**S Chand ICSE Mathematics Class 9 Solution Eighteenth Chapter Surface Area and Volume of 3D Solids Exercise 18A**

**(1) Find the surface area of the cuboids whose dimensions are:**

**(a) 3m, 4m, 5m**

**(b) 4 cm, 1.7 cm, 2.3 cm**

**Solution: **(a) Surface Area = 2(3×4 + 4×5 + 5×3) m^{2}

= 2 (12 + 20 + 15) m^{2}

= 2 (47) m^{2}

= 94 m^{2}

(b) Surface Area = 2 (4 x 1.7 + 1.7 x 2.3 + 2.3 x 4) cm^{2}

= 2 (6.8 + 3.91 + 9.2) cm^{2}

= 39.82 cm^{2}

**(2) (i) Find the surface area of the cubes whose dimensions are:**

**(a) 7 cm**

**(b) 10 m**

**(ii) To make a model of a cube 96 cm ^{2} of board was used. Find the length of one edge of the cube.**

**Solution (ii): **Let a be the length of edge,

∴ 6a^{2} = 96

Or, a^{2} = 16

Or, a = ±4,

Since, length is not negative So 4 cube the length of that cube.

**(iii) The total surface area of a cube is 726 cm ^{2}. Find its volume.**

**Solution (iii): **Let a be the length of one edge of the cube, then 6a^{2} = 726

Or, a^{2}= 121

Or, a = ±11

So, Since 11 cm be the length of its edge.

So, Volume is = 11^{3} = 1331 cm^{3}

**(3) Complete the following table. The measurements are in cm.**

(i) | (ii) | (iii) | (iv) | (v) | (vi) | (vii) | (viii) | |

length | 4 | 2 | 7 | 12 | 16 | 38.66 | 40 | 60 |

width | 3 | 6 | 3 | 9 | 14 | 28 | 24 | 18 |

height | 5 | 8 | 4 | 12 | 18 | 24 | 2.5 | 5 |

volume | 60 | 96 | 84 | 1296 | 4032 | 25 984 | 2400 | 5400 |

**(4) Find the length of the diagonal of the cuboid whose dimension are:**

**(i) 2, 3, 4 cm**

**(ii) 3, 4, 5 cm**

Solution:

**(5) (i) Find the length of the diagonal and the volume of the cubes whose each edge is –**

**(a) 2 m**

**(b) 5 m**

**(c) 8 cm**

Solution:

(a) Diagonal = √3a

= √3 x 2 m

= 2√3 m

(b) Diagonal is = 5√3 m

(c) Diagonal is = 8√3 m

**(ii) Find the volume & total surface area of a cube if its edge is 12 cm.**

Volume = (12)^{3}

= 12 x 12 x 12

= 1728 cm^{2 }

Total surface area = 6 x (12)^{2} cm^{2}

= 144 x 6

= 864 cm^{2}

**(6) Find each edge of the cubes whose volumes are:**

**(i) 216 m ^{3}**

**(ii) 2197 m ^{3}**

(i) Let each edge of the cube is ‘a’ m

(a) a^{3} = 216

a = 6

Each edge = 6 m

(b) Let each edge of a cube whose volume is 2197 m^{3} is, ‘l’ m

Therefore, l^{3} = 2197

l = 13

So, each edge = 13 m.

**(ii) Answer True or False:**

The volume of a rectangular solid measuring 1m by 50 cm by 0.5 m is 250 000 cm^{3}

Therefore, 1m = 100 cm,

50 cm,

0.5 m = 50 cm

Therefore Volume = 100 x 50 x 50 cm^{3}

= 250000 cm^{3}

So, statement is true.

**(iii) The diagonal of a rectangular solid is 5√2 dm. If its length and breadth are 5 dm and 4 dm, then find the height of the solid.**

**(7) The volume of a rectangular solid 3600 cm ^{3}. If it is 20 cm long & 9 cm high. Find its width.**

**Solution:** Let b be the width.

Therefore, l x b x h = 3600

Or, l x b x h = 3600

Or, 20 x b x 9 = 3600

Or, b = 400 / 20 = 20

So, the width is 20 cm.

**(8) The length and breadth of a rectangular solid are respectively 25 cm and 20 cm. If the volume is 7000. If the volume is 7000 cm ^{3}. Find its height.**

Let, h be the height,

25 x 20 x h = 7000

Or, h = 7000 / 500 = 14

So, height is 14 cm.

**(9) The perimeter of one face of a cube is 20 cm. Find (i) the total area of 6 faces (ii) The volume of the cube.**

**Solution:** Let a be the length of one edge of the cube,

(i) ∴ 4a = 20

Or, a = 5

∴ length of one edge is = 5 cm

So total area of 6 face

= 6 x a^{2} = 6 x 5^{2} = 150 cm^{2}

(ii) Volume of cube = a^{3} = 5^{3 }= 5125 cm^{3}

**(10) The area of a playground is 4800 m ^{2} Find the cost of covering it with gravel 1 cm deep, if the gravel cost Rs. 4.80 per-cubic metre.**

∴ Volume of the playground = Area x deep

= 4800 x 1/100 m^{3}

= 48 m^{3}

Cost of covering it with gravel = 48 x 4.80

= 230.40 Rs.

**(11) A rectangular water tank of base 7m x 6m contains water upto a depth of 5m. How many cubic metres of water are there in the tank?**

So, water in the tank = (7 x 6 x 5) m^{2}

= 42 x 5 m^{3}

= 210 m^{3}

**(12) The internal measurements of a box are 20 cm long, 16 cm wide, and 24 cm high. How many 4 cm cubes could be put into the box?**

**Solution: **Volume of the box = 20 x 16 x 24 cm^{3}

= 7680 cm^{3}

Volume of 4 cm cube = 4^{3} = 64 cm^{3}

So, total number of box = 7680 / 64 = 120

**(13) The length, breadth and height of a rectangular solid are in ratio 5 : 4 : 2. If the total surface area is 1216 cm ^{2}. Find the length, breadth and height.**

**(14) A lock in a canal is 40 m long, 7 m wide. When the sluices are opened the depth of water in the lock decreases from 5 m to 3m 80 cm. How many cubic metre of water runs out?**

**(16) A tank 72 cm long, 60 cm wide, 36 cm high contain water at the depth of 18 cm. A metal block of 48 cm by 36 cm by 15 cm is put into the tank and totally submerged. Find in cm the amount of water level rises.**

**Solution:** So, total quantity of water in the tank (72 x 60 x 18) cm^{3}

Let, h be the level of water rises, then,

72 x 60 x h = 48 x 36 x 15

Or, h = 48 x 36 x 15 / 72 x 60

= 6

So, the level of water rises 6 cm.

**(17) Find the volume of wood required for making the closed box with internal dimension, 20 cm by 12.5 cm by 9.5 cm, wood 1.25 cm thick.**

Solution: So enternal length of the box is = 20 + (1.25 xd 2)

= 22.5 c m

Breadth = 12.5 + (1.25 x 2)

= 15 cm

Height = 9.5 + 2.5

= 12 cm

So, enternal volume = (22.5 x 15 x 12) cm^{3}

= 2375 cm^{3}

∴ Volume of wood (4050 – 2375) cm^{3}

= 1,675 cm^{3}

**(17) Find the volume of wood required for making the open box with external dimensions 17.5 cm long, 14 cm wide, 10 cm high, wood 7.5 mm thick.**

**Solution:** External length = 17.5 cm

Breadth = 14 cm

Height = 10 cm

Internal length = (17.5 – 1.5) cm

= 16 cm

Internal breadth = (14 – 1.5) cm

= 12.5 cm

Internal height (10 – 0.75) cm

= 9.25 cm

∴ External volume (17.5 x 14 x 10) cm^{3}

= 2450 cm^{3}

Internal volume = (16 x 12.5 x 9.25)

= 1850 cm^{3}

∴ Volume of wood = (2450 – 1850) cm^{3}

= 600 cm^{3}

**(19) A field is a field is 30 M long and 18 metre broad .A pit 6 m long,4 m wide and 3 M deep is dug out from the middle of the field and the earth removed is evenly spread over the remaining area of the field find the rise in the level of the remaining part of the field in centimeters correct to one decimal place.**

**Solution:**

Area of the field = (30 x 18) m^{2}

= 540 m^{2}

Area of the pit = (6 x 4) m^{2}

= 24 m^{2}

Remaining area = (540 – 24) m^{2}

= 516 m^{2}

But, rise of remaining part be ‘h’

∴ volume of earth = (6 x 4 x 3) m^{3}

= 72 m^{3}

But ‘h’ be the rise in the level of remaining part,

∴ 516 x h = 72

Or, h = 0.139534

∴ rise in height = 0.1395 m

= 13.9 cm

**(20) An agricultural field is the farm of rectangle of length 20m and width 14m. A pit 6 m long, 3 m wide and 2.5 m deep is dug in the corner in the field and earth is taken out from the pit is spread uniformly over the remaining area of the field. Find the extent to which the level of the field has been raised.**

**Solution:** Area of agricultural field = (20 x 14) m^{2}

= 280 m^{2}

Area of the pit = (6 x 3) m^{2}

= 18 m^{2}

Remaining area (280 – 18) m^{2}

= 262 m^{2}

Volume of earth = (6 x 3 x 2.5) m^{3}

= 45 m^{3}

Let the level rise is ‘h’

262 x h = 45

Or, h = 0.178

∴ level rise = 0.178 m

= 17.8 cm

**(21) A certain quantity of wood costs Rs. 250 per m ^{2}. A solid cubical block of such wood is bought for Rs. 182.25. Calculate the volume of the block and use the method of factors to find the length of one edge of the block.**

**Solution: **

**23) Two cubes each with 8 cm edge and joined end to end. Find the surface area of the resulting cuboid.**

**Solution:** When two cubes are join together then they form a cuboid,

Length of cuboid = (8 + 8) = 16 cm

Breadth = 8 cm

Height = 8 cm.

∴ Surface area of cuboid = 2 (16 x 8 + 8 x 8 + 16 x 8)

= 2 (128 + 64 + 128)

= 2 (320)

= 640 m^{2}

**(24) The area of three adjacent faces of a cuboid are x, y, z. If the volume is V, prove that V ^{2} = xyz**

Let, the three edges of cuboid are A,B, C

∴ AB = x,

BC = y,

CA = 7

∴ V = ABC,

∴ (AB) x (BC) x (CA) = xyz

Or, A^{2}B^{2}C^{2} = xyz

Or, V^{2} = xyz

**(25) A metal cube of edge 12 cm is melted and farmed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm. Find the edge of third smaller cube.**

Solution: ∴ let, the edge of third smaller cube is = ‘l’ cm

∴ 6^{3} + 8^{3} + l^{3} = 12^{3}

Or, l^{3} = 1000

Or, l = 10

∴ Edge of the smaller cube is = 10 cm.

** (26) 50 students sit in a classroom, Each student requires 9 m ^{2} on floor and 108 m^{3} in space. If the length of the room is 25 m, find its breadth and height.**

∴ One student requires 9 m^{2} on the floor.

∴ 50 students require = 50 x 9

= 450 m^{2}

∴ Let the breadth of the room is ‘b’

∴ 25 x b = 450

Or, b = 450 / 25 = 18

∴breadth = 18 m

One student need 108 m^{3} volume in space

∴ 50 student requires = (50 x 108) m^{3}

= 5400 m^{3}

Let, the height of the room be ‘h’

∴ 25 x 18 x h = 5400

Or, h = 5400 / 25 x 18

= 12

∴ Height of the room be 12 metre.

**(27) A cardboard shut is rectangular shape with dimensions 42 x 36 cm. From each one of its corner a square of 6 cm cut off. An open box is made of the remaining shut. Find the volume of the box.**

**Solution:**

∴ A square of 6 cm cut off from each corner, then a open box is made,

Then length of the box = 42 – (2×6)

= 42 – 12

= 30 m

Breadth of the box = 36 – (2 x 6)

= 36 – 12

= 24 m

Height of the box = 6 m

So the volume of the open box is = (30 x 24 x 6) cm^{3}

= (180 xd 24) cm^{3}

= 4320 cm^{3}

**(28) Two cubes, each of volume 343 cm ^{3} are joined end to end. Find the surface area of the resulting cuboid.**

**Solution: **

Two cubes, each of volume 343 cm^{3} but ‘l’ be the edge of the cube,

∴ l^{3} = 343

L = 7

∴ Edge of cube = 7 cm

∴ Length of cuboid = (7 +7) cm

= 14 cm

Breadth of cuboid = 7 cm

Height of the cuboid = 7 cm

∴ Surface area = 2)14×7 + 7×7 + 7×14)

= 490 cm^{2}

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