Mix Examples - Surface Areas and Volumes Questions in English

(C) The dimensions of the cuboid are length $(l)$ = $50 \, cm$,breadth $(b)$ = $40 \, cm$,and height $(h)$ = $30 \, cm$.
The formula for the lateral surface area of a cuboid is $2h(l + b)$.
Substituting the given values into the formula:
Lateral Surface Area = $2 \times 30 \times (50 + 40)$
= $60 \times 90$
= $5400 \, cm^2$.

(D) When two cubes of edge $a = 8 \, cm$ are joined end to end,they form a cuboid.
The dimensions of the resulting cuboid are:
Length $(l)$ = $8 \, cm + 8 \, cm = 16 \, cm$
Breadth $(b)$ = $8 \, cm$
Height $(h)$ = $8 \, cm$
The total surface area of a cuboid is given by the formula: $TSA = 2(lb + bh + lh)$.
Substituting the values:
$TSA = 2(16 \times 8 + 8 \times 8 + 16 \times 8)$
$TSA = 2(128 + 64 + 128)$
$TSA = 2(320)$
$TSA = 640 \, cm^{2}$.

(A) The lateral surface area of a cube is given by the formula $4a^{2}$,where $a$ is the length of the edge of the cube.
Given,$a = 15 \, cm$.
Lateral Surface Area $= 4 \times (15)^{2} \, cm^{2}$.
Lateral Surface Area $= 4 \times 225 \, cm^{2}$.
Lateral Surface Area $= 900 \, cm^{2}$.

(B) The dimensions of the cuboidal tank are: length $(l)$ = $10 \, m$,breadth $(b)$ = $8 \, m$,and height $(h)$ = $3 \, m$.
The total surface area of a closed cuboidal tank is given by the formula: $2(lb + bh + lh)$.
Substituting the values: $2(10 \times 8 + 8 \times 3 + 10 \times 3) = 2(80 + 24 + 30) = 2(134) = 268 \, m^2$.
The cost of the iron sheet is calculated by multiplying the total surface area by the rate per $m^2$.
Cost = $268 \, m^2 \times ₹ 50/m^2 = ₹ 13,400$.

(C) $1$. Calculate the area of the four walls: $Area = 2(l + b) \times h = 2(7 + 5) \times 4 = 2(12) \times 4 = 96\, m^2$.
$2$. Calculate the area of the ceiling: $Area = l \times b = 7 \times 5 = 35\, m^2$.
$3$. Total area to be whitewashed (excluding doors and windows): $Total\, Area = (Area\, of\, walls + Area\, of\, ceiling) - (Area\, of\, 2\, doors + Area\, of\, 4\, windows)$.
$4$. $Area\, of\, 2\, doors = 2 \times (3 \times 2) = 12\, m^2$.
$5$. $Area\, of\, 4\, windows = 4 \times (2 \times 1) = 8\, m^2$.
$6$. $Total\, Area = (96 + 35) - (12 + 8) = 131 - 20 = 111\, m^2$.
$7$. $Cost = Total\, Area \times Rate = 111 \times 200 = ₹22,200$.

(N/A) Let the dimensions of the cuboidal box be $6x$,$4x$,and $3x$ meters.
The total surface area of the cuboid is $SA = 2(lb + bh + lh) = 2((6x)(4x) + (4x)(3x) + (6x)(3x)) = 2(24x^2 + 12x^2 + 18x^2) = 2(54x^2) = 108x^2 \, m^2$.
The cost of covering with white paper at $₹ 2.00/m^2$ is $2 \times 108x^2 = 216x^2$.
The cost of covering with coloured paper at $₹ 2.50/m^2$ is $2.50 \times 108x^2 = 270x^2$.
The difference in cost is $270x^2 - 216x^2 = 54x^2$.
Given that the difference is $₹ 1350$,we have $54x^2 = 1350$.
$x^2 = 1350 / 54 = 25$.
$x = 5$.
Thus,the dimensions are $6(5) = 30 \, m$,$4(5) = 20 \, m$,and $3(5) = 15 \, m$.

Given for the cylinder:
Radius $(r) = 28 \, cm$
Height $(h) = 90 \, cm$
Curved surface area of a cylinder $= 2 \pi r h$
$= 2 \times \frac{22}{7} \times 28 \times 90 \, cm^2$
$= 2 \times 22 \times 4 \times 90 \, cm^2$
$= 15,840 \, cm^2$
Total surface area of a cylinder $= 2 \pi r(h + r)$
$= 2 \times \frac{22}{7} \times 28 \times (90 + 28) \, cm^2$
$= 176 \times 118 \, cm^2$
$= 20,768 \, cm^2$

(B) For the given cylinder,height $(h) = 30 \, cm$ and curved surface area $= 1320 \, cm^2$.
The formula for the curved surface area of a cylinder is $A = \pi d h$,where $d$ is the diameter.
Substituting the given values: $1320 = \frac{22}{7} \times d \times 30$.
Solving for $d$: $d = \frac{1320 \times 7}{22 \times 30}$.
$d = \frac{1320 \times 7}{660} = 2 \times 7 = 14 \, cm$.
Thus,the diameter of the cylinder is $14 \, cm$.

$(440 \ cm^2)$ For a metal hollow cylinder: height $(h) = 70\, cm$.
Inner radius $(r) = 10\, cm$.
Outer radius $(R) = \text{Inner radius} + \text{Thickness of metal} = 10 + 1 = 11\, cm$.
The difference between the areas of the outer and inner curved surfaces is given by:
$\text{Difference} = 2\pi Rh - 2\pi rh = 2\pi h(R - r)$.
Substituting the values:
$\text{Difference} = 2 \times \frac{22}{7} \times 70 \times (11 - 10) = 2 \times 22 \times 10 \times 1 = 440\, cm^2$.

(D) The formula for the curved surface area $(CSA)$ of a cylinder is given by $CSA = 2 \pi rh$.
Given:
Height $(h)$ = $35 \, cm$
Radius $(r)$ = $15 \, cm$
Using $\pi = \frac{22}{7}$:
$CSA = 2 \times \frac{22}{7} \times 15 \times 35$
$CSA = 2 \times 22 \times 15 \times 5$
$CSA = 44 \times 75$
$CSA = 3300 \, cm^2$
Therefore,the correct option is $D$.

(A) The diameter of the cylinder is $d = 28 \text{ cm}$,so the radius is $r = d/2 = 14 \text{ cm}$.
The height of the cylinder is $h = 50 \text{ cm}$.
The formula for the total surface area of a cylinder is $A = 2\pi r(r + h)$.
Substituting the values: $A = 2 \times (22/7) \times 14 \times (14 + 50)$.
$A = 2 \times 22 \times 2 \times 64$.
$A = 88 \times 64 = 5632 \text{ cm}^2$.

(B) The formula for the curved surface area $(CSA)$ of a cylinder is $CSA = 2\pi rh$,where $r$ is the radius and $h$ is the height.
Given: $CSA = 3960\, cm^{2}$ and $r = 21\, cm$.
Using $\pi = \frac{22}{7}$,we have:
$3960 = 2 \times \frac{22}{7} \times 21 \times h$
$3960 = 2 \times 22 \times 3 \times h$
$3960 = 132 \times h$
$h = \frac{3960}{132}$
$h = 30\, cm$.
Therefore,the height of the cylinder is $30\, cm$.

(C) The curved surface area $(CSA)$ of a cylinder is given by the formula: $CSA = 2 \pi r h$,where $r$ is the radius and $h$ is the height.
Given: $CSA = 550 \, cm^2$ and $h = 50 \, cm$.
Substituting the values: $550 = 2 \times \frac{22}{7} \times r \times 50$.
$550 = \frac{2200}{7} \times r$.
$r = \frac{550 \times 7}{2200} = \frac{7}{4} = 1.75 \, cm$.
The diameter $(d)$ is $2 \times r = 2 \times 1.75 = 3.5 \, cm$.

(A) Given: Radius $(r)$ = $10\,cm$,Height $(h)$ = $30\,cm$,Rate = ₹ $3/cm^2$,$\pi = 3.14$.
The formula for the curved surface area $(CSA)$ of a cylinder is $2\pi rh$.
$CSA$ = $2 \times 3.14 \times 10 \times 30 = 1884\,cm^2$.
Total cost = $CSA$ $\times$ Rate = $1884 \times 3 = 5652$.
Thus,the cost of painting the outer curved surface is ₹ $5652$.

(A) The formula for the total surface area $(TSA)$ of a closed cylinder is given by $TSA = 2\pi r(r + h)$,where $r$ is the radius and $h$ is the height.
Given: $TSA = 748\, cm^2$ and $r = 7\, cm$.
Using $\pi = \frac{22}{7}$:
$748 = 2 \times \frac{22}{7} \times 7 \times (7 + h)$
$748 = 44 \times (7 + h)$
Divide both sides by $44$:
$7 + h = \frac{748}{44}$
$7 + h = 17$
$h = 17 - 7$
$h = 10\, cm$.

(B) The well is in the shape of a cylinder.
Given: Depth $(h)$ = $20 \, m$,Radius $(r)$ = $7 \, m$.
The inner curved surface area of the cylinder = $2 \pi r h$.
Area = $2 \times (22/7) \times 7 \times 20 = 2 \times 22 \times 20 = 880 \, m^2$.
The cost of cementing is ₹ $15$ per $m^2$.
Total cost = $880 \times 15 = ₹ 13,200$.

(C) Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$.
Given that $r_1 : r_2 = 3 : 7$ and $h_1 : h_2 = 28 : 15$.
The curved surface area $(CSA)$ of a cylinder is given by the formula $CSA = 2pi rh$.
Therefore,the ratio of their curved surface areas is:
$\frac{CSA_1}{CSA_2} = \frac{2pi r_1 h_1}{2pi r_2 h_2} = \frac{r_1}{r_2} \times \frac{h_1}{h_2}$
Substituting the given ratios:
$\frac{CSA_1}{CSA_2} = \frac{3}{7} \times \frac{28}{15}$
$\frac{CSA_1}{CSA_2} = \frac{3}{15} \times \frac{28}{7} = \frac{1}{5} \times 4 = \frac{4}{5}$
Thus,the ratio of their curved surface areas is $4 : 5$.

(D) Let the radius of the cylinder be $r$ and the height be $h$.
The curved surface area $(CSA)$ of a cylinder is $2\pi rh$.
The total surface area $(TSA)$ of a cylinder is $2\pi rh + 2\pi r^2 = 2\pi r(h + r)$.
Given the ratio $CSA : TSA = 2 : 5$,we have:
$\frac{2\pi rh}{2\pi r(h + r)} = \frac{2}{5}$
$\frac{h}{h + r} = \frac{2}{5}$
Cross-multiplying gives $5h = 2(h + r)$.
$5h = 2h + 2r$
$3h = 2r$
Therefore,the ratio of the radius to the height is $\frac{r}{h} = \frac{3}{2}$,which is $3: 2$.

(A) $1$. Convert the radius from $cm$ to $m$: $r = 14 \, cm = 0.14 \, m$.
$2$. The formula for the curved surface area $(CSA)$ of one cylinder is $2 \pi rh$.
$3$. $CSA = 2 \times \frac{22}{7} \times 0.14 \times 5 = 4.4 \, m^2$.
$4$. Total $CSA$ for $20$ pillars = $20 \times 4.4 = 88 \, m^2$.
$5$. Total cost = $\text{Total Area} \times \text{Rate} = 88 \times 120 = ₹ 10,560$.

(N/A) For a cone:
Radius $(r) = 15 \, cm$ and slant height $(l) = 25 \, cm$.
Curved surface area of a cone:
$= \pi r l$
$= 3.14 \times 15 \times 25 \, cm^2$
$= 1177.5 \, cm^2$
Total surface area of a cone:
$= \pi r(l + r)$
$= 3.14 \times 15 \times (25 + 15) \, cm^2$
$= 3.14 \times 15 \times 40 \, cm^2$
$= 1884 \, cm^2$

(N/A) Given: Slant height $(l) = 25 \,cm$,Curved Surface Area $(CSA) = 550 \,cm^{2}$.
$1$. Finding the radius $(r)$:
$CSA = \pi r l$
$550 = \frac{22}{7} \times r \times 25$
$r = \frac{550 \times 7}{22 \times 25} = 7 \,cm$.
$2$. Finding the height $(h)$:
Using the relation $l^{2} = h^{2} + r^{2}$:
$25^{2} = h^{2} + 7^{2}$
$625 = h^{2} + 49$
$h^{2} = 625 - 49 = 576$
$h = \sqrt{576} = 24 \,cm$.
$3$. Finding the total surface area $(TSA)$:
$TSA = \pi r(l + r)$
$TSA = \frac{22}{7} \times 7 \times (25 + 7)$
$TSA = 22 \times 32 = 704 \,cm^{2}$.

(A) For a given cone:
Radius $(r) = 7\, m$ and slant height $(l) = 12\, m$.
Area of tarpaulin used in the tent = Curved surface area of the cone.
Curved surface area $= \pi r l = \frac{22}{7} \times 7 \times 12 = 264\, m^2$.
Since the area of the tarpaulin is $l \times b$,where $b = 2\, m$:
$264 = l \times 2 \implies l = \frac{264}{2} = 132\, m$.
Thus,$132\, m$ of tarpaulin is required.
Cost of $1\, m$ tarpaulin = ₹ $48$.
Total cost = $132 \times 48 = ₹ 6336$.

(A) Given:
Diameter of the cone,$d = 6\, cm$.
Radius of the cone,$r = d / 2 = 6 / 2 = 3\, cm$.
Slant height of the cone,$l = 11\, cm$.
The formula for the total surface area of a cone is $TSA = \pi r(r + l)$.
Substituting the values:
$TSA = \pi \times 3 \times (3 + 11)$
$TSA = \pi \times 3 \times 14$
$TSA = 42\pi\, cm^2$.
Using $\pi \approx 22/7$:
$TSA = 42 \times (22 / 7) = 6 \times 22 = 132\, cm^2$.

(N/A) Given: Radius $(r) = 3.5 \, cm$,Height $(h) = 12 \, cm$.
First,find the slant height $(l)$ using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \, cm$.
Curved Surface Area $(CSA) = \pi rl = \frac{22}{7} \times 3.5 \times 12.5 = 22 \times 0.5 \times 12.5 = 137.5 \, cm^2$.
Total Surface Area $(TSA) = \pi r(r + l) = \frac{22}{7} \times 3.5 \times (3.5 + 12.5) = 22 \times 0.5 \times 16 = 11 \times 16 = 176 \, cm^2$.

(N/A) $1$. When revolved about $AB$: The height $h = AB = 28 \, cm$ and the radius $r = BC = 21 \, cm$. The slant height $l = \sqrt{r^2 + h^2} = \sqrt{21^2 + 28^2} = \sqrt{441 + 784} = \sqrt{1225} = 35 \, cm$. The curved surface area $CSA = \pi r l = \frac{22}{7} \times 21 \times 35 = 22 \times 3 \times 35 = 2310 \, cm^2$.
$2$. When revolved about $BC$: The height $h = BC = 21 \, cm$ and the radius $r = AB = 28 \, cm$. The slant height $l = \sqrt{r^2 + h^2} = \sqrt{28^2 + 21^2} = 35 \, cm$. The curved surface area $CSA = \pi r l = \frac{22}{7} \times 28 \times 35 = 22 \times 4 \times 35 = 3080 \, cm^2$.

(B) Given: Diameter of the cone $d = 42 \, m$,so radius $r = 21 \, m$. Height $h = 20 \, m$.
First,calculate the slant height $l$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29 \, m$.
The area of the canvas required is the curved surface area of the cone,given by $CSA = \pi r l$.
$CSA = \frac{22}{7} \times 21 \times 29 = 22 \times 3 \times 29 = 1914 \, m^2$.
The cost of the canvas is $1914 \, m^2 \times ₹ 75/m^2 = ₹ 143550$.

(N/A) Let the radius $r = 2x$ and the slant height $l = 7x$.
The curved surface area of a cone is given by $CSA = \pi rl$.
Given $CSA = 704 \, cm^2$ and taking $\pi = \frac{22}{7}$:
$704 = \frac{22}{7} \times (2x) \times (7x)$
$704 = 22 \times 2 \times x^2$
$704 = 44x^2$
$x^2 = \frac{704}{44} = 16$
$x = 4$.
Thus,radius $r = 2 \times 4 = 8 \, cm$ and slant height $l = 7 \times 4 = 28 \, cm$.
The height $h$ is given by $h = \sqrt{l^2 - r^2} = \sqrt{28^2 - 8^2} = \sqrt{(28-8)(28+8)} = \sqrt{20 \times 36} = \sqrt{720} = 12\sqrt{5} \, cm$.

(B) The total surface area of a cone is given by the formula: $TSA = \pi r(r + l)$,where $r$ is the radius and $l$ is the slant height.
Given: $r = 10\, cm$,$TSA = 880\, cm^2$,and $\pi = 22/7$.
Substituting the values into the formula:
$880 = (22/7) \times 10 \times (10 + l)$
$880 \times 7 = 220 \times (10 + l)$
$6160 = 220 \times (10 + l)$
$10 + l = 6160 / 220$
$10 + l = 28$
$l = 28 - 10$
$l = 18\, cm$.
Thus,the slant height of the cone is $18\, cm$.

(C) The total surface area $(TSA)$ of a cone is given by the formula: $TSA = \pi r(r + l) = \pi rl + \pi r^{2}$.
Given,$TSA = 3696 \, cm^{2}$ and the curved surface area $(CSA)$ is $\pi rl = 2310 \, cm^{2}$.
Substituting the value of $CSA$ into the $TSA$ formula:
$3696 = 2310 + \pi r^{2}$.
$\pi r^{2} = 3696 - 2310 = 1386$.
Using $\pi = \frac{22}{7}$,we get $\frac{22}{7} r^{2} = 1386$.
$r^{2} = \frac{1386 \times 7}{22} = 63 \times 7 = 441$.
$r = \sqrt{441} = 21 \, cm$.
Now,use the $CSA$ formula to find the slant height $(l)$:
$2310 = \pi rl = \frac{22}{7} \times 21 \times l$.
$2310 = 22 \times 3 \times l = 66l$.
$l = \frac{2310}{66} = 35 \, cm$.
The ratio of the radius $(r)$ to the slant height $(l)$ is $r:l = 21:35$.
Dividing both by $7$,we get $3:5$.

(D) Given,the radius of the sphere $(r) = 14 \, cm$.
The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Substituting the value of $r = 14 \, cm$ and $\pi = \frac{22}{7}$:
$A = 4 \times \frac{22}{7} \times 14 \times 14 \, cm^2$
$A = 4 \times 22 \times 2 \times 14 \, cm^2$
$A = 88 \times 28 \, cm^2$
$A = 2464 \, cm^2$.
Thus,the surface area of the sphere is $2464 \, cm^2$.

(A) Given,the diameter of the solid hemisphere is $7 \, cm$.
Radius $(r) = \frac{\text{diameter}}{2} = \frac{7}{2} \, cm = 3.5 \, cm$.
Curved surface area of a hemisphere $= 2 \pi r^2$.
$= 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \, cm^2 = 77 \, cm^2$.
Total surface area of a hemisphere $= 3 \pi r^2$.
$= 3 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \, cm^2 = 115.5 \, cm^2$.

(B) Given,the radius of the sphere $(r) = 3.5\, cm = \frac{7}{2}\, cm$.
The surface area of a sphere is given by the formula $A = 4\pi r^{2}$.
Substituting the values:
$A = 4 \times \frac{22}{7} \times \left(\frac{7}{2}\right)^{2}$
$A = 4 \times \frac{22}{7} \times \frac{49}{4} = 22 \times 7 = 154\, cm^{2}$.
The cost of plating gold is ₹ $150$ per $cm^{2}$.
Total cost = $\text{Surface area} \times \text{Rate}$
Total cost = $154 \times 150 = 23,100$.
Therefore,the cost of plating gold on its surface is ₹ $23,100$.

(C) Let the radius of the sphere be $r_1$ and the radius of the hemisphere be $r_2$.
Given that the ratio of their radii is $r_1 : r_2 = 3 : 2$,so $\frac{r_1}{r_2} = \frac{3}{2}$.
The surface area of a sphere is $4 \pi r_1^2$.
The total surface area of a hemisphere is $3 \pi r_2^2$.
Now,the ratio of the surface area of the sphere to the total surface area of the hemisphere is:
$\frac{4 \pi r_1^2}{3 \pi r_2^2} = \frac{4}{3} \times \left(\frac{r_1}{r_2}\right)^2$
$= \frac{4}{3} \times \left(\frac{3}{2}\right)^2$
$= \frac{4}{3} \times \frac{9}{4}$
$= \frac{3}{1}$
Therefore,the required ratio is $3:1$.

(C) The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
Given radius $r = 30 \, cm$ and $\pi = 3.14$.
Substituting the values into the formula:
$A = 4 \times 3.14 \times (30)^2$
$A = 4 \times 3.14 \times 900$
$A = 12.56 \times 900$
$A = 11304 \, cm^2$.

(A) The formula for the surface area of a sphere is $A = 4\pi r^2$.
Given,radius $r = 50\, cm$ and $\pi = 3.14$.
Substituting the values into the formula:
$A = 4 \times 3.14 \times (50)^2$
$A = 4 \times 3.14 \times 2500$
$A = 12.56 \times 2500$
$A = 31400\, cm^2$.

(B) The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
Given radius $r = 1.25 \, m$ and $\pi = 3.14$.
Substituting the values into the formula:
$A = 4 \times 3.14 \times (1.25)^2$
$A = 4 \times 3.14 \times 1.5625$
$A = 12.56 \times 1.5625$
$A = 19.625 \, m^2$.
Therefore,the correct option is $B$.

(C) Given,the diameter of the sphere $d = 21\, cm$.
Therefore,the radius $r = \frac{d}{2} = \frac{21}{2} = 10.5\, cm$.
The surface area of a sphere is given by the formula $A = 4\pi r^2$.
Substituting the values,we get $A = 4 \times \frac{22}{7} \times (10.5)^2$.
$A = 4 \times \frac{22}{7} \times 10.5 \times 10.5$.
$A = 4 \times 22 \times 1.5 \times 10.5$.
$A = 88 \times 15.75 = 1386\, cm^2$.

(D) The diameter of the sphere is $d = 49\, cm$.
Therefore,the radius $r = d / 2 = 49 / 2 = 24.5\, cm$.
The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Substituting the values,we get $A = 4 \times (22 / 7) \times (49 / 2) \times (49 / 2)$.
$A = 4 \times (22 / 7) \times (2401 / 4)$.
$A = (22 / 7) \times 2401$.
$A = 22 \times 343 = 7546\, cm^2$.

(A) Given,diameter of the sphere $d = 3.5\,m$.
Radius $r = \frac{d}{2} = \frac{3.5}{2} = 1.75\,m$.
The surface area of a sphere is given by the formula $A = 4\pi r^2$.
Substituting the values,$A = 4 \times \frac{22}{7} \times (1.75)^2$.
$A = 4 \times \frac{22}{7} \times 1.75 \times 1.75$.
$A = 4 \times 22 \times 0.25 \times 1.75$.
$A = 88 \times 0.4375 = 38.5\,m^2$.

(A) Given radius $r = 14\, cm$.
$1$. The curved surface area $(CSA)$ of a hemisphere is given by the formula $2\pi r^{2}$.
$CSA = 2 \times \frac{22}{7} \times 14 \times 14 = 2 \times 22 \times 2 \times 14 = 1232\, cm^{2}$.
$2$. The total surface area $(TSA)$ of a hemisphere is given by the formula $3\pi r^{2}$.
$TSA = 3 \times \frac{22}{7} \times 14 \times 14 = 3 \times 22 \times 2 \times 14 = 1848\, cm^{2}$.
Thus,the curved surface area is $1232\, cm^{2}$ and the total surface area is $1848\, cm^{2}$.

(N/A) Given,radius $r = 17.5 \, cm = \frac{35}{2} \, cm$.
Curved Surface Area of a hemisphere = $2\pi r^2$.
$= 2 \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 11 \times 5 \times 35 = 1925 \, cm^2$.
Total Surface Area of a hemisphere = $3\pi r^2$.
$= 3 \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 3 \times 11 \times 5 \times \frac{35}{2} = 165 \times 17.5 = 2887.5 \, cm^2$.

(A) Given radius $r = 2.1 \, m$.
Curved Surface Area of a hemisphere = $2 \pi r^2$.
$= 2 \times \frac{22}{7} \times 2.1 \times 2.1 = 2 \times 22 \times 0.3 \times 2.1 = 27.72 \, m^2$.
Total Surface Area of a hemisphere = $3 \pi r^2$.
$= 3 \times \frac{22}{7} \times 2.1 \times 2.1 = 3 \times 22 \times 0.3 \times 2.1 = 41.58 \, m^2$.

(A) Given: Diameter $(d)$ = $14 \, cm$.
Radius $(r)$ = $d / 2 = 14 / 2 = 7 \, cm$.
Curved Surface Area $(CSA)$ of a hemisphere = $2 \pi r^2$.
$CSA$ = $2 \times (22 / 7) \times 7 \times 7 = 2 \times 22 \times 7 = 308 \, cm^2$.
Total Surface Area $(TSA)$ of a hemisphere = $3 \pi r^2$.
$TSA$ = $3 \times (22 / 7) \times 7 \times 7 = 3 \times 22 \times 7 = 462 \, cm^2$.
Thus,the curved surface area is $308 \, cm^2$ and the total surface area is $462 \, cm^2$.

(N/A) Given diameter $d = 126 \, cm$,so radius $r = \frac{126}{2} = 63 \, cm$.
Curved Surface Area of a hemisphere $= 2 \pi r^2 = 2 \times \frac{22}{7} \times 63 \times 63 = 2 \times 22 \times 9 \times 63 = 24,948 \, cm^2$.
Total Surface Area of a hemisphere $= 3 \pi r^2 = 3 \times \frac{22}{7} \times 63 \times 63 = 3 \times 22 \times 9 \times 63 = 37,422 \, cm^2$.

(A) Given diameter $d = 2.8\,m$,so the radius $r = d/2 = 1.4\,m$.
The curved surface area $(CSA)$ of a hemisphere is given by $2\pi r^2$.
$CSA = 2 \times (22/7) \times (1.4)^2 = 2 \times (22/7) \times 1.96 = 2 \times 22 \times 0.28 = 12.32\,m^2$.
The total surface area $(TSA)$ of a hemisphere is given by $3\pi r^2$.
$TSA = 3 \times (22/7) \times (1.4)^2 = 3 \times (22/7) \times 1.96 = 3 \times 22 \times 0.28 = 18.48\,m^2$.
Thus,the curved surface area is $12.32\,m^2$ and the total surface area is $18.48\,m^2$.

(D) The dome is in the form of a hemisphere.
Radius $(r)$ = $4.2 \, m$.
Surface area of the hemispherical dome = Curved Surface Area $(CSA)$ = $2 \pi r^2$.
$CSA$ = $2 \times \frac{22}{7} \times 4.2 \times 4.2$.
$CSA$ = $2 \times 22 \times 0.6 \times 4.2 = 110.88 \, m^2$.
Cost of painting = Area $\times$ Rate.
Cost = $110.88 \times 18 = ₹ 1995.84$.

(A) The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Given that $A = 7850 \, cm^2$ and $\pi = 3.14$,we substitute these values into the formula:
$7850 = 4 \times 3.14 \times r^2$
$7850 = 12.56 \times r^2$
$r^2 = \frac{7850}{12.56}$
$r^2 = 625$
$r = \sqrt{625} = 25 \, cm$.
Therefore,the radius of the sphere is $25 \, cm$.

(A) The total surface area of a hemisphere is given by the formula $TSA = 3\pi r^2$,where $r$ is the radius.
Given $TSA = 4158 \, cm^2$.
So,$3 \times \frac{22}{7} \times r^2 = 4158$.
$r^2 = \frac{4158 \times 7}{3 \times 22}$.
$r^2 = \frac{4158 \times 7}{66} = 63 \times 7 = 441$.
$r = \sqrt{441} = 21 \, cm$.
The diameter $d = 2r = 2 \times 21 = 42 \, cm$.

(B) Given: Inner radius $(r)$ = $4 \, cm$,Thickness $(t)$ = $0.2 \, cm$.
The outer radius $(R)$ = $r + t = 4 + 0.2 = 4.2 \, cm$.
The formula for the curved surface area of a hemisphere is $2 \pi R^2$.
Outer curved surface area = $2 \times \frac{22}{7} \times (4.2)^2$.
$= 2 \times \frac{22}{7} \times 4.2 \times 4.2$.
$= 2 \times 22 \times 0.6 \times 4.2$.
$= 44 \times 2.52 = 110.88 \, cm^2$.

(C) Let the initial radius of the sphere be $r_1 = r$.
The initial surface area is $A_1 = 4 \pi r^2$.
The radius is decreased by $20 \%$,so the new radius $r_2 = r - 0.20r = 0.8r$.
The new surface area is $A_2 = 4 \pi (0.8r)^2 = 4 \pi (0.64r^2) = 0.64(4 \pi r^2) = 0.64 A_1$.
The decrease in surface area is $A_1 - A_2 = A_1 - 0.64 A_1 = 0.36 A_1$.
The percentage decrease is $\frac{0.36 A_1}{A_1} \times 100 \% = 36 \%$.