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As we know that the numerical coefficients of given numerical are 42 and 63.

Greatest common factor of 42, 63 is 21.

Common literals appearing in given numerical are x, y and z

Smallest power of x in two monomials is 2

Smallest power of y in two monomials is 1

Smallest power of z in two monomials is 1

Monomials of common literals with smallest power is x²yz

∴ The greatest common factor = 21x²yz

(i) We know that

Area of the floor of the tent = πr²

By substituting the values

Area of the floor of the tent = (22/7) × 5² = 550/7 m²

We know that the area required by one student is 5/7 m²

So the required number of students = (550/7)/ (5/7) = 110

(ii) We know that

Curved surface area of the tent = area of the cloth = 165 m²

So we get

πrl = 165

By substituting the values

(22/7) × 5 × l = 165

On further calculation

l = (165 × 7)/ (22 × 5) = 21/2 m

We know that

h = √ (l² – r²)

By substituting the values

h = √ ((21/2)² – 5²)

On further calculation

h = √ ((441/4) – 25) = √ (341/4)

So we get

h = 9.23 m

We know that

Volume of the tent = 1/3 πr²h

By substituting the values

Volume of the tent = 1/3 × (22/7) × 5² × 9.23

On further calculation

Volume of the tent = 241.7 m³

It is given that

Diameter of the cylinder = 140cm

Radius of the cylinder = 140/2 = 70cm

Height of the cylinder = 1m = 100cm

We know that

Area of sheet required = Total surface area of cylinder = 2 πr (h + r)

By substituting the values

Area of sheet required = 2 × (22/7) × 70 (100 + 70)

On further calculation

Area of sheet required = 2 × 22 × 10 × 170

So we get

Area of sheet required = 74800 cm² = 7.48 m²

Therefore, the area of sheet required is 7.48 m².

Given number= 2925

Resolve the given number into prime factors we get

2925 = 3 × 3 × 5 × 5 × 13

To get perfect square we have to divide by 13

Then we get 2925 = 3 × 3 × 5 × 5

New number = 2925 / 13 = 225

Therefore √225 = 3 × 5 = 15

(i) 1458 can be factorised as follows.

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, prime factor 2 does not have its pair. If 2 gets a pair, then the number will become a perfect square. Therefore, 1458 has to be multiplied with 2 to obtain a perfect square.

Therefore, 1458 × 2 = 2916 is a perfect square.

1458 × 2 = 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

∴ √2916 = 2 x 3 x 3 x 3 = 54

(ii) 768 can be factorised as follows.

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 768 has to be multiplied with 3 to obtain a perfect square.

Therefore, 768 × 3 = 2304 is a perfect square.

768 × 3 = 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

∴ √2304 = 2 x 2 x 2 x 2 x 3 = 48

It is given that

Radius of the conical tent = 24m

Height of conical tent = 10m

We know that

Slant height of conical tent can be written as

l = √(r² + h²)

By substituting the values

l = √(24² + 10²)

On further calculation

l = √(576 + 100) = √ 676

So we get

l = 26m

We know that

Curved surface area of conical tent = πrl

By substituting the values

Curved surface area of conical tent = (22/7) × 24 × 26

So we get

Curved surface area of conical tent = (13728/7) m²

It is given that the cost of 1m² canvas = ₹ 70

So the cost of (13728/7) m² canvas = ₹ 70 × (13728/7) = ₹ 137280

Therefore, the slant height of the tent is 26m and the cost of canvas required to make the tent is ₹ 137280.

It is given that

Diameter of the cone = 24m

Radius of the cone = 24/2 = 12m

Slant height of the cone = 21m

We know that

Total surface area of a cone = πr (l + r)

By substituting the values

Total surface area of a cone = (22/7) × 12 (21+12)

On further calculation

Total surface area of a cone = (22/7) × 12 × 33

So we get

Total surface area of a cone = 1244.57 m²

Therefore, the total surface area of a cone is 1244.57 m².

Given: 

volume of the frustum = 10459 \(\frac{3}7\) cm³ 

Radii of lower and upper ends resp.: r’ = 8 cm and r’’ = 20 cm

Volume = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\)

⇒ \(\frac{1}3π(8^2 + 20^2 + 8\times20)\times{h}\) = 10459 \(\frac{3}7\)

⇒ h = 16 cm 

Let slant height be l 

Slant height, l = \(\sqrt{(r' - r")^2 + h^2}\)

⇒ l = \(\sqrt{(20 - 8)^2 + 16^2}\)

⇒ l = 20 cm 

Total surface area of the frustum = π(r’ + r’’)l + πr’² + πr’’² 

= π(20 + 8)20 + π (20)² + π (8)² 

= 3218.29 cm² 

cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm² 

= 3218.28 × 1.40 

= Rs 4506

(d) 64 cm³

Explanation:

Diagonal of cube = a√3 = 4√3

a = 4

Volume of cube = a³ = 4³ = 64 cm³

Radius of lower end = r = 8 cm

Radius of upper end = R = 20 cm

Height of container frustum = h = 16 cm

Cost of 100 cm² metal sheet = Rs 10

So, Cost of 1 cm² metal sheet = 10/100 = Rs 0.1

Let l be the slant height.

l² = (R-r)² + h²

= (20-8)² + 16²

= 256 + 144

= 400

or l = 20 cm

Surface area of frustum of the cone = πr² + π(R + r)l cm²

= π[ (20 + 8 )20 + 8²]

= π[560 + 64]

= 624 X ( 22/7 )

= 1961.14 cm²

(i) Find cost of metal sheet used to make the container if it costs Rs. 10 per 100 cm²

Cost of metal sheet per 100 cm² = Rs. 10

Cost of metal for Rs. 1961.14 cm² = (1961.14 x10)/100 = Rs. 196.114

(ii) Find the volume of frustum:

Volume of frustum = 1/3 πh(r² + R² + rR)

= 1/3 x 22/7 x 16(8² + 20² + 8×20)

= 1/3 x 22/7 x 16(64 + 400 + 160)

= 10345.4208 cm³

= 10.345 liters

Using

1000 cm³ = 1 litre

1 cm³ = 1/1000 litre

Cost of 1 liter milk = Rs. 35

Cost of 10.345 liter milk = Rs. 35 x 10.345 = Rs. 363 (approx)

Answer : (b) = 2.25 π 

Let each side of the cube be a cm. 

Then DE = diagonal of square face = √2a 

⇒ DG = \(\frac{\sqrt{2}a}{2}\) = \(\frac{a}{\sqrt{2}}\) cm.

Let the radius of the cone, i.e., AF = FB = r cm. 

Height FC = h cm. 

Given, r = h√2 

In similar triangles, AFC and DGC

\(\frac{AF}{FC}=\frac{DG}{GC}\)  ⇒ \(\frac{r}{h} = \frac{\frac{a}{\sqrt{2}}}{(h-a)}\)  

⇒ \(\frac{a/\sqrt{2}}{h-a} = \sqrt{2} \)  

⇒ a = 2(h-a) ⇒ h = \(\frac{3a}{2}\) , r = \(\frac{3a}{2}\)  \(\times \,\sqrt{2}\) 

∴  Volume of cone : Volume of cube = \(\frac{\frac{1}{3}\pi \times \big(\frac{3a\sqrt{2}}{2}\big)^2 \times \frac{3a}{2}}{a^3}\)  

= \(\frac{9}{4} a^3\pi :a^3 = \frac{9}{4}\pi\) 

= 2.25 π 

(d) \(\frac{V}{6}\)

Volume of remaining metal = Volume of cylinder – Volume of cone 

= πr²h – \(\frac13πr^2h=\frac23πr^2h=\frac23V\)

Volume of 4 spheres = \(\frac23\) V 

⇒ Volume of one sphere = \(\frac{\frac23V}{4}=\frac{V}{6}.\)

We know that

Volume of solid hemisphere = Surface area of solid hemisphere

So we get

2/3 πr³ = 3 πr²

It can be written as

r³/ r² = (3 × π × 3)/ (2 × π)

We get

r = 9/2 units

So the diameter = 2 (9/2) = 9 units

Therefore, the diameter of the hemisphere is 9 units.

It is given that each edge of a cube = 9m

We know that

Volume of cube = a³

By substituting the values

Volume of cube = 9³

So we get

Volume of cube = 729 m³

We know that

Lateral surface area of cube = 4a²

By substituting the values

Lateral surface area of cube = 4 × 9²

So we get

Lateral surface area of cube = 4 × 81 = 324 m²

We know that

Total surface area of cube = 6a²

By substituting the values

Total surface area of cube = 6 × 9²

So we get

Total surface area of cube = 6 × 81 = 486 m²

We know that

Diagonal of cube = √3 a

By substituting the values

Diagonal of cube = 1.73 × 9 = 15.57 m

Therefore, the volume is 729 m³, lateral surface area is 324 m², total surface area is 486 m² and the diagonal of cube is 15.57 m.

(a) 50%

Volume of a prism = Area of base × height. 

Since, the base area is constant and the height is halved, so the volume will also be halved, i.e. volume will be reduced by 50%.

(a) 48 cm³

Explanation:

Let a be the length of smallest edge.

The edges are in the proportion a: 2a: 3a

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (2 a² + 3 a² + 6 a²)

88 = 22 a²

a = 2

2a=4

3a = 6

We know that volume of cuboid= length × breadth × height

V = 2 × 4 × 6

V = 48

For frustum: 

Base radius, r’ = \(\frac{20}2\) = 10 cm 

Top radius, r’’ = \(\frac{12}2\) = 6 cm 

Height, h = 3 cm 

Volume = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\)

= \(\frac{1}3π(10^2 + 6^2 + 10\times6)\times3\)

= 616 cub. cm 

Let l be the slant height of the cone, 

then 

⇒ l = \(\sqrt{(r' - r")^2 + h^2}\)

⇒ l = \(\sqrt{(10 - 6)^2 + 3^2}\)

⇒ l = 5 cm 

Total surface area of the frustum = π (r’ + r’’) × l + πr’² + πr’’² 

= π (20 + 10) × 15.620 + π (10)² + π (6)² 

= 678.85 cm²

Answer: (b) 5497.6 cm³ 

Remaining volume 

= (Volume of cube) – (Volume of cuboid formed by cutting the square hole) – (Volume of the cylinder formed by cutting the circular hole) + (Common volume of cuboid and cylinder)

= (20)³ – (10)² × 20 – π × (4)² × 20 + π × (4)² × 10 

= 8000 – 2000 – 160π 

= 6000 – 160 π 

= 5497.6 cm³.

Note: The common portion is a cylinder of diameter 8 cm and height 10 cm (side of square).

(c) 72

Let a be the length of an edge of a cube. Then, 

a³ = 12a            (∵ A cube has 12 edges) 

⇒ a² = 12 ⇒ a = 2√3 

∴ Total surface area of cube = 6a² = 6 x (2√3)² = 72.

It is given that

Inner radius of the bowl = 5cm

Thickness of the bowl = 0.25cm

External radius = 5 + 0.25 = 5.25cm

We know that

Outer curved surface area of the bowl = 2 πr²

By substituting the values

Outer curved surface area of the bowl = 2 × (22/7) × 5.25²

On further calculation

Outer curved surface area of the bowl = 2 × (22/7) × 27.5625

So we get

Outer curved surface area of the bowl = 173.25 cm²

Therefore, the outer curved surface area of the bowl is 173.25 cm².

Radius of top of the bucket, r’ = \(\frac{40}2\) = 20 cm 

Radius of bottom of the bucket, r’’ = \(\frac{20}2\) = 10 cm 

Depth, h = 12 cm 

Volume of the bucket = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\)

= \(\frac{1}3π(20^2 + 10^2 + 20\times10)\times12\)

= 8800 cub. cm 

Let l be slant height of the bucket 

⇒ l = \(\sqrt{(r'^2 - r"^2}) + h^2\)

⇒ l = \(\sqrt{(20^2 - 10^2) + 12^2}\)

⇒ l = 15.62 cm 

Total surface area of bucket = π (r’ + r’’) × l + πr’’² 

= π (20 + 10) × 15.620 + π (10)² 

= 17.81 dm² 

Cost of tin sheet = 1.20 × 17.87 

= Rs 21.40

Here, 

dimension of cuboid = 15 cm × 10 cm × 5 cm 

Radius of cylindrical hole = \(\frac{7}2\) cm 

Surface area of the remaining block = surface area of cuboidal solid metallic block – 2 × base area of cylindrical hole 

= 2(15 × 10 + 10 × 5 + 5 × 15) - 2 × \(\frac{22}7\times(\frac{7}2)^2\)

= 583 cm²

(c) 2 cm.

Let the internal radius of the shell be r cm.

∴ Internal volume of the shell = \(\frac43πr^3\) cu. cm

External radius of the shell = \(\frac{21}{2}\) cm.

External volume of the shell = \(\frac43\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}\) cu. cm

= 4851 cu. cm. 

Weight of 1 cu. cm of metal = 10g.

∴ Volume of the metal in the shell = \(22775\frac5{21}\times \frac{1}{10}\) cu. cm

= \(\frac{478280}{21}\times\frac{1}{10}\) cu. cm = \(\frac{47828}{21}\) cu. cm

∴ Internal vol. of the shell = \(4851-\frac{47828}{21}=\frac{54043}{21}\) cu. cm

⇒ \(\frac43πr^3=\frac{54043}{21}\) ⇒ r³ = \(\frac{54043\times3\times7}{21\times4\times22}\) = 614.125

⇒ \(r = 8.5 \) cm

∴ Thickness of shell = 10.5 cm – 8.5 cm = 2 cm

Answer is (B) the x-axis

Answer : (b) = 7 cm 

Let R be the radius of the ball. 

Then, 

Volume of water displaced = Volume of iron ball

⇒ \(\pi \times \big(\frac{14}{2}\big)^2 \times \frac{28}{3} = \frac{4}{3} \times\pi \times R^3\)  

⇒ R³ = 7³ 

⇒ R = 7.

Let z = x + iy. Then z – 2 – 3i = (x – 2) + i (y – 3)

Let θ be the amplitude of z – 2 – 3i. Then tanθ = (y-3)/(x-2)

⇒ tan π/4 =  (y-3)/(x-2)   [since  θ=π/4 ]

⇒ 1 =  (y-3)/(x-2) i.e. x – y + 1 = 0

Hence, the locus of z is a straight line.

We know that

Internal radius = 3/2 = 1.5cm

External radius = 1.5 + 1 = 2.5cm

We know that

Volume of cast iron = (π × (2.5)² × 100 – π × (1.5)² × 100)

Taking the common terms out

Volume of cast iron = π × 100 × (2.5² – 1.5²)

On further calculation

Volume of cast iron = (22/7) × 100 × (6.25 – 2.25)

So we get

Volume of cast iron = (22/7) × 100 × 4 = 1257.142 cm³

It is given that 1cm³ of cast iron weighs 21g

We know that 1kg = 1000g

So the weight of cast iron pipe = 1257.142 × (21/1000) = 26.4kg

Therefore, the weight of cast iron pipe is 26.4 kg.

Answer is (C),

|z +1|=1 ⇒ |(x +1)+iy| = 1

⇒ (x +1)² + y² = 1

which is a circle with centre (–1, 0) and radius 1.

Given: 

height of the tent, H = 77 dm = 7.7 m 

Height of the cylindrical portion, h = 44 dm = 4.4 m 

∴ Height of the conical portion, h’ = 7.7 – 4.4 = 3.3 m 

Radius of the cylinder, r = \(\frac{36}2\) = 18 m 

Curved surface area of the cylindrical portion of the tent 

= 2πr² h 

= 2 × π × (18)² × 4.4 

= 158.4 π m² 

Slant height of the conical portion of the tent = \(\sqrt{3.3^2 + 18^2}\)

= 18.3 m 

∴ Curved surface area of the conical portion = π × 18 × 18.3 

= 329.4π m² 

∴ Total surface area of the tent = 158.4 π + 329.4 π 

= 487.8 π 

= 1533.09 m² 

∴ Canvas required to make the tent = 1533.09 m² 

Total cost of the canvas = 1533.09 × 3.50 

= Rs 5365.80

Given,

Height of the tent = 77 dm

Height of a surmounted cone = 44 dm

Height of the Cylindrical Portion = Height of the tent – Height of the surmounted Cone

= 77 – 44

= 33 dm = 3.3 m

And, given diameter of the cylinder (d) = 36 m

So, its radius (r) of the cylinder = 36/2 = 18 m

Let’s consider L as the slant height of the cone.

Then, we know that

L² = r² + h²

L² = 18² + 3.3²

L² = 324 + 10.89

L² = 334.89

L = 18.3 m

Thus, slant height of the cone (L) = 18.3 m

Now, the Curved Surface area of the Cylinder (S₁) = 2πrh

S₁ = 2π (184.4) m² 

And, the Curved Surface area of the cone (S₂) = πrL

S₂ = π × 18 × 18.3 m² 

So, the total curved surface of the tent (S) = S₁ + S₂

S = S₁ + S₂

S = (2π18 × 4.4) + (π18 × 18.3)

S = 1533.08 m²

Hence, the total Curved Surface Area (S) = 1533.08 m²

Next,

The cost of 1 m² canvas = Rs 3.50

So, 1533.08 m² of canvas will cost = Rs (3.50 x 1533.08)

= Rs 5365.8

Given,

Radius of the cylindrical portion of the rocket (R) = 2.5 m

Height of the cylindrical portion of the rocket (H) = 21 m

Slant Height of the Conical surface of the rocket (L) = 8 m

Curved Surface Area of the Cone (S₁) = πRL = π(2.5)(8)= 20π

And,

Curved Surface Area of the Cone (S₂) = 2πRH + πR²

S₂ = (2π × 2.5 × 21) + π (2.5)²

S₂ = (π × 105) + (π × 6.25)

Thus, the total curved surface area S is

S = S₁ + S₂

S = (π20) + (π105) + (π6.25)

S = (22/7)(20 + 105 + 6.25) = 22/7 x 131.25

S = 412.5 m²

Therefore, the total Surface Area of the Conical Surface = 412.5 m²

Now, calculating the volume of the rocket

Volume of the conical part of the rocket (V₁) = 1/3 × 22/7 × R² × h

V₁ = 1/3 × 22/7 × (2.5)² × h

Let, h be the height of the conical portion in the rocket.

We know that,

L² = R² + h²

h² = L² – R² = 8² – 2.5²

h = 7.6 m

Using the value of h, we will get

Volume of the conical part (V₁) = 1/3 × 22/7 × 2.5² × 7.6 m²   = 49.67 m²      

Next,

Volume of the Cylindrical Portion (V₂) = πR²h

V₂ = 22/7 × 2.5² × 21 = 412.5 m²

Thus, the total volume of the rocket = V₁ + V₂

V = 412.5 + 49.67 = 462.17 m²

Hence, the total volume of the Rocket is 462.17 m² 

Height of cylinder (h) = 10 cm 

Height of conical part = 6 cm 

Slant height of cone (l) = \(\sqrt{r^2 + h^2}\)

⇒ l = \(\sqrt{3.5^2 + 6^2}\)

⇒ l = 48.25 cm 

Curved surface area of cone = πrl 

= π (3.5) (48.25) 

= 76.408 cm² …(1) 

Curved surface area of cylinder = 2πrh 

= 2π (3.5) (10) 

= 220 cm² …(2) 

Curved surface area of hemisphere = 2πr² 

= 2π (3.5)² 

= 77 cm² ….(3) 

∴ Total curved surface area = Curved surface area of(cone + cylinder + hemisphere) 

= 76.408 + 220 + 77 

= 373.408 

∴ Total surface area of solid = 373.408 cm²

Correct option is (C) Rocket

(A) Capsule is combination of two hemispheres and a cylinder.

(B) Ice cream is combination of a hemisphere and a cone.

(C) Rocket is combination of a cylinder and a cone.

Correct option is  C) Rocket

False

Explanation:

According to the question,

When one cylinder is placed over another, the base of first cylinder and top of other cylinder will not be covered in total surface area.

We know that,

Total surface area of cylinder = 2πrh + 2πr²h, where r = base radius and h = height

Total surface area of shape formed = 2(Total surface of single cylinder) – 2(Area of base of cylinder)

= 2(2πrh + 2πr²) – 2(πr²)

= 4πrh + 2πr²

False

Explanation:

Let the radius of sphere = r

When a solid ball is exactly fitted inside the cubical box of side a,

We get,

Diameter of ball = Edge length of cube

2r = a

Radius, r = a/2

We also know that,

Volume of sphere = 4/3πr³

Volume of ball = 4/3π(a/2)³ = 4/3π(a³/8) = 1/6πa³

Yes, the poet has used exaggerations such as the cat’s defiance of gravity and it being called a ‘monster of depravity’ and a ‘fiend’ in order to enhance the mystery surrounding the cat. Since the cat is shown to be super-fast as nobody from the Scotland Yard to the Flying Squad can catch it on the scene of crime, these exaggerations have been used by Eliot to lay stress on this monstrous s as well as surprising and mysterious nature of Macavity.

Examples:

(i) ‘He’s the bafflement of Scotland Yard, the Flying Squad’s despair’.

(ii) ‘He breaks the law of gravity’.

(iii) ‘His powers of levitation would make a fakir stare’.

(iv) ‘He’s a fiend in feline shape, a monster of depravity’.

Yes, it seems like the poet is fond of cats. He calls Macavity a ‘fiend’ and a ‘monster’ because he might have wanted to portray an evil side. He might have used a cat in order to create a negative character who is a criminal and escapes easily from police. The quick movements

Answer is D. Fakir

D. ‘Hidden Paw’

A.  A Ginger Cat

C. The Flying Squad

Macavity was a male cat. He was a clever criminal. He had no fear of law or the police. He escaped from the scene of crime before the police arrived there. He was tall and thin. He was careless about his clothes and whiskers. He moved like a snake and pretended to be asleep. He was, in fact, a devil in the shape of a cat. He was wicked and morally corrupt.

C. The Secret Service

B. ‘Doing complicated long division sums’.

The poet calls Macavity a master criminal. He could make a good escape before the police came to the scene of crime. He befooled the detectives and the Flying Squad. He was an outlaw. He was a master in his field. He was matchless, a devil in the shape of a cat.

Volume of sphere = \(\frac{4}{3}π R^3\)

= \(\frac{{4}\times{669} \times{8}}{{3}\times{213}}\)

= 33.50 cm³ 

Volume of 16 spheres = 16 × 33.50 

= 536 cm³ 

Volume of rectangle = 16 cm × 8 cm × 8 cm 

= 1024 cm³ 

Volume of this liquid = 1024 cm³ – 536 cm³ 

= 488 cm³

Macavity has “broken every human . law”. He even possesses several supernatural powers such as levitation which makes a person rise in the air. He is such an expert that when the police reach the scene of crime “Macavity’s not there !”

It is given that length = 24m, breadth = 25cm = 0.25m and height = 6m

We know that

Volume of cuboid = l × b × h

By substituting the values we get

Volume of cuboid = 24 × 0.25 × 6

By multiplication

Volume of cuboid = 36 m³

We know that

Lateral surface area of a cuboid = 2 (l + b) × h

By substituting the values

Lateral surface area of a cuboid = 2 (24 + 0.25) × 6

On further calculation

Lateral surface area of a cuboid = 2 × 24.25 × 6

By multiplication

Lateral surface area of a cuboid = 291 m²

We know that

Total surface area of cuboid = 2 (lb + bh + lh)

By substituting the values

Total surface area of cuboid = 2 (24 × 0.25 + 0.25 × 6 + 24 × 6)

On further calculation

Total surface area of cuboid = 2 (6 + 1.5 + 144)

So we get

Total surface area of cuboid = 2 × 151.5

By multiplication

Total surface area of cuboid = 303 m²

It is given that length = 12cm, breadth = 8cm and height = 4.5cm

We know that

Volume of cuboid = l × b × h

By substituting the values we get

Volume of cuboid = 12 × 8 × 4.5

By multiplication

Volume of cuboid = 432 cm³

We know that

Lateral surface area of a cuboid = 2 (l + b) × h

By substituting the values

Lateral surface area of a cuboid = 2 (12 + 8) × 4.5

On further calculation

Lateral surface area of a cuboid = 2 × 20 × 4.5

By multiplication

Lateral surface area of a cuboid = 180cm²

We know that

Total surface area of cuboid = 2 (lb + bh + lh)

By substituting the values

Total surface area of cuboid = 2 (12 × 8 + 8 × 4.5 + 12 × 4.5)

On further calculation

Total surface area of cuboid = 2 (96 + 36 + 54)

So we get

Total surface area of cuboid = 2 × 186

By multiplication

Total surface area of cuboid = 372 cm²

Given that height of the cylinder is 8cm and volume is 1232cm³.

Therefore volume of the cylinder = πr²h

1232 = 22/7 × r× r × 8

r²= 1232 × 7/ 8

r = 7cm

Also we know that curved surface area of cylinder= 2πrh

Curved surface area = 2 × 22/7 × 7 × 8

Curved surface area = 252 cm²

We know that total surface area of cylinder = 2πr(r + h)

Total surface area = 2 × 22/7 × 7 (7 + 8)

Total surface area = 2580 cm²