2 = 3.14×25/2 = 1.57×25 = 39.25 cm² The area of the semicircle is 39.25 cm² 5. Area of the given figure= Area of semicircle + Area of triangle = πr²/2 + ½ bh = 3.14×2 + 7×2 = 6.28+14 = 20. 28 cm² The area of the given figure is 20. 28 cm² 6. Total cost of painting= Area of square × unit cost = 15×20×10 = 3000 The cost of painting the wall is Rs. 3000. 7. Total cost of ploughing= Area of circle × Unit cost = πr² × 10 = 3.14 × 50 × 50 × 10 = 31.4 × 2500 = 78500 The cost of ploughing the land is Rs. 78500. 8. Area of trapezium= h × (sum of length of parallel sides)/2 =5(8+11)/2 =5(19)/2 =95/2 = 47.5 The area of the trapezium is 47.5 cm² Frank Solutions www.vedantu.com 8 Class VIII Maths Chapter 11 Mensuration 9. Area of rhombus = Product of two diagonals/2 = 15×12/2 =15×6 =90 The area of the rhombus is 90 cm² 10. Area of circle= πr² = 3.14×10×10 =314 The area of the given circle is 314 cm² 11. Total surface area of cube= 6×side×side =6×6×6 =216 The area of the cube is 216 cm² 12. Area of shaded region= Area of larger circle- Area of smaller circle = πR²- πr² =3.14 × 102 - 3.14 × 82 =3.14(100-64) =3.14(36) =113.04 The area of the shaded region is 113.04 cm² 13. Quantity of water that can be stored = Volume of cuboid =l×b×h =7×5×20 =35×20 =700 The quantity of water that can be stored in the tank is 700 m² 14. Area of sphere= 4 πr² Frank Solutions www.vedantu.com 9 Class VIII Maths Chapter 11 Mensuration =4×3.14×8×8 =4×3.14×64 =12.56×64 =803.84 The area of the given sphere is 803.84 cm² 15. No. of tiles required = Area of the floor/ Area of a tile = 1000/ (0.2 × 0.1) = 50000 A total of 50000 tiles are required to cover the rectangular floor. 16. Total surface area of cuboid= 2(lb+bh+hl) =2(10×4 + 4×5 + 5×10) =2(40 + 20 + 50) =2(110) =220 The area of the given cuboid is 220 cm². 17. Length of the other diagonal = Area of rhombus × 2 / Length of the known diagonal = 120 × 2 / 12 =240/12 =20 The length of the other diagonal is 20 cm. 18. Area of square= side×side =7×7 =49 The area of the given square is 49 cm² 19. Area of triangle= bh/2 =5×12/2 Frank Solutions www.vedantu.com 10 Class VIII Maths Chapter 11 Mensuration =30 The area of the triangle is 30 cm² 20. Area of given figure = bh/2 =10×5/2 =50/2 =25 The area of the figure is 25 cm² 21. Volume of cuboid = lbh =8×5×7 =40×7 =280 The volume of the cuboid is 280 cm³. 22. Area of given figure= Area of semicircle + Area of rectangle = πr²/2 + lb = (3.14×5×5)/2 + 10×3.5 =1.57×25 + 35 =39.25+35 =74.25 The area of the given figure is 74.25 cm² 23. Area of rhombus = Product of diagonals/ 2 =7×10/2 =70/2 =35 The area of the rhombus is 35 cm² 24. Area of the given figure= Area of square + 4 × Area of semicircle = side × side + 4 × πr²/2 = 5×5 + (4×3.14×2.5×2.5)/2 Frank Solutions www.vedantu.com 11 Class VIII Maths Chapter 11 Mensuration =25 +78.5/2 =25 + 39.25 =64.25 The area of the given figure is 64.25 cm² 25. Total cost of painting = Number of pillars × Area of a pillar × Unit cost = 12 × 2πrh × 50 = 12×2×3.14×2×20×50 =24×314×20 =314×480 =150720 The cost of painting is Rs.150720 Frank Solutions www.vedantu.com 12  CLASS-VIII SUB-MATHEMATICS MENSURATION(SURFACE AREA AND VOLUME) Worksheet-1(Basic) I) Multiple choice questions: (1 mark) 1. The volume of a cube of side 0.03m is a) 0.27 m3 b) 0.027 m3 c) 27 m3 d) 2.7 m3 2. If the heights of a cylinder is halved, its volume will be of its original volume 1 1 a) times b) times c) 2 times d) 3 times 2 3 3. The lateral surface area of a cuboid whose length, breadth and height are 2a, 2b and 2c respectively is. a) 2(ab+bc+ca) b) 4(ab+bc+ca) c) 8(a+b)c d) none of these 4. The sum of the areas of all faces (excluding top and bottom) of a cuboid is the _ of the cuboid. a) volume b)lateral surface area c) total surface area d)none of these 5. The volume of a cuboid whose length, breadth and height are 2a, 3a and 4a is. a)24a2 b)24a3 c)12a3 d)none of these 6. The curved surface area of a right circular cylinder of radius 2r and height 2h is a)2Πrh b)4Πrh c)8Πrh d)none of these 7. The lateral surface area of a right circular cylinder of radius 3cm is 94.2cm2.The height of the cylinder is a)5cm b)4.5cm c)5.5cm d)none of these 8. If each edge of a cube is doubled, how many times will its volume increase? a)2 times b)4 times c)8 times d)none of these 9. If the volume of a cube is v, then the side(in terms of v) 3 a)v3 b)√ c) √ d)v2 10. The volume of a cylinder 3 whose diameter is equal to its height is 3 a)Πr3h b) c) d)none of these 8 8 II) Short answer questions type I (2 mark) 11. Find the height of a cuboid whose volume is 756cm3 and base area is 63cm2? 12. The dimensions of a cuboid are in the ratio of 2:3:4 and its total surface area is 280m2. Find the dimensions. 13. If the volume of a cube is 512cm3.Find the total surface area of the cube 14. The circumference of the base of a right circular is 220cm. If the height of the cylinder is 2m, find the curved surface area of the cylinder. 15. A roller of diameter 70cm and the length 2m is rolling on the ground.What is the area covered by the roller in 50 revolutions? III) Short answer questions type II (3 mark) 16. A match box is of dimension 4cm by 2.5cm by 1.5cm.What will be the volume of a packet containing 12 such match boxes? How many such packets can be placed in card board box whose size is 60cm x 20cm x 24cm. 17. Two cubes, each of side 6cm, are joined end to end. Find the volume of the resulting cuboid. 18. How many 8cm cubes can be cut out from the cube whose edge is 32cm? 19. The radius and height of a cylinder are in the ratio 5:7 and the volume is 550 cm3.Find its total surface area. 20. A rectangular sheet of paper 88cm x 10cm is rolled along its length and a cylinder is formed. Find the volume of the cylinder. IV) Long answer questions (4 mark) 21. The students of DAV were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with the base, using cardboard. Each penholder was to be of radius 3cm and height 10.5cm.The school was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard w was required to be bought for the competition? 22. A cuboid is of dimensions 60cm x 54cm x 30cm.How many small cubes of side 6cm can be placed in the given cuboid. 23. The area of the base of a right circular cylinder is 15400 cm2 and its volume is 92400 cm3. Find the area of the curved surface. 24. A cube of side 5cm is cut into 1 cm cubes. Find the percentage increase in volume after such cutting. 25. The bottom of a tank measures 50m x 40m.Find its depth if it contains 4000m3 of water. 26. A closed cylindrical tank of radius 7m and height 3m is made from a sheet of metal. How much sheet of metal is required? 27. A suitcase with measures 80cm x 48cm x 24cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96cm is required to cover 100 such suitcase. 28. The lateral surface area of a hollow cylinder is 6600m2.It is cut along its height and formed a rectangular sheet of width 30cm.Find the perimeter of rectangular sheet. 29. A swimming pool is 40m in length, 20m in breadth and 5m in depth. Find enting its four walls and floor at the rate of 10 per m2. the cost of cementing 30. Two cylinders of same volume have their radii in the ratio 1:6.Find the ratio of their heights.  CLASS-VIII SUB-MATHEMATICS MENSURATION(SURFACE AREA AND VOLUME) Worksheet-2(Standard) I) Multiple choice questions (1 mark) 1. A cuboid is 40cm x 20cm x 10cm.What would be the side of a cube having the same volume? a) 20cm b)40cm c)10cm d)30cm 2. If each side of a cube is doubled then its volume a) is doubled b)becomes 4 times c) becomes 6 times d)becomes 8 times 3. The curved surface area of a cylindrical pillar is 264m2 and its volume is 924m3.The ratio of its diameter to its height is. a) 3:7 b)7:3 c)6:7 d)7:6 4. A square sheet of paper is converted into cylinder by rolling it along its length. What will be the ratio of the base radius to the side of a square? a)1:2Π b) 2Π:1 c)1:1 d)1: Π 5. If the radius of a right circular cylinder is decreased by 50% and its height is increased by 60%, its volume will be decreased by a)10% b)60% c)40% d)20% II) Short answer questions type I (2 mark) 6. A birthday cake has two tiers as shown in the figure below. Find the volume of the cake. 7. A solid cube with an edge 10 cm is melted to form two equal cubes. Find the ratio of the edge of the smaller cube to the edge of the bigger cube. 8. Two cubes have their volumes in the ratio 8:27. Find the ratio of their surface area. 9. If the capacity of a cylindrical tank is 1848m3 and the diameter of its base is 14m.Find the depth of the tank.  10. A school provides milk to the students daily in a cylindrical glasses of diameter 7cm.If the glass is filled with milk up to height of 12cm,find how many litres of milk is needed to serve 1600 students. III) Short answer questions type II (3 mark) 11. A cuboid has a volume of 3000cm3, with height 15cm and length 20cm. Find the area of the upper surface 12. An iron cube has each side equal to 5cm.Find its volume and also its weight in Kg if 1 cu.cm of iron weighs 50 grams. 13. The sum of length, breadth and depth of a cuboid is 19cm and the diagonal is 5√5.Find its surface area. 14. A lateral surface area of a hollow cylinder is 4224 cm2 it is cut along its height and formed a rectangular sheet of width 32cm.Find the perimeter of the rectangular sheet. 15. A boy is cycling such that the wheels of the cycle are making 140 revolutions per hour.If the diameter of the wheel is 60cm,then calculate the speed (in km/hr) with which the boy is cycling. IV) Long answer type questions ( 4 mark) 16. Eight identical cuboidal wooden blocks are stacked one on top of the other. The total volume of the solid formed is 128cm3 if the height of each block is 1cm and the base is a square.Find the dimension of each block. 17. A metallic sheet is of rectangular shape with dimensions 48m x 36m. From each of its corner, a square is cut off to make an open box. The length of the square is 8m.Find the volume of the box. 18. A hollow iron pipe is 21cm long and its external diameter is 8cm.If the thickness of the pipe is 1cm and iron weighs 8g/cm3, find the weight of the pipe. 19. The ratio of radii of two cylinders is 1:2 and the heights are in the ratio 2:3. Find the ratio of their volumes. 20. Four times the area of a curved surface area of a cylinder is equal to the six times the sum of the area of its base. If the height is 12cm , find its curved surface area.  CLASS-VIII SUB-MATHEMATICS MENSURATION(SURFACE AREA AND VOLUME) Worksheet-3(Advance) 1. How many bricks of size 22cm x 10 cm x 7 cm are required to construct a wall 11m long , 3.5 m high and 40 cm thick, if the cement and sand used in the construction occupy 1 th part of the wall. 10 2. The area of a side of a box is 120sq cm. The area of the other side of the box is 72 sq cm. If the area of the upper surface of the box is 60sq cm then find the volume of the box. 3. A cylindrical tank has a radius of 154cm.It is filled with water to a height of 3m.If water to a height of 4.5m is poured into it, what will be the increase in the volume of water in kilo litres. 4. A rectangular examination hall having seats for 500 candidates has to be built so, as to allow 4 cubic meters of air and 0.5 square meters of floor area per candidate. If the length of the hall is 25m.Find the height and breadth of the hall. 5. In the given figure R is the radius of the base of the hat. Find the total outer surface area of the hat. 6. A hollow garden roller of 42cm diameter and length 152cm is made of cast iron 2cm thick. Find the volume of iron used in the roller. 7. A rectangular tank is 225m by 162m at the base. With what speed must the water flow into it through an aperture 60cm by 45cm that the level may be raised 20cm in 5 hours. 8. A well with 20m diameter is dug 28m deep. Earth taken out of it spread all around to a width of 10m to form an embankment. Find the height of embankment. 9. A wooden box including the lid has external dimensions 40cm x 34cm x 30cm.If the wood is 1cm thick, how many cm3 of wood is used in it. 10. Three cubes with sides in the ratio 3:4:5 are melted to form a single cube whose diagonal is 12√3 cm. Find the sides of the cube.  St.Andr ewsScot sSr.Sec.School 9t hAv enue, I.P.Ext ensi on, Pat par ganj ,Del hi–110092 Sessi on:2022- 2023-Wor ksheet Cl ass:VI II Subj ect :Mat hemat ics Topi c:Mensur ati on Wor ksheet -29 DOASDI RECTED Q1:I fthelengthsoft hedi agonal sofar hombusar e16cm and12cm, findi tsar ea. (ans96cm. 2 ) Q2:Howmanybri ckseach25cm by15cm by8cm,ar erequi redf orawal l32m long, 3m hi ghand40cm t hick?( ans12800) Q3:Thear eaofat r apezi s400cm2, um i t hedi stancebetweent heparal lel sidesi s16 cm.Ifoneofthepar all elsi desi s20cm, findthelengt hoftheot herside?( ans 30cm) Q4:Thr eemetalcubesofsi des6cm, 8cm and10cm aremel tedandr ecasti ntoa bi gcube.Findit stot alsur facear ea. (ans864cm2) Q5:Thediamet erofar olleri s84cm andi t slengt his120cm.Ittakes500compl ete revoluti onstomov eonceov ert ol evelapl aygr ound.Fi ndtheareaofthe playgroundinm2.(ans1584m 2) Q6:Fi ndt heareaofar hombuswhoseonesi demeasur es5cm andonedi agonal is 8cm.( ans24sqcm) Q7:Theparal lel sidesofatr apezium ar e40cm and20cm.I fit snon-paral lelsi des arebothequal, eachbeing26cm, fi ndt heareaoft het rapezi um.(ans720cm2 ) Q8:Thelater alsur faceareaofahollowcyl i s4224cm2.I nderi tiscutalongi tshei ght andf ormedarect angularsheetofwi dth33cm.Fi ndt heperimeterof rectangul arsheet ? Q9:Acuboi disofdi mensi ons(60* 50*30) cm. Howmanysmal lcubeswi thsi de6cm canbepl acedi nthegi vencuboi d? Q10:I sitpossi bleforasol i dcuboi dofv olume500cm3i smel tedandf orm an10 cubesofside5cm  MENSURATION 169 CHAPTER Mensuration 11 11.1 Introduction We have learnt that for a closed plane figure, the perimeter is the distance around its boundary and its area is the region covered by it. We found the area and perimeter of various plane figures such as triangles, rectangles, circles etc. We have also learnt to find the area of pathways or borders in rectangular shapes. In this chapter, we will try to solve problems related to perimeter and area of other plane closed figures like quadrilaterals. We will also learn about surface area and volume of solids such as cube, cuboid and cylinder. 11.2 Let us Recall Let us take an example to review our previous knowledge. This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m. (i) What is the total length of the fence surrounding it? To find the length of the fence we need to find the perimeter of this park, which is 100 m. (Check it) (ii) How much land is occupied by the park? To find the land occupied by this park we need to find the area of this park which is 600 square meters (m2) (How?). (iii) There is a path of one metre width running inside along the perimeter of the park that has to be cemented. If 1 bag of cement is required to cement 4 m2 area, how many bags of cement would be required to construct the Fig 11.1 cemented path? area of the path We can say that the number of cement bags used =. area cemented by 1 bag Area of cemented path = Area of park – Area of park not cemented. Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m2. That is 28 × 18 m2. Hence number of cement bags used = ------------------ (iv) There are two rectangular flower beds of size 1.5 m × 2 m each in the park as shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered by grass. 170 MATHEMATICS Area of rectangular beds = ------------------ Area of park left after cementing the path = ------------------ Area covered by the grass = ------------------ We can find areas of geometrical shapes other than rectangles also if certain measurements are given to us. Try to recall and match the following: Diagram Shape Area rectangle a×a square b×h triangle πb 2 1 parallelogram b×h 2 circle a×b Can you write an expression for the perimeter of each of the above shapes? TRY THESE (a) Match the following figures with their respective areas in the box. 49 cm2 77 cm2 98 cm2 (b) Write the perimeter of each shape.  MENSURATION 171 EXERCISE 11.1 1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area? 2. Mrs. Kaushik has a square plot with the (a) (b) measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ` 55 per m2. 3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres]. 4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners). 5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle. (a) (b) (c) 11.3 Area of Trapezium Nazma owns a plot near a main road (Fig 11.2). Unlike some other rectangular plots in her neighbourhood, the plot has only one pair of parallel opposite sides. So, it is nearly a trapezium in shape. Can you find out its area? Let us name the vertices of this plot as shown in Fig 11.3. By drawing EC || AB, we can divide it into two parts, one of rectangular shape and the other of triangular shape, (which Fig 11.3 is right angled at C), as shown in Fig 11.3. Fig 11.2 (b = c + a = 30 m) 172 MATHEMATICS 1 1 Area of Δ ECD = h × c = × 12 × 10 = 60 m2. 2 2 Area of rectangle ABCE = h × a = 12 × 20 = 240 m2. Area of trapezium ABDE = area of Δ ECD + Area of rectangle ABCE = 60 + 240 = 300 m2. We can write the area by combining the two areas and write the area of trapezium as 1 ⎛c ⎞ area of ABDE = h ×c+h×a=h ⎜⎝ + a⎟⎠ 2 2 ⎛ c + 2a ⎞ ⎛ c + a + a⎞ = h ⎜⎝ ⎟⎠ = h ⎜⎝ ⎟⎠ 2 2 (b + a ) (sum of parallelsides) = h = height 2 2 (b + a ) By substituting the values of h, b and a in this expression, we find h = 300 m2. 2 TRY THESE 1. Nazma’s sister also has a trapezium shaped plot. Divide it into three parts as shown ( a + b) (Fig 11.4). Show that the area of trapezium WXYZ = h. 2 2. If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the h ( a + b) expression. 2 Fig 11.4 DO THIS 1. Draw any trapezium WXYZ on a piece of graph paper as shown in the figure and cut it out (Fig 11.5). Fig 11.5 2. Find the mid point of XY by folding the side and name it A (Fig 11.6). Fig 11.6  MENSURATION 173 3. Cut trapezium WXYZ into two pieces by cutting along ZA. Place Δ ZYA as shown in Fig 11.7, where AY is placed on AX. What is the length of the base of the larger triangle? Write an expression for the area of this triangle (Fig 11.7). Fig 11.7 4. The area of this triangle and the area of the trapezium WXYZ are same (How?). Get the expression for the area of trapezium by using the expression for the area of triangle. So to find the area of a trapezium we need to know the length of the parallel sides and the perpendicular distance between these two parallel sides. Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the area of trapezium. TRY THESE Find the area of the following trapeziums (Fig 11.8). (i) (ii) Fig 11.8 DO THIS In Class VII we learnt to draw parallelograms of equal areas with different perimeters. Can it be done for trapezium? Check if the following trapeziums are of equal areas but have different perimeters (Fig 11.9). Fig 11.9 174 MATHEMATICS We know that all congruent figures are equal in area. Can we say figures equal in area need to be congruent too? Are these figures congruent? Draw at least three trapeziums which have different areas but equal perimeters on a squared sheet. 11.4 Area of a General Quadrilateral A general quadrilateral can be split into two triangles by drawing one of its diagonals. This “triangulation” helps us to find a formula for any general quadrilateral. Study the Fig 11.10. Area of quadrilateral ABCD = (area of Δ ABC) + (area of Δ ADC) 1 1 = ( AC × h1) + ( AC × h2) 2 2 1 Fig 11.10 = AC × ( h1 + h2) 2 1 = d ( h1 + h2) where d denotes the length of diagonal AC. 2 Example 1: Find the area of quadrilateral PQRS shown in Fig 11.11. Solution: In this case, d = 5.5 cm, h1 = 2.5cm, h2 = 1.5 cm, 1 Area = d ( h1 + h2) 2 1 = × 5.5 × (2.5 + 1.5) cm2 2 1 Fig 11.11 = × 5.5 × 4 cm2 = 11 cm2 2 TRY THESE We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already? (Fig 11.12) Fig 11.12 11.4.1 Area of special quadrilaterals We can use the same method of splitting into triangles (which we called “triangulation”) to find a formula for the area of a rhombus. In Fig 11.13 ABCD is a rhombus. Therefore, its diagonals are perpendicular bisectors of each other. Area of rhombus ABCD = (area of Δ ACD) + (area of Δ ABC)  MENSURATION 175 1 1 1 = ( × AC × OD) + ( × AC × OB) = AC × (OD + OB) 2 2 2 1 1 = AC × BD = d1 × d2 where AC = d1 and BD = d2 2 2 Fig 11.13 In other words, area of a rhombus is half the product of its diagonals. Example 2: Find the area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm. 1 Solution: Area of the rhombus = d1 d2 where d1, d2 are lengths of diagonals. 2 1 = × 10 × 8.2 cm2 = 41 cm2. 2 THINK, DISCUSS AND WRITE A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles? TRY THESE Find the area of these quadrilaterals (Fig 11.14). (i) (iii) (ii) Fig 11.14 11.5 Area of a Polygon We split a quadrilateral into triangles and find its area. Similar methods can be used to find the area of a polygon. Observe the following for a pentagon: (Fig 11.15, 11.16) Fig 11.16 Fig 11.15 By constructing one diagonal AD and two perpendiculars BF By constructing two diagonals AC and AD the and CG on it, pentagon ABCDE is divided into four parts. So, pentagon ABCDE is divided into three parts. area of ABCDE = area of right angled Δ AFB + area of So, area ABCDE = area of Δ ABC + area of trapezium BFGC + area of right angled Δ CGD + area of Δ ACD + area of Δ AED. Δ AED. (Identify the parallel sides of trapezium BFGC.) 176 MATHEMATICS TRY THESE (i) Divide the following polygons (Fig 11.17) into parts (triangles and trapezium) to find out its area. Fig 11.17 FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR (ii) Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm. Area of Polygon ABCDE = area of Δ AFB +.... 1 1 Area of Δ AFB = × AF × BF = × 3 × 2 =.... 2 2 (BF + CH) Area of trapezium FBCH = FH × 2 (2 + 3) Fig 11.18 =3× [FH = AH – AF] 2 1 1 Area of ΔCHD = × HD× CH =....; Area of ΔADE = × AD × GE =.... 2 2 So, the area of polygon ABCDE =.... (iii) Find the area of polygon MNOPQR (Fig 11.19) if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm NA, OC, QD and RB are perpendiculars to diagonal MP. Fig 11.19 Example 1: The area of a trapezium shaped field is 480 m2, the distance between two parallel sides is 15 m and one of the parallel side is 20 m. Find the other parallel side. Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m. The given area of trapezium = 480 m2. 1 Area of a trapezium = h (a + b) 2 1 480 × 2 So 480 = × 15 × (20 + b) or = 20 + b 2 15 or 64 = 20 + b or b = 44 m Hence the other parallel side of the trapezium is 44 m.  MENSURATION 177 Example 2: The area of a rhombus is 240 cm2 and one of the diagonals is 16 cm. Find the other diagonal. Solution: Let length of one diagonal d1 = 16 cm and length of the other diagonal = d2 1 Area of the rhombus = d. d = 240 2 1 2 1 So, 16 ⋅ d 2 = 240 2 Therefore, d2 = 30 cm Hence the length of the second diagonal is 30 cm. Example 3: There is a hexagon MNOPQR of side 5 cm (Fig 11.20). Aman and Ridhima divided it in two different ways (Fig 11.21). Find the area of this hexagon using both ways. Ridhima’s method Aman’s method Fig 11.20 Fig 11.21 Solution: Aman’s method: Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. You can verify it by paper folding (Fig 11.22). (11 + 5) Now area of trapezium MNQR = 4 × = 2 × 16 = 32 cm2. 2 So the area of hexagon MNOPQR = 2 × 32 = 64 cm2. Ridhima’s method: Fig 11.22 Δ MNO and Δ RPQ are congruent triangles with altitude 3 cm (Fig 11.23). You can verify this by cutting off these two triangles and placing them on one another. Fig 11.23 1 Area of Δ MNO = × 8 × 3 = 12 cm2 = Area of Δ RPQ 2 Area of rectangle MOPR = 8 × 5 = 40 cm2. Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm2. EXERCISE 11.2 1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m. 178 MATHEMATICS 2. The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side. 3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC. 4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field. 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. 6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal. 7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ` 4. 8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. 9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface. 10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area? 11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same. 11.6 Solid Shapes In your earlier classes you have studied that two dimensional figures can be identified as the faces of three dimensional shapes. Observe the solids which we have discussed so far (Fig 11.24).  MENSURATION 179 Fig 11.24 Observe that some shapes have two or more than two identical (congruent) faces. Name them. Which solid has all congruent faces? DO THIS Soaps, toys, pastes, snacks etc. often come in the packing of cuboidal, cubical or cylindrical boxes. Collect, such boxes (Fig 11.25). Fig 11.25 Cuboidal Box Cubical Box All six faces are rectangular, and opposites faces are identical. So there are three pairs of identical faces. All six faces are squares Cylindrical Box and identical. One curved surface and two circular faces which are identical. Now take one type of box at a time. Cut out all the faces it has. Observe the shape of each face and find the number of faces of the box that are identical by placing them on each other. Write down your observations. 180 MATHEMATICS Did you notice the following: The cylinder has congruent circular faces that are parallel to each other (Fig 11.26). Observe that the line segment joining the center of circular faces is perpendicular to the base. Such cylinders are known as right circular cylinders. We are only going to study this type of cylinders, though there are other Fig 11.26 Fig 11.27 types of cylinders as well (Fig 11.27). (This is a right (This is not a right circular cylinder) circular cylinder) THINK, DISCUSS AND WRITE Why is it incorrect to call the solid shown here a cylinder? 11.7 Surface Area of Cube, Cuboid and Cylinder Imran, Monica and Jaspal are painting a cuboidal, cubical and a cylindrical box respectively of same height (Fig 11.28). Fig 11.28 They try to determine who has painted more area. Hari suggested that finding the surface area of each box would help them find it out. To find the total surface area, find the area of each face and then add. The surface area of a solid is the sum of the areas of its faces. To clarify further, we take each shape one by one. 11.7.1 Cuboid Suppose you cut open a cuboidal box and lay it flat (Fig 11.29). We can see a net as shown below (Fig 11.30). Write the dimension of each side. You know that a cuboid has three pairs of identical faces. What expression can you use to find the area of each face? Fig 11.29 Fig 11.30 Find the total area of all the faces of the box. We see that the total surface area of a cuboid is area I + area II + area III + area IV +area V + area VI =h×l+b×l+b×h+l×h+b×h+l×b  MENSURATION 181 So total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl ) where h, l and b are the height, length and width of the cuboid respectively. Suppose the height, length and width of the box shown above are 20 cm, 15 cm and 10 cm respectively. Then the total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15) = 2 ( 300 + 200 + 150) = 1300 m2. TRY THESE Find the total surface area of the following cuboids (Fig 11.31): Fig 11.31 The side walls (the faces excluding the top and bottom) make the lateral surface area of the cuboid. For example, the total area of all the four walls of the cuboidal room in which you are sitting is the lateral surface area of this room (Fig 11.32). Hence, the lateral surface area of a cuboid is given by 2(h × l + b × h) or 2h (l + b). Fig 11.32 DO THIS (i) Cover the lateral surface of a cuboidal duster (which your teacher uses in the class room) using a strip of brown sheet of paper, such that it just fits around the surface. Remove the paper. Measure the area of the paper. Is it the lateral surface area of the duster? (ii) Measure length, width and height of your classroom and find (a) the total surface area of the room, ignoring the area of windows and doors. (b) the lateral surface area of this room. (c) the total area of the room which is to be white washed. THINK, DISCUSS AND WRITE 1. Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base? 2. If we interchange the lengths of the base and the height of a cuboid (Fig 11.33(i)) to get another cuboid (Fig 11.33(ii)), will its lateral surface area change? (i) Fig 11.33 (ii) 182 MATHEMATICS 11.7.2 Cube DO THIS Draw the pattern shown on a squared paper and cut it out [Fig 11.34(i)]. (You know that this pattern is a net of a cube. Fold it along the lines [Fig 11.34(ii)] and tape the edges to form a cube [Fig 11.34(iii)]. (i) (ii) (iii) Fig 11.34 (i) (ii) Fig 11.35 (a) What is the length, width and height of the cube? Observe that all the faces of a cube are square in shape. This makes length, height and width of a cube equal (Fig 11.35(i)). (b) Write the area of each of the faces. Are they equal? (c) Write the total surface area of this cube. (d) If each side of the cube is l, what will be the area of each face? (Fig 11.35(ii)). Can we say that the total surface area of a cube of side l is 6l2 ? TRY THESE Find the surface area of cube A and lateral surface area of cube B (Fig 11.36). Fig 11.36  MENSURATION 183 THINK, DISCUSS AND WRITE (i) Two cubes each with side b are joined to form a cuboid (Fig 11.37). What is the surface area of this cuboid? Is it 12b2? Is the surface area of cuboid formed by joining three such cubes, 18b2? Why? Fig 11.37 (ii) How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area? (iii) After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions (Fig 11.38). How many have no face painted? 1 face painted? 2 faces painted? 3 faces painted? Fig 11.38 11.7.3 Cylinders Most of the cylinders we observe are right circular cylinders. For example, a tin, round pillars, tube lights, water pipes etc. DO THIS (i) Take a cylindrical can or box and trace the base of the can on graph paper and cut it [Fig 11.39(i)]. Take another graph paper in such a way that its width is equal to the height of the can. Wrap the strip around the can such that it just fits around the can (remove the excess paper) [Fig 11.39(ii)]. Tape the pieces [Fig 11.39(iii)] together to form a cylinder [Fig 11.39(iv)]. What is the shape of the paper that goes around the can? (i) (ii) (iii) (iv) Fig 11.39 184 MATHEMATICS Of course it is rectangular in shape. When you tape the parts of this cylinder together, the length of the rectangular strip is equal to the circumference of the circle. Record the radius (r) of the circular base, length (l ) and width (h) of the rectangular strip. Is 2πr = length of the strip. Check if the area of rectangular strip is 2πrh. Count how many square units of the squared paper are used to form the cylinder. Check if this count is approximately equal to 2πr (r + h). (ii) We can deduce the relation 2πr (r + h) as the surface area of a cylinder in another way. Imagine cutting up a cylinder as shown below (Fig 11.40). Fig 11.40 22 The lateral (or curved) surface area of a cylinder is 2πrh. Note: We take π to be 7 unless otherwise stated. The total surface area of a cylinder = πr2 + 2πrh + πr2 = 2πr2 + 2πrh or 2πr (r + h) TRY THESE Find total surface area of the following cylinders (Fig 11.41) Fig 11.41 THINK, DISCUSS AND WRITE Note that lateral surface area of a cylinder is the circumference of base × height of cylinder. Can we write lateral surface area of a cuboid as perimeter of base × height of cuboid? Example 4: An aquarium is in the form of a cuboid whose external measures are 80 cm × 30 cm × 40 cm. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed? Solution: The length of the aquarium = l = 80 cm Width of the aquarium = b = 30 cm  MENSURATION 185 Height of the aquarium = h = 40 cm Area of the base = l × b = 80 × 30 = 2400 cm2 Area of the side face = b × h = 30 × 40 = 1200 cm2 Area of the back face = l × h = 80 × 40 = 3200 cm2 Required area = Area of the base + area of the back face + (2 × area of a side face) = 2400 + 3200 + (2 × 1200) = 8000 cm2 Hence the area of the coloured paper required is 8000 cm2. Example 5: The internal measures of a cuboidal room are 12 m × 8 m × 4 m. Find the total cost of whitewashing all four walls of a room, if the cost of white washing is ` 5 per m2. What will be the cost of white washing if the ceiling of the room is also whitewashed. Solution: Let the length of the room = l = 12 m Width of the room = b = 8 m Height of the room = h = 4 m Area of the four walls of the room = Perimeter of the base × Height of the room = 2 (l + b) × h = 2 (12 + 8) × 4 = 2 × 20 × 4 = 160 m2. Cost of white washing per m2 = ` 5 Hence the total cost of white washing four walls of the room = ` (160 × 5) = ` 800 Area of ceiling is 12 × 8 = 96 m2 Cost of white washing the ceiling = ` (96 × 5) = ` 480 So the total cost of white washing = ` (800 + 480) = ` 1280 Example 6: In a building there are 24 cylindrical pillars. The radius of each pillar is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of all pillars at the rate of ` 8 per m2. Solution: Radius of cylindrical pillar, r = 28 cm = 0.28 m height, h = 4 m curved surface area of a cylinder = 2πrh 22 curved surface area of a pillar = 2 × × 0.28 × 4 = 7.04 m2 7 curved surface area of 24 such pillar = 7.04 × 24 = 168.96 m2 cost of painting an area of 1 m2 = ` 8 Therefore, cost of painting 1689.6 m2 = 168.96 × 8 = ` 1351.68 Example 7: Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm2. Solution: Let height of the cylinder = h, radius = r = 7cm Total surface area = 2πr (h + r) 186 MATHEMATICS 22 i.e., 2× × 7 × (7 + h) = 968 7 h = 15 cm Hence, the height of the cylinder is 15 cm. EXERCISE 11.3 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make? 2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases? 3. Find the side of a cube whose surface area is 600 cm2. 4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet. 5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room? 6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area? 7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required? 8. The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet? 9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m. 10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.  MENSURATION 187 11.8 Volume of Cube, Cuboid and Cylinder Amount of space occupied by a three dimensional object is called its volume. Try to compare the volume of objects surrounding you. For example, volume of a room is greater than the volume of an almirah kept inside it. Similarly, volume of your pencil box is greater than the volume of the pen and the eraser kept inside it. Can you measure volume of either of these objects? Remember, we use square units to find the area of a region. Here we will use cubic units to find the volume of a solid, as cube is the most convenient solid shape (just as square is the most convenient shape to measure area of a region). For finding the area we divide the region into square units, similarly, to find the volume of a solid we need to divide it into cubical units. Observe that the volume of each of the adjoining solids is 8 cubic units (Fig 11.42 ). Fig 11.42 We can say that the volume of a solid is measured by counting the number of unit cubes it contains. Cubic units which we generally use to measure volume are 1 cubic cm = 1 cm × 1 cm × 1 cm = 1 cm3 = 10 mm × 10 mm × 10 mm =............... mm3 1 cubic m = 1 m × 1 m × 1 m = 1 m3 =............................... cm3 1 cubic mm = 1 mm × 1 mm × 1 mm = 1 mm3 = 0.1 cm × 0.1 cm × 0.1 cm =...................... cm3 We now find some expressions to find volume of a cuboid, cube and cylinder. Let us take each solid one by one. 11.8.1 Cuboid Take 36 cubes of equal size (i.e., length of each cube is same). Arrange them to form a cuboid. You can arrange them in many ways. Observe the following table and fill in the blanks. cuboid length breadth height l×b×h=V (i) 12 3 1 12 × 3 × 1 = 36 (ii)............ 188 MATHEMATICS (iii)............ (iv)............ What do you observe? Since we have used 36 cubes to form these cuboids, volume of each cuboid is 36 cubic units. Also volume of each cuboid is equal to the product of length, breadth and height of the cuboid. From the above example we can say volume of cuboid = l × b × h. Since l × b is the area of its base we can also say that, Volume of cuboid = area of the base × height DO THIS Take a sheet of paper. Measure its area. Pile up such sheets of paper of same size to make a cuboid (Fig 11.43). Measure the height of this pile. Find the volume of the cuboid by finding the product of the area of the sheet and the height of this pile of sheets. Fig 11.43 This activity illustrates the idea that volume of a solid can be deduced by this method also (if the base and top of the solid are congruent and parallel to each other and its edges are perpendicular to the base). Can you think of such objects whose volume can be found by using this method? TRY THESE Find the volume of the following cuboids (Fig 11.44). (i) Fig 11.44  MENSURATION 189 11.8.2 Cube The cube is a special case of a cuboid, where l = b = h. Hence, volume of cube = l × l × l = l 3 TRY THESE Find the volume of the following cubes (a) with a side 4 cm (b) with a side 1.5 m DO THIS Arrange 64 cubes of equal size in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid shapes of same volume have same surface area? THINK, DISCUSS AND WRITE A company sells biscuits. For packing purpose they are using cuboidal boxes: box A →3 cm × 8 cm × 20 cm, box B → 4 cm × 12 cm × 10 cm. What size of the box will be economical for the company? Why? Can you suggest any other size (dimensions) which has the same volume but is more economical than these? 11.8.3 Cylinder We know that volume of a cuboid can be found by finding the product of area of base and its height. Can we find the volume of a cylinder in the same way? Just like cuboid, cylinder has got a top and a base which are congruent and parallel to each other. Its lateral surface is also perpendicular to the base, just like cuboid. So the Volume of a cuboid = area of base × height = l × b × h = lbh Volume of cylinder = area of base × height = πr2 × h = πr2h TRY THESE Find the volume of the following cylinders. (i) (ii) 190 MATHEMATICS 11.9 Volume and Capacity There is not much difference between these two words. (a) Volume refers to the amount of space occupied by an object. (b) Capacity refers to the quantity that a container holds. Note: If a water tin holds 100 cm3 of water then the capacity of the water tin is 100 cm3. Capacity is also measured in terms of litres. The relation between litre and cm3 is, 1 mL = 1 cm3,1 L = 1000 cm3. Thus, 1 m3 = 1000000 cm3 = 1000 L. Example 8: Find the height of a cuboid whose volume is 275 cm3 and base area is 25 cm2. Solution: Volume of a cuboid = Base area × Height Volume of cuboid Hence height of the cuboid = Base area 275 = = 11 cm 25 Height of the cuboid is 11 cm. Example 9: A godown is in the form of a cuboid of measures 60 m × 40 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m3 ? Solution: Volume of one box = 0.8 m3 Volume of godown = 60 × 40 × 30 = 72000 m3 Volume of the godown Number of boxes that can be stored in the godown = Volume of one box 60 × 40 × 30 = = 90,000 0.8 Hence the number of cuboidal boxes that can be stored in the godown is 90,000. Example 10: A rectangular paper of width 14 cm is rolled along its width and a cylinder 22 of radius 20 cm is formed. Find the volume of the cylinder (Fig 11.45). (Take for π) 7 Solution: A cylinder is formed by rolling a rectangle about its width. Hence the width of the paper becomes height and radius of the cylinder is 20 cm. Fig 11.45 Height of the cylinder = h = 14 cm Radius = r = 20 cm  MENSURATION 191 Volume of the cylinder = V = π r2 h 22 = × 20 × 20 × 14 = 17600 cm3 7 Hence, the volume of the cylinder is 17600 cm3. Example 11: A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder. Solution: Length of the paper becomes the perimeter of the base of the cylinder and width becomes height. Let radius of the cylinder = r and height = h Perimeter of the base of the cylinder = 2πr = 11 22 or 2× × r = 11 7 7 Therefore, r = cm 4 Volume of the cylinder = V = πr2h 22 7 7 = × × × 4 cm3 = 38.5 cm3. 7 4 4 Hence the volume of the cylinder is 38.5 cm3. EXERCISE 11.4 1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume. (a) To find how much it can hold. (b) Number of cement bags required to plaster it. (c) To find the number of smaller tanks that can be filled with water from it. 2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area? 3. Find the height of a cuboid whose base area is 180 cm2 and volume B A is 900 cm3? 4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid? 5. Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm ? 6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank? 7. If each edge of a cube is doubled, (i) how many times will its surface area increase? (ii) how many times will its volume increase? 192 MATHEMATICS 8. Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir. WHAT HAVE WE DISCUSSED? 1. Area of (i) a trapezium = half of the sum of the lengths of parallel sides × perpendicular distance between them. (ii) a rhombus = half the product of its diagonals. 2. Surface area of a solid is the sum of the areas of its faces. 3. Surface area of a cuboid = 2(lb + bh + hl) a cube = 6l 2 a cylinder = 2πr(r + h) 4. Amount of region occupied by a solid is called its volume. 5. Volume of a cuboid = l × b × h a cube = l 3 a cylinder = πr2h 6. (i) 1 cm3 = 1 mL (ii) 1L = 1000 cm3 (iii) 1 m3 = 1000000 cm3 = 1000L      Length of boundary of a simple closed figure is known as perimeter. Area is the measure of region enclosed in a simple closed curve. Perimeter of a rectangle = 2 (length + breadth). Area of a rectangle = length × breadth. Perimeter of a square = 4 × side. Area of a square = side × side. 1 Area of a triangle = × Base × Corresponding Height. 2 Area of a parallelogram = Base × Corresponding Height. Area of a circle = r 2 , where r is the radius. 1 Area of a trapezium = × (Sum of parallel sides) × Height. 2 1 Area of a rhombus = × Product of diagonals. 2 Lateral surface area of a cube = 4 (side)2. Total surface area of a cube = 6 (side)2. Lateral surface area of a cuboid = 2 × height × (length + breadth). Total surface area of a cuboid = 2(lb + bh + hl). Lateral (curved) surface area of a cylinder = 2πrh. Total surface area of a cylinder = 2πr (r + h), where r is the radius and h is the height. 12/04/18  Amount of space occupied by a solid is called its volume. Volume of a cube = (side)3. Volume of a cuboid = length × breadth × height. Volume of a cylinder = πr 2h. 1cm3 = 1ml 1L = 1000 cm3. 1m3 = 10,00,000 cm3 = 1,000 L.    In examples 1 and 2, there are four options out of which one is correct. Write the correct answer. Example 1 : What is the area of the triangle ADE in the following figure? A B E 8 cm D C 10 cm (a)45 cm2 (b) 50 cm2 (c) 55 cm2 (d) 40 cm2 Solution : The correct answer is (d). Example 2 : What will be the change in the volume of a cube when its side becomes 10 times the original side? (a) Volume becomes 1000 times. (b) Volume becomes 10 times. (c) Volume becomes 100 times. 1 (d) Volume becomes times. 1000 Solution : The correct answer is (a). In examples 3 and 4, fill in the blanks to make the statements true. Example 3 : Area of a rhombus is equal to __________ of its diagonals. Solution : Half the product.   12/04/18   Example 4 : If the area of a face of a cube is 10 cm2, then the total surface area of the cube is __________. Solution : 60 cm2. In examples 5 and 6, state whether the statements are true (T) or false (F). Example 5 : 1L = 1000 cm3 Solution : True. Example 6 : Amount of region occupied by a solid is called its surface area. Solution : False. Example 7 : 160 m3 of water is to be used to irrigate a rectangular field whose area is 800 m2. What will be the height of the water level in the field? Solution : Volume of water = 160 m3 Area of rectangular field = 800 m2 Let h be the height of water level in the field. Now, volume of water = volume of cuboid formed on the field by water. 160 = Area of base × height = 800 × h 160 h = = 0.2 800 So, required height = 0.2 m Example 8 : Find the area of a rhombus whose one side measures 5 cm and one diagonal as 8 cm. Solution : Let ABCD be the rhombus as shown below.   12/04/18  DO = OB = 4cm, since diagonals of a rhombus are perpendicular bisectors of each other. Therefore, using Pythagoras theorem in AOB, AO2 + OB2 = AB2 AO = AB2  OB2 = 52  42 = 3 cm So, AC = 2 × 3 = 6 cm 1 1 Thus, the area of the rhombus = × d1 × d2 = × 8 × 6 2 2 = 24 cm2. Example 9 : The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium. Solution : Let ABCD be the trapezium such that AB = 40 cm and CD = 20 cm and AD = BC = 26 cm.   To become familiar with some of the vocabulary terms in the chapter, consider the following: 1. The square root of a number is one of the two equal factors of the number. For example, 3 is a square root because 3  3 = 9. How might picturing plant roots help you remember the meaning of square root? 2. The word ‘perimeter’ comes from the Greek roots peri, meaning ‘all around,’ and metron, meaning ‘measure.’ What do the Greek roots tell you about the perimeter of a geometric figure? 3. To square a number means ‘to multiply the number by itself,’ as in 2  2. Keeping this idea of square in mind, what do you think a perfect square might be? 4. The word ‘circumference’ comes from the Latin word circumferre, meaning to “carry around”. How does the Latin meaning help you define the circumference of a circle?   12/04/18   Now, draw CL || AD Then ALCD is a parallelogram So AL = CD = 20 cm and CL = AD = 26 cm. In ∆CLB, we have CL = CB = 26 cm Therefore, ∆CLB is an isosceles triangle. Draw altitude CM of ∆CLB. Since ∆ CLB is an isosceles triangle. So, CM is also the median. 1 1 Then LM = MB = BL = × 20 cm = 10 cm 2 2 [as BL = AB – AL = (40 – 20) cm = 20 cm]. Applying Pythagoras theorem in ∆CLM, we have CL2 = CM2 + LM2 262 = CM2 + 102 CM2 = 262 – 102 = (26 – 10) (26 + 10) = 16 × 36 = 576 CM = 576 = 24 cm 1 Hence, the area of the trapezium = (sum of parallel 2 1 sides) × Height = (20 + 40) × 24 = 30 × 24 = 720 cm2. 2 Example 10 : Find the area of polygon ABCDEF, if AD = 18cm, AQ = 14 cm, AP = 12 cm, AN = 8 cm, AM = 4 cm, and FM, EP, QC and BN are perpendiculars to diagonal AD. Solution :   12/04/18  In the figure MP = AP – AM = (12 – 4) cm = 8 cm PD = AD – AP = (18 – 12) cm = 6 cm NQ = AQ – AN = (14 – 8) cm = 6 cm QD = AD – AQ = (18 – 14) cm = 4 cm Area of the polygon ABCDEF = area of ∆AFM + area of trapezium FMPE + area of ∆EPD + area of ∆ANB + area of trapezium NBCQ + area of ∆QCD. 1 1 1 1 = × AM × FM + (FM + EP) × MP + PD × EP + × 2 2 2 2 1 1 AN × NB + (NB + CQ) × NQ + QD × CQ 2 2 1 1 1 1 1 = ×4×5+ (5 + 6) × 8 + ×6×6+ ×8×5+ 2 2 2 2 2 1 (5 + 4) × 6 + × 4 × 4. 2 = 10 + 44 + 18 + 20 + 27 + 8 = 127 cm2      Example 11 : Horse stable is in the form of a cuboid, whose external dimensions are 70 m × 35 m × 40 m, surrounded by a cylinder halved vertically through diameter 35 m and it is open from one rectangular face 70 m × 40 m. Find the cost of painting the exterior of the stable at the rate of Rs 2/m2. Understand and Explore the problem What do you know? Here you know dimensions of cuboid, L = 70 m, B = 35 m, H = 40 m, diameter of cylinder 35 m and cost of painting Rs. 2 per m2. What fact do you need to solve the question and is not given? Height of cylinder.   12/04/18   Plan a Strategy Begin by visualising the shape of the stable and draw it (open from shaded part). Think of area to be painted in cuboidal part as well as in cylindrical part. Add the two areas calculated in step 2. Find cost. Solve Area of cylindrical top be painted 1 = [T.S.A] 2 1 = [2πR (R + H)] 2 1  22 35  35  = 2× 7 × 2  2 +70   2    = 4812.5 m2 Area of cuboid to be painted = area of three walls = lh + 2bh = 70 × 40 +2 × 40 × 35 = 2800 + 2800 = 5600 m2 Total area to be painted = 4812.5 + 5600 = 10412.5 m2 Cost of painting per m 2 = Rs 2 Cost of painting 10412.5 m2 = Rs 10412.5 × 2 = Rs 20825 Revise Verify your answer by adopting some other plan, i.e. here in this problem instead of taking area in two steps, let’s find in one step. Area to be painted = Area of three walls + Area of cylindrical part.   12/04/18  1 = 2bh + lh + [2πRH + 2πR2] 2 = h [2b + l ] + [πR (R + H)] 22 35  35  = 40 [2 × 35 + 70] + ×  + 70  7 2  2  = 40 + 55 × 87.5 = 5600 + 4812.5 Final cost (same as in previous method) Hence verified.    (a) What would be the cost of painting if cylindrical root is not to be painted? (b) What would be the cost if one face is not included. Is there any difference in the cost? A cube is a three-dimensional solid with six square faces. Its surface area is the total area of all Six of its faces. As each face is a square, the formula for surface area of a cube is A = 6s s2   In questions 1 to 28, there are four options out of which one is correct. Write the correct answer. 1. A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cubic centimetre cubes, how many 1 cubic centimetre cubes will have exactly one of their faces painted? (a) 27 (b) 42 (c) 54 (d) 142   12/04/18   2. A cube of side 4 cm is cut into 1 cm cubes. What is the ratio of the surface areas of the original cubes and cut-out cubes? (a) 1 : 2 (b) 1:3 (c) 1 : 4 (d) 1 : 6 3. A circle of maximum possible size is cut from a square sheet of board. Subsequently, a square of maximum possible size is cut from the resultant circle. What will be the area of the final square? 3 1 (a) of original square. (b) of original square. 4 2 1 2 (c) of original square. (d) of original square. 4 3 4. What is the area of the largest triangle that can be fitted into a rectangle of length l units and width w units? (a) lw/2 (b) lw/3 (c) lw/6 (d) lw/4 1 5. If the height of a cylinder becomes of the original height and the 4 radius is doubled, then which of the following will be true? (a) Volume of the cylinder will be doubled. (b) Volume of the cylinder will remain unchanged. (c) Volume of the cylinder will be halved. 1 (d) Volume of the cylinder will be of the original volume. 4 Volume is a measurement of the amount of space inside a three-dimensional object. It’s measured in cubic units and equals the number of unit cubes (cubes whose edges have length 1) that fit inside the object. In the diagram on the right, each side has a length of 2 units, so two unit cubes fit along each side. (One unit cube is shaded blue.) You can calculate the volume of cube using the formula. 3 V=S×S×S Or V = S   12/04/18  1 6. If the height of a cylinder becomes of the original height and the 4 radius is doubled, then which of the following will be true? (a) Curved surface area of the cylinder will be doubled. (b) Curved surface area of the cylinder will remain unchanged. (c) Curved surface area of the cylinder will be halved. 1 (d) Curved surface area will be of the original curved surface. 4 1 7. If the height of a cylinder becomes of the original height and the 4 radius is doubled, then which of the following will be true? (a) Total surface area of the cylinder will be doubled. (b) Total surface area of the cylinder will remain unchanged. (c) Total surface of the cylinder will be halved. (d) None of the above. 8. The surface area of the three coterminus faces of a cuboid are 6, 15 and 10 cm2 respectively. The volume of the cuboid is (a) 30 cm3 (b) 40 cm3 (c) 20 cm3 (d) 35 cm3 9. A regular hexagon is inscribed in a circle of radius r. The perimeter of the regular hexagon is (a) 3r (b) 6r (c) 9r (d) 12r 10. The dimensions of a godown are 40 m, 25 m and 10 m. If it is filled with cuboidal boxes each of dimensions 2 m × 1.25 m × 1 m, then the number of boxes will be (a) 1800 (b) 2000 (c) 4000 (d) 8000 Think about your answers to these questions. Discuss your ideas with other students and your teacher. Then write a summary of your findings in your notebook. 1. Explain how to find the total area of all the faces of a rectangular box. 2. Explain how to find the number of identical cubes it will take to fill a rectangular box. 3. Suppose several different nets are made for a given box. What do all of the nets have in common? What might be different?   12/04/18   11. The volume of a cube is 64 cm3. Its surface area is (a) 16 cm2 (b) 64 cm2 (c) 96 cm2 (d) 128 cm2 12. If the radius of a cylinder is tripled but its curved surface area is unchanged, then its height will be (a) tripled (b) constant (c) one sixth (d) one third 13. How many small cubes with edge of 20 cm each can be just accommodated in a cubical box of 2 m edge? (a) 10 (b) 100 (c) 1000 (d) 10000 14. The volume of a cylinder whose radius r is equal to its height is 1 3 πr 3 3 r3 (a) πr (b) (c) πr (d) 4 32 8 15. The volume of a cube whose edge is 3x is (a) 27x 3 (b) 9x3 (c) 6x3 (d) 3x3 16. The figure ABCD is a quadrilateral in which AB = CD and BC = AD. Its area is (a) 72 cm2 (b) 36 cm2 (c) 24 cm2 (d) 18 cm2 17. What is the area of the rhombus ABCD below if AC = 6 cm, and BE = 4cm? (a) 36 cm2 (b) 16 cm2 (c) 24 cm2 (d) 13 cm2   12/04/18  18. The area of a parallelogram is 60 cm2 and one of its altitude is 5 cm. The length of its corresponding side is (a) 12 cm (b) 6 cm (c) 4 cm (d) 2 cm 19. The perimeter of a trapezium is 52 cm and its each non-parallel side is equal to 10 cm with its height 8 cm. Its area is (a) 124 cm2 (b) 118 cm2 (c) 128 cm2 (d) 112 cm2 20. Area of a quadrilateral ABCD is 20 cm2 and perpendiculars on BD from opposite vertices are 1 cm and 1.5 cm. The length of BD is (a) 4 cm (b) 15 cm (c) 16 cm (d) 18 cm 21. A metal sheet 27 cm long, 8 cm broad and 1 cm thick is melted into a cube. The side of the cube is (a) 6 cm (b) 8 cm (c) 12 cm (d) 24 cm 22. Three cubes of metal whose edges are 6 cm, 8 cm and 1