Class 8 RD Sharma Solutions – Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.4 | Set 1

Question 1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.

Solution:

Given, the length of room = 12 m

The breadth of room = 9 m

The height of room = 8 m

We need to find the longest rod that can be placed in the room i.e we need to find the diagonal of the room

So, the diagonal of the room = √[l² + b² + h²]

= √[12² + 9² + 8²] = 17 m

Hence, the longest rod that can be placed in the room is 17 m

Question 2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that 1/V = 2/S (1/a + 1/b + 1/c)

Solution:

Given, the dimensions of the cuboid are a, b, c

V is the volume of the cuboid and S is the surface area of the cuboid

We know, the surface area of cuboid = 2(l × b + b × h + h × l)

So, S = (a × b + b × c + c × a)

And the volume of cuboid = lbh

So, V = a × b × c

S/V = 2 [a×b + b×c + c×a] / a×b×c

= 2[(a×b/a×b×c) + (b×c/a×b×c) + (c×a/a×b×c)]

Solving further we get,

1/V = 2/S (1/a + 1/b + 1/c)

Hence, proved

Question 3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V, prove that V² = xyz.

Solution:

Given, x, y, and z are the adjacent faces of the cuboid

Let l be the Length, b be breadth, h be the Height and V be the volume of the cuboid

So, x = l × b

y = b × h

z = l × h

Multiplying x, y, and z we get,

x × y × z = l × b × b × h × h × l

So, xyz = (l × b × h)²

xyz = V²

Hence, proved

Question 4. A rectangular water reservoir contains 105 m³ of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.

Solution:

Given, the volume of reservoir = 105 m³

The length of reservoir = 12 m

The breadth of reservoir = 3.5 m

Let h be the depth of the reservoir

So, Volume of reservoir = l × b × h

105 = 12 × 3.5 × h

So, h = 2.5 m

Hence, the depth of water in the reservoir is 2.5 m

Question 5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.

Solution:

Given, the edges of cube A, B, and C are 18 cm, 24 cm, and 30 cm

So, the Volume of cube A = (edge)³

= (18)³ = 5832 cm³

The Volume of cube B = (edge)³

= (24)³ = 13824 cm³

The Volume of cube C = (edge)²

= (30)³ = 27000 cm³

Let b be the edge of Cube D

The sum of volumes of cube A, B, and C will be equal to the volume of cube D

So, 5832 + 13824 + 27000 = a³

So, a = 36 cm

Hence, the edge of cube D is 36 cm

Question 6. The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. Dm. Find its dimensions.

Solution:

Let l, b, and h be the Length, Breadth, and Height of the room

Given, the volume of the room is 512 dm³

Also, the breadth = 2 × h and b = l/2

So, l = 2 × b

And h = b/2 

The volume of room = l × b × h

512 = 2b × b × (b/2)

So, b = 8 dm

Also, length = 2b = 16 dm

And, height = b/2 = 4 dm

Hence, the dimensions of the room are 16 dm, 8 dm, and 4 dm

Question 7. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs. 5 per meter sheet, sheet being 2 m wide.

Solution:

Given, the length of tank = 12 m

The breadth of tank = 9 m

The height of tank = 4 m

Also, the area of iron sheet will be equal to surface area of cuboid

= 2(length × breadth + breadth × height + height × length)

= 2(12 × 9 + 9 × 4 + 4 × 12) = 384 m²

Now, let the length of iron sheet is a m

And, breadth/width is 2 m 

So, length of sheet × width of sheet = 384 m²

a × 2 = 384

a = 192 m 

Cost of iron sheet will be 192 × 5 = Rs 960

Hence, the cost of iron sheet used is Rs 960

Question 8. A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12m×8m×6m, find the cost of iron sheet at Rs. 17.50 per meter.

Solution:

Given, the length of tank = 12 m

The breadth of tank = 8 m

The height of tank = 6 m

So, the area of sheet required = The surface area of tank with only one top 

= length × breadth + 2 (length×height + breadth×height)

= 12 × 8 + 2(12 × 6 + 8 × 6)

= 336 m²

Now, let the length of iron sheet is a m

And, breadth/width is 4 m 

So, length of sheet × width of sheet = 336 m²

a × 4 = 336

a = 84 m 

Cost of iron sheet will be 84 × 17.50 = Rs 1470

Hence, the cost of iron sheet used is Rs 1470

Question 9. Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

Solution:

Let a be the edges of three cubes placed adjacently

So, the sum of areas of 3 cubes will be 3 × 6 (edge)²

= 3 × 6a² = 18a²

Also, when these cubes are placed adjacently they form a cuboid 

The length of cuboid so formed = a + a + a = 3a m

And, the breadth of cuboid so formed = a m

And, the height of cuboid so formed = a m

We know surface area of cuboid = 2(length × breadth + breadth × height + height × length)

= 2 (3a × a + a × a + a × 3a)

= 14a²

And finally, the ration of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a²/18a²

= 14/18 = 7 : 9

Hence, the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes is 7 : 9

Question 10. The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs. 3.50 per square meter.

Solution:

Given, the length of room = 12.5 m

The breadth of room = 9 m

The height of room = 7 m

And, the dimensions of each door is 2.5 m × 1.2 m

And, the dimensions of each window is 1.5 m × 1 m

Now calculating area of four walls in which doors and windows are included,

= 2 (length×height + breadth×height)

= 2 (12.5×7 + 9×7) = 301 m²

Now calculating area of 2 doors and 4 windows,

= 2 [2.5 × 1.2] + 4 [1.5 × 1] = 12 m²

So, the area of four walls will be = 301 m² – 12 m²

= 289 m²

Now, the cost of painting four walls = Rs 3.50 × 289 = Rs 1011.50

Hence, the cost of painting four walls is Rs 1011.50

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