Mensuration — Exercise 19.3

The volume of a cube is 343 cm³, find the length of an edge of cube.

Answer

Given,

Volume of the cube = 343 cm³

Let the length of an edge of the cube be a cm.

By formula,

Volume of a cube = (edge)³

⇒ a³ = 343

⇒ a = $\sqrt[3]{343}$

⇒ a = 7 cm

Hence, the length of an edge of the cube is 7 cm.

Fill in the following blanks:

Answer

By formula,

Volume of a cuboid = length × breadth × height

(i) Volume of the cuboid = 90 cm³, breadth = 5 cm and height = 3 cm

Let the length of the cuboid be l cm.

⇒ l × 5 × 3 = 90

⇒ 15 × l = 90

⇒ l = $\dfrac{90}{15}$

⇒ l = 6 cm

Hence, the length of the cuboid is 6 cm.

(ii) Length = 15 cm, breadth = 8 cm and height = 7 cm

Volume of the cuboid = 15 × 8 × 7 cm³

= 840 cm³

Hence, the volume of the cuboid is 840 cm³.

(iii) Volume of the cuboid = 62.5 m³, length = 10 m and breadth = 5 m

Let the height of the cuboid be h m.

⇒ 10 × 5 × h = 62.5

⇒ 50 × h = 62.5

⇒ h = $\dfrac{62.5}{50}$

⇒ h = 1.25 m

Hence, the height of the cuboid is 1.25 m.

Find the height of a cuboid whose volume is 312 cm³ and base area is 26 cm².

Answer

Given,

Volume of the cuboid = 312 cm³

Base area of the cuboid = 26 cm²

Let the height of the cuboid be h cm.

By formula,

Volume of a cuboid = base area × height

⇒ 26 × h = 312

⇒ h = $\dfrac{312}{26}$

⇒ h = 12 cm

Hence, the height of the cuboid is 12 cm.

A godown is in the form of a cuboid of measures 55 m × 45 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 1.25 m³?

Answer

Given,

Dimensions of the godown = 55 m × 45 m × 30 m

Volume of one box = 1.25 m³

By formula,

Volume of a cuboid = length × breadth × height

Volume of the godown = 55 × 45 × 30 m³

= 74250 m³

$\text{ Number of boxes } = \dfrac{\text{Volume of the godown}}{\text{Volume of one box}} \\[1em] = \dfrac{74250}{1.25} \\[1em] = \dfrac{7425000}{125} \\[1em] = 59400$

Hence, 59400 cuboidal boxes can be stored in the godown.

A rectangular pit 1.4 m long, 90 cm broad and 70 cm deep was dug and 1000 bricks of base 21 cm by 10.5 cm were made from the earth dug out. Find the height of each brick.

Answer

Given,

Length of the rectangular pit = 1.4 m = 140 cm

Breadth of the rectangular pit = 90 cm

Depth of the rectangular pit = 70 cm

Number of bricks = 1000

Base of each brick = 21 cm × 10.5 cm

By formula,

Volume of a cuboid = length × breadth × height

Volume of the earth dug out = 140 × 90 × 70 cm³

= 882000 cm³

Volume of one brick = $\dfrac{\text{Volume of the earth dug out}}{\text{Number of bricks}}$

= $\dfrac{882000}{1000}$ cm³

= 882 cm³

Let the height of each brick be h cm.

Volume of one brick = base area × height

⇒ 21 × 10.5 × h = 882

⇒ 220.5 × h = 882

⇒ h = $\dfrac{882}{220.5}$

⇒ h = 4 cm

Hence, the height of each brick is 4 cm.

If each edge of a cube is tripled, then find how many times will its volume become?

Answer

Let the edge of the cube be a units.

By formula,

Volume of a cube = (edge)³

Volume of the given cube = a³ cubic units

When each edge is tripled, the edge of the new cube = 3a units

Volume of the new cube = (3a)³

= 27a³ cubic units

= 27 × Volume of the given cube

Hence, the volume of the cube becomes 27 times its original volume.

A milk tank is in the form of cylinder whose radius is 1.4 m and height is 8 m. Find the quantity of milk in litres that can be stored in the tank.

Answer

Given,

Radius of the cylindrical milk tank = 1.4 m

Height of the cylindrical milk tank = 8 m

By formula,

Volume of a cylinder = πr²h

$= \dfrac{22}{7} \times (1.4)^2 \times 8\\[1em] = \dfrac{22}{7} \times 1.96 \times 8\\[1em] = 22 \times 0.28 \times 8\\[1em] = 49.28 \text{ m}^3$

So, volume of the tank = 49.28 m³

Converting the volume into litres,

Quantity of milk = 49.28 × 1000 litres

[∵ 1 m³ = 1000 litres]

= 49280 litres

Hence, 49280 litres of milk can be stored in the tank.

A closed box is made of 2 cm thick wood with external dimension 84 cm × 75 cm × 64 cm. Find the volume of the wood required to make the box.

Answer

Given,

External dimensions of the closed box = 84 cm × 75 cm × 64 cm

Thickness of the wood = 2 cm

Since the box is closed, the wood is present on both the opposite faces in each direction. So, each external dimension decreases by 2 × 2 = 4 cm.

Internal length = 84 - 4 cm = 80 cm

Internal breadth = 75 - 4 cm = 71 cm

Internal height = 64 - 4 cm = 60 cm

By formula,

Volume of a cuboid = length × breadth × height

External volume of the box = 84 × 75 × 64 cm³

= 403200 cm³

Internal volume of the box = 80 × 71 × 60 cm³

= 340800 cm³

Volume of the wood = External volume - Internal volume

= 403200 - 340800 cm³

= 62400 cm³

Hence, the volume of the wood required to make the box is 62400 cm³.

Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.

Answer

Given,

The diameters of the two cylindrical jars are in the ratio 3 : 4.

Both the jars contain the same amount of milk.

Let the diameters of the two jars be 3a and 4a.

By formula,

Radius of a circle = $\dfrac{\text{Diameter}}{2}$

Radius of the first jar (r₁) = $\dfrac{3a}{2}$

Radius of the second jar (r₂) = $\dfrac{4a}{2}$ = 2a

Let the heights of the two jars be h₁ and h₂ respectively.

By formula,

Volume of a cylinder = πr²h

Since both the jars contain the same amount of milk, their volumes are equal.

$\Rightarrow \pi \times \left(\dfrac{3a}{2}\right)^2 \times h_1 = \pi \times (2a)^2 \times h_2\\[1em] \Rightarrow \dfrac{9a^2}{4} \times h_1 = 4a^2 \times h_2\\[1em] \Rightarrow \dfrac{h_1}{h_2} = 4a^2 \times \dfrac{4}{9a^2}\\[1em] \Rightarrow \dfrac{h_1}{h_2} = \dfrac{16}{9}$

Hence, the ratio of the heights of the two jars is 16 : 9.

The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?

Answer

Let the radius of the base of the original cylinder be r and its height be h.

By formula,

Volume of a cylinder = πr²h

Volume of the original cylinder = πr²h

For the new cylinder, the radius is halved and the height is doubled.

Radius of the new cylinder = $\dfrac{r}{2}$

Height of the new cylinder = 2h

$\text{Volume of the new cylinder} = \pi \times \left(\dfrac{r}{2}\right)^2 \times 2h\\[1em] = \pi \times \dfrac{r^2}{4} \times 2h\\[1em] = \dfrac{\pi r^2 h}{2}$

Ratio of the volumes = $\dfrac{\text{Volume of the new cylinder}}{\text{Volume of the original cylinder}}$

$= \dfrac{\pi r^2 h}{2} \div \pi r^2 h\\[1em] = \dfrac{\pi r^2 h}{2} \times \dfrac{1}{\pi r^2 h}\\[1em] = \dfrac{1}{2}$

Hence, the ratio of the volume of the new cylinder to that of the original cylinder is 1 : 2.

A rectangular piece of tin of size 30 cm × 18 cm is rolled in two ways, once along its length (30 cm) and once along its breadth. Find the ratio of volumes of two cylinders so formed.

Answer

Given,

Size of the rectangular piece of tin = 30 cm × 18 cm

Case 1: Rolled along its length (30 cm)

Here, the length of the sheet becomes the circumference of the base of the cylinder and the breadth becomes its height.

Let the radius of the base of this cylinder be r₁ cm.

⇒ 2πr₁ = 30

⇒ r₁ = $\dfrac{30}{2\pi}$

Height of this cylinder, h₁ = 18 cm

Case 2: Rolled along its breadth (18 cm)

Here, the breadth of the sheet becomes the circumference of the base of the cylinder and the length becomes its height.

Let the radius of the base of this cylinder be r₂ cm.

⇒ 2πr₂ = 18

⇒ r₂ = $\dfrac{18}{2\pi}$

Height of this cylinder, h₂ = 30 cm

By formula,

Volume of a cylinder = πr²h

$\text{Ratio of the volumes} = \dfrac{\pi r_1^2 h_1}{\pi r_2^2 h_2}\\[1em] = \dfrac{\pi \times \left(\dfrac{30}{2\pi}\right)^2 \times 18}{\pi \times \left(\dfrac{18}{2\pi}\right)^2 \times 30}\\[1em] = \dfrac{30^2 \times 18}{18^2 \times 30}\\[1em] = \dfrac{30}{18}\\[1em] = \dfrac{5}{3}$

Hence, the ratio of the volumes of the two cylinders so formed is 5 : 3.

Water flows through a cylindrical pipe of internal diameter 7 cm at 5 m per sec. Calculate

(i) the volume in litres of water discharged by the pipe in one minute.

(ii) the time in minutes, the pipe would take to fill an empty rectangular tank of size 4 m × 3 m × 2.31 m

Answer

Given,

Internal diameter of the cylindrical pipe = 7 cm

Speed of the water = 5 m per sec

Radius of the pipe = $\dfrac{7}{2}$ cm = 3.5 cm

(i) Length of the column of water discharged in one second = 5 m = 500 cm

Length of the column of water discharged in one minute = 500 × 60 cm

= 30000 cm

By formula,

Volume of a cylinder = πr²h

$\text{Volume of water discharged in one minute} = \dfrac{22}{7} \times (3.5)^2 \times 30000\\[1em] = \dfrac{22}{7} \times 12.25 \times 30000\\[1em] = 22 \times 1.75 \times 30000\\[1em] = 1155000 \text{ cm}^3$

So, volume of water discharged in one minute = 1155000 cm³

Converting the volume into litres,

Volume of water = $\dfrac{1155000}{1000}$ litres

[∵ 1 litre = 1000 cm³]

= 1155 litres

Hence, 1155 litres of water is discharged by the pipe in one minute.

(ii) Dimensions of the rectangular tank = 4 m × 3 m × 2.31 m

= 400 cm × 300 cm × 231 cm

By formula,

Volume of a cuboid = length × breadth × height

Volume of the tank = 400 × 300 × 231 cm³

= 27720000 cm³

Time taken to fill the tank = $\dfrac{\text{Volume of the tank}}{\text{Volume of water discharged in one minute}}$

= $\dfrac{27720000}{1155000}$ minutes

= 24 minutes

Hence, the pipe would take 24 minutes to fill the empty rectangular tank.

Two cylindrical vessels are filled with milk. The radius of one vessel is 15 cm and height is 40 cm, and the radius of other vessel is 20 cm and height is 45 cm. Find the radius of another cylindrical vessel of height 30 cm which may just contain the milk which is in the two given vessels.

Answer

Given,

Radius of the first vessel = 15 cm and its height = 40 cm

Radius of the second vessel = 20 cm and its height = 45 cm

Height of the third vessel = 30 cm

By formula,

Volume of a cylinder = πr²h

Volume of the first vessel = π × (15)² × 40

= 9000π cm³

Volume of the second vessel = π × (20)² × 45

= 18000π cm³

Total volume of the milk = 9000π + 18000π cm³

= 27000π cm³

Let the radius of the third cylindrical vessel be r cm.

Since the third vessel just contains all the milk,

⇒ π × r² × 30 = 27000π

⇒ r² = $\dfrac{27000\pi}{30\pi}$

⇒ r² = 900

⇒ r = $\sqrt{900}$

⇒ r = 30

Hence, the radius of the other cylindrical vessel is 30 cm.

A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m³

Answer

Given,

Height of the wooden pole = 7 m

Diameter of the wooden pole = 20 cm

Weight of the wood = 225 kg per m³

Radius of the pole = $\dfrac{20}{2}$ cm = 10 cm = 0.1 m

By formula,

Volume of a cylinder = πr²h

$= \dfrac{22}{7} \times (0.1)^2 \times 7\\[1em] = \dfrac{22}{\cancel{7}} \times 0.01 \times \cancel{7}\\[1em] = 22 \times 0.01\\[1em] = 0.22 \text{ m}^3$

So, volume of the wooden pole = 0.22 m³

Weight of the pole = Volume of the pole × Weight of 1 m³ of wood

= 0.22 × 225 kg

= 49.5 kg

Hence, the weight of the wooden pole is 49.5 kg.

A cylinder of maximum volume is cut from a wooden cuboid of length 30 cm and cross-section a square of side 14 cm. Find the volume of the cylinder and the volume of the wood wasted.

Answer

Given,

Length of the wooden cuboid = 30 cm

Side of the square cross-section = 14 cm

The cylinder of maximum volume that can be cut from this cuboid has its height equal to the length of the cuboid and the diameter of its base equal to the side of the square cross-section.

Height of the cylinder = 30 cm

Diameter of the base of the cylinder = 14 cm

Radius of the base of the cylinder = $\dfrac{14}{2}$ cm = 7 cm

By formula,

Volume of a cylinder = πr²h

$= \dfrac{22}{7} \times 7^2 \times 30\\[1em] = \dfrac{22}{\cancel{7}} \times \cancel{7} \times 7 \times 30\\[1em] = 22 \times 7 \times 30\\[1em] = 4620 \text{ cm}^3$

So, volume of the cylinder = 4620 cm³

By formula,

Volume of a cuboid = length × breadth × height

Volume of the wooden cuboid = 14 × 14 × 30 cm³

= 5880 cm³

Volume of the wood wasted = Volume of the cuboid - Volume of the cylinder

= 5880 - 4620 cm³

= 1260 cm³

Hence, the volume of the cylinder is 4620 cm³ and the volume of the wood wasted is 1260 cm³.