EDULABZ INTERNATIONAL PERIMETER AND AREA, VOLUME AND SURFACE AREA OF SOLIDS

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1 18 PERIMETER AND AREA, VOLUME AND SURFACE AREA OF SOLIDS 1. Find the area and the perimeter of the following figures. All angles are 90 and all sides are in cm. (i) (ii) Ans. (i) Perimeter of fig = ( ) cm = 36 cm Area of figure = ( ) cm = (4 + 14) cm = 38 cm (ii) Area of figure = ( ) cm = ( ) = 480 cm Perimeter of figure (ii) = ( ) cm = 10 cm. Find the area of the (i) shaded part (ii) unshaded part of the following figures, given that all adjacent sides are at right angles : 1 m 8 m 16 m m 8 m Ans. (a) Length of outer rectangle = m Breadth of outer rectangle = 16 m 4 m.5m Area of outer rectangle = ( 16) m = 35 m Length of inner rectangle = 1 m Breadth of inner rectangle = 8 m Area of inner rectangle = area of unshaded part = 1 m 8 m = 96 m m ICSE Math Class VII 1 Question Bank

2 4.5m m m 1m 4.5m.5m Area of shaded part = Area of outer rectangle Area of inner rectangle = (35 96) m = 56 m (b) Length of rectangle = 8 m Breadth of rectangle = 15 m Area of rectangle = (8 15)m = 40 m Area of shaded part = ( ) m = ( ) m = m Hence, area of unshaded part = ( ) m = 80.5m 3. Find the area and perimeter of the shaded part in each of the following figures : (i) 9m.5m 1.5m 18m A B (ii) m m m m 3m 3m m m m.5m E F 1.5m G 18m H 9m DJ I C 9m L Ans. (i) Area of the shaded part = Area of rect angle ABCD + Area of rectangle EFKL + Area of rectangle GHIJ. = (9 ) m + [( ).5] m + (18 1) m K 1m = 18 m + (7.5) m + 18 m = ( ) m = 53.5 m Perimeter of the shaded part = AB + BC + CL + LK + KJ + JI + IH + HG + GF + FE + ED + DA ( ) m = 73 m. (ii) By partitioning the given figure into suitable rectangles and squares as shown below, we get the rectangle AFGR, GHIJ, KLMN, OPQR and square BCDE. ICSE Math Class VII Question Bank

3 Q m P M m L I m H R A O m m N K J 3m m m 3m 9m C B Area of the shaded part = Area of rectangle AFGR + Area of rectangle GHIJ + Area of rectangle of KLMN Area of rectangle OPQR + Area of square BCDE E D F G = [( ) 3] m + [(9 3) ] m +[(9 3 ) ] m + [(9 3) ] m + ( ) m = (10 3) m + (6 ) m + (6 ) m + (6 ) m + 4 m = ( ) m = 70 m Perimeter of the shaded part = AB + BC + CD + DE + EF + FH + HI + IJ + JK + KC + LM + MN + ON + OP + PQ + QA = ( + ) m + m+ m+ m + ( + ) m + 9 m + m + (9 3) m + m + (9 3) m + m + (9 3) m+ m + (9 3) m + m + 9 m = ( ) m = 66 m. 4. ABCD is a square of side 4 cm. EF is parallel to BC and AE = 15 cm. By how much does D F C A E B (i) the permieter of AEFD exceed the perimeter of EBCF? (ii) the area of AEFD exceed the area of EBCF? ICSE Math Class VII 3 Question Bank

4 Ans. (i) Perimeter of AEFD = (15 + 4) cm = 39 = 78 cm Perimeter of EBCF = (4 + 9) cm = 33 = 66 cm D 15cm F 9cm C 4cm 4cm 4cm A 15cm Perimeter of AEFD exceeds perimeter of EBCF by = (78 66) cm = 1 cm (ii) Area of AEFD = L B = (15 4) = 360 cm x = 189 ICSE Math Class VII 4 Question Bank E 9cm Area of EBCF = L B = (4 9) = 16 cm Hence, area of AEFD exceeds the area of EBCF by = (360 16) = 144 cm 5. The perimeter of a rectangular plot of land is 40 m and its length is 63m. Find the breadth and area of the plot. Ans. Perimeter of the rectangular plot = 40 m Length of the plot = 63 m Let the breadth of the plot be x m (x + 63) = 40 (length + breadth) = perimeter 40 x + 63 = = 10 x = (10 63) = 57 m Breadth of the plot = 57 m Area of the plot = (length breadth) = (63 57) m = 3591 m 6. The perimeter of a rectangular grassy plot is 189 m and its breadth is 10.5 m. Find the length and area of the plot. Ans. Perimeter of the rectangular plot = 189 m Beadth of the plot = 10.5 m Let the length of the plot be x m (x ) = 189 B

5 x = 94.5 x = = 84 Length of the plot = 84 m Area of the plot = (length breadth) = = 88 m 7. A rectangular graden is 175 m long and 96 m broad. find the cost of fencing it at Rs 1.50 per metre. Also, find the cost of ploughing it at 50 paise per square metre. Ans. Length of the graden = 175 m Breadth of the graden = 96 m Perimeter of the graden = (l + b) = ( ) = (71) = 54 m Length of the wire required for fencing = 54 m Cost of fencing = Rs ( ) = Rs 813 Also area of the graden = (l b) = (175 96) m = m Cost, of ploughing = paise = rupees = Rs Square tiles of side 0 cm are to be laid on the floor of a room 10 m by 4.5 m. How many tiles will be needed? Find the cost of putting the tiles at Rs 1.40 per tile. Ans. Length of the floor = 10 m = (10 100) cm = 1000 cm Breadth of the floor = 4.5 m = ( ) cm = 450 cm Area of the floor ( ) cm Area of one tile = (0 0) cm Number of tiles needed = Area of thefloor Area of one t ile = = 5 45 = Cost of putting the tile = Rs ( ) = Rs The perimeter of a rectangular field is 151 m. If its breadth is 3 m, find its length and area. Ans. Perimeter of rectangular field = 151 m, Breadth = 3 m Let length = x m (x + 3) = 151 x + 64 = 151 x = = x = = 43.5 Length of the rectangular field = 43.5 m Area = L B = (43.5 3) m = 139 m ICSE Math Class VII 5 Question Bank

6 10. The area of a rectangular plot is 340 m and its breaeth is 17 m. Find the cost of surrounding the plot with a fence at Rs 5.70 per metre. Ans. Area of rectangular plot = 340 m, Breadth = 17 m Area 340 Length = = = 0 m Breadth 17 Perimeter = (L + B) = (0 + 17) m = ( 37) = 74 m Cost of fencing 1 metre = Rs 5.70 Cost of fencing 74 metre = ( ) = Rs A rectangular room is 10 m long and 7.5 m wide. Find the cost of covering the floor with carpet 1.5 m wide at Rs 5 per metre. Ans. Length of room = 10 m Breadth of room = 7.5 m Area of room = (10 7.5) = 75 m Width of carpet = 1.5 m 75 Length of carpet = = m Cost of 1 metre = Rs 5 Cost of 60 metre of carpet = 60 Rs 5 = Rs Find the cost of flooring a room 6.5 m by 5 m with square tiles of side 5 cm at the rate of Rs per tile. Ans. Area of room = L B = = 3.5 m Area of square tile = 5 cm 5 cm Number of tiles required = 65 = 65 cm = m Area of room Area of tile = = = Cost of 1 tile = Rs 4.70 Cost of 50 tiles = ( ) = Rs The length of the base of a triangle is 1 cm and its area is 108 cm ; find the height of the triangle. If the height of this triangle is halved and the length of the base is doubled then find : (i) area of the new triangle; (ii) increase or decrease in area of the triangle ICSE Math Class VII 6 Question Bank

7 Ans. Length of base of a triangle (b) = 1 cm and its area = 108 cm Let h be the height of the triangle We know that 1 bh = Area of triangle 1 1 h = 108 6h = 108 h = = Thus, height of the triangle = 18 cm As per condition Height of triangle = 1 18 cm = 9 cm and base = 1 = 4 cm (i) area of new triangle = 1 base height = = 108 cm (ii) there is no change in the area of the triangle 14. The area of a triangular field is 34 m and its base is 18 m. Find the corresponding height (length of altitude) of the triangle. If this triangular field is exchanged with a rectangular field of the same area and of length 4 ; what is : (i) the area of this rectangular field? (ii) the breadth of it? (iii) the perimeter of the rectangular field? Ans.Case I Area of triangular field = 34 m, Base (b) = 18 m Let length of its height be h Then, Area of triangle = 1 base height 1 18 h = 34 9h = h = = 36 9 Height = 36 m Case II (i) Area of rectangular field = 34 m (ii) Length of the field (l) = 4 m ICSE Math Class VII 7 Question Bank

8 m then breadth (b) = Area 34 7 = = m = 13.5 m. Length 4 (iii) Perimeter of the rectangular field = (l + b) = ( ) m = 37.5 = 75 m 15. The adjoining figure shows a square ABCD and a triangle ABE. If each side of the square is 4 cm, find : (i) the area of the square ABCD ; (ii) the area of the triangle ABE ; (iii) the area of the unshaded protion of the figure ; (iv) ratio between the areas of the square ABCD and the triangle ABE. Ans.In square ABCD. (i) Its each side (a) = 4 cm Area of the square = side a = (4) cm = 576 cm (ii) Base of the triangle ABE = AB = 4 cm and height (altitude h = AD = 4 cm A D Area of triangle ABE = 1 bh = = 88 cm (iii) Area of unshaded portion = Area of square Area of shaded triangle = = 88 cm E (iv) Ratio between the area of square ABCD and ABE = 576 : 88 = : A room is 10 m long and 6 m broad. It is surrounded by a verandah which is m wide all around it. Find the cost of flooring the verandah with marble at Rs 36 per m. Ans. Length of room (l) = 10 m Breadth (b) = 6 m Width of verandah = m Outer length (L) = 10 + m = = 14 m C B 10m m 14m 6m 10m ICSE Math Class VII 8 Question Bank

9 Outer width (B) = 6 + m = = 10 m Thus, Area of verandah = L B l b = = = 80 m Cost of 1 m = Rs 36 Hence, total cost = Rs = Rs A hall is 16 m long and 1 m broad. Find the cost of carpeting it at Rs. 15 per m, after leaving a margin of 1 metre all around. Ans. Length of Hall (L) = 16 m, width (B) = 1 m Width of margin left = 1 m Inner length (l) = 16 1 = 16 = 14 m Inner width = 1 1 = 1 = 10 m Thus, inner area leaving margin = l b = = 140 m Rate of carpeting = Rs 15 per m Hence, total cost = Rs = Rs A path 3 m wide is running along the inside of the boundary of a rectangular field 116 m by 76 m. How much money is needed to gravel the path at Rs.50 per m? Ans. Length of field (L) = 116 m and breadth of field (B) = 76 m Width of Path (inside) = 3 m Inner length (l) = = = 110 m Linner breadth (b) = 76 3 = 76 6 = 70 m Thus, area of path L B l b = = ( ) m = 1116 m Rate of gravelling the path = Rs.50 per m Hence, total cost Rs = Rs A path.5 m wide is running around a rectangular grassy plot 40 m by 35 m. Find the area of the path and the money needed for tilling it at Rs per m. Ans. Length of plot of (l) = 35 m and breadth of plot (b) = 35 m width of path =.5 m Thus, outer length (L) = = = 45 m and outer breadth (B) = = = 40 m Thus, area of path = L B l b = ( ) m = ( ) = 400 m Rate of tilling it = Ra 5.60 per m ICSE Math Class VII 9 Question Bank

10 Hence, total cost = Rs = Rs A rectangular lawn 115 m long and 65 m broad has two cross-paths at right angles, one m wide running parallel to its length and the other.5 wide running parallel to its breadth. Find the cost of gravelling the paths at Rs. 14 per m. Ans. Length of lawn (l) = 115 m Breadth of lawn (b) = 64 m Width of length wise path = m and width of breadth wise path =.5 m m 64m Area of path = m = = = 385 m Rate of gravelling = Ra 14 per m Hence, total cost = Rs = Rs The central hall of a school is m long and 15. m wide. A carpet is to be laid on the floor leaving a strip of 75 cm width frm lthe walls uncovered. Find the area of the carpet and the area of the strip left uncovered. Ans. Length of Hall (L) = m and breadth of hall (B) = 15.5 m Width of strip = 75 cm = 3 4 m Inner length (l) = 3 = 1.5 = 0.5 m 4 and inner breadth (b) = = 14 m Thus, area of strip = L B l b = ( ) m = (341 87) = 54 m and area of inner part = area of the carpet = l b = = 87 m. The floor of a rectangular hall has a perimeter 36 m. Its height is 4.5 m. Find the cost of painting its four walls (doors and windows be ignored) at the rate of Rs per square metre. Ans. Perimeter = (l + b) = 36 m Area of four walls = (l + b) h = = 106 m Cost of painting l m = Rs 8.40 Cost of painting 106 m = Rs = Rs A rectangular fish tank has a length of 30 cm, a breadth of 0 cm and a height of 0 cm. tank is placed on a horizontal table and it is three-quarters full of water. Find the area of the tank which is in contact with water. Ans. Length of tank = 30 cm ICSE Math Class VII 10 Question Bank.5m

11 Breadth of tank = 0 cm Height of tank = 0 cm, As the tank is three-quarters full of water Height of water in the tank = 0 3 = 15cm 4 Area of the tank in contact with the water = Area of floor of tank + Area of 4 walls upto 15 cm. = (30 + 0) 15 = = ( ) = 100 cm 4. If cost Rs. 936 to fence a square field at Rs per metre. Find the area of the square field. Ans. Total cost = Rs 936, Rate per metre = Rs 7.80 Perimeter of field = Total cost of fencing Rate per metre 936 = = m. Perimeter of square field = 4 side = 10 m Thus, side = = m Hence, Area of square field = 30m 30 m = 900 m 5. A person walks at 3km/hr. How long will he take to go round a square gorund 5 times, the area of which being 05m? Ans. Area of square ground = 05 m (side) = 05 m = (45m) side = 45 m Perimeter of ground = 45 4 = 180 m. To go 5 rounds, total distance = = 900 m. Speed of person = 3 km/hr. Time taken to walk 3 km = 1 hr Time taken to walk to walk 900 m = = 18 minutes Water is to be transferred from one tank to the other tank of dimensions.1 m 1.5 m 0.6 m such that the second tank is completely filled with this water the length and breadth of the first tank are m and 1 m respectively. Find the height of the water when it was in the first tank. ICSE Math Class VII 11 Question Bank

12 Ans. Volume of water in the second tank =.1 m 1.5 m 0.6 m = 1.89 m 3 Volume of the water in the first tank = 1.89m 3 Length (l) = m and width (b) = 1 m Let h be the hight of the water, then 1 h = 1.89 h = 1.89 m Hence, height = 1.89 m 7. How many bricks, each measuring 18 cm 1 cm 6 cm, will be needed to build a wall of length = 6 m, width = 4 cm and height = 3.6 m? Ans. Measure of each brick = 18 cm 1 cm 6 cm Volume of each brick = 196 cm 3 Length of wall (l) = 6 m = 600 cm width (b) = 4 cm and height (h) = 3.6 m = 360 cm Volume of wall = l b h = cm 3 Number of bricks used in the wall = Volume of wall Volume of one brick = = Find the area of four walls of a room whose length = 3. m, breadth =.5 m and height = 3m Ans. Length of the room (l) = 3. m, breadth (b) =.5 m and height (h) = 3m Area of 4 walls = (l b) h = (3. +.5) 3m = m = 34. m 9. A rectangular water reservoir contains 4000 litres of water. Find the depth of the water in the reservoir if its base measures 6 m by 3.5 m. Ans. Volume of reservoir = = m [ 1000 litres = 1 m 3 ] Area = l b = 6 m 3.5 m Height of reservoir = Volume 4 = Area = m Depth of reservoir = m. ICSE Math Class VII 1 Question Bank

13 30. A cuboidal tea packet measures 10 cm by 6 cm by 4 cm. How many such tea packets can be placed in a cardboard box of dimensions 50 cm by 30 cm by 0. m? Ans. Volume of 1 cuboidal tea packet = 10 cm 6 cm 4 cm = 40 cm 3 Volume of cardboard box = 50 cm 30 cm 0. m = 50 cm 30 cm 0 cm = cm 3 Number of tea packets = Volume of Box Volume of 1 tea packet = = Number of tea packet = The dimensions of a rectangular tin oil are 6 cm by 6 cm by 45 cm. Find the area of the tin sheet required for making 40 such tins. If 1 square metre of the tin sheet costs Rs 0, find the cost of the tin sheet used for these 40 tins. Ans. Dimensions of rectangular tin oil = 6 cm 6 cm 45 cm Area of tin sheet = lb + (l + b) h = ( 6 6) + (6 + 6) 45 = ( ) cm = 603 cm Area of tin sheet for these 40 tins = 603 cm 40 = 4180 cm Cost of 1sq m of tin sheet = Rs 0 Rs Cost of 4180 sq cm of tin sheet = = Rs The volume of a cuboid is 144 cm 3, its height is 4 cm and the base is a square. Find (i) a side of the square base (ii) surface area of the cuboid. Ans. (i) Volume of cuboid= 144 cm 3 height of cuboid = 4 cm Since base is a square length and breadth are equal side height = volume or side = volume height 144 = = = Hence, side of square = 6 cm (ii) surface area of cuboid = (lb + lh + bh) ICSE Math Class VII 13 Question Bank = ( ) = ( ) = 84 = 168 cm (i)

14 33. The external dimensions of a closed rectangular wooden box are 94 cm by 68 cm by 0.46 m. If the wood is cm thick all around. Find. (i) the internal dimensions of the box. (ii) the capacity of the box (iii) the volume of the wood used in making the box. Ans. (i) External dimensions of wooden box = 94 cm 68 cm 0.46 m = 94 cm 68 cm 46 cm Thickness of wood = cm Deducting thickness of wood of both sides i.e. = 4 cm from the external dimensions, the internal dimensions of the box are (94 4) cm by (68 4) cm by (46 4) i.e. 90 cm by 64 cm by 4 cm Capacity of the box = Volume of internal box = 90 cm 64 cm 4 cm = 4190 cm 3 Volume of outer wooden box 94 cm 68 cm 46 cm = 9403 cm 3 Hence, Volume of wood used in making the box = Volume of outer box Volume of inner box = ( ) cm 3 = 511 cm A rectangular room is 6 m long 5 m wide and 3.5 m hights. It has one door 1 m by m and two windows 1.5 m m each. Find the cost of white washing the walls and the ceiling of the room at the rate of Rs 4.60 per square metre. Ans. Length (l) = 6 m Breadth (b) = 5 m Height of (h) = 3.5 m Area of ceiling = 6 5 = 30 m Area of four walls = ( (6 + 5) 3.5) m = 77 m Area of one door = 1 m m = m Area of two windows = ( 1.5 ) = 6 m Area to be white washed = Area of ceiling + Area of 4 walls Area of one door Area of windows = = 99 m Cost of white washing 1 sq. m = Rs 4.60 Hence, Cost of white washing 99 sq. m = Rs = Rs ICSE Math Class VII 14 Question Bank

15 35. If the total surface area of a cubical box is 16 cm find the length of its one side. Also, find the area of four walls. Total surface area of the box = 16 cm Total surface area of a cube = 6(side) 16 6 side = 16 side = 36 (6) 6 = = Therefore, side = 6 Area of four walls of cubical box = 4 (side) = 4 (6) = 4 36 cm = 144 cm 36. If the total surface area of a cube is 96 cm, find the length of its each side and its volume. Ans. Total surface area of a cube = 96 cm Let a be its each side, then 6a = a = = 16 6 a = (4) a = 4 cm Volume of cube = (side) 3 = (4) 3 cm 3 = 64 cm If the rainfall on a certain day was 3.5 cm, how many litres of water fell on 1 hectare land on that day? Ans. Area of land = 1 hectare = = m l b = m and height = rainfall = 3.5 cm = m = m volume of water that fall on, 1 hectare of land = ( ) m 3 = 350 m 3 = litres = litres. 38. What is the weight of a cubical block of ice edge 60 cm in length, if one cubic metre of ice weighs 900 kilograms? Ans. One cubic metre of ice weighs = 900 kg, edge of block = 60 cm. Volume of cubical block = (edge) 3 = (60 cm) 3 60 = 100 m = (.6.6.6) m 3 =.16 m 3 Weight of ice cube = kg = kg. 39. A rectangular pit 1.4 m long, 90 cm broad and 70 cm deep was dugs. and 1000 bricks each of base 1 cm by 10.5 cm were made from the earth dug out. Find the height of each brick. ICSE Math Class VII 15 Question Bank 3

16 Ans. Length of rectangular pit = 1.4 m = 140 cm Breadth = 90 cm Depth = 70 cm Volume of cuboidal pit = ( ) cm 3 = cm Bricks were made out of it So,volume of 1000 bricks = cm 3 Volume of brick = cm3 = 88 cm 3 Length of brick = 1 cm. Breadth of brick = 10.5 cm 88 Height of brick = cm = 4 cm How many wooden cubes of an edge 5 cm in length can be cut out from a cuboidal log of wood of size.5 m by 0.5 m by 50 cm, assuming that there is no wastage? Ans. Volume of cuboidal log of wood =.5 m 0.5 m 50 cm = 50 cm 50 cm 50 cm = cm 3 Edge of wooden cube = 5 cm Volume of one wooden cube = (5 cm) 3 Number of wooden cubes = = cm 3 Volume of log of wood Volume of 1wooden cube = = 41. How many soap cakes can be placed in a box of size 56 cm by 40 cm by 5 cm, if the size of a soap cake is 7 cm by 5 cm by.5 cm? Ans. Volume of box = 56 cm 40 cm 5 cm = cm 3 size of a soap cake is 7cm by 5 cm by.5 cm volume of soap cake = cm 3 = 87.5 cm 3 Number of soap cakes = Volume of box Volume of 1 soap cake = = ICSE Math Class VII 16 Question Bank

17 4. A room is 4.8 m long, 3.6 m broad and.4 m high. Find the cost of laying tiles on its floor and on its four walls at the rate of Rs. 80 per m. Ans. Length of room (l) = 4.8 m, Breadth (b) = 3.6 m and height (h) =.4 m Area of floor = l b = = 17.8 m Area of four walls = (l + b) h = ( ).4 m = m = 40.3 m Total area = = 57.6 m Rate of laying tiles = Rs 80 per m Hence, total cost = Rs = Rs The inside of a room has a square base of side 3.6 m. The inside height of the room is 4 m. Find : (i) the space (internal volume) of the room. (ii) how many boxes each of dimension 0.9 m 40 cm 5 cm can be placed in the room? Ans.Each side of square base of room = 3.6 m and inside height (h) = 4 m (i) Internal volume of the room = Area of base height = (3.6) 4 = m 3 = m 3 (ii) Dimensions of each box = 0.9 m 40 cm 5 cm = 0.9 m 0.4 m 0.5 m = 0.09 m 3 Number of boxes to be placed in the room = Volumeof room Volume of each box = = 100 = How many children can be accommodated in a hall of length 16 m, breadth 1.5 m and height 1.45 m assuming 3.6 m 3 of air is required for each child? Ans. Length of the hall (l) = 16 m, breadth (b) = 1.5 m height (h) = 4.5 m Volume of the hall = l b h = ( ) m 3 = 900 m 3 Volume of air required for one child = 3.6 m 3 Hence, number of children can be accommodated in the hall = = 10 = 50 ICSE Math Class VII Question Bank

18 45. A path of length 35 m and width 30 m is laid with concreate upto a depth of 40 cm. Find the cost of laying the concrete at Rs.0 per cubic metre. Ans. Length of the path = 35 m, width of the path = 30 m and depth of the path = 40 cm = m = 0.4 m Volume of the path = l b h = ( ) m3 = 3900 m 3 Cost of laying the concrete of 1 m 3 = Rs 0 Cost of laying the concrete of 3900 m 3 = Rs = Rs Water is to be transferred from a tank of length 3 m and breadth 80 cm to another tank of dimensions.4m by 1.5m by 60 cm to fill it completely. What was the height of the water in the first tank? Ans. Length of the first tank (l) = 3 m Breadth of the first tank (b) = 80 cm = 0.8 m Let h be the hight of water in the first tank Hence, volume of water in the first tank = l b h = h m 3 =.4h m 3 Volume of water in the other tank =.16 m 3.4h =.16 =.4 m 1.5 m 0.6 m ( 60 cm = m = 0.6) h = = m = m = 0.9 m = cm = 90 cm Hence, height of water in the first tank is 90 cm. 47. There cubes each of side 8 m are joined together side by side to form a cuboid. Find the surface area of the resulting cuboid. Ans. Side of each cube = 8 cm By joining three cubes, a cuboid is formed Length of cuboid (l) = 8 3 = 4 cm breadth of cuboid (b) = 8 cm and height of coboid (h) = 8 cm Surface area of cuboid = (lb + bh + hl) = ( ) cm = ( ) cm = 448 = 896 cm ICSE Math Class VII 18 Question Bank

19 48. A cuboidal block of 6 cm 9 cm 1 cm is cut up into exact number of equal cubes. Find the least possible number of cubes. Ans. Length of cuboid (l) = 1 cm Breadth (b) = 9 cm Height (h) = 6 cm HCF of 1, 9, 6 = 3 cm Hence, side of each cube = 3 cm Volume of cuboid = l b h = cm 3 = 648 cm 3 and volume of each cube = (side) 3 = (3) 3 cm 3 = 7 cm 3 Hence, number of cube to be cut out = Volume of cuboid Volume of cube = = 49. How many cubes of 3 cm edge can be cut out of a cube of 18 cm edge? Ans. Edge of big cube = 18 cm Volume of big cube = cm 3 Edge of small cube = 3 cm Volume of small cube = cm 3 Number of total cubes cut out = = = ICSE Math Class VII 19 Question Bank

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