Exercise Q.1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Transcription

1 11 MENSURATION Exercise 11.1 Q.1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area? (a) Side = 60 m (Given) Perimeter of square = 4 side = 4 60 = 240 m Area of square = (Side) 2 = (60) 2 = 3600 m (i) (b) Here, length (l) = 80 m (Given) breadth (b) =? Perimeter of rectangle = 2(l + b) 240 m = 2(80 + b) or, = 80 + b 120 = 80 + b = b 40 = b b = 40 m Area of rectangle = l b = 80 m 40 m = 3200 m (ii) 1

2 From eqn. (i) and (ii), we have Area of square > Area of rectangle Hence, field (a) has larger area. Q.2. Mrs. Kaushik has a square plot with the 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 Rs 55 per m 2. We know that area of a square = (Side) 2 Area of rectangular plot = l b Subtracting eqn. (ii) from (i), we get = (25) 2 = 625 m (i) = = 300 m (ii) Area of garden = Area of square plot Area of middle plot (Rectangular plot) = ( ) m 2 = 325 m 2 Hence, the total cost of developing the garden = 325 m 2 Rs 55 = Rs 17,875 Q.3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 ( ) metres]. 2

3 Length of rectangle = 20 ( ) m = (20 7) m = 13 m Breadth of rectangle = 7 m Also, breadth of rectangle = Diameter of semi-circle = 7 m As radius Diameter Radius of semicircular ends = 3.5 m Hence, perimeter of semicircular ends = 2 πr = = 22 m (i) Perimeter of the rectangular portion = 2(l + b) Adding eqn. (i) and (ii), we get = 2 (13 + 7) = 2 20 = 40 m (ii) Perimeter of the garden = (22 m + 40 m) = 62 m Area of the rectangular portion = l b = 13 7 = 91 m 2 Area of semicircular ends = πr2 = = 38.5 m 2 Area of the garden = 38.5 cm cm 2 = m 2 Hence, perimeter of the garden = 62 m Area of the garden = m 2 3

4 Q.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 m 2? (If required you can split the tiles in whatever way you want to fill up the corners). Area of a tile = base height = b h We know, 1 m = 100 cm = = 240 cm 2 or, 1 m 2 = cm 2 Area of the floor = 1080 m 2 Number of tiles required = = cm 2 = cm 2 Area of floor Area of flooring tiles = Hence, tiles are required. = Q.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) (a) Given diameter of semi-circle = 2.8 cm 4

5 So, radius of semi-circle = 1.4 cm Circumference of semi-circle = 4.4 cm (b) Perimeter of the given shape = Circumference of semicircle + Perimeter of square = 4.4 cm cm = 10.4 cm (c) Perimeter of the given shape = Circumference of semicircle + 2 cm + 2 cm. = 4.4 cm + 2 cm + 2 cm = 8.4 cm Hence, the ant have to take a longer round for the (b) food-piece. Exercise 11.2 Q.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. Area of the trapezium h(a + b) [Where h is height and = 1 a and b are parallel sides] ( ) = 0.88 m2 5

6 Q.2. The area of a trapezium is 34 cm 2 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. Area of trapezium h(a + b) 34 4 (10 + b) 34 = 2 (10 + b) 34 = b 2b = b = 14 b = 14 2 or, b = 7 cm Hence, another parallel side is 7 cm Q.3. Length of the fence of a trapezium shaped filed 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. Perimeter of the trapezium ABCD = AB + BC + CD + AD 120 = AB = AB AB = AB = 15 m Area of trapezium ABCD (a + b) h 6

7 ( ) = 660 m 2 Q.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. Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC AC MB AC ND AC (MB + ND) 24(13 + 8) = 12(21) = = 252 m 2 Q.5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Area of rhombus d 1 d 2 Length of one diagonal (d 1 ) = 7.5 cm Length of other diagonal (d 2 ) = 12 cm 7

8 Area of rhombus = = = 45.0 = 45 cm 2 Q.6. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal. Area of rhombus d 1 d 2 Also, area of the given rhombus = 6 4 = 24 cm 2 So, 1 2 d 1 d 2 = d 2 = d 2 = 6 4 d 2 = 6 cm Q.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 m 2 is Rs 4. Area of each tiles = Area of rhombus 8

9 d 1 d = = 675 cm 2 Area of 1 tile = 675 cm 2 Area of 3000 tiles = cm cm = = m 2 Total cost of polishing the floor = = Rs = Rs 810 Hence, the cost of polishing the floor = Rs 810 Q.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 m 2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. Let the length of one of the parallel sides (a) = xm Length of the other parallel side (b) = 2xm 2 Area of trapezium (a + b) h (x + 2x) 100 9

10 3x m 2 = 150x Hence, = x x = 70 m a = 70 m b = 2 70 = 140 m Q.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. Area of trapeziumabch = Area of trapezium GDEF (a + b) h (11 + 5) = 32 m2 Area of rectangle HCDG = length breadth = 11 5 = 55 m 2 Hence, area of the octagonal surface = Area of trapezium ABCH + Area of rectangle HCDG + Area of trapezium GDEF. = 32 m m m 2 = 119 m 2 10

11 Q.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? Jyoti s way : We know area of trapezium (a + b) h ( ) = m2 Hence, area of figure using Jyoti s way = = m 2 Kavita s way : Area of the given figure = Area of triangle + Area of square We know area of triangle base height Area of square = (side) 2 11

12 Area of the given figure (15)2 = = = 675 = m2 2 Hence, area of figure using Kavita s way = m 2 Q.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. Given KE = FL = HI = GJ = 4 cm Area of section AEFB = 1 h (a + b) 2 4 ( ) 4 40 = 80 cm Similarly, Area of section DCGH 4( ) 12

13 4 40 = 80 cm 2 Area of section DHEA = Area of trapezium = 1 h(a + b) 2 4( ) = 96 cm2 Similarly, Area of section BFGC = 1 h(a + b) 2 4( ) 4 48 = 96 cm 2 Exercise 11.3 Q.1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make? (a) l = Length = 60 cm b = breadth = 40 cm h = height = 50 cm 13

14 Total surface area of cuboidal box = 2(lb + bh + hl) (b) = 2 ( ) = 2 ( ) = 2 (7400) = cm 2 Here, l = b = h = 50 cm Total surface area of cube = 6 (side 2 ) = 6 (50) 2 = = cm 2 Hence, box (a) requires the lesser amount of material to make. Q.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? Total surface area of suitcase = 2(lb + bh + hl) Length of required tarpaulin = = 2( ) = 2 ( ) = = cm 2 Surface area of suitcase Width of tarpaulin = = 144 cm Q.3. Find the side of a cube whose surface area is 600 cm 2. Surface area of cube = 6 side 2 (l = side) 600 cm 2 = 6l 2 14

15 600 6 = l = l = l (Taking square root on both side) 10 = l Hence, the required side of the cube = 10 cm. Q.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. Here, l = 2 m, b = 1 m and h = 1.5 m Lateral surface area of the cabinet = 2h(l + b) = (2 + 1) = = 9 m 2 Area of the top of the cabinet = 1 2 m 2 = 2 m 2 Hence, required covered surface area = 9 m m 2 = 11 m 2 Q.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 m 2 of area is painted. How many cans of paint will she need to paint the room? Here, Length = (l) = 15 m Breadth = (b) = 10 m Height = (h) = 7 m 15

16 Lateral surface area of the hall = 2h(l + b) = 2 7( ) = = 350 m 2 Area of ceiling = l b = = 150 m 2 Hence, the total area to be painted = 350 m m 2 = 500 m 2 Since, each can of paint covers 100 m 2 of area So, number of cans that will be required to paint = = 5 Hence, 5 cans will be needed to paint the room. Q.6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area? One of the given figure is cylinder while other is cube edge having an edge of 7 cm. The height of both figures are same. Lateral surface area of cylinder = 2πrh = = 154 cm 2 where d = 7 cm 7 r = cm 2 h = 7 cm Lateral surface area of cube = 4 (side 2 ) = 4 (7) 2 [ side = 7 cm] 16

17 = 4 49 = 196 cm 2 Hence, the cube has larger lateral surface area. Q.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? Total surface area of cylindrical tank = 2πr (r + h) = (7 + 3) 7 = = 440 m 2 Hence, the sheet of metal required = 440 m 2. Q.8. The lateral surface area of a hollow cylinder is 4224 cm 2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet? Surface area of cylinder = 4224 cm 2 2πrh = 4224 cm 2 33 h = 4224 cm 2 ( 2πr = 33 cm) h = 128 cm Perimeter of rectangle = 2(l + b) = 2( ) = = 322 cm Q.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. 17

18 Diameter of road roller = 84 cm Radius of road roller (r) = 84 2 = 42 cm Here, Surface area of road roller = 2πrh l = h = 1 m = 100 cm = So, area of the road = = cm 2 Hence, area of the road = 1980 cm = cm2 = 1980 m 2 ( 1 m 2 = cm 2 ) Q.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. Let r cm be the radius of the label. We know diameter (d) = 2 radius (r) r = 2 d = 14 2 = 7 cm The label is placed 2 cm from top and bottom, then required height = 20 4 = 16 cm Lateral surface area of the label = 2πrh = = = 704 cm 2 Hence, area of the label = 704 cm 2 18

19 Exercise 11.4 Q.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. (a) Volume (b) Surface area (c) Volume Q.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? (A)Diameter of cylinder A = 7 cm d 7 Radius = r = = cm 2 2 Also, height = h = 14 cm Volume of cylinder = πr 2 h =

20 (B) Diameter of cylinder B = 14 cm = = 539 cm (i) Radius = r = 14 2 = 7 cm Also, height = h = 7 cm Volume of cylinder = πr 2 h = 22 7 (7)2 7 = = 1078 cm (ii) Hence, the volume of cylinder (B) is greater than the volume of cylinder (A) Curved surface area of cylinder A = 2πrh = = 308 cm 2 Curved surface area of cylinder B = = 308 cm 2 Hence, surface area of cylinder A and cylinder B are equal. Q.3. Find the height of a cuboid whose base area is 180 cm 2 and volume is 900 cm 3? Let h be the height of cuboid. Volume of cuboid = l b h 20

21 Base area = l b = 180 cm 2 Volume of cuboid = 180 h 180 h = 900 cm 3 h = = 5 cm Hence, Height of the cuboid = 5 cm Q.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? Volume of cuboid = = cm 3 Volume of cube = (6 cm) 3 = 216 cm 3 Number of small cubes = = 450 cubes Hence, 450 cubes of side 6 cm can be placed in the given cuboid. Q.5. Find the height of the cylinder whose volume is 1.54 m 3 and diameter of the base is 140 cm? Let h be the height of the cylinder Volume of cylinder = 1.54 m 3 Diameter of cylinder = 140 cm Radius = r = Volume of cylinder = πr 2 h = 70 cm = m 1.54 m 3 = h 21

22 1.54 = h = h 1 = h Hence, height of cylinder = 1 m Q.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? Radius of milk tank = 1.5 m Height of milk tank = 7m Volume of milk tank = πr 2 h = Volume of cylinder Since, 1 m 3 = 22 7 = 49.5 cm 3 = 1000 litres m 3 = m 3 = litres Hence, the quantity of milk in tank = litres. Q.7. If each edge of a cube is doubled, (a) how many times will its surface area increase? (b) how many times will its volume increase? (a) Let the edge of a cube be x units 22

23 If each edge of a cube is doubled then the new edge will becomes 2x units Surface area of cube = 6 (side) 2 = 6x 2 sq. units If edge is doubled, then surface area of cube = 6 (2x) 2 = 6 4x 2 = 24x 2 sq. units Hence, its surface area will becomes four times. (b) Volume of cube = (l) 3 = x 3 cube units If edge is doubled, then volume of cube = (2x) 3 = 8x 3 cubic units = x 3 8 = 8x 3 cubic units Hence, its volume will become eight times. Q.8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m 3, find the number of hours it will take to fill the reservoir. Volume of cuboidal reservoir = 108 m 3 Volume of cuboidal reservoir in litres = litres ( 1 m 3 = 1000l) = litres Required time = = 1800 minutes = 1800 = 30 hours ( 1 hr = 60 minutes) 60 23