1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Solution:
(a)
Given,
Length of cuboidal box (l) = 60 cm
Breadth of cuboidal box (b) = 40 cm
Height of cuboidal box (h) = 50 cm
According to question:
Total surface area of cuboidal box = 2×(lb + bh + hl)
= 2×(60×40 +40×50 + 50×60)
= 2×(2400+2000+3000)
= 14800 cm2
(b) Given,
Length of cubical box (l) = 50 cm
Breadth of cubical box (b) = 50 cm
Height of cubical box (h) = 50 cm
Total surface area of cubical box = 6(side)2
= 6(50×50)
= 6×2500
= 15000 cm2
Hence, according to the result of (a) and (b), we get that cuboidal box requires the lesser amount of material to make.
2. A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Given,
Length of suitcase (l) = 80 cm,
Breadth of suitcase (b)= 48 cm
Height of cuboidal (h) = 24 cm
Then,
Total surface area of suitcase box = 2(lb+bh+hl)
= 2(80×48+48×24+24×80)
= 2 (3840+1152+1920)
= 2×6912
= 13824 cm2
Now,
Area of Tarpaulin cloth = Surface area of suitcase
l×b = 13824
l ×96 = 13824
l = 144
Required tarpaulin for 100 suitcases = 144×100
= 14400 cm = 144 m
Hence, required tarpaulin cloth to cover 100 suitcases is 144 m.
3. Find the side of a cube whose surface area is 600cm2.
Solution:
Given, surface area of cube = 600 cm2
From formula we know that,
Surface area of a cube = 6(side)2
So,
6(side)2 = 600
(side)2 = 100
side = ±10
Hence, side cannot be negative. So, the measure of each side of a cube is 10 cm.
4. Rukshar painted the outside of the cabinet of measure 1 m ×2 m ×1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?
Solution:
Given,
Length of cabinet (l) = 2 m
Breadth of cabinet (b) = 1 m
Height of cabinet (h) = 1.5 m
Then,
Surface area of cabinet except the bottom of the cabinet = lb+2(bh+hl)
= 2×1+2(1×1.5+1.5×2)
= 2+2(1.5+3.0)
= 2+9.0
= 11 m2
Hence, required surface area of cabinet is 11m2.
5. Daniel is paining the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room?
Solution:
Given,
Length of wall (l) = 15 m
Breadth of wall (b) = 10 m
Height of wall (h) = 7 m
Then,
Total Surface area of classroom except floor = lb+2(bh+hl )
= 15×10+2(10×7+7×15)
= 150+2(70+105)
= 150+350
= 500 m2
Now,
Total number of required cans = Area of classroom/Area of one can
= 500/100 = 5
Hence, 5 cans are required to paint the room.
6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface areas?
Solution:
Given,
Diameter of cylinder (d) = 7 cm
So, radius of cylinder (r) = 7/2 cm [ r = d/2]
Height of cylinder (h) = 7 cm
Now,
Lateral surface area of cylinder = 2πrh
= 2 × (22/7) × (7/2) × 7
= 154 cm2
Now,
Lateral surface area of cube = 4 (side)2
= 4×72
= 4×49
= 196 cm2
Hence, the cube has larger lateral surface area.
7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Solution:
Given,
Radius of cylindrical tank (r) = 7 m
Height of cylindrical tank (h) = 3 m
Then,
Total surface area of cylindrical tank = 2πr(h+r)
= 2×(22/7)×7(3+7)
= 44×10 = 440 m2
Hence, 440 m2 metal sheet is required.
8. The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Solution:
Given, lateral surface area of hollow cylinder = 4224 cm2
Width of rectangular sheet (b) = 33 cm
Let l be the length of rectangular sheet.
Now,
Lateral surface area of cylinder = Area of rectangular sheet
4224 = b × l
4224 = 33 × l
l = 4224/33 = 128 cm
Hence, the length of the rectangular sheet is 128 cm.
Now,
Perimeter of rectangular sheet = 2(l+b)
= 2(128+33)
= 322 cm
9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length 1 m.
Solution:
Given,
Diameter of road roller (d) = 84 cm
So, radius of road roller (r) = 84/2 = 42 cm [r = d/2]
Length of road roller (h) = 1 m = 100 cm
Now,
Curved surface area of road roller = 2πrh
= 2×(22/7)×42×100
= 26400 cm2
Now,
Area covered by road roller in 750 revolutions = 26400×750cm2
= 1,98,00,000cm2
= 1980 m2 [1 m2= 10,000 cm2]
Hence, the area of the road is 1980 m2.
10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label?
Solution:
Given,
Diameter of cylindrical container (d) = 14 cm
So, radius of cylindrical container (r) = 14/2 = 7 cm [r = d/2]
Height of cylindrical container (H) = 20 cm
According to the figure,
Height of the label, say (h) = 20–2–2 ()
= 16 cm
Now,
Curved surface area of label = 2πrh
= 2×(22/7)×7×16
= 704
Hence, the area of the label is 704 cm2.
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