VOLUME AND SURFACE AREAS OF SOLIDS


 Chad Phelps
 7 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
CHAPTER 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationCSU Fresno Problem Solving Session. Geometry, 17 March 2012
CSU Fresno Problem Solving Session Problem Solving Sessions website: http://zimmer.csufresno.edu/ mnogin/mfdprep.html Math Field Day date: Saturday, April 21, 2012 Math Field Day website: http://www.csufresno.edu/math/news
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 informationBasic Math for the Small Public Water Systems Operator
Basic Math for the Small Public Water Systems Operator Small Public Water Systems Technology Assistance Center Penn State Harrisburg Introduction Area In this module we will learn how to calculate the
More informationPythagoras Theorem. Page I can... 1... identify and label rightangled triangles. 2... explain Pythagoras Theorem. 4... calculate the hypotenuse
Pythagoras Theorem Page I can... 1... identify and label rightangled triangles 2... eplain Pythagoras Theorem 4... calculate the hypotenuse 5... calculate a shorter side 6... determine whether a triangle
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 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 informationCalculating the Surface Area of a Cylinder
Calculating the Measurement Calculating The Surface Area of a Cylinder PRESENTED BY CANADA GOOSE Mathematics, Grade 8 Introduction Welcome to today s topic Parts of Presentation, questions, Q&A Housekeeping
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 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 informationSandia High School Geometry Second Semester FINAL EXAM. Mark the letter to the single, correct (or most accurate) answer to each problem.
Sandia High School Geometry Second Semester FINL EXM Name: Mark the letter to the single, correct (or most accurate) answer to each problem.. What is the value of in the triangle on the right?.. 6. D.
More informationGeometry Unit 6 Areas and Perimeters
Geometry Unit 6 Areas and Perimeters Name Lesson 8.1: Areas of Rectangle (and Square) and Parallelograms How do we measure areas? Area is measured in square units. The type of the square unit you choose
More informationHiker. A hiker sets off at 10am and walks at a steady speed for 2 hours due north, then turns and walks for a further 5 hours due west.
Hiker A hiker sets off at 10am and walks at a steady speed for hours due north, then turns and walks for a further 5 hours due west. If he continues at the same speed, what s the earliest time he could
More informationGeometry and Measurement
The student will be able to: Geometry and Measurement 1. Demonstrate an understanding of the principles of geometry and measurement and operations using measurements Use the US system of measurement for
More information1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?
CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 33, with
More informationMathematical Modeling and Optimization Problems Answers
MATH& 141 Mathematical Modeling and Optimization Problems Answers 1. You are designing a rectangular poster which is to have 150 square inches of tet with inch margins at the top and bottom of the poster
More informationPlatonic Solids. Some solids have curved surfaces or a mix of curved and flat surfaces (so they aren't polyhedra). Examples:
Solid Geometry Solid Geometry is the geometry of threedimensional space, the kind of space we live in. Three Dimensions It is called threedimensional or 3D because there are three dimensions: width,
More informationSurface Area Quick Review: CH 5
I hope you had an exceptional Christmas Break.. Now it's time to learn some more math!! :) Surface Area Quick Review: CH 5 Find the surface area of each of these shapes: 8 cm 12 cm 4cm 11 cm 7 cm Find
More informationMECHANICS OF SOLIDS  BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.
MECHANICS OF SOLIDS  BEAMS TUTOIAL 1 STESSES IN BEAMS DUE TO BENDING This is the first tutorial on bending of beams designed for anyone wishing to study it at a fairly advanced level. You should judge
More informationWednesday 15 January 2014 Morning Time: 2 hours
Write your name here Surname Other names Pearson Edexcel Certificate Pearson Edexcel International GCSE Mathematics A Paper 4H Centre Number Wednesday 15 January 2014 Morning Time: 2 hours Candidate Number
More information2. A painted 2 x 2 x 2 cube is cut into 8 unit cubes. What fraction of the total surface area of the 8 small cubes is painted?
Black Surface Area and Volume (Note: when converting between length, volume, and mass, 1 cm 3 equals 1 ml 3, and 1 ml 3 equals 1 gram) 1. A rectangular container, 25 cm long, 18 cm wide, and 10 cm high,
More informationMaximum and minimum problems. Information sheet. Think about
Maximum and minimum problems This activity is about using graphs to solve some maximum and minimum problems which occur in industry and in working life. The graphs can be drawn using a graphic calculator
More informationLesson 22. Circumference and Area of a Circle. Circumference. Chapter 2: Perimeter, Area & Volume. Radius and Diameter. Name of Lecturer: Mr. J.
Lesson 22 Chapter 2: Perimeter, Area & Volume Circumference and Area of a Circle Circumference The distance around the edge of a circle (or any curvy shape). It is a kind of perimeter. Radius and Diameter
More informationMath 0306 Final Exam Review
Math 006 Final Exam Review Problem Section Answers Whole Numbers 1. According to the 1990 census, the population of Nebraska is 1,8,8, the population of Nevada is 1,01,8, the population of New Hampshire
More information121 Representations of ThreeDimensional Figures
Connect the dots on the isometric dot paper to represent the edges of the solid. Shade the tops of 121 Representations of ThreeDimensional Figures Use isometric dot paper to sketch each prism. 1. triangular
More information12 Surface Area and Volume
12 Surface Area and Volume 12.1 ThreeDimensional Figures 12.2 Surface Areas of Prisms and Cylinders 12.3 Surface Areas of Pyramids and Cones 12.4 Volumes of Prisms and Cylinders 12.5 Volumes of Pyramids
More information43 Perimeter and Area
43 Perimeter and Area Perimeters of figures are encountered in real life situations. For example, one might want to know what length of fence will enclose a rectangular field. In this section we will study
More informationME 111: Engineering Drawing
ME 111: Engineering Drawing Lecture # 14 (10/10/2011) Development of Surfaces http://www.iitg.ernet.in/arindam.dey/me111.htm http://www.iitg.ernet.in/rkbc/me111.htm http://shilloi.iitg.ernet.in/~psr/ Indian
More informationAlgebra Geometry Glossary. 90 angle
lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example:
More informationTeacher Page Key. Geometry / Day # 13 Composite Figures 45 Min.
Teacher Page Key Geometry / Day # 13 Composite Figures 45 Min. 91.G.1. Find the area and perimeter of a geometric figure composed of a combination of two or more rectangles, triangles, and/or semicircles
More informationShow that when a circle is inscribed inside a square the diameter of the circle is the same length as the side of the square.
Week & Day Week 6 Day 1 Concept/Skill Perimeter of a square when given the radius of an inscribed circle Standard 7.MG:2.1 Use formulas routinely for finding the perimeter and area of basic twodimensional
More informationFinding Volume of Rectangular Prisms
MA.FL.7.G.2.1 Justify and apply formulas for surface area and volume of pyramids, prisms, cylinders, and cones. MA.7.G.2.2 Use formulas to find surface areas and volume of threedimensional composite shapes.
More informationCHAPTER 8, GEOMETRY. 4. A circular cylinder has a circumference of 33 in. Use 22 as the approximate value of π and find the radius of this cylinder.
TEST A CHAPTER 8, GEOMETRY 1. A rectangular plot of ground is to be enclosed with 180 yd of fencing. If the plot is twice as long as it is wide, what are its dimensions? 2. A 4 cm by 6 cm rectangle has
More informationCylinder Volume Lesson Plan
Cylinder Volume Lesson Plan Concept/principle to be demonstrated: This lesson will demonstrate the relationship between the diameter of a circle and its circumference, and impact on area. The simplest
More informationFCAT FLORIDA COMPREHENSIVE ASSESSMENT TEST. Mathematics Reference Sheets. Copyright Statement for this Assessment and Evaluation Services Publication
FCAT FLORIDA COMPREHENSIVE ASSESSMENT TEST Mathematics Reference Sheets Copyright Statement for this Assessment and Evaluation Services Publication Authorization for reproduction of this document is hereby
More informationMCB4UW Optimization Problems Handout 4.6
MCB4UW Optimization Problems Handout 4.6 1. A rectangular field along a straight river is to be divided into smaller fields by one fence parallel to the river and 4 fences perpendicular to the river. Find
More informationDŵr y Felin Comprehensive School. Perimeter, Area and Volume Methodology Booklet
Dŵr y Felin Comprehensive School Perimeter, Area and Volume Methodology Booklet Perimeter, Area & Volume Perimeters, Area & Volume are key concepts within the Shape & Space aspect of Mathematics. Pupils
More informationThe GED math test gives you a page of math formulas that
Math Smart 643 The GED Math Formulas The GED math test gives you a page of math formulas that you can use on the test, but just seeing the formulas doesn t do you any good. The important thing is understanding
More informationWhat You ll Learn. Why It s Important
These students are setting up a tent. How do the students know how to set up the tent? How is the shape of the tent created? How could students find the amount of material needed to make the tent? Why
More informationUnit 15: Measurement Length, Area and Volume
Unit 15: Measurement Length, Area and Volume Section A: Circumference measuring the circumference of a round object. You will to show all of your working out. C 5 d where: C 5 circumference 5 3.14 d 5
More informationEDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES. Maths Level 2. Chapter 5. Shape and space
Shape and space 5 EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES Maths Level 2 Chapter 5 Shape and space SECTION H 1 Perimeter 2 Area 3 Volume 4 2D Representations of 3D Objects 5 Remember what you
More informationWhat You ll Learn. Why It s Important
What is a circle? Where do you see circles? What do you know about a circle? What might be useful to know about a circle? What You ll Learn Measure the radius, diameter, and circumference of a circle.
More informationWEIGHTS AND MEASURES. Linear Measure. 1 Foot12 inches. 1 Yard 3 feet  36 inches. 1 Rod 5 1/2 yards  16 1/2 feet
WEIGHTS AND MEASURES Linear Measure 1 Foot12 inches 1 Yard 3 feet  36 inches 1 Rod 5 1/2 yards  16 1/2 feet 1 Furlong 40 rods  220 yards  660 feet 1 Mile 8 furlongs  320 rods  1,760 yards 5,280 feet
More informationPostulate 17 The area of a square is the square of the length of a. Postulate 18 If two figures are congruent, then they have the same.
Chapter 11: Areas of Plane Figures (page 422) 111: Areas of Rectangles (page 423) Rectangle Rectangular Region Area is measured in units. Postulate 17 The area of a square is the square of the length
More informationVolume of a Cylinder
Volume of a Cylinder Focus on After this lesson, you will be able to φ determine the volume of a cylinder How much water do you use? You might be surprised. The water storage tank shown has a height of
More informationPIZZA! PIZZA! TEACHER S GUIDE and ANSWER KEY
PIZZA! PIZZA! TEACHER S GUIDE and ANSWER KEY The Student Handout is page 11. Give this page to students as a separate sheet. Area of Circles and Squares Circumference and Perimeters Volume of Cylinders
More informationCharacteristics of the Four Main Geometrical Figures
Math 40 9.7 & 9.8: The Big Four Square, Rectangle, Triangle, Circle Pre Algebra We will be focusing our attention on the formulas for the area and perimeter of a square, rectangle, triangle, and a circle.
More informationThe teacher gives the student a ruler, shows her the shape below and asks the student to calculate the shape s area.
Complex area Georgia is able to calculate the area of a complex shape by mentally separating the shape into familiar shapes. She is able to use her knowledge of the formula for the area of a rectangle
More informationChapter 4: Area, Perimeter, and Volume. Geometry Assessments
Chapter 4: Area, Perimeter, and Volume Geometry Assessments Area, Perimeter, and Volume Introduction The performance tasks in this chapter focus on applying the properties of triangles and polygons to
More informationFilling and Wrapping: Homework Examples from ACE
Filling and Wrapping: Homework Examples from ACE Investigation 1: Building Smart Boxes: Rectangular Prisms, ACE #3 Investigation 2: Polygonal Prisms, ACE #12 Investigation 3: Area and Circumference of
More informationMeasurement Length, Area and Volume
Length, Area and Volume Data from international studies consistently indicate that students are weaker in the area of measurement than any other topic in the mathematics curriculum Thompson & Preston,
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Santa Monica College COMPASS Geometry Sample Test MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the area of the shaded region. 1) 5 yd 6 yd
More informationArea and Circumference
4.4 Area and Circumference 4.4 OBJECTIVES 1. Use p to find the circumference of a circle 2. Use p to find the area of a circle 3. Find the area of a parallelogram 4. Find the area of a triangle 5. Convert
More informationAngles that are between parallel lines, but on opposite sides of a transversal.
GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,
More informationMECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA. Define and calculate 1st. moments of areas. Define and calculate 2nd moments of areas.
MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA The concepts of first and second moments of area fundamental to several areas of engineering including solid mechanics and fluid mechanics. Students who are
More informationChapter 8 Geometry We will discuss following concepts in this chapter.
Mat College Mathematics Updated on Nov 5, 009 Chapter 8 Geometry We will discuss following concepts in this chapter. Two Dimensional Geometry: Straight lines (parallel and perpendicular), Rays, Angles
More informationA Resource for Freestanding Mathematics Qualifications
To find a maximum or minimum: Find an expression for the quantity you are trying to maximise/minimise (y say) in terms of one other variable (x). dy Find an expression for and put it equal to 0. Solve
More informationACTIVITY: Finding a Formula Experimentally. Work with a partner. Use a paper cup that is shaped like a cone.
8. Volumes of Cones How can you find the volume of a cone? You already know how the volume of a pyramid relates to the volume of a prism. In this activity, you will discover how the volume of a cone relates
More informationMultiply circumference by 0.3183. Or divide circumference by 3.1416. Multiply diameter by 3.1416. Or divide diameter by 0.3183.
RULES RELATIVE TO THE CIRCLE TO FIND DIAMETER TO FIND CIRCUMFERENCE TO FIND RADIUS TO FIND SIDE OF AN INSCRIBED SQUARE TO FIND SIDE OF AN EQUAL SQUARE Multiply circumference by 0.383. Or divide circumference
More informationAttachment G1: Pit Latrine Diagram. Fig E.1a: Pit Latrine. Fig E.1b: Plan View of Twin Pits
Attachment G1: Pit Latrine Diagram Fig E.1a: Pit Latrine Fig E.1b: Plan View of Twin Pits Fig E.1c: Section of a watersealed pan Fig E.1d: 3D view of Overflow Pipe Fig E.1e: 2D view of Overflow Pipe
More informationFinding Areas of Shapes
Baking Math Learning Centre Finding Areas of Shapes Bakers often need to know the area of a shape in order to plan their work. A few formulas are required to find area. First, some vocabulary: Diameter
More informationGAP CLOSING. 2D Measurement GAP CLOSING. Intermeditate / Senior Facilitator s Guide. 2D Measurement
GAP CLOSING 2D Measurement GAP CLOSING 2D Measurement Intermeditate / Senior Facilitator s Guide 2D Measurement Diagnostic...4 Administer the diagnostic...4 Using diagnostic results to personalize interventions...4
More informationChapter 22: Electric Flux and Gauss s Law
22.1 ntroduction We have seen in chapter 21 that determining the electric field of a continuous charge distribution can become very complicated for some charge distributions. t would be desirable if we
More information