# 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 = âˆš[l

^{2}+ b^{2}+ h^{2}]= âˆš[12

^{2}+ 9^{2}+ 8^{2}] = 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 Ã— b + b Ã— h + h Ã— l)

So, S = (a Ã— b + b Ã— c + c Ã— a)

And the volume of cuboid = lbh

So, V = a Ã— b Ã— c

S/V = 2 [aÃ—b + bÃ—c + cÃ—a] / aÃ—bÃ—c

= 2[(aÃ—b/aÃ—bÃ—c) + (bÃ—c/aÃ—bÃ—c) + (cÃ—a/aÃ—bÃ—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 V**^{2} = xyz.

^{2}= 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 Ã— b

y = b Ã— h

z = l Ã— h

Multiplying x, y, and z we get,

x Ã— y Ã— z = l Ã— b Ã— b Ã— h Ã— h Ã— l

So, xyz = (l Ã— b Ã— h)

^{2}xyz = V

^{2}

Hence, proved

**Question 4. A rectangular water reservoir contains 105 m**^{3} of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.

^{3}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 m

^{3}The length of reservoir = 12 m

The breadth of reservoir = 3.5 m

Let

hbe the depth of the reservoirSo, Volume of reservoir = l Ã— b Ã— h

105 = 12 Ã— 3.5 Ã— 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 cm^{3}The Volume of cube B = (edge)

^{3}= (24)

^{3}= 13824 cm^{3}The Volume of cube C = (edge)

^{2}= (30)

^{3}= 27000 cm^{3}Let

bbe the edge of Cube DThe sum of volumes of cube A, B, and C will be equal to the volume of cube D

So, 5832 + 13824 + 27000 = a

^{3}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 dm

^{3}Also, the breadth = 2 Ã— h and b = l/2

So, l = 2 Ã— b

And h = b/2

The volume of room = l Ã— b Ã— h

512 = 2b Ã— b Ã— (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 Ã— breadth + breadth Ã— height + height Ã— length)

= 2(12 Ã— 9 + 9 Ã— 4 + 4 Ã— 12) = 384 m

^{2}Now, let the length of iron sheet is

amAnd, breadth/width is 2 m

So, length of sheet Ã— width of sheet = 384 m

^{2}a Ã— 2 = 384

a = 192 m

Cost of iron sheet will be 192 Ã— 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 12mÃ—8mÃ—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 Ã— breadth + 2 (lengthÃ—height + breadthÃ—height)

= 12 Ã— 8 + 2(12 Ã— 6 + 8 Ã— 6)

= 336 m

^{2}Now, let the length of iron sheet is

amAnd, breadth/width is 4 m

So, length of sheet Ã— width of sheet = 336 m

^{2}a Ã— 4 = 336

a = 84 m

Cost of iron sheet will be 84 Ã— 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

abe the edges of three cubes placed adjacentlySo, the sum of areas of 3 cubes will be 3 Ã— 6 (edge)

^{2}= 3 Ã— 6a

^{2}= 18a^{2}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 Ã— breadth + breadth Ã— height + height Ã— length)

= 2 (3a Ã— a + a Ã— a + a Ã— 3a)

= 14a

^{2}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 = 14a

^{2}/18a^{2}= 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 Ã— 1.2 m

And, the dimensions of each window is 1.5 m Ã— 1 m

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

= 2 (lengthÃ—height + breadthÃ—height)

= 2 (12.5Ã—7 + 9Ã—7) = 301 m

^{2}Now calculating area of 2 doors and 4 windows,

= 2 [2.5 Ã— 1.2] + 4 [1.5 Ã— 1] = 12 m

^{2}So, the area of four walls will be = 301 m

^{2}â€“ 12 m^{2}= 289 m

^{2}Now, the cost of painting four walls = Rs 3.50 Ã— 289 = Rs 1011.50

Hence, the cost of painting four walls is Rs 1011.50

## Please

Loginto comment...