For resulting cuboid
Length (l)    = 6 + 6 = 12 cm
Breadth (b) = 6 cm
Height (h)   = 6 cm
\(\therefore \) Surface area 
= 2(lb + bh + hl) 
= 2(12 \(\times\) 6 + 6 \(\times\) 6 + 6 \(\times\) 12)
= 360 cm²    

Let the height of the hall be h m.
Area of the floor   = l \(\times\) b
= 20  16 = 320 m²
Area of the flat roof   = l \(\times\) b
= 20 \(\times\) 16 = 320 m²
Sum of the areas of the four walls
= 2(l + b) h
= 2(20 +16) h = 72 h m²
According to the question,
72 h = 320 + 320
\(\Rightarrow \) 72h = 640  \(\Rightarrow\)  \(h=\frac { 640 }{ 72 } m\)
\(\Rightarrow \)   \(h=\frac { 80 }{ 9 } m\)
Hence, the height of the hall is \(\frac { 80 }{ 9 } m.\) 

Let the dimensions of the box be 2k, 3k and 4k.
\(\therefore \)  Total surface area = 2(lb + bh + hl)
= 2(2k.3k + 3k.4k + 4k.2k)
= 52k²m²
Cost of covering at the rate of Rs 8 per m²
= 52k² \(\times\) 8 = Rs 416k²
Cost of covering at the rate of Rs 9.50 per m²
= 52k² \(\times\) 9.50 = Rs 494k²
Difference between the costs
= Rs 494k² - Rs 416k² = Rs 78k²
According to the question,
78k² = 1248
\(\Rightarrow\)  k² = 16 \(\Rightarrow\) k = 4
Hence, the dimensions of the box are 8 m, 12 m and 16 m.

Radius of the base (r) \(=\frac { 1 }{ 2 } m\)
\(=\frac { 1 }{ 2 } \times 100cm=50cm\)
Height (h) = 1.4 m
= 1.4 \(\times\) 100 cm = 140 cm
Surface area to be tin-coated \(=2\left( 2\pi rh+\pi { r }^{ 2 } \right) \)
 \(=2\left[ 2\times 3.14\times 50\times 140+3.14\times { \left( 50 \right) }^{ 2 } \right] \)
\(\\ =2\left[ 43960+7850 \right] =2\left( 51810 \right)\)
\( \\ =103620{ cm }^{ 2 }\)
\(\therefore \)   Cost of tin-coating at the rate of Rs 50 per 1000 cm²
= Rs \(\frac { 50 }{ 1000 } \times 103620\) = Rs 5181.

r = 5 cm, h = 12 cm
\(\therefore l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(\\ =\sqrt { { \left( 5 \right) }^{ 2 }+{ \left( 12 \right) }^{ 2 } } =\sqrt { 25+144 }\)
\( \\ =\sqrt { 169 } =13cm\)
Curved surface area \(=\pi rl\)
\(=\pi \left( 5 \right) \left( 13 \right) =65\pi { cm }^{ 2 }\)

Let the base radius of the conical tent be r m. Then,
\(2\pi r=44\)
\(\Rightarrow 2.\frac { 22 }{ 7 } .r=44\)
\(\\ \Rightarrow r=7m\)
\(\\ h=10m\)
\(\\ \ \therefore \ l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(\\=\sqrt { 49+100 } =\sqrt { 149 } \)
\(\therefore\)   Curved surface area = \(\pi rl\)
 \(=\frac { 22 }{ 7 } .7.\sqrt { 149 } =22\sqrt { 149 } m\)
= Area of the canvas used
\(\therefore\)   Length of the canvas used
\(=\frac { 22\sqrt { 149 } }{ 2 } m=11\sqrt { 149 } m\)
\(\\ =11\times 12.2m=134.2m\)

For cone
Base radius (r) = 8 m
Height (h)        = 6 m
\(\therefore \)  Slant height (l) = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { \left( 8 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 } } =10m\)
\(\therefore \)  Curved surface area = \(\pi rl\)
\(=\pi \left( 8 \right) \left( 10 \right) \)
\(\\ =80\pi { m }^{ 2 }\)
For cylinder
Base radius (R) = 8 m
Height (H) = 14 m
\(\therefore \)  Curved surface area = \(2\pi RH\)
 \(=2\pi \left( 8 \right) \left( 14 \right) \)
\(\\ =224\pi { m }^{ 2 }\)
\(\therefore \)  Total curved surface area = Curved surface area of the cone + Curved surface area of the cylinder
\(=80\pi +224\pi =304\pi { m }^{ 2 }\)
\(\\ =304\times 3.14{ m }^{ 2 }=954.56{ m }^{ 2 }\ \)
\(\therefore \)  Cost of canvas = 954.56 \(\times\) 50
= Rs 47728

Let the base radius and height of the cylinder be r cm and h cm respectively.
Then,
\(2 \pi r=22\)
\(\Rightarrow 2\times \frac{22}{7}\times r=22\)
\(\Rightarrow r=\frac{7}{2}cm\)
h = 22 cm
ஃ  Volume of the cylinder 
\(=\pi r^2h\)
\(=\frac{22}{7}. \frac{7}{2}.\frac{7}{2}.22\)
= 847 cm³

0.31 km/hour
= 0.31 x 1000 m/hour
= 310 m/ hour
=\(\frac{310}{60}\) m/minute=\(\frac{31}{6}\) m/minute
ஃ  Volume of water that falls into the sea in a minute
\(=4 \times60\times\frac{31}{6}=1240 \ m^3\)

l:b = 4:3, h = 550 cm,
l = 4y, b = 3y (taking y as constant)
h = 550 cm = 5.5 m
Total cost = L.S.A \(\times\)Rate per square metres
5082 = 2h(l + b)\(\times\)6.60
=2 \(\times\)5.5(4y + 3y) \(\times\)6.60
5082 = 2\(\times\)5.5\(\times\)7y\(\times\)6.60
10 = y
l = 4y = 4\(\times\)10 = 40 m
b = 3y = 3\(\times\)10 = 30 m

Height of the cone(h) = 24 cm
Let r cm be the radius of the base and l cm an can be the slant height of the cone, then
\(l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+{ 24 }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+576 } \)
Now, curved surface area=\(\pi\)rl
\(\Rightarrow \ \frac { 22 }{ 7 } \times r\times \sqrt { { r }^{ 2 }+576 } =550\)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =550\times \frac { 7 }{ 22 } \)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =175\)
Squaring both the sides we get
r² (r² + 576) = 30625
(r²)² + 576r²-30625 = 0
Let r² = x
x² + 576x - 30625 = 0
\(\Rightarrow\) x² + 625x - 49x - 30625 = 0
\(\Rightarrow\) x(x + 625)-  49(x + 625) = 0
\(\Rightarrow\) (x + 625)(x - 49) = 0
\(\Rightarrow\) x + 625=0 or x - 49 = 0
\(\Rightarrow\) x = 625 or x = 49
not possible x = 49
\(\therefore\) r²= 49
\(\Rightarrow\) r = 7 cm
Volume & the cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+24\)
= 1232 cm³.

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.