Base diameter = 40 cm
\(\therefore\) Base radius (r) = \(\frac { 40 }{ 2 } cm=20cm\)
\(=\frac { 20 }{ 100 } m=0.2m\)
Height (h) = 1 m
 \(\therefore l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(\\ =\sqrt { { \left( 0.2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 0.04+1 } \)
\(\\ =\sqrt { 1.04 } =1.02m\)
\(\therefore\)  Curved surface area = \(\pi rl\)
= 3.14 \(\times\) 0.2 \(\times\) 1.02
= 0.64056 m²
\(\therefore\)  Curved surface area of 50 cones
= 0.64056 \(\times\) 50 m²
= 32.028 m²
\(\therefore\)  Cost of painting all these cones
= 32.028 \(\times\) 12
= 384.336 = Rs 384.34 (approximately).

Base radius (r) = 7 cm
Height (h)  = 24 cm
\(\therefore \)  Slant height (l) = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
 \(=\sqrt { { \left( 7 \right) }^{ 2 }+{ \left( 24 \right) }^{ 2 } } =\sqrt { 49+576 } \)
\(\\ =\sqrt { 625 } =25cm\)
\(\therefore \)  Curved surface area of a cap = \(\pi rl\)
\(=\frac { 22 }{ 7 } \times 7\times 25=550{ cm }^{ 2 }\)
\(\therefore \)  Curved surface area of 10 caps
= 550 \(\times\) 10 = 5500 cm²
 Hence, the area of the sheet required to make 10 such caps is 5500 cm² .

Slant height (l) = 25 m
Base diameter (d) = 14 m
\(\therefore \) Base radius (r) = \(\frac { 14 }{ 2 } m=7m\)
\(\therefore \) Curved surface area of the tomb =\(\pi rl\)
\(=\frac { 22 }{ 7 } \times 7\times 25=550{ m }^{ 2 }\)
\(\therefore \)  Cost of white-washing the curved surface of the tomb at the rate of Rs 210 per 100 m²
= Rs \(\frac { 210 }{ 100 } \times 550\) = Rs 1155.

For conical tent
h = 8 m, r = 6 m
\(\therefore l=\sqrt { { r }^{ 2 }+{ h }^{ 2 }\ }\)
\( \\ =\sqrt { { \left( 6 \right) }^{ 2 }+{ \left( 8 \right) }^{ 2 } } =\sqrt { 36+64 }\)
\( \\ =\sqrt { 100 } =10m\)
\(\therefore \) Curved surface area = \(\pi rl\)
= 3.14 \(\times\) 6 \(\times\) 10 = 188.4 m².
Width of tarpaulin = 3 m
\(\therefore \) Length of tarpaulin 
Extra length of the material required
= 20 cm = 0.2 m
\(\therefore \) Actual length of tarpaulin required
= 62.8 m + 0.2 m = 63 m.  

\(\because \) Diameter of the base   = 10.5 cm
\(\therefore \) Radius of the base (r) \(=\frac { 10.5 }{ 2 } cm\)
= 5.25 cm
Slant height (l) = 10 cm
\(\therefore \) Curved surface area of the cone = \(\pi rl\)
 \(=\frac { 22 }{ 7 } \times 5.25\times 10=165{ cm }^{ 2 }.\)

2r = 20 cm
\(\Rightarrow\)  r = 10 cm
h = 30 cm
\(\therefore \)   Cloth required = \(2\pi r\left( h+2.5+2.5 \right) \)
 \(=2\pi r\left( h+5 \right) =2\times \frac { 22 }{ 7 } \times 10\times \left( 30+5 \right) \)
\(\\ =2200{ cm }^{ 2 }.\)

(i)    2r = 4.2 m
\(\therefore \)    \(r=\frac { 4.2 }{ 2 } m=2.1\quad m\)
h = 4.5 m
\(\therefore \)    Lateral or curved surface area = \(2\pi rh\)
 \(=2\times \frac { 22 }{ 7 } \times 2.1\times 4.5=59.4{ m }^{ 2 }\)
(ii) Total surface area = \(2\pi r\left( h+r \right) \)
 \(=2\times \frac { 22 }{ 7 } \times 2.1\times \left( 4.5+2.1 \right) \)
\(\\ =2\times \frac { 22 }{ 7 } \times 2.1\times 6.6=87.12{ m }^{ 2 }\)
Let the actual area of steel used be x m².
Since \(\frac { 1 }{ 12 } \) of the actual steel used was wasted, the area of the steel which has gone into the tank \(=\frac { 11 }{ 12 } \)  of x.
\(\therefore \)   \(\frac { 11 }{ 12 } x=87.12\)
\(\therefore \)   \(x=\frac { 87.12\times 12 }{ 11 } =95.04{ m }^{ 2 }\)
\(\therefore \)  Steel actually used = 95.04 m².

h = 28 m
2r = 5 cm
\(\therefore \)  \(r=\frac { 5 }{ 2 } cm=\frac { 5 }{ 2\times 100 } m\)
\(=\frac { 5 }{ 200 } m=\frac { 1 }{ 40 } m\)
\(\therefore \) Total radiating surface in the system 
\(=2\pi rh\)
\(\\ =2\times \frac { 22 }{ 7 } \times \frac { 1 }{ 40 } \times 28=4.4{ m }^{ 2 }.\)

h = 77 cm
2r = 4 cm
r = 2 cm
2R = 4.4 cm
R = 2.2 cm
(i) Inner curved surface area \(=2\pi rh\)
 \(=2\times \frac { 22 }{ 7 } \times 2\times 77=968{ cm }^{ 2 }\)
(ii) Outer curved surface area =
 \(=2\times \frac { 22 }{ 7 } \times 2.2\times 77=1064.8{ cm }^{ 2 }\)
(iii) Total surface area
 \(=2\pi Rh+2\pi rh+2\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) \)
\(\\ =1064.8+2\times \frac { 22 }{ 7 } \times 2\times 77+2\times \frac { 22 }{ 7 } \times \left\{ { \left( 2.2 \right) }^{ 2 }-{ \left( 2 \right) }^{ 2 } \right\} \)
\(\\ =1064.8+968+2\times \frac { 22 }{ 7 } \times \left( 4.84-4 \right) \)
\(\\ =1064.8+968+2\times \frac { 22 }{ 7 } \times 0.84\)
\(\\ =1064.8+968+5.28=2038.08{ cm }^{ 2 }.\)

h = 1 m = 100 cm
2r = 140 cm
\(\Rightarrow\)  \(r=\frac { 140 }{ 2 } cm=70cm\)
 \(\therefore\)  Total surface area of the closed cylindrical tank   
 \(=2\pi r\left( h+r \right) \)
\(\\ =2\times \frac { 22 }{ 7 } \times 70\left( 100+70 \right) \)
\(\\ =74800{ cm }^{ 2 }=\frac { 74800 }{ 100\times 100 } { m }^{ 2 }\)
\(\\ =7.48{ m }^{ 2 }\)
Hence, 7.48 square metres of the sheet are required.

Let the radius of the base of the cylinder be r cm.
h = 14 cm
 Curved surface area = 88 cm²   Given
\(\Rightarrow\)   \(2\pi rh=88\)
\(\Rightarrow\)   \(2\times \frac { 22 }{ 7 } \times r\times 14=88\)
\(\Rightarrow\)   \(r=\frac { 88\times 7 }{ 2\times 22\times 14 } \)
\(\Rightarrow\)   r = 1
\(\Rightarrow\)  2r = 2
Hence, the diameter of the base of the cylinder is 2 cm.

For a brick
l = 22.5 cm, b = 10 cm,
h = 7.5 cm
\(\therefore \)  Total surface area of a brick 
= 2 (lb + bh + hl)
= 2 (22.5 \(\times\) 10 + 10 \(\times\) 7.5 + 7.5 \(\times\) 22.5)
= 2 (225 + 75 + 168.75)
= 2(468.75) = 937.5 cm² = .09375 m²
\(\therefore \)  Number of bricks that can be painted out 
\(=\frac { 9.375 }{ .09375 } =100.\)

Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Perimeter = 250 m
\(\Rightarrow\) 2(l + b) = 250
\(\Rightarrow\)  l + b = 125                ...(1)
Area of the four walls
\(=\frac { 15000 }{ 10 } =1500\)m²      
\(\Rightarrow\) 2(l + b)h = 1500
\(\Rightarrow\)  (l + b)h  = 750                  
\(\Rightarrow\)  125 h  = 750                Using (1)
\(\Rightarrow\)   \(h=\frac { 750 }{ 125 } \)
\(\Rightarrow\)   h = 6 m
Hence, the height of the hall is 6 m.            

l = 5 m, b = 4 m,
h = 3 m
Area of the walls of the room = 2(l + b)h
= 2(5 + 4)3 = 54 m²
Area of the ceiling = lb
= (5) (4) = 20 m2
\(\therefore \) Total area of the walls of the room and the ceiling = 54 m² + 20 m² = 74 m²
\(\therefore \) Cost of white washing the walls of the room and the ceiling = 74 \(\times\) 7.50 = Rs 555.