# ML Aggarwal Mensuration Exe-18.4 Class 8 ICSE Ch-18 Maths Solutions

ML Aggarwal Mensuration Exe-18.4 Class 8 ICSE Ch-18 Maths Solutions. We Provide Step by Step Answer of Exe-18.4 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Mensuration Exe-18.4 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-18 | Mensuration |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-18.4 Questions |

Edition | 2023-2024 |

**Mensuration Exe-18.4**

ML Aggarwal Class 8 ICSE Maths Solutions

**Page-312**

**Question 1. The surface area of a cube is 384 cm**^{2}. Find

^{2}. Find

(i) the length of an edge

(ii) volume of the cube.

**Answer:**

Surface area of a cube = 384 cm^{2}

(i) Surface area of cube = 6(side)^{2}

Hence, edge (side) = √(surface area/6)

= √(384/6)

= √64

= 8 cm

(ii) Volume = (Edge)^{3} = (8)^{3 }= 8 × 8 × 8 cm^{3} = 512 cm^{3}

**Question 2. Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of n.**

**Answer:**

Radius of a solid cylinder (r) = 5 cm

Height (h) = 10 cm

Total surface area = 2πrh + 2πr^{2}

= 2rπ(h + r)

= 2π × 5(10 + 5)

= π × 10 × 15 = 150π cm^{2}

**Question 3. An aquarium is in the form of a cuboid whose external measures are 70 cm × 28 cm × 35 cm. The base, side faces and back face are to be covered with coloured paper. Find the area of the paper needed.**

**Answer:**

A cuboid shaped aquarium,

Length (l) = 70 cm

Breadth (b) = 28 cm

and height (h) = 35 cm

Area of base = 70 × 28 cm^{3} = 1960 cm^{3}

Area of side face = (28 × 35) × 2 cm^{2 }= 1960 cm^{2}

Area of back face = 70 × 35 cm^{2} = 2450 cm

∴. Total area = 1960 + 1960 + 2450 = 6370 cm^{2}

∴ Area of paper required = 6370 cm^{2}

**Question 4. The internal dimensions of rectangular hall are 15 m × 12 m × 4 m. There are 4 windows each of dimension 2 m × 1.5 m and 2 doors each of dimension 1.5 m × 2.5 m. Find the cost of white washing all four walls of the hall, if the cost of white washing is ₹5 per m**^{2}. What will be the cost of white washing if the ceiling of the hall is also white washed?

^{2}. What will be the cost of white washing if the ceiling of the hall is also white washed?

**Answer:**

Internal dimension of rectangular hall = 15m × 12 m × 4 m

Area of 4-walls = 2(l + b) × h

= 2(15 + 12) × 4

= 2 × 27 × 4 m^{2}

= 216 m^{2}

Area of 4 windows of size = 2 × 1.5 = 2 × 1.5 × 4 = 12 m^{2}

Area of 2 door of size = 1.5 × 2.5 = 2 × 1.5 × 2.5 = 7.5 m^{2}

∴ Area of remaining hall = 216 – (12 + 7.5) = 216 – 19.5 m^{2} = 196.5 m^{2}

Area of ceiling = l × b = 15 × 12= 180 m^{2}

Cost of white washing the walls at the rate of ₹5 per m^{2
}= 196.5 × 5 = ₹982.50

Area of ceiling = l × b = 15 × 12= 180 m^{2}

Cost of white washing = 180 × 5 = ₹900

∴ Total cost = ₹982.50 + 900.00 = ₹1882.50

**Question 5. A swimming pool is 50 m in length, 30 m in breadth and 2·5 m in depth. Find the cost of cementing its floor and walls at the rate of ₹27 per square metre.**

**Answer:**

Length of swimming pool = 50 m

Breadth of swimming pool = 30 m

Depth (Height) of swimming pool = 2·5 m

Area of floor = 50 × 30 = 1500 m^{2}

Area of four walls = 2 (50 + 30) × 2·5 = 160 × 2·5 = 400 m^{2}

Area to be cemented = 1500 m^{2} + 400 m^{2} = 1900 m^{2}

Cost of cementing 1m2 = ₹27

Cost of cementing 1900m^{2 }= ₹27 × 1900 = ₹51300

**Question 6. The floor of a rectangular hall has a perimeter 236 m. Its height is 4·5 m. Find the cost of painting its four walls (doors and windows be ignored) at the rate of Rs. 8.40 per square metre.**

**Answer:**

Perimeter of Hall = 236 m.

Height = 4·5 m

Perimeter = 2 (l + b) = 236 m

Area of four walls = 2 (l + b) × h = 236 × 4·5 = 1062 m^{2}

Cost of painting 1 m^{2} = ₹8·40

Cost of painting 1062 m^{2} = ₹8·40 × 1062 = ₹8920·80

**Question 7. A cuboidal fish tank has a length of 30 cm, a breadth of 20 cm and a height of 20 cm. The tank is placed on a horizontal table and it is three-quarters full of water. Find the area of the tank which is in contact with water.**

**Answer:**

Length of tank = 30 cm

Breadth of tank = 20 cm

Height of tank = 20 cm

As the tank is three-quarters full of water

∴ Height of water in the tank = = 15 cm

Area of the tank in contact with the water = Area of floor of Tank

+ Area of 4 walls upto 15 cm

= 30 × 20 + 2 (30 + 20) × 15

= 600 + 2 × 50 × 15

= 600 + 1500 = 2100 cm^{2}

**Question 8. The volume of a cuboid is 448 cm**^{3}. Its height is 7 cm and the base is a square. Find

^{3}. Its height is 7 cm and the base is a square. Find

(i) a side of the square base

(ii) surface area of the cuboid.

**Answer:**

Volume of a cuboid = 448 cm^{3}

Height = 7 cm

∴ Area of base = 448/7 = 64 cm^{2}

∵ Base is a square.

(i) ∴ Side of square base = √64 = 8 cm

(ii) Surface area of the cuboid = 2 [lb + bh + hl]
= 2[8 × 8 + 8 × 7 + 7 × 8] cm^{2}

= 2[64 + 56 + 56]
= 2 × 176 = 352 cm^{2}

**(ML Aggarwal Mensuration Exe-18.4 Class 8 ICSE Maths)**

**Question 9. The length, breadth and height of a rectangular solid are in the ratio 5 : 4 : 2. If its total surface area is 1216 cm**^{2}, find the volume of the solid.

^{2}, find the volume of the solid.

**Answer:**

Ratio in length, breadth and height of a rectangular solid = 5 : 4 : 2

Total surface area =1216 cm^{2}

Let Length = 5x,

Breadth = 4x

and height = 2x

Total surface area = 2[5x × 4x + 4x × 2x + 2x × 5x]
= 2[20x^{2} + 8x^{2} + 10x^{2} ]
= 2 × 38x^{2} = 76x^{2}

∴ 94x^{2} = 1216

⇒ x^{2} = 1216/76 = 16 = (4)^{2}

∴ x = 4

∴ Length = 5 × 4 = 20 cm

Breadth = 4 × 4 = 16 cm

Height = 2 × 4 = 8 cm

and volume = lbh = 20 × 16 × 8 = 2560 cm^{3}

**Question 10. A rectangular room is 6 m long, 5 m wide and 3·5 m high. It has 2 doors of size 1·1 m by 2 m and 3 windows of size 1·5 m by 1·4 m. Find the cost of whitewashing the walls and the ceiling of the room at the rate of ₹5·30 per square metre.**

**Answer:**

Length of room = 6 m

Breadth of room = 5 m

Height of room = 3·5 m

Area of four walls = 2 (l + b) × h

= 2 (6 + 5) × 3·5 = 77 m^{2}

Area of 2 doors and 3 windows

= (2 × 1·1 × 2 + 3 × 1·5 × 1·4)

= (44 + 6·3) m^{2} = 10·7 m^{2}

Area of ceiling = l × b = 6 × 5 = 30 m^{2}

Total area for white washing

= (77 – 10·7 + 30) m^{2} = 96·3 m^{2}

Cost of white washing = ₹(96·3 × 5·30) = ₹510·39

**Question 11. A cuboidal block of metal has dimensions 36 cm by 32 cm by 0·25 m. It is melted and recast into cubes with an edge of 4 cm.**

(i) How many such cubes can be made?

(ii) What is the cost of silver coating the surfaces of the cubes at the rate of ₹0·75 per square centimetre?

**Answer:**

(i) Length of cuboid = 36 cm

Breadth of cuboid = 32 cm

Height of cuboid = 0·25 × 100 = 25 cm

Volume of cubiod = lbh = (36 × 32 × 25) cm^{3} = 28800 cm^{3}.

Volume of cube = (side)^{2 }= (4)^{2} = 64 cm^{2}

Number of cubes recasting from cubiod = 28800/64 = 450

(ii) Surface area of 1 cube = 6 × a^{2 }= 6 × 16 = 96 cm^{2}

Surface area of 450 cubes = 96 × 450 = 43200 cm^{2}

Cost of silver coating on cubes = ₹0·75 × 43200 = ₹32400

**Mensuration Exe-18.4**

**ML Aggarwal Class 8 ICSE Maths Solutions**

**Page-313**

**Question 12. Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube, find the cost of coating the surface of the new cube with gold at the rate of ₹3·50 per square centimetre?**

**Answer:**

Let a cm be the edge of new cube

∴ According to question,

a^{3} = 3^{3} + 4^{3} + 5^{3} = 27 + 64 + 125 = 216 cm^{3}

a = 3√216

⇒ a = 6 cm.

Surface area of new cube = 6 × (side)^{2 }= 6 × (6)^{2} = 216 cm^{2}

Cost of coating the surface of new cube = ₹3·50 × 216 = ₹156

**Question 13. The curved surface area of a hollow cylinder is 4375 cm**^{2}, it is cut along its height and formed a rectangular sheet of width 35 cm. Find the perimeter of the rectangular sheet.

^{2}, it is cut along its height and formed a rectangular sheet of width 35 cm. Find the perimeter of the rectangular sheet.

**Answer:**

Curved surface area of a hollow cylinder = 4375 cm^{2}

By cutting it from the height,

It becomes a rectangular sheet whose width = 35 cm

So, the height of cylinder = 35 cm

And, length of sheet = Area/Height

= 4375/35

= 125 cm

Perimeter of the sheet = 2(l + b)

= 2 × (125 + 35)

= 2 × 160

= 320 cm

**Question 14. A road roller has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must take in order to level a playground of size 120 m × 44 m.**

**Answer:**

Diameter of a road roller = 0.7 m = 70 cm

So, radius (r) = 70/2 cm = 35 cm = 35/100 m

and width (h) = 1.2 m

Now,

Curved surface area = 2πrh

= (2 × 22/7 × 35/100 × 1.2) m^{2}

= 264/100 m^{2}

Area of playground = 120 m × 44 m

= 120 × 44 m^{2}

= 5280 m^{2}

Hence, the number of revolution made by the road roller = (5280/264) × 100

= 2000 revolutions

**Question 15. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?**

**Answer:**

Diameter of cylindrical container = 14 cm

∴ Radius (r) = 14/2 = 7 cm

and height (h) = 20 cm

Width of lable = 2 0 – (2 + 2) cm = 20 – 4 = 16 cm

∴ Area of lable = 2πrh = 2 × 22/7 × 7 × 16 = 704 cm^{2}

**(ML Aggarwal Mensuration Exe-18.4 Class 8 ICSE Maths)**

**Question 16. The sum of the radius and height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm**^{2}. Find the height and the volume of the cylinder.

^{2}. Find the height and the volume of the cylinder.

**Answer:**

Sum of height and radius of a cylinder = 37 cm

Total surface area = 1628 cm^{2}

Let’s consider the radius to be r

Then, height = (37 – r) cm

Total surface area = 2π(h + r)

⇒ r = 7 cm

Height = 37 – 7 = 30 cm

Volume = πr^{2}h

= (22/7) × 7 × 7 × 30 cm^{3 }

= 4620 cm^{3}

**Question 17. The ratio between the curved surface and total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm**^{3}.

^{3}.

**Answer:**

Given that ratio between curved surface and total surface area of a cylinder = 1 : 2

Total surface area = 616 cm^{2}

So, curved surface area = 616/2 = 308 cm^{2}

Area of two circular faces = 616 – 308 = 308 cm^{2}

Area of one circular face = 308/2 = 154 cm^{2}

Now, let’s consider the radius to be r

πr^{2} = 154

(22/7) × r^{2} = 154

r^{2} = (154 × 7)/ 22 = 49

⇒ r = 49 = √7 cm

the volume = πr^{2}h = (22/7) × 7 × 7 × 7 = 1078 cm^{2}

**Question 18. The given figure shown a metal pipe 77 cm long. The inner diameter of cross section is 4 cm and the outer one is 4.4 cm. Find its**

(i) inner curved surface area

(ii) outer curved surface area

(iii) total surface area.

**Answer:**

Length of metal pipe (h) = 77 cm

Inner diameter = 4 cm

and outer diameter = 4.4 cm

So, inner radius (r) = 4/2 = 2 cm

And outer radius (R) = 4.4/2 = 2.2 cm

**(i) Inner curved surface area = 2πrh**

= 2 × 22/7 × 2 × 77 cm^{2}

= 968 cm^{2}

**(ii) Outer surface area = 2πRH**

= 2 × 22/7 × 2.2 × 77 cm^{2}

= 1064.8 cm^{2}

**(iii) Surface area of upper and lower rings = 2[πR ^{2} – πr^{2}]**

= 2 × 22/7 (2.2^{2} – 2^{2}) cm^{2}

= 44/7 × 4.2 × 0.2

= 5.28 cm^{2}

Total surface area = (968 + 1064.8 + 5.28) cm^{2} = 2038.08 cm^{2}

— End of Mensuration ** Exe-18.4 **Class 8 ICSE Maths Solutions :–

Return to : **– **ML Aggarwal Maths Solutions for ICSE Class -8

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