Chapter 18: Surface Area and Volume of a Cuboid and Cube Exercise – 18.2

Question: 1

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold?

Solution:

Given data:

Length (l) = 6 m

Breadth (b) = 5 m

Height (h) = 4.5 m

Volume of the tank = l * b * h

= 6 * 5 * 4.5

= 135 m³

It is given that,

1m³ = 1000 liters

Therefore, 135m³ = (135 ∗ 1000) liters

= 135000 liters

The tank can hold 1, 35, 000 liters of water.

Question: 2

A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?

Solution:

Given that

Length of the vessel (l) = 10 m

Width of the Cuboidal vessel = 8 m

Let 'h' be the height of the cuboidal vessel.

Volume of the vessel = 380 m³

Therefore, l ∗ b ∗ h = 380 m³

⇒ 10 ∗ 8 ∗ h = 380

⇒ h = 4.75 m

Therefore, height of the vessel should be 4.75 m.

Question: 3

Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m³

Solution:

Given,

Length of the cuboidal pit (l) = 8 m

Breadth of the cuboidal pit (b) = 6 m

Depth of the cuboidal pit (h) = 3 m

Volume of the Cuboidal pit = l * b * h

= 8 * 6 * 3

= 144 m³

Cost of digging 1m³ = Rs. 30

Cost of digging 144m³ = 144 * 30 = Rs.4320

Question: 4

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 Data:

Length of the cube (l) = a

Breadth of the cube (b) = b

Height of the cube (h) = c

Volume of the cube (V) = l * b * h

= a * b * c

= abc

Surface area of the cube (S) = 2 (lb + bh + hl)

= 2(ab + bc + ca)

Hence Proved.

Question: 5

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

Solution:

Let a, b and d be the length, breadth, and height of the cuboid.

Then, x = ab

y = bc

z = ca

and V = abc [V = l * b * h]

= xyz = ab * bc * ca = (abc)²

and V = abc

V² = (abc)²

Therefore,V² = (xyz)

Question: 6

If the areas of three adjacent face of a cuboid are 8 cm³, 18 cm³ and 25 cm³. Find the volume of the cuboid.

Solution:

WKT, if x, y, z denote the areas of three adjacent faces of a cuboid.

= x = l * b, y = b * h, z = l * h

Volume (V) is given by

V = l * b * h

Now, xyz = lb * bh * hl = V²

Here x = 8

y = 18

z = 25

Therefore, V² = 8 ∗ 18 ∗ 25 = 3600

⇒ V = 60 cm³

Question: 7

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:

Consider l, b and h are the length, breadth and height of the room.

So, b = 2h and b = (1/2)l

⇒ l/2 = 2h

⇒ l = 4h

⇒ l = 4h, b = 2h

Now, Volume = 512 dm³

⇒ 4h ∗ 2h ∗ h = 512

⇒ h³ = 64

⇒ h = 4

So, Length of the room (l) = 4h = 4 * 4 = 16 dm

Breadth of the room (b) = 2h = 2 * 4 = 8 dm

And Height of the room (h) = 4 dm.

Question: 8

A river 3m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Solution:

Radius of the water flow = 2 km per hour = (2000/60) m/min

= (100/3) m/min

Depth of the river (h) = 3 m

Width of the river (b) = 40 m

Volume of the water flowing in 1 min = 100/3 ∗ 40 ∗ 3 = 4000³

Thus, 1 minute 4000m³ = 4000000 litres of water will fall in the sea.

Question: 9

Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km every hour. What much area will it irrigate in 30 minutes if 8 cm of standing water is desired?

Solution:

Given that,

Water in the canal forms a cuboid of Width (b) = 30 dm = 3 m

Height (h) = 12 dm = 1.2 m

Cuboid length is equal to the distance traveled in 30 min with the speed of 100 km per hour.

Therefore, Length of the cuboid = 100∗30/60 = 60 km = 50000 metres

So, volume of water to be used for irrigation = 5000 * 3 * 1.2 m³

Water accumulated in the field forms a cuboid of base area equal to the area of the field and height equal to 8/100 metres

Therefore, Area of field∗8/100 = 50000 ∗ 3 ∗ 1.2

⇒ 22, 50, 000 metres.

Question: 10

Three metal cubes with edges 6cm, 8cm, 10cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.

Solution:

Let ‘a’ be the length of each edge of the new cube.

Then a³ = (6³ + 8³ + 10³) cm³

⇒ a³ = 1728

⇒ a = 12

Therefore, Volume of the new cube = a³ = 1728 cm³

Surface area of the new cube = 6a² = 6 ∗ (12)² = 864 cm²

Diagonal of the newly formed cube = √3 a = 12√3 cm

Question: 11

Two cubes, each of volume 512 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Given that,

Volume of the cube = 512 cm³

⇒ side³ = 512

⇒ side³ = 8³

⇒ side = 8 cm

Dimensions of the new cuboid formed

Length (l) = 8 + 8 = 16 cm, Breadth (b) = 8 cm, Height (h) = 8 cm

Surface area = 2(lb + bh + hl)

= 2(16 * 8 + 8 * 8 + 16 * 8)

= 640 cm²

Therefore, Surface area is 640 cm².

Question: 12

Half cubic meter of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold-sheet.

Solution: