VOLUME AND SURFACE AREAS OF SOLIDS


 Chad Phelps
 2 years ago
 Views:
Transcription
1 VOLUME AND SURFACE AREAS OF SOLIDS Q.1. Find the total surface area and volume of a rectangular solid (cuboid) measuring 1 m by 50 cm by 0.5 m Ans. Length of cuboid l = 1 m, Breadth of cuboid, b = 50 cm = = m Height of cuboid, h = 0.5 m = m= m Volume of cuboid = l b h = 1 = m = 0.5 m Total surface area of cuboid = ( lb + bh + hl) = = + + = = = =.50 m 4 Hence, volume of cuboid is 0.5 m and total surface area of cuboid is.50 m Q.. The length, breadth and height of a rectangular solid are in the ratio 5 : 4 :. If the total surface area is 116 cm, find the length, breadth and height of the solid. Ans. Let length, l = 5 x, breadth, b = 4x and height, h = x. Total surface area = 116 cm ( lb + bh + hl) = 116 (5x 4x + 4x x + x 5 x) = 116 (0x + 8x + 10 x ) = 116 (8 x ) = 116 6x = x = = 16 x = 16 x = 4 l = 5x = 5 4 = 0 cm 6 b = 4x = 4 4 = 16 cm, and h = x = 4 = 8 cm Hence, length, l = 0 cm, breadth, b = 16 cm, and height h = 8 cm. Math Class IX 1 Question Bank
2 Q.. The volume of a rectangular wall is 8 m, find the width of the wall. m. If its length is 16.5 m and height Ans. Volume of rectangular wall = m Length of wall ( l ) = 16.5 m, Height of wall ( h ) = 8 m Let width of wall = b m, then volume of rectangular wall = l b h b = 1 b = = m=0.5 m Hence, width of wall is 0.5 m. Q.4. A class room is 1.5 m long, 6.4 m broad and 5 m high. How many students can accommodate if each student needs 1.6 m of floor area? How many cubic metres of air would each student get? Ans. Length of room ( l ) = 1.5 m, width of room ( b ) = 6.4 m and height of room ( h ) = 5 m Volume of air inside the room = l b h = m = 400 m Area of floor of the room = l b = m = 80 m For each student area required = 1.6 m Number of students = = = Volume of air 400 The required air received by each student = = = 8 m Number of students 50 Q.5. Find the length of the longest rod that can be placed in a room measuring 1 m 9 m 8 m. Ans. Length of room ( l ) = 1 m, breadth of room ( b ) = 9 m, and height of room ( h ) = 8 m Hence, the longest rod required to place in the room = l + b + h = (1) + (9) + (8) = m = 89 = 1 m Q.6. The volume of a cuboid is cm and its height is 15 cm. The crosssection of the cuboid is a rectangle having its sides in the ratio 5 :. Find the perimeter of the crosssection. Ans. Volume of cuboid = cm, and Height of cuboid ( h ) = 15 cm volume Length Breadth cm = = = 960 cm h 15 Ratio of remaining sides = 5 : Thus, length = 5x and breadth = x Math Class IX Question Bank
3 5x x = x = 960 x = 64 = (8) x = 8 Length = 5x = 8 5 = 40 cm and breadth = x = 8 = 4 cm Hence, perimeter of rectangular crosssection = ( l + b) = (40 + 4) cm = 64 = 18 cm. Q.. The area of path is 6500 m. Find the cost of covering it with gravel 14 cm deep at the rate of Rs 5.60 per cubic metre. 14 Ans. Area of path = 6500 m and Depth of gravel = 14 cm = m Volume of gravel = Area Depth = 6500 = 910 m 100 Rate of covering the gravel = Rs 5.60 per m Total cost = Rs = Rs = Rs Q.8. The cost of papering the four walls of a room 1 m long at Rs 6.50 per square metre is Rs 168 and the cost of matting the floor at Rs.50 per square metre is Rs 8. Find the height of the room. Ans. Rate of papering the walls = Rs 6.50 per m and Total cost = Rs Area of four walls = = m = 5 m Rate of matting the floor = Rs.50 per m and Total cost = Rs Area of floor = = m = 108 m Area of floor 108 Length of room = 1 m Breadth of room = = = 9 m Length 1 Area of four walls = ( l + b) h = 5 (1 + 9) h = 5 1h = 5 5 h = h = 6 1 Hence, height of the room is 6 m. Math Class IX Question Bank
4 Q.9. The dimensions of a field are 15 m 1 m. A pit.5 m 6 m 1.5 m is dug in one corner of the field and the earth removed from it, is evenly spread over the remaining area of the field, calculate, by how much the level of the field is raised? Ans. Length of field ( l ) = 15 m and breadth of field ( b ) = 1 m Area of total field = l b = 15 1 = 180 m Length of pit =.5 m breadth of pit = 6 m and depth of pit = 1.5 m Area of pit = l b =.5 6 = 45 m and volume of earth removed = l b h = m = 6.5 m Area of field leaving pit = (180 45) m = 15 m and volume of earth removed = l b h = m = 6.5 m Area of field leaving pit = ) m = 15 m Volume of earth Level (height) of the earth in the field = Area of remaining part = = m = 0.5 m = 50 cm Q.10. The sum of length, breadth and depth of a cuboid is 19 cm, and the length of its diagonal is 11 cm. Find the surface area of the cuboid. Ans. Let l, b and h be the length, breadth and depth of the cuboid, then l + b + h = 19 cm and l + b + h = 11 cm l + b + h = (11) = 11 The surface area of the cuboid = ( lb + bh + hl) We know that ( l + b + h) = l + b + h + ( lb + bh + hl) (19) = 11+ ( lb + bh + hl) ( lb + bh + hl) = (19) 11 = = 40 cm Hence, surface area of the cuboid = 40 cm Q.11. Find the number of bricks, each measuring 5 cm 1.5 cm.5 cm, required to construct a wall 6 m long, 5 m high and 50 cm thick, while the cement and the sand mixture occupies 1 th of the volume of the wall. 0 Ans. Volume of one brick = 5 cm 1.5 cm.5 cm m 1 1 = m = m = Math Class IX 4 Question Bank
5 Length of wall ( l ) = 6 m and Height of wall ( h ) = 5 m 50 1 and thickness of wall ( b ) = m = m Volume of wall = l b h = 6 5 = 15 m 1 Volume of cement and sand = of 15 m = m 0 4 Volume of bricks = Volume of wall Volume of cement and sand 60 5 = m = m 4 4 Volume of total bricks Number of bricks = Volume of one brick = = = 19 0 = Q.1. The total surface area of a cube is 6 cm. Find its volume. Ans. Total surface area of cube = 6 cm Let edge of cube = a cm Total surface area of cube = 6a = 6 6 a = = 11 a = 11 = 11 cm 6 Edge of cube = 11 cm Volume of cube Q.1. The volume of a cube is 9 Ans. Volume of cube = 9 cm Let edge of cube = a = (side) = (11 cm) = cm = 11 cm cm. Find its total surface area. a = 9 = a = 9 cm Total surface area of cube = 6a = 6 9 cm 9 cm = 486 cm. Q.14. The edges of three cubes of metal are cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube. Ans. The edges of three cubes are cm, 4 cm and 5 cm. Volume of three cubes are (), (4) and (5) i.e. cm, 64 cm and 15 cm. Volume of new cube = = 16 cm Let edge of new cube be a Math Class IX 5 Question Bank
6 a = 16 = 6 6 6, a = 6 cm Hence, edge of new cube = 6 cm Q.15. Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 1 cm. Find x. Ans. Edges of three cubes are x cm, 8 cm and 10 cm respectively Volume of these cubes are x, (8) and (10) i.e. x cm, 51 cm and 1000 cm Edge of new cube formed = 1 cm Volume of new cube = (side) = (1) = = 18 cm According to the given problem x = 18 x = 18 x = = 16 = x = 6 cm Hence, edge of cube = x = 6 cm. Q.16. Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes. Ans. Let edge of each of three cubes be x when three cubes are placed end to end, a cuboid is formed whose dimensions are : l = x + x + x = x, b = x, h = x Total surface area of a resulting cuboid = ( lb + bh + hl) = (x x + x x + x x) Sum of total surface areas of three cubes = (x + x + x ) = x = 14x = 6x = 18x 14x 14 Ratio of suface areas of cubes = = = 18x 18 9 Hence, the required ratio is : 9. Q.1. A metal cube of edge 1 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, find the edge of third smaller cube. Ans. Edge of metal cube = 1 cm Volume of metal cube = (1) = 18 cm Edge of first smaller cube = 6 cm Volume of smaller cube = (6) = 16 cm Edge of second smaller cube = 8 cm Math Class IX 6 Question Bank
7 Volume of second smaller cube = (8) = 51 cm Volume of third smaller cube = 18 ( ) = 18 8 = 1000 cm 1 1 Hence, edge of third cube = (1000) cm = [(10) ] cm = 10 cm Q.18. The dimensions of a metallic cuboid are 100 cm 80 cm 64 cm. It is melted and recast into a cube. Find (i) the edge of the cube (ii) the surface area of the cube. Ans. Length of the metallic cuboid ( l ) = 100 cm breadth of the metallic cuboid ( b ) = 80 cm and height of metallic cuboid ( h ) = 64 cm Volume of metallic cuboid = l b h = = cm Volume of cube so formed = cm (i) Edge of the cube (ii) Surface area of cube ( a ) = (51000) 1/ = 6a = 6 (80) 1/ = [(80) ] = 80 cm = cm Q.19. The square on the diagonal of cube has an area of 185 the side of the cube (ii) the total surface area of the cube. Ans. (Diagonal) of cube = 185 cm (i) Let side of cube = a ( a ) = 185 a = 65 = (5) a = 5 Hence, side of the cube is 5 cm. (ii) Total surface area of the cube a = a = = 8400 cm cm, Calculate (i) = 6 side = 6a = 6 (5) = 6 65 cm = 50 cm Q.0. Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid is 648 cm. Find the length of edge of each cube. Also find the ratio between the surface area of resulting cuboid and the surface area of a cube. Ans. Surface area of cuboid = 648 cm, let edge of each cube = x cm Length of cuboid = 4 x cm, width of cuboid = x cm and height of cuboid = x cm Surface area of cuboid = ( lb + bh + hl) 648 = (4x x + x x + x 4 x) = (4x + x + 4 x ) 648 = 18x 648 x = = 6 = (6) x = 6 18 Math Class IX Question Bank
8 Thus, length of edge of each cube = 6 cm Surface area of cube = 6 (side) = 6x = 6 (6) = 6 6 = 16 cm Hence, ratio between the surface area of cuboid and that of cube = 648 : 16 = :1 Q.1. A rectangular container has base of length 1 cm and width 9 cm. A cube of edge 6 cm is placed in the container and then sufficient water is filled into it, so that the cube is just submerged. Find the fall in level of water in the container, when the cube is removed. Ans. Length of the base of container = 1 cm and width of container = 9 cm Let fall in water level be x cm. Side of cube = 6 cm Volume of cube = = 16 cm Volume of water = 16 cm According to the given problem, we get x = 16 x = = 1 9 Hence, fall in water level is cm. Q.. A rectangular container whose base is a square of side 1 cm, contains sufficient water to submerge a rectangular solid 8 cm 6 cm cm. Find the rise in level of water in the container when the solid is in it. Ans. Let the rise in water level = x cm Length of solid cuboid = 8 cm, width of solid cuboid = 6 cm Height of solid cuboid = cm Volume of cuboid = l b h = 8 6 = 144 cm Volume of water = 144 cm Side of the square base = 1 cm Area of base = 1 1 = 144 cm According to given problem, we get 144 x = 144 x = 1 Hence, rise in water level is 1 cm. Q.. A field is 10 m long and 50 m broad. A tank 4 m long, 10 m broad and 6 m deep is dug any where in the field and the earth taken out of the tank is evenly spread over the remaining part of the field. Find the rise in level of the field. Ans. Length of rectangular field = 10 m, Breadth of rectangular field = 50 m Area of of the total field = m = 6000 m Length of tank = 4 m, breadth of tank = 10 m and depth of tank = 6 m Volume of earth dug out = m = 1440 m Math Class IX 8 Question Bank
9 Area of surface of tank = l b = 4 10 = 40 m Area of remaining part of the field = Area of total field Surface area of tank = = 560 m Let height of rise of field = x m 560 x = x = = = 0.5. Hence, rise in level of the field = 0.5 m or 5 cm Q.4. The following figure shows a solid of uniform crosssection. Find the volume of the solid. All measurements are in centimetres. Assume that all angles in the figure are rightangles. Ans. The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, cm and 9 cm respectively. Hence, volume of solid = = = 180 cm Q.5. A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and m deep respectively if the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool. Ans. The crosssection of the swimming pool is a trapezium with parallel sides 1.5 m and m (i.e. depths of both ends) and 40 m (i.e. length of pool) as the between sides. The width of pool i.e. 15 m can be taken as the length of crosssection. Length of swimming pool ( l ) = 40 m Breadth of swimming pool ( b ) = 15 m Depth of its ends = 1.5 m and.5 m 1 Volume of water [(1.5.0) 40] 15 m = m = = 150 m Capacity of water in litres = litres = litres (1 m = 1000 litres) r distance Math Class IX 9 Question Bank
10 Q.6. The crosssection of a piece of metal m in length is shown in the adjoining figure. Calculate : (i) The area of its crosssection. (ii) The volume of piece of metal. (iii) The weight of piece of metal to the nearest kg, if 1 cm of the metal weighs 6.5 g. Ans. From C, draw CF AB then CF = AB = 1 cm AF = CB = 8 cm EF = EA FA = 1 8 = 4 cm (i) Now area of figure ABCDE (crosssection) = ar. (Rectangle ABCF) + ar. (Trapezium FCDE) 1 = 1 cm 8 cm + (1 + 16) 4 cm = = = 16 cm (ii) Length of the piece = m = 00 cm Volume of the metal piece = Area Length = cm = 400 cm 400 = =.4 m (iii) Weight of 1 cm metal = 6.5 g Total weight of the metal piece = g = g = 11 kg (approx) Q.. A square brass plate of side x cm is 1 mm thick and weighs 5.44 kg. If 1 cm of brass weighs 8.5 g, find the value of x. Ans. Side of square plate = x cm Thickness of square plate = 1 mm Total weight of square plate = 5.44 kg Weight of 1 cm = 8.5 g Total volume of the plate cm = 8.5 = 640 cm Math Class IX 10 Question Bank
11 According to the given problem x = 6400 x = 6400 = 80 cm 1 = x x = Q.8. The area of crosssection of a rectangular pipe is 5.4 cm and water is pumped out of it at the rate of kmph. Find, in litres, the volume of water which flows out of the pipe in 1 minute. Ans. Area of crosssection of rectangular pipe = 5.4 cm Speed of water pumped out = kmph Time = 1 minute Length of water flow = 1000 m = 450 m 60 Volume of water = Area Length 5.4 = 450 m = m m = m = Volume of water in litres 1000 = (1 m = 1000 litres) 1000 = 4 litres Q.9. The crosssection of a tunnel, perpendicular to its length is a trapezium ABCD in which AB = 8 m, DC = 6 m and AL = BM. The height of the tunnel is.4 m and its length is 40 m. Find : (i) The cost of paving the floor of the tunnel at Rs 16 per m. (ii) The cost of painting the internal surface of the tunnel, excluding the floor at the rate of Rs 5 per m. Ans. The crosssection of a tunnel is of the trapezium shaped ABCD in which AB = 8 m, DC = 6 m and AL = BM. The height is.4 m and its length is 40 m. (i) Area of floor of tunnel = l b = 40 8 = 0 m Math Class IX 11 Question Bank
12 Rate of cost of paving = Rs 16 per m Total cost = 0 16 = Rs (ii) AL = BM = = = 1 m In ADL, AD = AL + DL (Using Pythagoras Theorem) = (1) + (.4) = = 6.6 = (.6) AD =.6 m Perimeter of the crosssection of the tunnel = ( ) m = 18.8 m Length = 40 m Internal surface area of the tunnel (except floor) = ( ) m = (5 0) m = 4 m Rate of painting = Rs 5 per m Hence, total cost of painting = Rs 5 4 = Rs 160 Q.0. A stream, which flows at a uniform rate of 4 km/hr, is 10 metres wide and 1. m deep at a certain point. If its crosssection is rectangular is shape find, in litres, the volume of water that flows in a minute Ans. Speed of stream = 4 km/hr = 4 m/s = m/s 18 9 Area of crosssection of stream = = 1 m Volume of water discharged in 1 second = 1 = m 9 But 1 m = 1000 litres m = 1000 = litres Volume of water discharged in 1 minutes i.e. 60 sec. = = litres. Math Class IX 1 Question Bank
13 Q.1. If cubic metres of sand be thrown into a rectangular tank 14 m long and 5 m wide; find the rise in level of the water. Ans. Length of rectangular tank = 14 m Width of rectangular tank = 5 m Let rise in level = x m Volume = 14 5 x = 0 x m But volume = m (Given) 0x = x = x = x = x = 1.44 m 100 Q.. A rectangular cardboard sheet has length cm and breadth 6 cm. Squares of each side cm are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed. Ans. Length of sheet = cm Breadth of sheet = 6 cm Side of each square = cm Inner length = = 6 = 6 cm Inner breadth = 6 = 6 6 = 0 cm By folding the sheet, the length of the container = 6 cm Breadth of the container = 0 cm and height of container = cm Volume of the container = l b h = 6 cm 0 cm cm = 1560 cm Q.. A swimming pool is 18 m long and 8 m wide its deep and shallow ends are m and 1. m respectively find the capacity of the pool, assuming that the bottom of the pool slopes uniformly. Ans. Length of pool = 18 m Breadth of pool = 8 m Height of one side = m Height on second side = 1. m ( + 1.) Volume of pool = 18 8 m = = 0.4 m Math Class IX 1 Question Bank
14 Q.4. A plot is 8 m long and 5 m broad. A tank 15 m long, m broad and 4 m deep is dug anywhere in the plot and the earth so removed is evenly spread over the remaining part of the plot. Find the rise in level of the plot. Ans. Length of plot = 8 m and width of plot = 5 m Area of plot = 8 5 m = 905 m Area of tank = l b = 15 m = 105 m Area of remaining plot = ( ) m = 800 m Volume of earth dug out = l b d = 15 4 m = 40 m Volume of earth Height of level of earth spread over the remaining part = Area of plot 40 = m = m = 100 = 15 cm Q.5. Find the volume of the cylinder in which : (i) Height = 1 cm and Base radius = 5 cm (ii) Diameter = 8 cm and Height = 40 cm. Ans. (i) Height of cylinder ( h ) = 1 cm and radius of the base ( r ) = 5 cm Volume of cylinder = π r h cm = = 1650 cm 8 (ii) Radius of the base of cylinder, r = = 14 cm Height of cylinder ( h ) = 40 cm Volume of cylinder = π r h cm = = 8 40 = 4640 cm Q.6. Find the weight of the solid cylinder of radius 10.5 cm and height 60 cm, if the material of the cylinder weighs, 5 grams per cu. cm. Ans. Radius of solid cylinder ( r ) = 10.5 cm Height of cylinder ( h ) = 60 cm Volume of cylinder = π r h cm = = = 090 cm Weight of 1 cubic cm (i.e. 1 cm ) = 5 g Math Class IX 14 Question Bank
15 10950 Total weight of the cylinder = g = g = kg = kg 1000 Q.. A cylindrical tank has a capacity of 6160 m. Find its depth if its radius is 14 m. Also find the cost of painting its curved surface at Rs per m. Ans. Volume of cylinder = 6160 m and radius of cylinder ( r ) = 14 m Let depth of cylinder be h m. Volume of cylinder = π r h = h = 6160 h = = 10 m Curved surface area of cylinder = π rh = m = 880 m Rate of painting its curved surface = Rs per m Total cost = 880 = Rs 640. Q.8. The curved surface area of a cylinder is 4400 cm and the circumference of its base is 110 cm. Find the height and the volume of the cylinder. Ans. Curved surface area = 4400 cm and circumference of its base = 110 cm Let radius be r and height be h Circumference of base = π r = r = 110 r = = 5 cm Curved surface area of cylinder = π rh = h = h = = 40 cm Volume of the cylinder = π r h 40 cm = = 8500 cm Q.9. The total surface area of a solid cylinder is 46 cm and its curved surface area is onethird of its total surface area. Find the volume of the cylinder. Ans. Total surface area of cylinder = 46 cm 1 1 and curved surface area = of total surface area cm = = Let r be the radius and h be its height, then π rh = rh = 49 rh = 154 =...(i) 44 and Area of two bases π r = = 08 r = r = = 49 = () r = cm Math Class IX 15 Question Bank
16 Now putting the value of r in (i), we get h = h = = cm Hence, volume of cylinder = π r h = cm = 59 cm Q.40. The sum of the radius of the base and the height of a solid cylinder is m. If the total surface area of the cylinder be 168 m, find its volume. Ans. Let r be the radius and h be the height of the cylinder. Then total surface area of cylinder = 168 m π r ( r + h) = r = 168 r = = h = = 0 m. But r + h = + h = Volume of the cylinder = π r h 0 m = = 460 m Q.41. A road roller is cylindrical in shape. Its circular end has a diameter 0. m and its width is m. Find the least number of complete revolutions that the roller must make in order to level a playground measuring 80 m by 44 m. 0. Ans. Diameter of cylinder = 0. m Radius of cylinder, r = = 0.5 m Width of roller, h = m Curved surface area of roller = π rh = 0.5 = 4.40 m In 1 revolution, area covered by roller = 4.40 m Area of playground = = 50 m Number of revolutions = = 100 = = Q.4. How many cubic metres of earth must be dug out to make a well 8 m deep and.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq. metre. Ans. Depth of well, h = 8 m and diameter of well =.8 m.8 Radius of well, r = m = 1.4 m Volume of earth dug out = π r h m = = Math Class IX 16 Question Bank
17 Curved surface of well = π rh = = 46.4 m Cost of platering at the rate of Rs 4.50 per sq. metre = = Rs Q.4. Find the length of 11 kg copper wire of diameter 0.4 cm. Given one cubic cm of copper weighs 8.4 gram. Ans. Let length of copper wire be h cm. Diameter of copper wire = 0.4 cm then radius of copper wire ( r ) = 0. cm Volume of copper wire (cylinder) = π r h = h 88 = h = h cm Weigth of 1 cm of wire = 8.4 grams 88 Total weight of wire = h = h grams But, weight of wire = 11 kg = = gm h = h = h = = h = cm = m = m h = m Q.44. A cylinder has a diameter of 0 cm. The area of curved surface is 100 (sq. cm). Find : (i) the height of the cylinder correct to one decimal place. (ii) the volume of the cylinder correct to one decimal place. 0 Ans. (i) Diameter of cylinder = 0 cm then radius r = = 10 cm Let height of cylinder be h cm. Curved surface area of cylinder = 100 cm π rh = h = h = = cm = cm Math Class IX 1 Question Bank
18 (ii) If h is taken as 1.6 cm Then volume of cylinder = π r h cm = = Hence, (i) height of cylinder = 1.6 cm (ii) volume of cylinder 50.9 cm. Q.45. A metal pipe has a bore (inner diameter) of 5 cm. The pipe is mm thick all round. Find the weight in kilogram, of metres of the pipe, if 1 cm of the metal weighs. g. 5 Ans. Inner diameter = 5 cm Inner radius r 1 = =.5 cm Thickness = mm = 0. cm Outer radius, r =.5 cm + 0. cm =. cm Length of pipe = metres = 00 cm π 1 Volume of metal in the pipe = π r h r h = π h [ r r1 ] = 00 [(.) (.5) ] 104 = = 00 = 100 = 00 [.9 6.5] cm 1 cm of metal weighs =. gm of metal weighs =. = = 50.6 gm 50.6 = kg = 5.06 kg = 5.04 kg 1000 Q.46. Find the total surface area of a hollow cylinder open at both ends, if its length is 1 cm, external diameter is 8 cm and the thickness is cm. Ans. Length of hollow cylinder ( h ) = 1 cm External diameter of cylinder = 8 cm 8 External radius of cylinder ( r ) = cm = 4 cm Thickness of the cylinder = cm Inner radius of cylinder = 4 = cm Thus, total surface area of cylinder = [π Rh + π rh + ( π R π r )] cm = cm Math Class IX 18 Question Bank
19 = + + cm = + cm = + = = 58 cm Q.4. Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is cm, into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in 0 minutes. Ans. Diameter of pipe = cm, Radius of pipe = 1 cm Rate of water flow = 8 m/sec Time taken = 0 minutes Length of water flow into the pipe ( h ) = = m Volume of water = π r h = m m = Radius of the base of the cylindrical tank = 40 cm = m = m Let h be the water increased in the tank π r h = h = h = = 9 m 00 Hence, height of water is 9 m. Q.48. The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm, and the volume of material in it is 1056 cm. Find the internal and external radii. Given that the height of the cylinder is 1 cm. Ans. Let r and R be the inner and outer radii and h be their height π Rh + π rh = 1056 π h ( R + r) = ( R + r) = ( R + r) = 1056 R + r = = 8...(i) 1 Again π R h π r h = 1056 π h ( R r ) = ( ) = 1056 R 66 ( R r ) = 1056 r Math Class IX 19 Question Bank
20 1056 R r = = ( R + r) ( R r) = 16 8 ( R r) = R r = = 8 R + r = 8...(ii) R r =...(iii) Adding (ii) and (iii), we get R = 10 R = 5 Subtracting eqn. (iii) from (ii), we get r = 6 r = Hence, outer radius of cylinder = 5 cm and inner radius of cylinder = cm. Math Class IX 0 Question Bank
Total Time: 90 mins Practise Exercise no. 1 Total no. Of Qs: 51
Mensuration (3D) Total Time: 90 mins Practise Exercise no. 1 Total no. Of Qs: 51 Q1. A wooden box of dimensions 8m x 7m x 6m is to carry rectangular boxes of dimensions 8cm x 7cm x 6cm. The maximum number
More informationCHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON SOLIDS
CHAPTER 9 VOLUMES AND SURFACE AREAS OF COMMON EXERCISE 14 Page 9 SOLIDS 1. Change a volume of 1 00 000 cm to cubic metres. 1m = 10 cm or 1cm = 10 6m 6 Hence, 1 00 000 cm = 1 00 000 10 6m = 1. m. Change
More informationExercise 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?
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
More informationChapter 16. Mensuration of Cylinder
335 Chapter 16 16.1 Cylinder: A solid surface generated by a line moving parallel to a fixed line, while its end describes a closed figure in a plane is called a cylinder. A cylinder is the limiting case
More informationPERIMETER AND AREA OF PLANE FIGURES
PERIMETER AND AREA OF PLANE FIGURES Q.. Find the area of a triangle whose sides are 8 cm, 4 cm and 30 cm. Also, find the length of altitude corresponding to the largest side of the triangle. Ans. Let ABC
More informationSURFACE AREAS AND VOLUMES
CHAPTER 1 SURFACE AREAS AND VOLUMES (A) Main Concepts and Results Cuboid whose length l, breadth b and height h (a) Volume of cuboid lbh (b) Total surface area of cuboid 2 ( lb + bh + hl ) (c) Lateral
More informationMENSURATION. Base Corresponding Height. Area of a parallelogram = Base Corresponding Height. (Sum of parallel sides) Height. Product of diagonals.
UNIT 11 MENSURATION (A) Main Concepts and Results 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
More informationPerimeter and Area. Chapter 11 11.1 INTRODUCTION 11.2 SQUARES AND RECTANGLES TRY THESE
PERIMETER AND AREA 205 Perimeter and Area Chapter 11 11.1 INTRODUCTION In Class VI, you have already learnt perimeters of plane figures and areas of squares and rectangles. Perimeter is the distance around
More information9 Area, Perimeter and Volume
9 Area, Perimeter and Volume 9.1 2D Shapes The following table gives the names of some 2D shapes. In this section we will consider the properties of some of these shapes. Rectangle All angles are right
More informationGCSE Revision Notes Mathematics. Volume and Cylinders
GCSE Revision Notes Mathematics Volume and Cylinders irevise.com 2014. All revision notes have been produced by mockness ltd for irevise.com. Email: info@irevise.com Copyrighted material. All rights reserved;
More informationGCSE Exam Questions on Volume Question 1. (AQA June 2003 Intermediate Paper 2 Calculator OK) A large carton contains 4 litres of orange juice.
Question 1. (AQA June 2003 Intermediate Paper 2 Calculator OK) A large carton contains 4 litres of orange juice. Cylindrical glasses of height 10 cm and radius 3 cm are to be filled from the carton. How
More informationChapter 11 Area of Quadrilateral
75 Chapter 11 11.1 Quadrilateral A plane figure bounded by four sides is known as a quadrilateral. The straight line joining the opposite corners is called its diagonal. The diagonal divides the quadrilateral
More informationSOLID SHAPES M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier
Mathematics Revision Guides Solid Shapes Page 1 of 19 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier SOLID SHAPES Version: 2.1 Date: 10112015 Mathematics Revision Guides Solid
More informationNCERT. Area of the circular path formed by two concentric circles of radii. Area of the sector of a circle of radius r with central angle θ =
AREA RELATED TO CIRCLES (A) Main Concepts and Results CHAPTER 11 Perimeters and areas of simple closed figures. Circumference and area of a circle. Area of a circular path (i.e., ring). Sector of a circle
More informationCHAPTER 27 AREAS OF COMMON SHAPES
EXERCISE 113 Page 65 CHAPTER 7 AREAS OF COMMON SHAPES 1. Find the angles p and q in the diagram below: p = 180 75 = 105 (interior opposite angles of a parallelogram are equal) q = 180 105 0 = 35. Find
More informationChapter 15 Mensuration of Solid
3 Chapter 15 15.1 Solid: It is a body occupying a portion of threedimensional space and therefore bounded by a closed surface which may be curved (e.g., sphere), curved and planer (e.g., cylinder) or
More informationBy the end of this set of exercises, you should be able to:
BASIC GEOMETRIC PROPERTIES By the end of this set of exercises, you should be able to: find the area of a simple composite shape find the volume of a cube or a cuboid find the area and circumference of
More information1 cm 3. 1 cm. 1 cubic centimetre. height or Volume = area of crosssection length length
Volume Many things are made in the shape of a cuboid, such as drink cartons and cereal boxes. This activity is about finding the volumes of cuboids. Information sheet The volume of an object is the amount
More informationThe formulae for calculating the areas of quadrilaterals, circles and triangles should already be known : Area = 1 2 D x d CIRCLE.
Revision  Areas Chapter 8 Volumes The formulae for calculating the areas of quadrilaterals, circles and triangles should already be known : SQUARE RECTANGE RHOMBUS KITE B dd d D D Area = 2 Area = x B
More informationMENSURATION. Definition
MENSURATION Definition 1. Mensuration : It is a branch of mathematics which deals with the lengths of lines, areas of surfaces and volumes of solids. 2. Plane Mensuration : It deals with the sides, perimeters
More informationAREA QUESTIONS AND ANSWERS
AREA QUESTIONS AND ANSWERS 1. The length and breadth of a square are increased by 40$ and 30% respectively. The area of the resulting rectangle exceeds the area of the square by: (a) 42% (b) 62% (c) 82%
More informationLESSON SUMMARY. Measuring Shapes
LESSON SUMMARY CXC CSEC MATHEMATICS UNIT SIX: Measurement Lesson 11 Measuring Shapes Textbook: Mathematics, A Complete Course by Raymond Toolsie, Volume 1 (Some helpful exercises and page numbers are given
More informationPerimeter, Area, and Volume
Perimeter, Area, and Volume Perimeter of Common Geometric Figures The perimeter of a geometric figure is defined as the distance around the outside of the figure. Perimeter is calculated by adding all
More informationCONNECT: Volume, Surface Area
CONNECT: Volume, Surface Area 1. VOLUMES OF SOLIDS A solid is a threedimensional (3D) object, that is, it has length, width and height. One of these dimensions is sometimes called thickness or depth.
More informationGeometry Chapter 12. Volume. Surface Area. Similar shapes ratio area & volume
Geometry Chapter 12 Volume Surface Area Similar shapes ratio area & volume Date Due Section Topics Assignment Written Exercises 12.1 Prisms Altitude Lateral Faces/Edges Right vs. Oblique Cylinders 12.3
More informationThe Area is the width times the height: Area = w h
Geometry Handout Rectangle and Square Area of a Rectangle and Square (square has all sides equal) The Area is the width times the height: Area = w h Example: A rectangle is 6 m wide and 3 m high; what
More informationArea is a measure of how much space is occupied by a figure. 1cm 1cm
Area Area is a measure of how much space is occupied by a figure. Area is measured in square units. For example, one square centimeter (cm ) is 1cm wide and 1cm tall. 1cm 1cm A figure s area is the number
More informationCIRCUMFERENCE AND AREA OF A CIRCLE
CIRCUMFERENCE AND AREA OF A CIRCLE 1. AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take = 3.14) 2. In the given
More informationWorking in 2 & 3 dimensions Revision Guide
Tips for Revising Working in 2 & 3 dimensions Make sure you know what you will be tested on. The main topics are listed below. The examples show you what to do. List the topics and plan a revision timetable.
More informationMaximum / Minimum Problems
171 CHAPTER 6 Maximum / Minimum Problems Methods for solving practical maximum or minimum problems will be examined by examples. Example Question: The material for the square base of a rectangular box
More informationCONNECT: Volume, Surface Area
CONNECT: Volume, Surface Area 2. SURFACE AREAS OF SOLIDS If you need to know more about plane shapes, areas, perimeters, solids or volumes of solids, please refer to CONNECT: Areas, Perimeters 1. AREAS
More informationB = 1 14 12 = 84 in2. Since h = 20 in then the total volume is. V = 84 20 = 1680 in 3
45 Volume Surface area measures the area of the twodimensional boundary of a threedimensional figure; it is the area of the outside surface of a solid. Volume, on the other hand, is a measure of the space
More information25.5 m Upper bound. maximum error 0.5 m
Errors When something is measured, the measurement is subject to uncertainty. This uncertainty is called error even though it does not mean that a mistake has been made. The size of the error depends on
More informationChapter 19. Mensuration of Sphere
8 Chapter 19 19.1 Sphere: A sphere is a solid bounded by a closed surface every point of which is equidistant from a fixed point called the centre. Most familiar examples of a sphere are baseball, tennis
More informationGrade 9 Mathematics Unit 3: Shape and Space Sub Unit #1: Surface Area. Determine the area of various shapes Circumference
1 P a g e Grade 9 Mathematics Unit 3: Shape and Space Sub Unit #1: Surface Area Lesson Topic I Can 1 Area, Perimeter, and Determine the area of various shapes Circumference Determine the perimeter of various
More informationDIFFERENTIATION OPTIMIZATION PROBLEMS
DIFFERENTIATION OPTIMIZATION PROBLEMS Question 1 (***) 4cm 64cm figure 1 figure An open bo is to be made out of a rectangular piece of card measuring 64 cm by 4 cm. Figure 1 shows how a square of side
More informationArea of Parallelograms, Triangles, and Trapezoids (pages 314 318)
Area of Parallelograms, Triangles, and Trapezoids (pages 34 38) Any side of a parallelogram or triangle can be used as a base. The altitude of a parallelogram is a line segment perpendicular to the base
More information3Surface Area, Volume, and Capacity
124 Chapter 3Surface Area, Volume, and Capacity How much water do you think this water tank can hold? What would you need to know to calculate the exact amount? 3.1 Surface Area of Prisms REVIEW: WORKING
More informationYimin Math Centre. Year 6 Problem Solving Part 2. 2.1 Measurements... 1. 2.2 Practical Exam Questions... 8. 2.3 Quiz... 10. Page: 10 11 12 Total
Year 6 Problem Solving Part 2 Student Name: Grade: Date: Score: Table of Contents 2 Problem Solving Part 2 1 2.1 Measurements....................................... 1 2.2 Practical Exam Questions.................................
More information3D Geometry: Chapter Questions
3D Geometry: Chapter Questions 1. What are the similarities and differences between prisms and pyramids? 2. How are polyhedrons named? 3. How do you find the crosssection of 3Dimensional figures? 4.
More informationSURFACE AREA AND VOLUME
SURFACE AREA AND VOLUME In this unit, we will learn to find the surface area and volume of the following threedimensional solids:. Prisms. Pyramids 3. Cylinders 4. Cones It is assumed that the reader has
More informationMensuration. The shapes covered are 2dimensional square circle sector 3dimensional cube cylinder sphere
Mensuration This a mixed selection of worksheets on a standard mathematical topic. A glance at each will be sufficient to determine its purpose and usefulness in any given situation. These notes are intended
More informationALPERTON COMMUNITY SCHOOL MATHS FACULTY ACHIEVING GRADE A/A* EXAM PRACTICE BY TOPIC
ALPERTON COMMUNITY SCHOOL MATHS FACULTY ACHIEVING GRADE A/A* EXAM PRACTICE BY TOPIC WEEK Calculator paper Each set of questions is followed by solutions so you can check & mark your own work CONTENTS TOPIC
More informationEDEXCEL FUNCTIONAL SKILLS PILOT. Maths Level 2. Chapter 5. Working with shape and space
EDEXCEL FUNCTIONAL SKILLS PILOT Maths Level 2 Chapter 5 Working with shape and space SECTION H 1 Perimeter 75 2 Area 77 3 Volume 79 4 2D Representations of 3D Objects 81 5 Remember what you have learned
More informationA Resource for Freestanding Mathematics Qualifications
When something is measured, the measurement is subject to error. The size of the error depends on the sensitivity of the measuring instrument and how carefully it is used. Often when measurements are given
More informationPizza! Pizza! Assessment
Pizza! Pizza! Assessment 1. A local pizza restaurant sends pizzas to the high school twelve to a carton. If the pizzas are one inch thick, what is the volume of the cylindrical shipping carton for the
More informationName: School Team: X = 5 X = 25 X = 40 X = 0.09 X = 15
7th/8th grade Math Meet Name: School Team: Event : Problem Solving (no calculators) Part : Computation ( pts. each) ) / + /x + /0 = X = 5 ) 0% of 5 = x % of X = 5 ) 00  x = ()()(4) + 6 X = 40 4) 0.6 x
More informationIntegrated Algebra: Geometry
Integrated Algebra: Geometry Topics of Study: o Perimeter and Circumference o Area Shaded Area Composite Area o Volume o Surface Area o Relative Error Links to Useful Websites & Videos: o Perimeter and
More informationPerfume Packaging. Ch 5 1. Chapter 5: Solids and Nets. Chapter 5: Solids and Nets 279. The Charles A. Dana Center. Geometry Assessments Through
Perfume Packaging Gina would like to package her newest fragrance, Persuasive, in an eyecatching yet costefficient box. The Persuasive perfume bottle is in the shape of a regular hexagonal prism 10 centimeters
More informationMathematics and steel roll pressing
Mathematics and steel roll pressing Part 1 Making cylinders and bends by roll pressing steel Outcomes for Part 1(Old syllabus): Maths: NS4.3, NS5.2.1, PAS4.1, PAS4.4, MS4.1, MS4.2, WMS4.2, WMS4.3, WMS4.4
More informationSolids. Objective A: Volume of a Solids
Solids Math00 Objective A: Volume of a Solids Geometric solids are figures in space. Five common geometric solids are the rectangular solid, the sphere, the cylinder, the cone and the pyramid. A rectangular
More information124 Volumes of Prisms and Cylinders. Find the volume of each prism. The volume V of a prism is V = Bh, where B is the area of a base and h
Find the volume of each prism. The volume V of a prism is V = Bh, where B is the area of a base and h The volume is 108 cm 3. The volume V of a prism is V = Bh, where B is the area of a base and h the
More informationUnit 04: Fundamentals of Solid Geometry  Shapes and Volumes
Unit 04: Fundamentals of Solid Geometry  Shapes and Volumes Introduction. Skills you will learn: a. Classify simple 3dimensional geometrical figures. b. Calculate surface areas of simple 3dimensional
More informationPractice Tests Answer Keys
Practice Tests Answer Keys COURSE OUTLINE: Module # Name Practice Test included Module 1: Basic Math Refresher Module 2: Fractions, Decimals and Percents Module 3: Measurement Conversions Module 4: Linear,
More informationRight Prisms Let s find the surface area of the right prism given in Figure 44.1. Figure 44.1
44 Surface Area The surface area of a space figure is the total area of all the faces of the figure. In this section, we discuss the surface areas of some of the space figures introduced in Section 41.
More informationQ1. Here is a flag. Calculate the area of the shaded cross. Q2. The diagram shows a rightangled triangle inside a circle.
Q1. Here is a flag. Calculate the area of the shaded cross. 2 marks Q2. The diagram shows a rightangled triangle inside a circle. The circle has a radius of 5 centimetres. Calculate the area of the triangle.
More informationFunctional Skills Mathematics
Functional Skills Mathematics Level Learning Resource Perimeter and Area MSS1/L.7 Contents Perimeter and Circumference MSS1/L.7 Pages 36 Finding the Area of Regular Shapes MSS1/L.7 Page 710 Finding the
More informationIWCF United Kingdom Branch
IWCF United Kingdom Branch Drilling Calculations Distance Learning Programme Part 2 Areas and Volumes IWCF UK Branch Distance Learning Programme  DRILLING CALCULATIONS Contents Introduction Training objectives
More informationIntroduction. Rectangular Volume. Module #5: GeometryVolume Bus 130 1:15
Module #5: Geometryolume Bus 0 :5 Introduction In this module, we are going to review the formulas for calculating volumes, rectangular, cylindrical, and irregular shaped. To set the stage, let s suppose
More informationLength, perimeter and area 3.1. Example
3.1 Length, perimeter and area Kno and use the names and abbreviations for units of length and area Be able to measure and make a sensible estimate of length and area Find out and use the formula for the
More informationGeometry Notes VOLUME AND SURFACE AREA
Volume and Surface Area Page 1 of 19 VOLUME AND SURFACE AREA Objectives: After completing this section, you should be able to do the following: Calculate the volume of given geometric figures. Calculate
More informationMATHEMATICS FOR ENGINEERING BASIC ALGEBRA
MATHEMATICS FOR ENGINEERING BASIC ALGEBRA TUTORIAL 4 AREAS AND VOLUMES This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals.
More informationGAP CLOSING. Volume and Surface Area. Intermediate / Senior Student Book
GAP CLOSING Volume and Surface Area Intermediate / Senior Student Book Volume and Surface Area Diagnostic...3 Volumes of Prisms...6 Volumes of Cylinders...13 Surface Areas of Prisms and Cylinders...18
More informationVolume of Prisms, Cones, Pyramids & Spheres (H)
Volume of Prisms, Cones, Pyramids & Spheres (H) A collection of 91 Maths GCSE Sample and Specimen questions from AQA, OCR, PearsonEdexcel and WJEC Eduqas. Name: Total Marks: 1. A cylinder is made of
More informationGeometry Vocabulary Booklet
Geometry Vocabulary Booklet Geometry Vocabulary Word Everyday Expression Example Acute An angle less than 90 degrees. Adjacent Lying next to each other. Array Numbers, letter or shapes arranged in a rectangular
More information16 Circles and Cylinders
16 Circles and Cylinders 16.1 Introduction to Circles In this section we consider the circle, looking at drawing circles and at the lines that split circles into different parts. A chord joins any two
More informationModule: Fundamental Construction Math
Module: Fundamental Construction Math Overview: A background review of fundamental math principles, concepts, and calculation used in common construction activities including estimating, load calculations,
More informationJunior Math Circles March 10, D Geometry II
1 University of Waterloo Faculty of Mathematics Junior Math Circles March 10, 2010 3D Geometry II Centre for Education in Mathematics and Computing Opening Problem Three tennis ball are packed in a cylinder.
More information124 Volumes of Prisms and Cylinders. Find the volume of each prism.
Find the volume of each prism. 3. the oblique rectangular prism shown at the right 1. The volume V of a prism is V = Bh, where B is the area of a base and h is the height of the prism. If two solids have
More informationGeometry Notes PERIMETER AND AREA
Perimeter and Area Page 1 of 57 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter
More informationIn Problems #1  #4, find the surface area and volume of each prism.
Geometry Unit Seven: Surface Area & Volume, Practice In Problems #1  #4, find the surface area and volume of each prism. 1. CUBE. RECTANGULAR PRISM 9 cm 5 mm 11 mm mm 9 cm 9 cm. TRIANGULAR PRISM 4. TRIANGULAR
More informationMEASUREMENTS. U.S. CUSTOMARY SYSTEM OF MEASUREMENT LENGTH The standard U.S. Customary System units of length are inch, foot, yard, and mile.
MEASUREMENTS A measurement includes a number and a unit. 3 feet 7 minutes 12 gallons Standard units of measurement have been established to simplify trade and commerce. TIME Equivalences between units
More informationYOU MUST BE ABLE TO DO THE FOLLOWING PROBLEMS WITHOUT A CALCULATOR!
DETAILED SOLUTIONS AND CONCEPTS  SIMPLE GEOMETRIC FIGURES Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. Thank you! YOU MUST
More informationDraft copy. Circles, cylinders and prisms. Circles
12 Circles, cylinders and prisms You are familiar with formulae for area and volume of some plane shapes and solids. In this chapter you will build on what you learnt in Mathematics for Common Entrance
More informationArea of a triangle: The area of a triangle can be found with the following formula: 1. 2. 3. 12in
Area Review Area of a triangle: The area of a triangle can be found with the following formula: 1 A 2 bh or A bh 2 Solve: Find the area of each triangle. 1. 2. 3. 5in4in 11in 12in 9in 21in 14in 19in 13in
More informationArea, Perimeter, Volume and Pythagorean Theorem Assessment
Area, Perimeter, Volume and Pythagorean Theorem Assessment Name: 1. Find the perimeter of a right triangle with legs measuring 10 inches and 24 inches a. 34 inches b. 60 inches c. 120 inches d. 240 inches
More informationCalculating Area, Perimeter and Volume
Calculating Area, Perimeter and Volume You will be given a formula table to complete your math assessment; however, we strongly recommend that you memorize the following formulae which will be used regularly
More informationVOLUME of Rectangular Prisms Volume is the measure of occupied by a solid region.
Math 6 NOTES 7.5 Name VOLUME of Rectangular Prisms Volume is the measure of occupied by a solid region. **The formula for the volume of a rectangular prism is:** l = length w = width h = height Study Tip:
More informationPERIMETERS AND AREAS
PERIMETERS AND AREAS 1. PERIMETER OF POLYGONS The Perimeter of a polygon is the distance around the outside of the polygon. It is the sum of the lengths of all the sides. Examples: The perimeter of this
More informationEDEXCEL FUNCTIONAL SKILLS PILOT. Maths Level 1. Chapter 5. Working with shape and space
EDEXCEL FUNCTIONAL SKILLS PILOT Maths Level 1 Chapter 5 Working with shape and space SECTION H 1 Calculating perimeter 86 2 Calculating area 87 3 Calculating volume 89 4 Angles 91 5 Line symmetry 92 6
More informationCircle Theorems. Angles at the circumference are equal. The angle in a semicircle is x The angle at the centre. Cyclic Quadrilateral
The angle in a semicircle is 90 0 Angles at the circumference are equal. A B They must come from the same arc. Look out for a diameter. 2x Cyclic Quadrilateral Opposite angles add up to 180 0 A They must
More information24HourAnswers.com. Online Homework. Focused Exercises for Math SAT. Skill Set 12: Geometry  2D Figures and 3D Solids
24HourAnsers.com Online Homeork Focused Exercises for Math SAT Skill Set 12: Geometry  2D Figures and 3D Solids Many of the problems in this exercise set came from The College Board, riters of the SAT
More informationShape Dictionary YR to Y6
Shape Dictionary YR to Y6 Guidance Notes The terms in this dictionary are taken from the booklet Mathematical Vocabulary produced by the National Numeracy Strategy. Children need to understand and use
More information1. Kyle stacks 30 sheets of paper as shown to the right. Each sheet weighs about 5 g. How can you find the weight of the whole stack?
Prisms and Cylinders Answer Key Vocabulary: cylinder, height (of a cylinder or prism), prism, volume Prior Knowledge Questions (Do these BEFORE using the Gizmo.) [Note: The purpose of these questions is
More informationMath 115 Extra Problems for 5.5
Math 115 Extra Problems for 5.5 1. The sum of two positive numbers is 48. What is the smallest possible value of the sum of their squares? Solution. Let x and y denote the two numbers, so that x + y 48.
More informationRevision Notes Adult Numeracy Level 2
Revision Notes Adult Numeracy Level 2 Place Value The use of place value from earlier levels applies but is extended to all sizes of numbers. The values of columns are: Millions Hundred thousands Ten thousands
More information1. Determine how much paper Ramiro and Danielle need to make one of each style and size of shade.
Surface Area  Blue Problems Making Lampshades In Exercise 17, use the following information. Ramiro and Danielle are making lampshades using wire and different colors of heavyweight tissue paper. They
More informationNational 4: Expressions and Formulae
Biggar High School Mathematics Department National 4 Pupil Booklet National 4: Expressions and Formulae Learning Intention Success Criteria I can simplify algebraic expressions. I can simplify and carry
More informationAssessing pupils progress in mathematics at Key Stage 3. Year 9 assessment package Shape, space and measures Teacher pack
Assessing pupils progress in mathematics at Key Stage 3 Year 9 assessment package Shape, space and measures Teacher pack Year 9 Shape task: Wrap around and Prismatic Levels ( 3/ )4/5/6 Note that for classes
More informationPerimeter, Area and Volume of Regular Shapes
Perimeter, Area and Volume of Regular Sapes Perimeter of Regular Polygons Perimeter means te total lengt of all sides, or distance around te edge of a polygon. For a polygon wit straigt sides tis is te
More informationAPPLICATION OF DERIVATIVES
6. Overview 6.. Rate of change of quantities For the function y f (x), d (f (x)) represents the rate of change of y with respect to x. dx Thus if s represents the distance and t the time, then ds represents
More informationArea of Parallelograms (pages 546 549)
A Area of Parallelograms (pages 546 549) A parallelogram is a quadrilateral with two pairs of parallel sides. The base is any one of the sides and the height is the shortest distance (the length of a perpendicular
More informationChapter 17. Mensuration of Pyramid
44 Chapter 7 7. Pyramid: A pyramid is a solid whose base is a plane polygon and sides are triangles that meet in a common vertex. The triangular sides are called lateral faces. The common vertex is also
More informationCircles & Pythagoras
National 5 Mathematics Exam Questions by Topic Circles & Pythagoras stepbystep worked solutions included The SQA material contained in this Practice Paper is copyright Scottish Qualifications Authority
More informationPRACTICAL GEOMETRY SYMMETRY AND VISUALISING SOLID SHAPES NCERT
UNIT 12 PRACTICAL GEOMETRY SYMMETRY AND VISUALISING SOLID SHAPES (A) Main Concepts and Results Let a line l and a point P not lying on it be given. By using properties of a transversal and parallel lines,
More informationBegin recognition in EYFS Age related expectation at Y1 (secure use of language)
For more information  http://www.mathsisfun.com/geometry Begin recognition in EYFS Age related expectation at Y1 (secure use of language) shape, flat, curved, straight, round, hollow, solid, vertexvertices
More informationHeight. Right Prism. Dates, assignments, and quizzes subject to change without advance notice.
Name: Period GL UNIT 11: SOLIDS I can define, identify and illustrate the following terms: Face Isometric View Net Edge Polyhedron Volume Vertex Cylinder Hemisphere Cone Cross section Height Pyramid Prism
More informationChapter 6. Volume. Volume by Counting Cubes. Exercise 6 1. 1 cm 3. The volume of a shape is the amount of space it takes up.
Chapter 6 Volume Volume by Counting Cubes The volume of a shape is the amount of space it takes up. 3 The basic unit of volume is the cubic centimetre. A small cube which measures by by is said to have
More informationTerm Definition Example Dimension Size, distance The dimensions of the plate were given in metres. amount an object can hold
Term Definition Example Dimension Size, distance The dimensions of the plate were given in metres. Capacity Volume  the amount an object can hold The capacity of an icecream container is 2 litres. Mass
More informationName: Date: Geometry Solid Geometry. Name: Teacher: Pd:
Name: Date: Geometry 20122013 Solid Geometry Name: Teacher: Pd: Table of Contents DAY 1: SWBAT: Calculate the Volume of Prisms and Cylinders Pgs: 17 HW: Pgs: 810 DAY 2: SWBAT: Calculate the Volume of
More information