1300 cm²

21.6 m

(i) 968 cm²    
(ii) 1210 cm²

5 : 6

130 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.

For shelter
l = 4 m, b = 3 m,
h = 2.5 m
\(\therefore\) Total surface area of the shelter
= lb + 2(bh + hl)
= (4)(3) + 2[(3)(2.5) + (2.5)(4)]
= 12 + 2[7.5 + 10]
= 47 m²
Hence, 47 m² of tarpaulin will be 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.

7 cm, 10.5 cm

\(9\pi { a }^{ 2 }{ unit }^{ 2 },2\sqrt { 3 } \pi { a }^{ 3 }{ unit }^{ 3 }\)

Diameter = 1 mm
ஃ  Radius (r) = \(\frac{1}{2}mm\)
Length of the pencil (h) =14 cm =140 mm 
ஃ  Volume of the graphite =\(\pi r^2h\)
\(=\frac{22}{7}\times(\frac{1}{2})^2\times 140=110mm^3\)
\(=\frac{110}{10\times10\times10}cm^3=0.11cm^2\)
Diameter = 7 mm
ஃ Radius (R) = \(\frac{7}{2}mm\)
Length of the pencil (h)  = 14 cm = 140 mm
ஃ Volume of the wood
\(=\pi (R^2-r^2)h=\frac{22}{7}(\frac{7}{2})^2-(\frac{1}{2})^2140\)
\(5280 mm^3=\frac{5280}{10\times10\times10}cm^3=5.28cm^3\)

Let the length and the breadth of the room be 4x m and x m respectively.
 \(\therefore\)  Area of the four walls of the room
= 2(l + b) h
= 2 (4x + x) 5 = 50 x m²
 \(\therefore\)  Cost of papering the four walls of the room @ 70 paise per square metre
= Rs 50x \(\times\) 0.70 = Rs 35 x
According to the question,
35x = 157.50
\(\Rightarrow\)  \(x=\frac { 157.50 }{ 35 } =4.5\)
\(\therefore\)  Breadth of the room = 4.5 m
\(\therefore\)  Length of the room = 4 \(\times\) 4.5 m = 18 m.

It r be the radius of cone, then slant height
\(l=\sqrt { { h }^{ 2 }+{ r }^{ 2 } } \)
\(\therefore\) c = \(\pi\)rl
\(V=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
L.H.S.= \(3\pi V{ h }^{ 3 }-{ c }^{ 2 }{ h }^{ 2 }+{ 9V }^{ 2 }\)
\(=3\pi \left( \frac { 1 }{ 3 } \pi { r }^{ 2 }h \right) { h }^{ 3 }-(\pi rl)^{ 2 }{ h }^{ 2 }+9{ \left( \frac { 1 }{ 3 } \pi { r }^{ 2 }h \right) }^{ 2 }\)
\(={ \pi }^{ 2 }{ r }^{ 2 }{ h }^{ 4 }-{ \pi }^{ 2 }{ r }^{ 2 }{ l }^{ 2 }{ h }^{ 2 }+{ \pi }^{ 2 }{ r }^{ 4 }{ h }^{ 2 }\)
\(={ \pi }^{ 2 }{ r }^{ 2 }{ h }^{ 4 }-{ \pi }^{ 2 }{ r }^{ 2 }({ h }^{ 2 }+{ r }^{ 2 }){ h }^{ 2 }+{ \pi }^{ 2 }{ r }^{ 4 }{ h }^{ 2 }\)
= 0.