Textbook - Surface Areas and Volumes Questions in English

(A) Since Mary wants to paste the paper on the outer surface of the box,the quantity of paper required is equal to the total surface area of the box,which is in the shape of a cuboid.
The dimensions of the box are: Length $(l)$ = $80 \, cm$,Breadth $(b)$ = $40 \, cm$,Height $(h)$ = $20 \, cm$.
The total surface area of the cuboid = $2(lb + bh + hl)$.
Surface area = $2[(80 \times 40) + (40 \times 20) + (20 \times 80)] \, cm^2$.
Surface area = $2[3200 + 800 + 1600] \, cm^2$.
Surface area = $2 \times 5600 \, cm^2 = 11200 \, cm^2$.
The area of each square sheet of paper = side $\times$ side = $40 \, cm \times 40 \, cm = 1600 \, cm^2$.
The number of sheets required = (Total surface area of the box) / (Area of one sheet of paper).
Number of sheets = $11200 / 1600 = 7$.
Therefore,she would require $7$ sheets.

(B) Since Hameed is getting the five outer faces of the tank covered with tiles,he needs to calculate the surface area of these five faces to determine the number of tiles required.
Edge of the cubical tank $(a)$ $= 1.5\, m = 150\, cm$.
Surface area of the five faces $= 5 \times a^2 = 5 \times 150 \times 150\, cm^2 = 112500\, cm^2$.
Area of each square tile $= 25\, cm \times 25\, cm = 625\, cm^2$.
Number of tiles required $= \frac{\text{Total surface area}}{\text{Area of one tile}} = \frac{112500}{625} = 180$.
Cost of $1$ dozen $(12)$ tiles $= Rs. 360$.
Cost of $1$ tile $= \frac{360}{12} = Rs. 30$.
Total cost for $180$ tiles $= 180 \times 30 = Rs. 5400$.

(109) Given:
Length $(l) = 1.5\, m$
Breadth $(b) = 1.25\, m$
Depth $(h) = 65\, cm = 0.65\, m$
$(i)$ Since the box is open at the top,the area of the sheet required is the sum of the area of the four walls and the base:
Area $= 2lh + 2bh + lb$
$= [2 \times 1.5 \times 0.65 + 2 \times 1.25 \times 0.65 + 1.5 \times 1.25]\, m^2$
$= (1.95 + 1.625 + 1.875)\, m^2 = 5.45\, m^2$
$(ii)$ Cost of sheet per $m^2 = Rs. 20$
Total cost for $5.45\, m^2 = 5.45 \times 20 = Rs. 109$

(D) Given: Length $(l) = 5\, m$,Breadth $(b) = 4\, m$,Height $(h) = 3\, m$.
We need to whitewash the four walls and the ceiling.
Area to be whitewashed = Area of four walls + Area of ceiling
$= 2(l + b)h + (l \times b)$
$= 2(5 + 4) \times 3 + (5 \times 4)$
$= 2(9) \times 3 + 20$
$= 54 + 20 = 74\, m^2$.
Cost of whitewashing per $m^2 = Rs. 7.50$.
Total cost = $74 \times 7.50 = Rs. 555$.

(A) Let the length,breadth,and height of the rectangular hall be $l\, m, b\, m,$ and $h\, m$ respectively.
The area of the four walls is given by the formula: $\text{Area} = 2lh + 2bh = 2(l + b)h$.
The perimeter of the floor of the hall is $2(l + b) = 250\, m$.
Substituting the perimeter into the area formula,we get: $\text{Area of four walls} = 250 \times h = 250h\, m^2$.
The cost of painting is given as $Rs. 10$ per $m^2$.
Total cost of painting = $\text{Area} \times \text{Rate} = 250h \times 10 = 2500h$.
Given that the total cost is $Rs. 15000$,we set up the equation: $2500h = 15000$.
Solving for $h$: $h = \frac{15000}{2500} = 6\, m$.
Therefore,the height of the hall is $6\, m$.

(B) The total surface area of one brick is given by the formula $2(lb + bh + lh)$.
Substituting the given dimensions: $l = 22.5 \,cm$,$b = 10 \,cm$,$h = 7.5 \,cm$.
Surface area of one brick $= 2(22.5 \times 10 + 10 \times 7.5 + 22.5 \times 7.5) \,cm^2$.
$= 2(225 + 75 + 168.75) \,cm^2 = 2(468.75) \,cm^2 = 937.5 \,cm^2$.
Convert the total paintable area from $m^2$ to $cm^2$: $9.375 \,m^2 = 9.375 \times 10000 \,cm^2 = 93750 \,cm^2$.
Let $n$ be the number of bricks that can be painted.
$n \times 937.5 = 93750$.
$n = \frac{93750}{937.5} = 100$.
Thus,$100$ bricks can be painted.

(C) Lateral surface area of a cube $= 4 \times (\text{edge})^2$.
Given edge of the cube $= 10\, cm$.
Lateral surface area of the cubical box $= 4 \times (10\, cm)^2 = 4 \times 100\, cm^2 = 400\, cm^2$.
Lateral surface area of a cuboid $= 2 \times h \times (l + b)$.
Given length $(l) = 12.5\, cm$,breadth $(b) = 10\, cm$,and height $(h) = 8\, cm$.
Lateral surface area of the cuboidal box $= 2 \times 8\, cm \times (12.5\, cm + 10\, cm) = 16\, cm \times 22.5\, cm = 360\, cm^2$.
Comparing the two areas,$400\, cm^2 > 360\, cm^2$.
The cubical box has a greater lateral surface area.
The difference is $400\, cm^2 - 360\, cm^2 = 40\, cm^2$.

(D) Total surface area of the cubical box $= 6 \times (\text{edge})^2 = 6 \times (10\, cm)^2 = 600\, cm^2$.
Total surface area of the cuboidal box $= 2(lh + bh + lb) = 2(12.5 \times 8 + 10 \times 8 + 12.5 \times 10) = 2(100 + 80 + 125) = 2(305) = 610\, cm^2$.
Comparing the two, the total surface area of the cubical box $(600\, cm^2)$ is smaller than that of the cuboidal box $(610\, cm^2)$.
The difference in surface area $= 610\, cm^2 - 600\, cm^2 = 10\, cm^2$.
Thus, the cubical box has a smaller total surface area by $10\, cm^2$.

(A) The greenhouse is in the shape of a cuboid.
Length $(l) = 30 \, cm$
Breadth $(b) = 25 \, cm$
Height $(h) = 25 \, cm$
The total surface area of the glass is equal to the total surface area of the cuboid.
Total Surface Area $= 2(lb + lh + bh)$
$= 2(30 \times 25 + 30 \times 25 + 25 \times 25) \, cm^2$
$= 2(750 + 750 + 625) \, cm^2$
$= 2(2125) \, cm^2$
$= 4250 \, cm^2$
Thus,the total area of the glass is $4250 \, cm^2$.

(B) cuboid has $12$ edges,consisting of $4$ lengths $(l)$,$4$ widths $(b)$,and $4$ heights $(h)$.
Given dimensions are length $(l)$ = $30 \, cm$,width $(b)$ = $25 \, cm$,and height $(h)$ = $25 \, cm$.
The total length of tape required for all $12$ edges is given by the formula:
Total length $= 4(l + b + h)$
$= 4(30 + 25 + 25) \, cm$
$= 4(80) \, cm$
$= 320 \, cm$
Therefore,$320 \, cm$ of tape is required.

(C) Length $(l_1)$ of bigger box $= 25 \,cm$,Breadth $(b_1) = 20 \,cm$,Height $(h_1) = 5 \,cm$.
Total surface area of bigger box $= 2(l_1b_1 + l_1h_1 + b_1h_1) = 2(25 \times 20 + 25 \times 5 + 20 \times 5) \,cm^2 = 2(500 + 125 + 100) \,cm^2 = 1450 \,cm^2$.
Extra area for overlaps $= 5\% \text{ of } 1450 \,cm^2 = (1450 \times 0.05) \,cm^2 = 72.5 \,cm^2$.
Total area for $1$ bigger box $= 1450 + 72.5 = 1522.5 \,cm^2$.
Area for $250$ bigger boxes $= 250 \times 1522.5 = 380625 \,cm^2$.
Length $(l_2) = 15 \,cm$,Breadth $(b_2) = 12 \,cm$,Height $(h_2) = 5 \,cm$.
Total surface area of smaller box $= 2(15 \times 12 + 15 \times 5 + 12 \times 5) \,cm^2 = 2(180 + 75 + 60) \,cm^2 = 630 \,cm^2$.
Extra area for overlaps $= 5\% \text{ of } 630 \,cm^2 = (630 \times 0.05) \,cm^2 = 31.5 \,cm^2$.
Total area for $1$ smaller box $= 630 + 31.5 = 661.5 \,cm^2$.
Area for $250$ smaller boxes $= 250 \times 661.5 = 165375 \,cm^2$.
Total cardboard area $= 380625 + 165375 = 546000 \,cm^2$.
Cost $= (546000 / 1000) \times 4 = 546 \times 4 = Rs. 2184$.

(D) The shelter is a cuboidal structure without a base.
Length $(l) = 4 \, m$,Breadth $(b) = 3 \, m$,and Height $(h) = 2.5 \, m$.
The tarpaulin is required for the four walls and the top.
Area of tarpaulin required $= \text{Lateral Surface Area} + \text{Area of the top}$
$= 2(lh + bh) + (l \times b)$
$= [2(4 \times 2.5 + 3 \times 2.5) + (4 \times 3)] \, m^2$
$= [2(10 + 7.5) + 12] \, m^2$
$= [2(17.5) + 12] \, m^2$
$= (35 + 12) \, m^2 = 47 \, m^2$.
Thus,$47 \, m^2$ of tarpaulin is required.

(A) The kaleidoscope is cylindrical in shape.
Radius of the base of the cylinder $(r) = 3.5\, cm$.
Height (length) of the cylinder $(h) = 25\, cm$.
The area of the chart paper required is equal to the curved surface area of the cylinder.
Curved Surface Area $= 2\pi rh$
$= 2 \times \frac{22}{7} \times 3.5 \times 25\, cm^{2}$
$= 2 \times 22 \times 0.5 \times 25\, cm^{2}$
$= 44 \times 12.5\, cm^{2}$
$= 550\, cm^{2}$

(B) Given:
Height $(h) = 14\, cm$
Curved surface area $(CSA) = 88\, cm^2$
Let the radius of the base be $r$ and the diameter be $d = 2r$.
The formula for the curved surface area of a cylinder is $CSA = 2\pi rh$.
Since $d = 2r$,we can write $CSA = \pi dh$.
Substituting the given values:
$88 = \frac{22}{7} \times d \times 14$
$88 = 22 \times d \times 2$
$88 = 44 \times d$
$d = \frac{88}{44} = 2\, cm$.
Thus,the diameter of the base of the cylinder is $2\, cm$.

(C) The height $(h)$ of the cylindrical tank is $1 \,m$.
The base diameter is $140 \,cm$,so the base radius $(r)$ is $\frac{140}{2} \,cm = 70 \,cm = 0.7 \,m$.
The area of the sheet required is equal to the total surface area of the closed cylindrical tank,which is given by the formula $2\pi r(r + h)$.
Substituting the values: $\text{Area} = 2 \times \frac{22}{7} \times 0.7 \times (0.7 + 1) \,m^2$.
$\text{Area} = 2 \times 22 \times 0.1 \times 1.7 \,m^2$.
$\text{Area} = 4.4 \times 1.7 \,m^2 = 7.48 \,m^2$.
Thus,$7.48 \,m^2$ of the metal sheet is required.

(D) Given:
Length (height) of the cylindrical pipe,$h = 77\, cm$.
Inner diameter,$d_{1} = 4\, cm$.
Inner radius,$r_{1} = \frac{d_{1}}{2} = \frac{4}{2} = 2\, cm$.
The formula for the curved surface area $(CSA)$ of a cylinder is $2\pi rh$.
Inner curved surface area $= 2 \times \pi \times r_{1} \times h$
$= 2 \times \frac{22}{7} \times 2 \times 77$
$= 2 \times 22 \times 2 \times 11$
$= 968\, cm^{2}$.
Thus,the inner curved surface area of the metal pipe is $968\, cm^{2}$.

(A) Given: Length of the pipe $(h)$ $= 77 \,cm$.
Outer diameter $(D)$ $= 4.4 \,cm$.
Outer radius $(R)$ $= \frac{D}{2} = \frac{4.4}{2} = 2.2 \,cm$.
The outer curved surface area of a cylinder is given by the formula $2 \pi R h$.
Outer curved surface area $= 2 \times \frac{22}{7} \times 2.2 \times 77 \,cm^{2}$.
$= 2 \times 22 \times 2.2 \times 11 \,cm^{2}$.
$= 44 \times 24.2 \,cm^{2}$.
$= 1064.8 \,cm^{2}$.

(B) Given:
Length of the pipe $(h)$ = $77 \, cm$
Inner diameter $(d_1)$ = $4 \, cm$,so inner radius $(r_1)$ = $2 \, cm$
Outer diameter $(d_2)$ = $4.4 \, cm$,so outer radius $(r_2)$ = $2.2 \, cm$
Total surface area of the pipe = (Inner curved surface area) + (Outer curved surface area) + (Area of two circular rings at the ends)
Inner curved surface area = $2 \pi r_1 h = 2 \times \frac{22}{7} \times 2 \times 77 = 2 \times 22 \times 2 \times 11 = 968 \, cm^2$
Outer curved surface area = $2 \pi r_2 h = 2 \times \frac{22}{7} \times 2.2 \times 77 = 2 \times 22 \times 2.2 \times 11 = 1064.8 \, cm^2$
Area of two circular rings = $2 \times \pi (r_2^2 - r_1^2) = 2 \times \frac{22}{7} \times (2.2^2 - 2^2) = 2 \times \frac{22}{7} \times (4.84 - 4) = 2 \times \frac{22}{7} \times 0.84 = 2 \times 22 \times 0.12 = 5.28 \, cm^2$
Total surface area = $968 + 1064.8 + 5.28 = 2038.08 \, cm^2$.

(C) The roller is cylindrical in shape.
Height $(h)$ of the cylindrical roller = Length of the roller = $120\, cm$.
Radius $(r)$ of the circular end of the roller = $\frac{84}{2}\, cm = 42\, cm$.
Curved Surface Area $(CSA)$ of the roller = $2\pi rh$.
$CSA = 2 \times \frac{22}{7} \times 42 \times 120\, cm^2$.
$CSA = 2 \times 22 \times 6 \times 120\, cm^2 = 31680\, cm^2$.
The area of the playground is equal to the total area covered in $500$ revolutions.
Area of the playground = $500 \times CSA$ of the roller.
Area = $500 \times 31680\, cm^2 = 15840000\, cm^2$.
Since $1\, m^2 = 10000\, cm^2$,we convert the area to $m^2$:
Area = $\frac{15840000}{10000}\, m^2 = 1584\, m^2$.

(D) Given: Diameter of the cylindrical pillar $= 50 \, cm = 0.5 \, m$.
Radius $(r) = \frac{0.5}{2} = 0.25 \, m$.
Height $(h) = 3.5 \, m$.
Curved Surface Area $(CSA)$ of the cylinder $= 2 \pi rh$.
$CSA = 2 \times \frac{22}{7} \times 0.25 \times 3.5 \, m^2$.
$CSA = 2 \times \frac{22}{7} \times 0.25 \times \frac{35}{10} = 2 \times 22 \times 0.25 \times 0.5 = 5.5 \, m^2$.
Cost of painting per $m^2 = Rs. \, 12.50$.
Total cost of painting $= 5.5 \times 12.50 = Rs. \, 68.75$.

(A) Let the height of the right circular cylinder be $h$.
The radius $(r)$ of the base of the cylinder is given as $0.7 \, m$.
The curved surface area $(CSA)$ of the cylinder is given as $4.4 \, m^2$.
The formula for the curved surface area of a cylinder is $CSA = 2 \pi r h$.
Substituting the given values into the formula:
$2 \times \frac{22}{7} \times 0.7 \times h = 4.4$
Simplify the expression:
$2 \times 22 \times 0.1 \times h = 4.4$
$4.4 \times h = 4.4$
Solving for $h$:
$h = \frac{4.4}{4.4} = 1 \, m$.
Therefore,the height of the cylinder is $1 \, m$.

(B) The inner diameter of the circular well is $d = 3.5 \, m$.
The inner radius $(r)$ is given by $r = \frac{d}{2} = \frac{3.5}{2} \, m = 1.75 \, m$.
The depth $(h)$ of the circular well is $10 \, m$.
The inner curved surface area of a cylinder is given by the formula $2 \pi r h$.
Substituting the values,we get:
Inner curved surface area $= 2 \times \frac{22}{7} \times 1.75 \times 10 \, m^2$.
$= 2 \times \frac{22}{7} \times \frac{175}{100} \times 10 \, m^2$.
$= 2 \times 22 \times \frac{1}{7} \times \frac{7}{4} \times 10 \, m^2$.
$= 44 \times 0.25 \times 10 \, m^2 = 110 \, m^2$.
Thus,the inner curved surface area of the circular well is $110 \, m^2$.

$(C)$ Given: Diameter of the well $(d)$ $= 3.5\, m$, Depth $(h)$ $= 10\, m$.
Radius $(r)$ $= d/2 = 3.5/2 = 1.75\, m$.
The curved surface area $(CSA)$ of the cylindrical well $= 2\pi rh$.
$CSA = 2 \times (22/7) \times 1.75 \times 10$.
$CSA = 2 \times (22/7) \times 17.5 = 44 \times 2.5 = 110\, m^2$.
Cost of plastering $= \text{Area} \times \text{Rate} = 110\, m^2 \times Rs. 40/m^2 = Rs. 4400$.

(D) The height $(h)$ of the cylindrical pipe is equal to its length,so $h = 28 \, m$.
The radius $(r)$ of the circular end of the pipe is half of the diameter,so $r = \frac{5}{2} \, cm = 2.5 \, cm$.
To maintain consistent units,convert the radius into meters: $r = \frac{2.5}{100} \, m = 0.025 \, m$.
The total radiating surface of the pipe is its Curved Surface Area $(CSA)$,which is given by the formula $CSA = 2 \pi r h$.
Substituting the values: $CSA = 2 \times \frac{22}{7} \times 0.025 \times 28$.
$CSA = 2 \times 22 \times 0.025 \times 4 = 44 \times 0.1 = 4.4 \, m^2$.
Thus,the total radiating surface in the system is $4.4 \, m^2$.

(A) Height $(h)$ of the cylindrical tank $= 4.5\, m$.
Radius $(r)$ of the circular end of the cylindrical tank $= \frac{4.2}{2}\, m = 2.1\, m$.
Lateral or curved surface area $(CSA)$ of the cylinder is given by the formula $2\pi rh$.
Substituting the values: $CSA = 2 \times \frac{22}{7} \times 2.1 \times 4.5\, m^2$.
$CSA = 2 \times 22 \times 0.3 \times 4.5\, m^2$.
$CSA = 44 \times 1.35\, m^2$.
$CSA = 59.4\, m^2$.
Therefore,the lateral or curved surface area of the tank is $59.4\, m^2$.

(B) First, calculate the Total Surface Area $(TSA)$ of the closed cylindrical tank:
$TSA = 2\pi r(r + h)$
Given $r = 2.1 \text{ m}$ and $h = 4.5 \text{ m}$,
$TSA = 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5) = 2 \times \frac{22}{7} \times 2.1 \times 6.6 = 87.12 \text{ m}^2$.
Let $A$ be the actual steel used. Since $1/12$ of the steel was wasted, the steel used for the tank is $A - \frac{1}{12}A = \frac{11}{12}A$.
Equating this to the surface area of the tank:
$\frac{11}{12}A = 87.12$
$A = \frac{87.12 \times 12}{11} = 7.92 \times 12 = 95.04 \text{ m}^2$.
Thus, the actual steel used is $95.04 \text{ m}^2$.

(C) The frame of the lampshade is cylindrical in shape.
Given:
Diameter of the base $= 20 \, cm$,so the radius $(r) = \frac{20}{2} = 10 \, cm$.
Height of the frame $= 30 \, cm$.
Margin for folding at the top and bottom $= 2.5 \, cm$ each.
Total height $(h)$ of the cloth required $= 30 \, cm + 2.5 \, cm + 2.5 \, cm = 35 \, cm$.
The cloth required to cover the lampshade is equal to the curved surface area of the cylinder.
Curved Surface Area $= 2 \pi rh$
$= 2 \times \frac{22}{7} \times 10 \times 35 \, cm^2$
$= 2 \times 22 \times 10 \times 5 \, cm^2$
$= 2200 \, cm^2$.
Thus,$2200 \, cm^2$ of cloth is required to cover the lampshade.

(D) The penholder is a cylinder with a base,so its surface area is the sum of the curved surface area and the area of the circular base.
Radius $(r) = 3 \,cm$
Height $(h) = 10.5 \,cm$
Surface area of $1$ penholder $= 2\pi rh + \pi r^2 = \pi r(2h + r)$
$= \frac{22}{7} \times 3 \times (2 \times 10.5 + 3) \,cm^2$
$= \frac{22}{7} \times 3 \times (21 + 3) \,cm^2$
$= \frac{22}{7} \times 3 \times 24 \,cm^2 = \frac{1584}{7} \,cm^2$
Total cardboard required for $35$ competitors $= 35 \times \left(\frac{1584}{7}\right) \,cm^2$
$= 5 \times 1584 \,cm^2 = 7920 \,cm^2$
Thus,$7920 \,cm^2$ of cardboard is required.

(A) The formula for the curved surface area of a right circular cone is given by $CSA = \pi r l$,where $r$ is the base radius and $l$ is the slant height.
Given:
Base radius $r = 7\, cm$
Slant height $l = 10\, cm$
Substituting the values into the formula:
$CSA = \frac{22}{7} \times 7 \times 10\, cm^{2}$
$CSA = 22 \times 10\, cm^{2}$
$CSA = 220\, cm^{2}$

(B) Given: height $h = 16 \, cm$ and radius $r = 12 \, cm$.
First,calculate the slant height $l$ using the formula $l = \sqrt{h^2 + r^2}$.
$l = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \, cm$.
Curved surface area of the cone $= \pi rl = 3.14 \times 12 \times 20 = 753.6 \, cm^2$.
Total surface area of the cone $= \pi rl + \pi r^2 = \pi r(l + r) = 3.14 \times 12 \times (20 + 12) = 3.14 \times 12 \times 32 = 1205.76 \, cm^2$.

(C) Since the grains of corn are found only on the curved surface of the corn cob,we need to calculate the curved surface area of the cone.
Given:
Radius $(r)$ = $2.1\, cm$
Height $(h)$ = $20\, cm$
First,we find the slant height $(l)$:
$l = \sqrt{r^2 + h^2} = \sqrt{(2.1)^2 + (20)^2} = \sqrt{4.41 + 400} = \sqrt{404.41} \approx 20.11\, cm$
Curved Surface Area of the cone = $\pi rl$
$= \frac{22}{7} \times 2.1 \times 20.11 = 22 \times 0.3 \times 20.11 = 6.6 \times 20.11 = 132.726\, cm^2$
Given that there are $4$ grains per $1\, cm^2$:
Total number of grains = $132.726 \times 4 = 530.904$
Rounding to the nearest whole number,we get $531$ grains.

(A) Given,diameter of the base of the cone $= 10.5\, cm$.
Radius $(r) = \frac{10.5}{2}\, cm = 5.25\, cm$.
Slant height $(l) = 10\, cm$.
The formula for the curved surface area of a cone is $CSA = \pi rl$.
Substituting the values:
$CSA = \frac{22}{7} \times \frac{10.5}{2} \times 10\, cm^2$.
$CSA = \frac{22}{7} \times 5.25 \times 10\, cm^2$.
$CSA = \frac{22}{7} \times 52.5\, cm^2$.
$CSA = 22 \times 7.5\, cm^2 = 165\, cm^2$.
Therefore,the curved surface area of the cone is $165\, cm^2$.

(B) Given:
Diameter of the base $(d) = 24 \, m$
Radius of the base $(r) = \frac{d}{2} = \frac{24}{2} = 12 \, m$
Slant height $(l) = 21 \, m$
The formula for the total surface area of a cone is given by:
Total Surface Area $= \pi r(r + l)$
Substituting the values:
Total Surface Area $= \frac{22}{7} \times 12 \times (12 + 21) \, m^2$
$= \frac{22}{7} \times 12 \times 33 \, m^2$
$= \frac{8712}{7} \, m^2$
$\approx 1244.57 \, m^2$

(N/A) Given: Curved surface area $= 308 \, cm^2$,Slant height $(l) = 14 \, cm$.
$(i)$ Let the radius of the base be $r \, cm$.
We know that the curved surface area of a cone is given by $\pi r l$.
$\therefore \pi r l = 308$
$\Rightarrow \frac{22}{7} \times r \times 14 = 308$
$\Rightarrow 44 \times r = 308$
$\Rightarrow r = \frac{308}{44} = 7 \, cm$.
Thus,the radius of the base is $7 \, cm$.
$(ii)$ Total surface area of a cone $= \text{Curved surface area} + \text{Base area}$.
Base area $= \pi r^2 = \frac{22}{7} \times 7^2 = 154 \, cm^2$.
Total surface area $= 308 \, cm^2 + 154 \, cm^2 = 462 \, cm^2$.

(D) Given:
Height of the tent $(h) = 10 \, m$
Radius of the base $(r) = 24 \, m$
The slant height $(l)$ of a cone is given by the formula:
$l = \sqrt{h^2 + r^2}$
Substituting the given values:
$l = \sqrt{10^2 + 24^2} \, m$
$l = \sqrt{100 + 576} \, m$
$l = \sqrt{676} \, m$
$l = 26 \, m$
Thus,the slant height of the tent is $26 \, m$.

(A) Given: Height of the cone $(h) = 10 \, m$,Radius of the base $(r) = 24 \, m$.
First,calculate the slant height $(l)$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \, m$.
The curved surface area of the cone is given by $\pi r l$.
Area of the canvas required $= \frac{22}{7} \times 24 \times 26 \, m^2 = \frac{13728}{7} \, m^2$.
The cost of $1 \, m^2$ canvas is $Rs. 70$.
Total cost of the canvas $= \text{Area} \times \text{Rate} = \frac{13728}{7} \times 70 = 13728 \times 10 = Rs. 137280$.

(B) Given: Base radius $(r) = 6\, m$,Height $(h) = 8\, m$.
Slant height $(l) = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\, m$.
Curved surface area of the tent $= \pi rl = 3.14 \times 6 \times 10 = 188.4\, m^2$.
Let the length of the tarpaulin be $L$. Since the width is $3\, m$,the area is $L \times 3 = 188.4\, m^2$.
$L = \frac{188.4}{3} = 62.8\, m$.
Extra length for margins and wastage $= 20\, cm = 0.2\, m$.
Total length required $= 62.8\, m + 0.2\, m = 63.0\, m$.

(A) Given: Slant height $(l) = 25 \, m$ and base diameter $(d) = 14 \, m$.
First,calculate the radius $(r)$ of the base:
$r = \frac{d}{2} = \frac{14}{2} = 7 \, m$.
Next,calculate the curved surface area $(CSA)$ of the conical tomb:
$CSA = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \, m^2$.
The rate of white-washing is $Rs. 210$ per $100 \, m^2$.
Therefore,the total cost of white-washing is:
$\text{Cost} = \left( \frac{210}{100} \right) \times 550 = 2.1 \times 550 = Rs. 1155$.

(B) Given: Radius of the base $(r) = 7\, cm$ and height $(h) = 24\, cm$.
First,calculate the slant height $(l)$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\, cm$.
The lateral surface area of one cone is given by $\pi rl$.
Lateral surface area $= \frac{22}{7} \times 7 \times 25 = 550\, cm^2$.
To find the area of the sheet required for $10$ such caps,multiply the area of one cap by $10$.
Total area $= 10 \times 550 = 5500\, cm^2$.

(C) Given:
Diameter of the base $= 40 \, cm$
Radius $(r) = \frac{40}{2} \, cm = 20 \, cm = 0.2 \, m$
Height $(h) = 1 \, m$
Slant height $(l) = \sqrt{r^2 + h^2} = \sqrt{(0.2)^2 + (1)^2} \, m = \sqrt{0.04 + 1} \, m = \sqrt{1.04} \, m = 1.02 \, m$
Curved surface area of one cone $= \pi r l = 3.14 \times 0.2 \times 1.02 \, m^2$
Curved surface area of $50$ cones $= 50 \times 3.14 \times 0.2 \times 1.02 \, m^2$
$= 50 \times 0.2 \times 3.14 \times 1.02 \, m^2 = 10 \times 3.14 \times 1.02 \, m^2 = 32.028 \, m^2$
Cost of painting $= \text{Total Area} \times \text{Rate} = 32.028 \times 12 = Rs. 384.336$
Rounding to two decimal places,the cost is $Rs. 384.34$.

(A) The surface area of a sphere is given by the formula $4 \pi r^{2}$.
Given the radius $r = 7 \, cm$.
Substituting the value of $r$ in the formula:
Surface Area $= 4 \times \frac{22}{7} \times (7 \, cm)^{2}$
$= 4 \times \frac{22}{7} \times 49 \, cm^{2}$
$= 4 \times 22 \times 7 \, cm^{2}$
$= 88 \times 7 \, cm^{2}$
$= 616 \, cm^{2}$.

(B) Given radius $r = 21\, cm$.
$(i)$ The curved surface area of a hemisphere is given by the formula $2\pi r^2$.
$= 2 \times \frac{22}{7} \times 21 \times 21\, cm^2$
$= 2 \times 22 \times 3 \times 21\, cm^2 = 2772\, cm^2$.
$(ii)$ The total surface area of a hemisphere is given by the formula $3\pi r^2$.
$= 3 \times \frac{22}{7} \times 21 \times 21\, cm^2$
$= 3 \times 22 \times 3 \times 21\, cm^2 = 4158\, cm^2$.

(C) The diameter of the hollow sphere is $d = 7 \,m$.
The radius $r$ of the sphere is given by $r = \frac{d}{2} = \frac{7}{2} = 3.5 \,m$.
The area available to the motorcyclist for riding is the surface area of the sphere,which is calculated using the formula $A = 4 \pi r^2$.
Substituting the values:
$A = 4 \times \frac{22}{7} \times 3.5 \times 3.5$
$A = 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$A = 4 \times 22 \times \frac{7}{4}$
$A = 22 \times 7 = 154 \,m^2$.
Thus,the area available for riding is $154 \,m^2$.

(D) Since only the rounded surface of the dome is to be painted,we need to find the curved surface area of the hemisphere.
Circumference of the base of the dome $= 17.6 \, m$.
Let the radius be $r$. Then,$2\pi r = 17.6$.
$r = \frac{17.6}{2 \times \pi} = \frac{17.6 \times 7}{2 \times 22} = 0.8 \times 3.5 = 2.8 \, m$.
Curved surface area of the hemisphere $= 2\pi r^2 = 2 \times \frac{22}{7} \times 2.8 \times 2.8 = 49.28 \, m^2$.
Cost of painting $100 \, cm^2 = Rs. 5$.
Since $1 \, m^2 = 10000 \, cm^2$,the cost of painting $1 \, m^2 = \frac{5}{100} \times 10000 = Rs. 500$.
Total cost of painting $= 49.28 \times 500 = Rs. 24640$.

(A) The radius $(r)$ of the sphere is $10.5 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^{2}$.
Substituting the given values:
Surface area $= 4 \times \frac{22}{7} \times (10.5)^{2} \, cm^{2}$
$= 4 \times \frac{22}{7} \times 10.5 \times 10.5 \, cm^{2}$
$= 4 \times 22 \times 1.5 \times 10.5 \, cm^{2}$
$= 88 \times 15.75 \, cm^{2}$
$= 1386 \, cm^{2}$
Therefore,the surface area of the sphere is $1386 \, cm^{2}$.

(B) The radius $(r)$ of the sphere is $5.6 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^{2}$.
Substituting the given values:
Surface area $= 4 \times \frac{22}{7} \times (5.6)^{2} \, cm^{2}$
$= 4 \times \frac{22}{7} \times 5.6 \times 5.6 \, cm^{2}$
$= 4 \times 22 \times 0.8 \times 5.6 \, cm^{2}$
$= 88 \times 4.48 \, cm^{2}$
$= 394.24 \, cm^{2}$.
Therefore,the surface area of the sphere is $394.24 \, cm^{2}$.

(C) The radius $(r)$ of the sphere is $14 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^2$.
Substituting the given values:
Surface area $= 4 \times \frac{22}{7} \times (14)^2 \, cm^2$
$= 4 \times \frac{22}{7} \times 14 \times 14 \, cm^2$
$= 4 \times 22 \times 2 \times 14 \, cm^2$
$= 88 \times 28 \, cm^2$
$= 2464 \, cm^2$.
Therefore,the surface area of the sphere is $2464 \, cm^2$.

(D) The radius $(r)$ of the sphere is given by:
$r = \frac{\text{Diameter}}{2} = \frac{14 \, cm}{2} = 7 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^2$.
Substituting the values:
Surface Area $= 4 \times \frac{22}{7} \times (7 \, cm)^2$
$= 4 \times \frac{22}{7} \times 49 \, cm^2$
$= 4 \times 22 \times 7 \, cm^2$
$= 88 \times 7 \, cm^2$
$= 616 \, cm^2$.
Therefore,the surface area of the sphere is $616 \, cm^2$.

(A) Given,diameter of the sphere $d = 21 \, cm$.
Radius $(r) = \frac{d}{2} = \frac{21}{2} \, cm = 10.5 \, cm$.
The surface area of a sphere is given by the formula $4 \pi r^{2}$.
Substituting the values: $\text{Surface Area} = 4 \times \frac{22}{7} \times (10.5)^{2} \, cm^{2}$.
$= 4 \times \frac{22}{7} \times 10.5 \times 10.5 \, cm^{2}$.
$= 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \, cm^{2}$.
$= 22 \times 3 \times 21 \, cm^{2} = 1386 \, cm^{2}$.
Thus,the surface area of the sphere is $1386 \, cm^{2}$.

(B) Given: Diameter $(d) = 3.5 \, cm$.
Radius $(r) = \frac{d}{2} = \frac{3.5}{2} = 1.75 \, cm$.
The surface area of a sphere is given by the formula: $A = 4 \pi r^2$.
Substituting the values: $A = 4 \times \frac{22}{7} \times (1.75)^2$.
$A = 4 \times \frac{22}{7} \times 1.75 \times 1.75$.
$A = 4 \times \frac{22}{7} \times \frac{7}{4} \times 1.75$.
$A = 22 \times 1.75 = 38.5 \, cm^2$.
Thus,the surface area of the sphere is $38.5 \, cm^2$.