Transcript
PERIMETER AND AREA, VOLUME AND SURFACE AREA OF SOLIDS
18
1. Find the area and the perimeter of the following figures. All angles are 90° and all sides are in cm. 2 16
10
(i)
(ii)
3 8
Z B
16
10
40
L A A N L IO U AT D N © E ER T IN
Ans. (i) Perimeter of fig = (10 + 8 + 3 + 6 + 7 + 2) cm = 36 cm Area of figure = (8 × 3 + 7 × 2) cm2 = (24 + 14) cm2 = 38 cm2 (ii) Area of figure = (40 × 10 + 10 × 8) cm2 = (400 + 80) = 480 cm2 Perimeter of figure (ii) = (40 + 10 + 16 + 10 + 8 + 10 + 16 + 10) cm = 120 cm 2. Find the area of the (i) shaded part (ii) unshaded part of the following figures, given that all adjacent sides are at right angles : 2.5m
4m
8m
16 m
15 m
12 m
22 m
28 m
Ans. (a) Length of outer rectangle = 22 m Breadth of outer rectangle = 16 m Area of outer rectangle = (22 × 16) m2 = 352 m2 Length of inner rectangle = 12 m Breadth of inner rectangle = 8 m Area of inner rectangle = area of unshaded part = 12 m × 8 m = 96 m2 ICSE Math Class VII
1
Question Bank
Area of shaded part = Area of outer rectangle – Area of inner rectangle = (352 – 96) m2 = 256 m2 (b) Length of rectangle = 28 m Breadth of rectangle = 15 m Area of rectangle = (28 × 15)m2 = 420 m2 Area of shaded part = (28 × 4 + 15 × 2.5 – 4 × 2.5) m2 = (112 + 37.5 – 10) m2 = 139.5 m2 Hence, area of unshaded part = (420 – 139.5) m2 = 280.5m2 3. Find the area and perimeter of the shaded part in each of the following figures :
(ii)
Z B
3m
2m
2m
2m
9m
2m
3m
18m
4.5m
9m
2m 1m
(i)
2m
2.5m 1.5m
2m
L A A N L IO U AT D N © E ER T IN 2m
F 1.5m 18m H G
B
L
4.5m
2m
DJ C
1m
A 9m
2.5m
2.5m
E
I
K
Ans. (i) Area of the shaded part = Area of rect angle ABCD + Area of rectangle EFKL + Area of rectangle GHIJ. = (9 × 2) m2 + [(2.5 + 4.5) × 2.5] m2 + (18 × 1) m2 = 18 m2 + (7 × 2.5) m2 + 18 m2 = (18 + 17.5 + 18) m2 = 53.5 m2 Perimeter of the shaded part = AB + BC + CL + LK + KJ + JI + IH + HG + GF + FE + ED + DA (2 + 9 + 2.5 + 2.5 + 4.5 + 18 + 1 + 18 + 1.5 + 2.5 + 2.5 + 9) 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
2
Question Bank
2m
P
M 2m L
I 2m H
9m
Q
B E
A
3m
2m K
G
J
F
2m
2m N O 3m
R
C 2m D
∴ 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 = [(2 + 2 + 2 + 2 + 2) × 3] m2 + [(9 – 3) × 2] m2 +[(9 – 3 ) × 2] m2 + [(9 – 3) × 2] m2 + (2 × 2) m2 = (10 × 3) m2 + (6 × 2) m2 + (6 × 2) m2 + (6 × 2) m2 + 4 m2 = (30 + 12 + 12 + 12 + 4) m2 = 70 m2 Perimeter of the shaded part = AB + BC + CD + DE + EF + FH + HI + IJ + JK + KC + LM + MN + ON + OP + PQ + QA = (2 + 2) m + 2 m+ 2 m+ 2 m + (2 + 2) m + 9 m + 2 m + (9 – 3) m + 2 m + (9 – 3) m +2 m + (9 – 3) m+ 2 m + (9 – 3) m + 2 m + 9 m = (4 + 2 + 2 + 2 + 4 + 9 + 2 + 6 + 2 + 6 + 2 + 6 + 2 + 6 + 2 + 9) m = 66 m. 4. ABCD is a square of side 24 cm. EF is parallel to BC and AE = 15 cm. By how much does
Z B
L A A N L IO U AT D N © E ER T IN D
F
A
E
C
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
Ans. (i) Perimeter of AEFD = 2 (15 + 24) cm = 2 × 39 = 78 cm Perimeter of EBCF = 2 × (24 + 9) cm = 2 × 33 = 66 cm D
24cm
A
15cm
F 9cm
24cm
24cm
E 9cm
15cm
C
B
∴ Perimeter of AEFD exceeds perimeter of EBCF by = (78 – 66) cm = 12 cm (ii) Area of AEFD = L × B = (15 × 24) = 360 cm2 Area of EBCF = L × B = (24 × 9) = 216 cm2 Hence, area of AEFD exceeds the area of EBCF by = (360 – 216) = 144 cm2 5. The perimeter of a rectangular plot of land is 240 m and its length is 63m. Find the breadth and area of the plot. Ans. Perimeter of the rectangular plot = 240 m Length of the plot = 63 m Let the breadth of the plot be x m ∵ 2 (length + breadth) = perimeter 2 (x + 63) = 240 ⇒
Z B
⇒
L A A N L IO U AT D N © E ER T IN x + 63 =
240 = 120 2
x = (120 – 63) = 57 m ⇒ ∴ Breadth of the plot = 57 m Area of the plot = (length × breadth) = (63 × 57) m2 = 3591 m2 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 2 (x + 10.5) = 189 ∴ ⇒ ICSE Math Class VII
x + 10.5 =
189 2 4
Question Bank
x + 10.5 = 94.5 ⇒ x = 94.5 – 10.5 = 84 ⇒ ∴ Length of the plot = 84 m Area of the plot = (length × breadth) = 84 × 10.5 = 882 m2 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 = 2 (l + b) = 2 (175 + 96) = 2 (271) = 542 m ∴ Length of the wire required for fencing = 542 m Cost of fencing = Rs (542 × 1.50) = Rs 813 Also area of the graden = (l × b) = (175 × 96) m2 = 16800 m2
Z B
16800 × 50 rupees = Rs 8400 100
L A A N L IO U AT D N © E ER T IN
Cost, of ploughing = 16800 × 50 paise =
8. Square tiles of side 20 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 = (4.5 × 100) cm = 450 cm Area of the floor (1000 × 450) cm2 Area of one tile = (20 × 20) cm2 Number of tiles needed =
Area of the floor 1000 × 450 = = 25 × 45 = 1125 20 × 20 Area of one t ile
Cost of putting the tile = Rs (1125 × 1.40) = Rs 1575 9. The perimeter of a rectangular field is 151 m. If its breadth is 32 m, find its length and area. Ans. Perimeter of rectangular field = 151 m, Breadth = 32 m Let length = x m 2(x + 32) = 151 ∴ 2x + 64 = 151 ⇒ 2x = 151 – 64 = 87 ⇒ 87 = 43.5 2 ∴ Length of the rectangular field = 43.5 m Area = L × B = (43.5 × 32) m2 = 1392 m2
⇒
ICSE Math Class VII
x=
5
Question Bank
10. The area of a rectangular plot is 340 m2 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 m2, Breadth = 17 m Area 340 = = 20 m Breadth 17 Perimeter = 2 (L + B) = 2 × (20 + 17) m = (2 × 37) = 74 m Cost of fencing 1 metre = Rs 5.70 Cost of fencing 74 metre = (74 × 5.70) = Rs 421.80 11. A rectangular room is 10 m long and 7.5 m wide. Find the cost of covering the floor with carpet 1.25 m wide at Rs 25 per metre. Ans. Length of room = 10 m Breadth of room = 7.5 m Area of room = (10 × 7.5) = 75 m2 Width of carpet = 1.25 m
Length =
Z B
L A A N L IO U AT D N © E ER T IN
75 = 60 m Length of carpet = ∴ 1.25 Cost of 1 metre = Rs 25 ∴ ∴ Cost of 60 metre of carpet = 60 × Rs 25 = Rs 1500 12. Find the cost of flooring a room 6.5 m by 5 m with square tiles of side 25 cm at the rate of Rs. 4.70 per tile. Ans. Area of room = L × B = 6.5 × 25 = 32.5 m2 Area of square tile = 25 cm × 25 cm
= 625 cm2 =
Number of tiles required =
625 2 m 10000
Area of room Area of tile
32.5 325 10000 = × = 520 625 10 625 10000 Cost of 1 tile = Rs 4.70 ∴ Cost of 520 tiles = (520 × 4.70) = Rs 2444 ∴ 13. The length of the base of a triangle is 12 cm and its area is 108 cm2 ; 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
Length of base of a triangle (b) = 12 cm and its area = 108 cm2 Let h be the height of the triangle We know that
Ans.
1 bh = Area of triangle 2
⇒
1 × 12 × h = 108 2 6h = 108
⇒ ⇒ ⇒
108 = 18 6 Thus, height of the triangle = 18 cm As per condition
h=
Z B
L A A N L IO U AT D N © E ER T IN
1 Height of triangle = × 18 cm = 9 cm and base = 12 × 2 = 24 cm 2 1 1 × base × height = × 24 × 9 = 108 cm2 2 2 (ii) there is no change in the area of the triangle 14. The area of a triangular field is 324 m2 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 24 ; 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 = 324 m2, Base (b) = 18 m Let length of its height be h
(i)
area of new triangle =
Then, Area of triangle = ⇒ ⇒
1 × base × height 2
1 × 18 × h = 324 ⇒ 9h = 324 2 324 h= = 36 9 Height = 36 m
⇒ Case II (i) Area of rectangular field = 324 m2 (ii) Length of the field (l) = 24 m
ICSE Math Class VII
7
Question Bank
then breadth (b) =
Area 324 27 = = m = 13.5 m. Length 24 2
(iii) Perimeter of the rectangular field = 2 (l + b) = 2 (24 + 13.5) m = 2 × 37.5 = 75 m 15. The adjoining figure shows a square ABCD and a triangle ABE. If each side of the square is 24 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. E
D
Z B C
L A A N L IO U AT D N © E ER T IN B
A
Ans.In square ABCD. (i) Its each side (a) = 24 cm ∴ Area of the square = side a2 = (24)2 cm2 = 576 cm2 (ii) Base of the triangle ABE = AB = 24 cm and height (altitude h = AD = 24 cm
1 1 bh = × 24 × 24 = 288 cm2 2 2 (iii) Area of unshaded portion = Area of square – Area of shaded triangle = 576 – 288 = 288 cm2
⇒ Area of triangle ABE =
2m
(iv) Ratio between the area of square ABCD and ∆ ABE = 576 : 288 = 2 : 1 16. A room is 10 m long and 6 m broad. It is surrounded by a verandah which is 2 m wide all around it. Find the cost of flooring the verandah with marble at Rs 36 per m2. Ans. Length of room (l) = 10 m 6m 10m Breadth (b) = 6 m 10m Width of verandah = 2 m 2m ∴ Outer length (L) = 10 + 2 × 2 m = 10 + 4 = 14 m 14m ICSE Math Class VII
8
Question Bank
Outer width (B) = 6 + 2 × 2 m = 6 + 4 = 10 m Thus, Area of verandah = L × B –l × b = 14 × 10 – 10 × 6 = 140 – 60 = 80 m2 Cost of 1 m2 = Rs 36 Hence, total cost = Rs 36 × 80 = Rs 2880 17. A hall is 16 m long and 12 m broad. Find the cost of carpeting it at Rs. 15 per m2, after leaving a margin of 1 metre all around. Ans. Length of Hall (L) = 16 m, width (B) = 12 m Width of margin left = 1 m Inner length (l) = 16 – 1 × 2 = 16 – 2 = 14 m ∴ Inner width = 12 – 1 × 2 = 12 – 2= 10 m Thus, inner area leaving margin = l × b = 14 × 10 = 140 m2
Z B
18. Ans.
∴
19. Ans.
L A A N L IO U AT D N © E ER T IN
Rate of carpeting = Rs 15 per m2 Hence, total cost = Rs 15 × 140 = Rs 2100 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 2.50 per m2 ? Length of field (L) = 116 m and breadth of field (B) = 76 m Width of Path (inside) = 3 m Inner length (l) = 116 – 3 × 2 = 116 – 6 = 110 m Linner breadth (b) = 76 – 3 × 2 = 76 – 6 = 70 m Thus, area of path L × B – l × b = 116 × 76 – 110 × 70 = (8816 – 7700) m2 = 1116 m2 Rate of gravelling the path = Rs 2.50 per m2 Hence, total cost Rs 2.50 × 1116 = Rs 2790 A path 2.5 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. 5.60 per m2. Length of plot of (l) = 35 m and breadth of plot (b) = 35 m width of path = 2.5 m Thus, outer length (L) = 40 + 2 × 2.5 = 40 + 5 = 45 m and outer breadth (B) = 35 + 2 × 2.5 = 35 + 5 = 40 m Thus, area of path = L × B – l × b = (45 × 40 – 40 × 35) m2 = (1800 – 1400) = 400 m2 Rate of tilling it = Ra 5.60 per m2
ICSE Math Class VII
9
Question Bank
Ans.
21.
Ans.
2.5m
20.
Hence, total cost = Rs 5.60 × 400 = Rs 2240 A rectangular lawn 115 m long and 65 m broad has two cross-paths at right angles, one 2 m wide running parallel to its length and the other 2.5 wide running parallel to its breadth. Find the cost of gravelling the paths at Rs. 14 per m2. Length of lawn (l) = 115 m Breadth of lawn (b) = 64 m 2m 64m Width of length wise path = 2m and width of breadth wise path = 2.5 m 115m ∴ Area of path = 2 × 115 + 2.5 × 64 – 2 × 2.5 2 = 230 + 160 – 5 = 390 – 5 = 385 m Rate of gravelling = Ra 14 per m2 Hence, total cost = Rs 14 × 385 = Rs 5390 The central hall of a school is 22 m long and 15.2 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. Length of Hall (L) = 22 m and breadth of hall (B) = 15.5 m
Z B
L A A N L IO U AT D N © E ER T IN Width of strip = 75 cm =
∴
22.
Ans.
23.
Ans.
3 m 4
3 = 22 – 1.5 = 20.5 m 4 and inner breadth (b) = 15.5 – 1.5 = 14 m Thus, area of strip = L × B – l × b = (22 × 15.5 – 20.5 × 14) m2 = (341 – 287) = 54 m2 and area of inner part = area of the carpet = l × b = 20.5 × 14 = 287 m2 The floor of a rectangular hall has a perimeter 236 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. 8.40 per square metre. Perimeter = 2 (l + b) = 236 m Area of four walls = 2 (l + b) × h = 236 × 4.5 = 1062 m2 Cost of painting l m2 = Rs 8.40 Cost of painting 1062 m2 = Rs 8.40 × 1062 = Rs 8920.80 A rectangular fish tank has a length of 30 cm, a breadth of 20 cm and a height of 20 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. Length of tank = 30 cm
ICSE Math Class VII
Inner length (l) = 22 – 2 ×
10
Question Bank
Breadth of tank = 20 cm Height of tank = 20 cm, As the tank is three-quarters full of water 20 × 3 = 15 cm 4 Area of the tank in contact with the water = Area of floor of tank + Area of 4 walls upto 15 cm. = 30 × 20 + 2 (30 + 20) × 15 = 600 + 2 × 50 × 15 = (600 + 1500) = 2100 cm2 24. If cost Rs. 936 to fence a square field at Rs. 7.80 per metre. Find the area of the square field. Ans. Total cost = Rs 936, Rate per metre = Rs 7.80
∴
∴
Height of water in the tank =
Z B
L A A N L IO U AT D N © E ER T IN Perimeter of field =
Total cost of fencing Rate per metre
936 = 120 m. 7.80 Perimeter of square field = 4 × side = 120 m
=
120 = 30 m 4 Hence, Area of square field = 30m × 30 m = 900 m2 25. A person walks at 3km/hr. How long will he take to go round a square gorund 5 times, the area of which being 2025m2 ? Ans. Area of square ground = 2025 m2 ⇒ (side)2 = 2025 m2 = (45m)2 ⇒ side = 45 m Perimeter of ground = 45 × 4 = 180 m. To go 5 rounds, total distance = 180 × 5 = 900 m. Speed of person = 3 km/hr. Time taken to walk 3 km = 1 hr
Thus, side =
1× 60 × 900 = 18 minutes. 3 × 1000 26. Water is to be transferred from one tank to the other tank of dimensions 2.1 m × 1.5 m × 0.6 m such that the second tank is completely filled with this water the length 1 and breadth of the first tank are 2 m and m respectively. Find the height of the 2 water when it was in the first tank.
Time taken to walk to walk 900 m =
ICSE Math Class VII
11
Question Bank
Ans. Volume of water in the second tank = 2.1 × m ×1.5 m× 0.6 m = 1.89 m3 ∴ Volume of the water in the first tank = 1.89m3 Length (l) = 2 m and width (b) =
1 m 2
1 × h = 1.89 2 h = 1.89 m
Let h be the hight of the water, then 2 ×
Hence, height = 1.89 m 27. How many bricks, each measuring 18 cm × 12 cm × 6 cm, will be needed to build a wall of length = 6 m, width = 24 cm and height = 3.6 m ? Ans. Measure of each brick = 18 cm × 12 cm × 6 cm Volume of each brick = 1296 cm3 ∴ Length of wall (l) = 6 m = 600 cm width (b) = 24 cm and height (h) = 3.6 m = 360 cm Volume of wall = l× b × h = 600 × 24 × 360 cm3 ∴
Z B
L A A N L IO U AT D N © E ER T IN
Number of bricks used in the wall =
∴
Volume of wall Volume of one brick
600×24×360 = 4000 1296 Find the area of four walls of a room whose length = 3.2 m, breadth = 2.5 m and height = 3m Length of the room (l) = 3.2 m, breadth (b) = 2.5 m and height (h) = 3m Area of 4 walls = 2 (l × b) × h = 2 (3.2 + 2.5) × 3m2 = 2 × 5.7 × 3 m2 = 34.2 m2 A rectangular water reservoir contains 42000 litres of water. Find the depth of the water in the reservoir if its base measures 6 m by 3.5 m. Volume of reservoir
=
28. Ans.
∴
29. Ans.
=
42000 = 42 m2 [∵ 1000 litres = 1 m3] 1000 Area = l × b = 6 m × 3.5 m 42 Volume = 2m = 6 × 3.5 Area Depth of reservoir = 2 m.
∴ Height of reservoir = ∴
ICSE Math Class VII
12
Question Bank
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 0.2 m ? Ans. Volume of 1 cuboidal tea packet = 10 cm × 6 cm × 4 cm = 240 cm3 Volume of cardboard box = 50 cm × 30 cm × 0.2 m = 50 cm × 30 cm × 20 cm = 30000 cm3
∴ Number of tea packets =
30000 Volume of Box = 125. = 240 Volume of 1 tea packet
∴ Number of tea packet = 125. 31. The dimensions of a rectangular tin oil are 26 cm by 26 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 20, find the cost of the tin sheet used for these 40 tins. Ans. Dimensions of rectangular tin oil = 26 cm × 26 cm × 45 cm Area of tin sheet = 2lb + 2 (l + b) h ∴ = (2 × 26 × 26) + 2 (26 + 26) × 45 = (1352 + 4680) cm2 = 6032 cm2 ∴ Area of tin sheet for these 40 tins = 6032 cm2 × 40 = 241280 cm2 Cost of 1sq m of tin sheet = Rs 20
Z B
L A A N L IO U AT D N © E ER T IN
Rs20 × 241280 = Rs 482.56 10000 32. The volume of a cuboid is 144 cm3, 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 cm3 height of cuboid = 4 cm Since base is a square ∵ length and breadth are equal side2 × height = volume
Cost of 241280 sq cm of tin sheet =
or
side2 =
=
volume height
144 = 36 = 6 2 4
Hence, side of square = 6 cm (ii) surface area of cuboid = 2 (lb + lh + bh) = 2 (6 × 6 + 6 × 4 × 6 × 4) = 2 (36 + 24 + 24) = 2 × 84 = 168 cm2 ICSE Math Class VII
13
…(i)
Question Bank
33. The external dimensions of a closed rectangular wooden box are 94 cm by 68 cm by 0.46 m. If the wood is 2 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 = 2 cm Deducting thickness of wood of both sides i.e. 2 × 2 = 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 42 cm Capacity of the box = Volume of internal box = 90 cm × 64 cm × 42 cm = 241920 cm3 Volume of outer wooden box 94 cm × 68 cm × 46 cm = 294032 cm3 Hence, Volume of wood used in making the box = Volume of outer box – Volume of inner box = (294032 – 241920) cm3 = 52112 cm3. 34. A rectangular room is 6 m long 5 m wide and 3.5 m hights. It has one door 1 m by 2 m and two windows 1.5 m × 2 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 m2 Area of four walls = (2 (6 + 5) × 3.5) m2 = 77 m2 Area of one door = 1 m × 2 m = 2 m2 Area of two windows = (2 × 1.5 × 2) = 6 m2 ∴ Area to be white washed = Area of ceiling + Area of 4 walls Area of one door – Area of 2 windows = 30 + 77 – 2 – 6 = 99 m2 Cost of white washing 1 sq. m = Rs 4.60 Hence, Cost of white washing 99 sq. m = Rs 4.60 × 99 = Rs 455.40
Z B
L A A N L IO U AT D N © E ER T IN
ICSE Math Class VII
14
Question Bank
35. If the total surface area of a cubical box is 216 cm2 find the length of its one side. Also, find the area of four walls. Total surface area of the box = 216 cm2 ∵ Total surface area of a cube = 6(side)2 6 side2 = 216 ⇒ side2 =
216 = 36 = (6)2 6
Therefore, side = 6 ∴ Area of four walls of cubical box = 4 (side)2 = 4 × (6)2 = 4 × 36 cm2 = 144 cm2 36. If the total surface area of a cube is 96 cm2, find the length of its each side and its volume. Ans. Total surface area of a cube = 96 cm2 Let a be its each side, then
Z B
96 = 16 6 a2 = (4)2 ⇒ a = 4 cm ⇒ ∴ Volume of cube = (side)3 = (4)3 cm3 = 64 cm3 37. 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 = 10000 = 10000 m2
6a2 = 96 ⇒
a2 =
L A A N L IO U AT D N © E ER T IN
⇒ l × b = 10000 m2 and height = rainfall = 3.5 cm =
∴
3.5 m = 0.035 m 100
volume of water that fall on, 1 hectare of land = (10000 × 0.035) m3 = 350 m3 = 350 × 1000 litres = 350000 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 3
60 m = 100 = (.6 × .6 × .6) m3 = .216 m3 ∴ Weight of ice cube = .216 × 900 kg = 194.4 kg. 39. A rectangular pit 1.4 m long, 90 cm broad and 70 cm deep was dugs. and 1000 bricks each of base 21 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
Ans. Length of rectangular pit = 1.4 m = 140 cm Breadth = 90 cm Depth = 70 cm Volume of cuboidal pit = (140 × 90 × 70) cm3 = 882000 cm3 1000 Bricks were made out of it So,volume of 1000 bricks = 882000 cm3 882000 cm3 = 882 cm3 1000 Length of brick = 21 cm. Breadth of brick = 10.5 cm
Volume of brick =
Z B
882 cm = 4 cm 21× 10.5 40. How many wooden cubes of an edge 25 cm in length can be cut out from a cuboidal log of wood of size 2.5 m by 0.5 m by 50 cm, assuming that there is no wastage ? Ans. Volume of cuboidal log of wood = 2.5 m × 0.5 m × 50 cm = 250 cm × 50 cm × 50 cm = 250 × 50 × 50 cm3 Edge of wooden cube = 25 cm Volume of one wooden cube = (25 cm)3 = 25 × 25 × 25 cm3
Height of brick =
L A A N L IO U AT D N © E ER T IN
∴ Number of wooden cubes =
Volume of log of wood Volume of 1wooden cube
250×50×50 = 40 25×25×25 41. How many soap cakes can be placed in a box of size 56 cm by 40 cm by 25 cm, if the size of a soap cake is 7 cm by 5 cm by 2.5 cm ? Ans. Volume of box = 56 cm × 40 cm × 25 cm = 56000 cm3 size of a soap cake is 7cm by 5 cm by 2.5 cm volume of soap cake = 7 × 5 × 2.5 cm3 = 87.5 cm3 ∴
=
∴
Number of soap cakes =
= ICSE Math Class VII
Volume of box Volume of 1 soap cake
56000 = 640. 87.5 16
Question Bank
42. A room is 4.8 m long, 3.6 m broad and 2.4 m high. Find the cost of laying tiles on its floor and on its four walls at the rate of Rs. 80 per m2. Ans. Length of room (l) = 4.8 m, Breadth (b) = 3.6 m and height (h) = 2.4 m Area of floor = l × b = 4.8 × 3.6 = 17.28 m2 Area of four walls = 2 (l + b) h = 2 (4.8 + 3.6) × 2.4 m2 = 2 × 8.4 × 2.4 m2 = 40.32 m2 ∴ Total area = 17.28 + 40.32 = 57.6 m2 Rate of laying tiles = Rs 80 per m2 Hence, total cost = Rs 80 × 57.6 = Rs 4608 43. 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 × 25 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)2 × 4 = 3.6 × 3.6 × 4m3 = 51.84 m3 (ii) Dimensions of each box = 0.9 m × 40 cm × 25 cm = 0.9 m × 0.4 m × 0.25 m = 0.09 m3
Z B
L A A N L IO U AT D N © E ER T IN
∴ Number of boxes to be placed in the room =
Volumeof room Volume of each box
51.84 5184 × 100 = = 576 0.09 100 × 9 44. How many children can be accommodated in a hall of length 16 m, breadth 12.5 m and height 1.45 m assuming 3.6 m3 of air is required for each child ? Ans. Length of the hall (l) = 16 m, breadth (b) = 12.5 m height (h) = 4.5 m Volume of the hall = l × b × h = (16 × 12.5 × 4.5) m3 = 900 m3 Volume of air required for one child = 3.6 m3 Hence, number of children can be accommodated in the hall
=
= ICSE Math Class VII
900 900 × 10 = = 250 3.6 36 17
Question Bank
45. A path of length 325 m and width 30 m is laid with concreate upto a depth of 40 cm. Find the cost of laying the concrete at Rs.20 per cubic metre. Ans. Length of the path = 325 m, width of the path = 30 m and depth of the path = 40 cm = Volume of the path = l × b × h = (325 × 30 ×
40 m = 0.4 m 100
4 ) m3 = 3900 m3 10
Cost of laying the concrete of 1 m3 = Rs 20 Cost of laying the concrete of 3900 m3 = Rs 3900 × 20 = Rs 78000. 46. Water is to be transferred from a tank of length 3 m and breadth 80 cm to another tank of dimensions 2.4m 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 = 3 × 0.8 × h m3 = 2.4h m3 Volume of water in the other tank 6 m = 0.6) = 2.4 m × 1.5 m × 0.6 m (∵ 60 cm = 100 = 2.16 m3 ∵ 2.4h = 2.16
Z B
L A A N L IO U AT D N © E ER T IN
216 2.16 126 10 = × m= m = 0.9 m 240 2.4 100 24 = 0.9 × 100 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 = 24 cm breadth of cuboid (b) = 8 cm and height of coboid (h) = 8 cm ∴ Surface area of cuboid = 2 (lb + bh + hl) = (24 × 8 + 8 × 8 × 8 × 24) cm2 = 2(192 + 64 + 192) cm2 = 2 × 448 = 896 cm2
h=
ICSE Math Class VII
18
Question Bank
48. A cuboidal block of 6 cm × 9 cm × 12 cm is cut up into exact number of equal cubes. Find the least possible number of cubes. Ans. Length of cuboid (l) = 12 cm Breadth (b) = 9 cm Height (h) = 6 cm HCF of 12, 9, 6 = 3 cm Hence, side of each cube = 3 cm Volume of cuboid = l × b × h = 12 × 9 × 6 cm3 = 648 cm3 and volume of each cube = (side)3 = (3)3 cm3 = 27 cm3 Hence, number of cube to be cut out =
Z B
Volume of cuboid Volume of cube
L A A N L IO U AT D N © E ER T IN
648 = 24 27 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 = 18 × 18 × 18 cm3 ∴ Edge of small cube = 3 cm Volume of small cube = 3 × 3 × 3 cm3 ∴ ∴ Number of total cubes cut out
=
=
ICSE Math Class VII
18 ×18 × 18 = 6 × 6 × 6 = 216 3× 3× 3
19
Question Bank
