1. A box made of plastic The dimensions are to be 1.5 m long, 1.25 m wide, and 65 cm deep. The top is supposed to be open. Regardless of the plastic sheet's thickness, calculate
(I) The portion of the sheet needed for the box's construction.
(ii) The price of the sheet for it, if a sheet with a surface area of 1 m2 costs Rs.
1. A box made of plastic The dimensions are to be 1.5 m long, 1.25 m wide, and 65 cm deep. The top is supposed to be open. Regardless of the plastic sheet's thickness, calculate
(I) The portion of the sheet needed for the box's construction.
(ii) The price of the sheet for it, if a sheet with a surface area of 1 m2 costs Rs.
Explanation:
Given: 1.5m is the box's length (l).
Box width (b) is 1.25 metres.
Box height (h) = 0.65 metres.
(i) The top of the box must be open.
Required sheet area.
2lh + 2bh + lb
= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m2
= (1.95+1.625+1.875) m2 = 5.45 m2
(ii) Sheets cost Rs. 20 per square metre.
5.45 m2 sheet costs Rs. (5.45 x 20).
= Rs.109.
2. A room is 5 metres long, 4 metres wide, and 3 metres tall, accordingly. Calculate the price of whitewashing the room's walls and ceiling at a rate of Rs. 7.50 per m2.
2. A room is 5 metres long, 4 metres wide, and 3 metres tall, accordingly. Calculate the price of whitewashing the room's walls and ceiling at a rate of Rs. 7.50 per m2.
Explanation:
Room length (l) equals 5m
Room width (b) equals 4m
Room height (h) equals 3m
It is clear that the room's four walls as well as its ceiling must be painted white.
Total area to be whitewashed equals the sum of the area of the room's walls and ceiling.
2lh + 2bh + lb
= [2×5×3+2×4×3+5×4]
= (30+24+20)
= 74
Area = 74 m2
Also,
Price of whitewash per square metre: Rs. 7.50 (Given)
Whitewashing a 74 m2 area costs Rs. (74 x 7.50).
= Rs. 555
3. A rectangular hall's floor has a 250 m circumference. Find the height of the hall if painting the four walls will cost Rs. 15,000 at a rate of Rs. 10 per m2.
[Hint: Lateral surface area = area of the four walls.]
3. A rectangular hall's floor has a 250 m circumference. Find the height of the hall if painting the four walls will cost Rs. 15,000 at a rate of Rs. 10 per m2.
[Hint: Lateral surface area = area of the four walls.]
Explanation:
The rectangular hall should have the following dimensions: l, b, and h, correspondingly.
2lh+2bh is the area of four walls.
= 2(l+b)h
The perimeter of the hall's floor equals 2(l+b).
= 250 m
Area of four walls is equal to 2(l+b)h, or 250h m2.
Painting costs ten rupees per square metre.
A 250-hour painting job would cost (250-hours times 10) or $2500.
However, it is acknowledged that painting the walls will cost Rs. 15,000 in total.
15000 = 2500h
Or h = 6
As a result, the hall has a 6 m height.
4. There is enough paint in a certain container to cover a 9.375 m2 area. How many 22.5 cm x 10 cm x 7.5 cm painted bricks can you get into this box?
4. There is enough paint in a certain container to cover a 9.375 m2 area. How many 22.5 cm x 10 cm x 7.5 cm painted bricks can you get into this box?
Explanation:
A brick's total surface area is equal to 2(lb + bh + lb).
= [2(22.5×10+10×7.5+22.5×7.5)] cm2
= 2(225+75+168.75) cm2
= (2×468.75) cm2
= 937.5 cm2
The container's paint can be used to cover n bricks.
Area of n bricks is equal to n 937.5 cm2 937.5 n cm2.
The area that the paint in the bottle may paint is 9.375 m2 (or 93750 cm2), according to the specifications.
Thus, 93750 = 937.5n.
n = 100
Therefore, the container paint has the ability to cover 100 blocks.
5. One cuboidal box is 12.5 cm long, 10 cm broad, and 8 cm high, while the edges of a cubical box are each 10 cm long.
(i).Which box, and by how much, has the larger lateral surface area?
(ii).Which box, and by how much, has the smallest total surface area?
5. One cuboidal box is 12.5 cm long, 10 cm broad, and 8 cm high, while the edges of a cubical box are each 10 cm long.
(i).Which box, and by how much, has the larger lateral surface area?
(ii).Which box, and by how much, has the smallest total surface area?
Explanation:
The question statement gives us
A cube's edge is 10 cm.
The length is 12.5 cm.
B = 10 cm for breadth.
H = 8 cm for height
For both figures, determine the lateral surface area (i).
Cubical box's lateral surface area is 4 (edge)2
= 4(10)2
= 400 cm2 …(1)
Cuboidal box's lateral surface area is equal to 2[lh+bh].
= [2(12.5×8+10×8)]
= (2×180) = 360
Consequently, the cuboidal box's lateral surface area is 360 cm2. …(2)
According to (1) and (2), the cubical box's lateral surface area is greater than the cuboidal box's lateral surface area. There is a 40 cm2 difference between the two lateral surfaces.
(Lateral surface area of the cubic box - Lateral surface area of the cuboidal box = 400 cm2 - 360 cm2 - 40 cm2)
Find the combined surface area of both figures in (ii).
The cubical box's overall surface area is 6 square feet.(edge)2 = 6(10 cm)2 = 600 cm2…(3)
The cuboidal box's entire surface area
= 2[bh+lb+lh]
= [2(12.5×8+10×8+12.5×100)]
= 610
Accordingly, the cuboidal box must have a total surface area of 610 cm2.(4)
The cubical box's total surface area is less than the cuboidal box's according to (3) and (4). Their disparity is 10 cm2.
In light of this, the cubical box's total surface area is 10 cm2 less than the cuboidal box's.
6. A small indoor greenhouse (herbarium) is constructed entirely of glass panes that are taped together, including the base. It measures 25 cm in width, 30 cm in length, and 25 cm in height.
(i) How big is the glass?
(ii) How much tape is required for each of the 12 edges?
6. A small indoor greenhouse (herbarium) is constructed entirely of glass panes that are taped together, including the base. It measures 25 cm in width, 30 cm in length, and 25 cm in height.
(i) How big is the glass?
(ii) How much tape is required for each of the 12 edges?
Explanation:
length of the greenhouse, suppose l = 30 cm
The greenhouse's width, suppose b = 25 cm
Height of the greenhouse, suppose h = 25 cm
(i) The area of the glass divided by the total surface area of the greenhouse is 2[lb+lh+bh].
= [2(30×25+30×25+25×25)]
= [2(750+750+625)]
= (2×2125) = 4250
4250 cm2 is the total glass surface area.
(ii)
The tape is needed along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF, according to the figure.
Total tape length equals 4(l+b+h).
= [4(30+25+25)] (after substituting the values)
= 320
As a result, 320 cm of tape is needed for each of the 12 edges.
7. Shanti Sweets The stall was ordering cardboard boxes to pack its confections. There were two sizes of boxes needed. 25 cm x 20 cm x 5 cm, and 15 cm x 12 cm x 5 cm, are the larger and smaller dimensions, respectively. 5% of the total surface area must be added to account for all overlaps. Find the cost of cardboard needed to supply 250 boxes of each type if the price of cardboard is Rs. 4 for 1000 cm2
7. Shanti Sweets The stall was ordering cardboard boxes to pack its confections. There were two sizes of boxes needed. 25 cm x 20 cm x 5 cm, and 15 cm x 12 cm x 5 cm, are the larger and smaller dimensions, respectively. 5% of the total surface area must be added to account for all overlaps. Find the cost of cardboard needed to supply 250 boxes of each type if the price of cardboard is Rs. 4 for 1000 cm2
Explanation:
The box's length, width, and height should be denoted by l, b, and h.
a larger box
l = 25cm
b = 20 cm
h = 5 cm
Surface area of the larger box is equal to 2(lb+lh+bh).
= [2(25×20+25×5+20×5)]
= [2(500+125+100)]
= 1450 cm2
14505/100 cm2 more space is needed for overlapping
= 72.5 cm2
The greater box's overall surface area, taking into account any overlaps.
= (1450+72.5) cm2 = 1522.5 cm2
For 250 of these larger boxes, a certain amount of cardboard is needed.
= (1522.5×250) cm2 = 380625 cm2
Littler Box
The smaller box's total surface area is equal to [2(1512+155+125)]. cm2
= [2(180+75+60)] cm2
= (2×315) cm2
= 630 cm2
As a result, the additional space needed for overlapping is 6305/100 cm2 = 31.5 cm2.
the surface area of a single smaller box after taking into account all overlaps
= (630+31.5) cm2 = 661.5 cm2
250 smaller boxes require a cardboard sheet with a surface area of (250 x 661.5) cm2 or 165375 cm2.
The total amount of cardboard needed is (380625+165375) cm2.
= 546000 cm2
Considering that a 1000 cm2 cardboard sheet costs Rs.
Due to this, the price of a 546000 cm2 cardboard sheet is equal to (546000 4)/1000, or Rs. 2184.
As a result, Rs. 2,184 worth of cardboard will be needed to deliver 250 boxes of each type.
8. Praveen intended to construct a box-like structure with tarpaulin that covers the top of the car and all four sides as a temporary shelter for her car.How many tarpaulins would be needed to construct a shelter with a height of 2.5 metres and a base measurement of 4 metres by 3 meters, assuming that the stitching margins are minimal and very small?
8. Praveen intended to construct a box-like structure with tarpaulin that covers the top of the car and all four sides as a temporary shelter for her car.How many tarpaulins would be needed to construct a shelter with a height of 2.5 metres and a base measurement of 4 metres by 3 meters, assuming that the stitching margins are minimal and very small?
Explanation:
Let l, b, and h represent the shelter's length, width, and height.
Given:
l = 4m
b = 3m
h = 2.5m
The top of the shelter and its four wall sides will both need tarpaulins.
The formula for the amount of tarpaulin needed is 2(lh+bh)+lb.
The result of combining the values of l, b, and h is
= [2(4×2.5+3×2.5)+4×3] m2
= [2(10+7.5)+12]m2
= 47m2
There will therefore be a need for 47 m2 of tarpaulin.
9. An upright circular cylinder with a 14 cm height has an 88 cm2 curved surface area. Assuming = 22/7, determine the cylinder's base's diameter.
9. An upright circular cylinder with a 14 cm height has an 88 cm2 curved surface area. Assuming = 22/7, determine the cylinder's base's diameter.
Explanation:
Cylinder height, 14 cm in height
The cylinder's diameter should be d.
88 cm2 is the size of the cylinder's curved surface.
We are aware that the curved surface area of a cylinder can be calculated using the formula 2rh.
therefore 2 rh = 88 cm2 (r is the radius of the base of the cylinder)
2×(22/7)×r×14 = 88 cm2
2r = 2 cm
d =2 cm
As a result, the base of the cylinder has a 2 cm diameter.
10. A closed cylindrical tank with a base diameter of 140 cm and a height of 1 m must be constructed out of metal sheet. What percentage of the sheet's total square metres is needed for the same? Suppose = 22/7
10. A closed cylindrical tank with a base diameter of 140 cm and a height of 1 m must be constructed out of metal sheet. What percentage of the sheet's total square metres is needed for the same? Suppose = 22/7
Explanation:
Let the dimensions of a cylindrical tank be h for height and r for radius.
Tank's cylindrical height, h = 1m
Radius = diameter half, or (140/2) cm = 70cm = 0.7m
Required sheet area equals tank's total surface area, which is 2r(r+h) unit square.
= [2×(22/7)×0.7(0.7+1)]
equals 7.48 square metres
The sheet must therefore be 7.48 square metres in size.
11. A 77 cm long metal pipe. A cross-section has an inner diameter of 4 cm and an outside diameter of 4.4 cm. (see fig. 13.11). Get its
the area of the inner curved surface
(ii) region of curved outer surface
Total surface area (iii)
(Presume 22/7).
11. A 77 cm long metal pipe. A cross-section has an inner diameter of 4 cm and an outside diameter of 4.4 cm. (see fig. 13.11). Get its
the area of the inner curved surface
(ii) region of curved outer surface
Total surface area (iii)
(Presume 22/7).
Explanation:
Inner and outer radii of the cylindrical pipe, r1 and r2, respectively.
r1 = 4/2 cm = 2 cm
r2 = 4.4/2 cm = 2.2 cm
Height of cylindrical pipe, h = cylindrical pipe's length, which is 77 cm
(i) The pipe's outer surface's curved surface area is equal to 2r1h.
= 2×(22/7)×2×77 cm2
= 968 cm2
(ii) The pipe's outer surface's curved surface area is equal to 2r2h.
= 2×(22/7)×2.2×77 cm2
= (22×22×2.2) cm2
= 1064.8 cm2
(iii) The pipe's total surface area is calculated as the sum of its inner and outer curves as well as the areas of its two round ends.
= 2r1h+2r2h+2π(r12-r22)
= 9668+1064.8+2×(22/7)×(2.22-22)
= 2031.8+5.28
= 2038.08 cm2
Consequently, the cylindrical pipe has a total surface area of 2038.08 cm2.
12. rA oller has a diameter of 84 cm and a length of 120 cm. 500 full revolutions are necessary to
To level a playground, move once more. Determine the playground's square footage in m2 (presume = 22/7).
12. rA oller has a diameter of 84 cm and a length of 120 cm. 500 full revolutions are necessary to
To level a playground, move once more. Determine the playground's square footage in m2 (presume = 22/7).
Explanation:
The form of a roller resembles a cylinder.
Let r be the roller's radius and h its height.
H = Roller Length = 120 cm
The roller's round end has a radius of r = (84/2) cm, or 42 cm.
Therefore, CSA of roller = 2rh.
= 2×(22/7)×42×120
= 31680 cm2
Field area = 500 CSA of roller
= (500×31680) cm2
= 15840000 cm2
= 1584 m2.
Consequently, the playground has a surface area of 1584 m2.
13. A cylindrical pillar is 3.5 metres tall and has a diameter of 50 cm. Calculate the price of painting the pillar's curving surface at a rate of Rs. 12.50 per m2.
(Assume 1/22 = )
13. A cylindrical pillar is 3.5 metres tall and has a diameter of 50 cm. Calculate the price of painting the pillar's curving surface at a rate of Rs. 12.50 per m2.
(Assume 1/22 = )
Explanation:
Let h be the cylindrical pillar's height and r be its radius.
Given:
cylindrical pillar with a height of 3.5 metres
Radius of the pillar's circular end is calculated as follows: r = diameter/2 = 50/2 = 25 cm = 0.25 m
CSA of the pillar = 2rh.
= 2×(22/7)×0.25×3.5
= 5.5 m2
The price of painting Area of 1 m2: Rs. 12.50
Painting a 5.5 m2 space costs Rs. (5.5 x 12.50).
= Rs.68.75
As a result, painting the curved surface of the pillar will cost Rs. 68.75 at a rate of Rs. 12.50 per m2.
14. A right circular cylinder has a surface area that is curved of 4.4 m2. Find the height of the cylinder if the cylinder's base has a radius of 0.7 m. (Pretend = 22/7)
14. A right circular cylinder has a surface area that is curved of 4.4 m2. Find the height of the cylinder if the cylinder's base has a radius of 0.7 m. (Pretend = 22/7)
Explanation:
Let h represent the cylinder's height and r represent its radius.
r = 0.7m is the radius of the cylinder's base.
Cylinder CSA equals two rh
cylinder's CSA equals 4.4m2
Equating both formulas, we obtain
2×(22/7)×0.7×h = 4.4
Or h = 1
The cylinder is 1 m tall as a result.
15. A round well has a 3.5-meter inner diameter. It's 10 metres deep. Find
(i) The curved interior surface area.
(ii) the price of plastering this curved surface, which is Rs. 40 per square metre.
(Pretend = 22/7)
15. A round well has a 3.5-meter inner diameter. It's 10 metres deep. Find
(i) The curved interior surface area.
(ii) the price of plastering this curved surface, which is Rs. 40 per square metre.
(Pretend = 22/7)
Explanation:
R = 3.5/2m = 1.75m is the inner radius of a round well.
Say h = 10m for the circular well's depth.
(i) The inner curved surface area equals 2rh
= (2×(22/7 )×1.75×10)
= 110 m2
The round well's inner curved surface area is 110 m2, as a result.
(ii) Plastering one square metre costs 40 rupees.
Plastering a 110 m2 space will cost you Rs (110 x 40).
= Rs.4400
Therefore, it will cost Rs. 4,400 to plaster the well's curving surface.
16. A cylindrical pipe with a diameter of 5 cm and a length of 28 m exists in a hot water heating system. Find the system's entire radiating surface. (Assume 1/22 = )
16. A cylindrical pipe with a diameter of 5 cm and a length of 28 m exists in a hot water heating system. Find the system's entire radiating surface. (Assume 1/22 = )
Explanation:
28 metres is equal to the height of a cylindrical pipe.
diameter/2 = 5/2 cm = 2.5 cm = 0.025m, which is the radius of the pipe's round end.
Now, the CSA of a cylindrical pipe is equal to 2rh, where r is the cylinder's radius and h its height.
= 2×(22/7)×0.025×28 m2
= 4.4m2
The system's radiating surface measures 4.4 m2.
17. Find
(i) a closed cylindrical, 4.2-meter-diameter, 4.5-meter-tall fuel storage tank's lateral or curved surface area.
If 1/12 of the steel actually used was squandered in the tank's construction, how much steel was actually used? Suppose = 22/7.
17. Find
(i) a closed cylindrical, 4.2-meter-diameter, 4.5-meter-tall fuel storage tank's lateral or curved surface area.
If 1/12 of the steel actually used was squandered in the tank's construction, how much steel was actually used? Suppose = 22/7.
Explanation:
Tank's cylindrical height is 4.5 metres high.
The circular end's radius is r, which equals (4.2/2)m = 2.1m.
(i) The cylindrical tank has a surface area that is 2 rh on the lateral or curved side.
= 2×(22/7)×2.1×4.5 m2
= (44×0.3×4.5) m2
= 59.4 m2
Accordingly, the tank's CSA is 59.4 m2.
(ii) The tank's total surface area is 2r(r+h).
= 2×(22/7)×2.1×(2.1+4.5)
= 44×0.3×6.6
= 87.12 m2
Let's say that S m2 of steel sheet was actually used to construct the tank.
S(1 -1/12) = 87.12 m2
This suggests that S = 95.04 m2.
As a result, 95.04m2 of steel were actually used to construct this tank.
18. You can notice a lampshade's frame in fig. 13.12. It should be draped with a pretty material.
The frame is 30 cm tall with a 20 cm base diameter. To fold it over the top and bottom of the frame, provide a gap of 2.5 cm. Learn how much fabric is needed to completely cover the lampshade. (Assume 1/22 = )
18. You can notice a lampshade's frame in fig. 13.12. It should be draped with a pretty material.
The frame is 30 cm tall with a 20 cm base diameter. To fold it over the top and bottom of the frame, provide a gap of 2.5 cm. Learn how much fabric is needed to completely cover the lampshade. (Assume 1/22 = )
Explanation:
Say that h is equal to the height of the lampshade's frame, which has a cylindrical appearance.
R stands for radius.
H = (2.5+30+2.5) cm = 35 cm is the total height.
r = (20/2) cm = 10cm
The cloth needed to cover the lampshade is 2rh, which can be calculated using the curved surface area formula.
= (2×(22/7)×10×35) cm2
= 2200 cm2
19. The Vidyalaya students were asked to take part in a competition for creating and embellishing cardboard penholders in the form of a cylinder with a base. Each pen holder had to have a 3 cm radius and a 10.5 cm height. Cardboard was to be provided by the Vidyalaya to the participants. How much cardboard would need to be purchased for the tournament if there were 35 competitors? (Presume = 22/7.
19. The Vidyalaya students were asked to take part in a competition for creating and embellishing cardboard penholders in the form of a cylinder with a base. Each pen holder had to have a 3 cm radius and a 10.5 cm height. Cardboard was to be provided by the Vidyalaya to the participants. How much cardboard would need to be purchased for the tournament if there were 35 competitors? (Presume = 22/7.
Explanation:
The cylindrical penholder's circular end has a 3 cm radius.
Holder's height is 10.5 centimetres.
A penholder's surface area is calculated by adding its CSA and base area.
= 2πrh+πr2
= 2×(22/7)×3×10.5+(22/7)×32= 1584/7
In light of this, one competitor uses a cardboard sheet with an area of 1584/7 cm2.
Therefore, the area of the cardboard sheet used by 35 competitors is 351584.7, which equals 7920 cm2.
The competition will therefore require a cardboard sheet of 7920 cm2.
Also Read: Surface Area and Volume Extra Questions
