In this chapter, we provide RD Sharma Solutions for Class 8 Chapter 21 Mensuration for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Chapter 21 Mensuration pdf, free RD Sharma Solutions for Class 8 Chapter 21 Mensuration book pdf download. Now you will get step by step solution to each question.

### Access answers to RD Sharma Maths Solutions For Class 8 Exercise 21.1 Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)

**1. Find the volume of a cuboid whose(i) length = 12 cm, breadth = 8 cm, height = 6 cm(ii) length = 1.2 m, breadth = 30 cm, height = 15 cm(iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm.**

**Solution:**

**(i)** The given details are:

Length of a cuboid = 12 cm

Breadth of a cuboid = 8 cm

Height of a cuboid = 6 cm

By using the formula

Volume of a cuboid = length × breadth × height

= 12 × 8 ×6

= 576 cm^{3}

**(ii)** The given details are:

Length of a cuboid = 1.2 m = 120 cm

Breadth of a cuboid = 30 cm

Height of a cuboid = 15 cm

By using the formula

Volume of a cuboid = length × breadth × height

= 120 × 30 × 15

= 54000 cm^{3}

**(iii)** The given details are:

Length of a cuboid = 15 cm

Breadth of a cuboid = 2.5 dm = 25 cm

Height of a cuboid = 8 cm

By using the formula

Volume of a cuboid = length × breadth × height

= 15 × 25 × 8

= 3000 cm^{3}

**2. Find the volume of a cube whose side is(i) 4 cm (ii) 8 cm(iii) 1.5 dm (iv) 1.2 m(v) 25 mm**

**Solution:**

**(i)** Given details are,

Side of cube = 4 cm

Volume of cube = (side)^{ 3}

= 4^{3} = 64 cm^{3}

**(ii)** Given details are,

Side of cube = 8 cm

Volume of cube = (side)^{ 3}

= 8^{3} = 512 cm^{3}

**(iii)** Given details are,

Side of cube = 1.5 dm

Volume of cube = (side)^{ 3}

= 1.5^{3} = 3.375 dm^{3} = 3375 cm^{3}

**(iv)** Given details are,

Side of cube = 1.2 m

Volume of cube = (side)^{ 3}

= 1.2^{3} = 1.728 m^{3}

**(v)** Given details are,

Side of cube = 25 mm

Volume of cube = (side)^{ 3}

= 25^{3} = 15625 mm^{3}= 15.625 cm^{3}

**3. Find the height of a cuboid of volume 100 cm ^{3}, whose length and breadth are 5 cm and 4 cm respectively.**

**Solution:**

Given details are,

Volume of a cuboid = 100 cm^{3}

Length of a cuboid = 5 cm

Breadth of a cuboid = 4 cm

Let height of cuboid be ‘h’ cm

We know that, l × b × h = 100cm

h = 100/( l × b)

= 100/(5×4)

= 5cm

**4. A cuboidal vessel is 10 cm long and 5 cm wide. How high it must be made to hold 300 cm ^{3} of a liquid?**

**Solution:**

Given details are,

Volume of a liquid in the vessel = 300 cm^{3}

Length of a cuboidal vessel = 10 cm

Breadth of a cuboidal vessel = 5 cm

Let height of cuboidal vessel be ‘h’ cm

We know that, l × b × h = 300 cm^{3}

h = 300/(l × b)

= 300/(10×5)

= 6cm

**5. A milk container is 8 cm long and 50 cm wide. What should be its height so that it can hold 4 litres of milk?**

**Solution:**

Given details are,

Volume = 4 litres = 4000 cm^{3}

Length of a milk container = 8 cm

Breadth of a milk container = 50 cm

Let height of milk container be ‘h’ cm

We know that, l × b × h = 4000 cm^{3}

h = 4000/ (l × b)

= 4000/ (50×8)

= 10cm

**6. A cuboidal wooden block contains 36 cm ^{3} wood. If it be 4 cm long and 3 cm wide, find its height.**

**Solution:**

Given details are,

Volume of wooden block = 36 cm^{3}

Length of the wooden block = 4 cm

Breadth of a wooden block = 3 cm

Let height of wooden block be ‘h’ cm

We know that, l × b × h = 36 cm^{3}

h = 36/( l × b)

= 36/(4×3)

= 3cm

**7.** **What will happen to the volume of a cube, if its edge is**

(i) halved (ii) trebled?

**Solution:**

Let us consider edge of a cube be ‘a’ cm

Volume of a cube will be ‘a^{3}’cm

**(i) **When halved

Edge = a/2

Volume = (a/2)^{3} = a^{3}/2^{3} = a^{3}/8 = 1/8times

**(ii) **When trebled

Edge = 3a

Volume = (3a)^{3} = 27a^{3} = 27times

**8. What will happen to the volume of a cuboid if its:(i) Length is doubled, height is same and breadth is halved?(ii) Length is doubled, height is doubled and breadth is same?**

**Solution:**

Let us consider,

Length of a cuboid be ‘l’

Breadth of a cuboid be ‘b’

Height of a cuboid be ‘h’

So, Volume of a cuboid = l × b × h

Now,

(i) Length of a cuboid becomes = 2l

Breadth = b/2

Height = h

Volume of cuboid = 2l × b/2 × h = l × b × h (remains same)

(ii) Length of a cuboid becomes = 2l

Breadth = b

Height = 2h

Volume of cuboid = 2l × b × 2h = 4lbh (four times)

**9. Three cuboids of dimensions 5 cm × 6cm × 7cm, 4cm × 7cm × 8cm and 2 cm × 3 cm × 13 cm are melted and a cube is made. Find the side of cube.**

**Solution:**

Given details are,

Volume of First cuboid = 5 × 6 × 7 = 210 cm^{3}

Volume of second cuboid = 4 × 7 × 8 = 224 cm^{3}

Volume of third cuboid = 2 × 3 × 13 = 78 cm^{3}

So, Volume of a cube = 210 + 224 + 78 = 512 cm^{3}

Let side of a cube be ‘a’

a^{3} = 512

∴ a = 8 cm

**10. Find the weight of solid rectangular iron piece of size 50 cm×40cm × 10 cm, if 1 cm ^{3} of iron weights 8 gm.**

**Solution:**

Given details are,

Dimension of rectangular iron piece = 50cm × 40cm × 10cm

Volume of solid rectangular = 50 × 40 × 10 = 20000 cm^{3}

Weight of 1 cm^{3} iron = 8 gm.

∴ Weight of 20000 cm^{3} iron = 8 × 20000

= 160000 gm.

= 160 kg

**11. How many wooden cubical blocks of side 25 cm can be cut from a log of wood of size 3 m by 75 cm by 50 cm, assuming that there is no wastage?**

**Solution:**

Given details are,

Dimensions of log of wood = 3m × 75cm × 50cm

Side of cubical block = 25cm

We know that,

Number of cubical block that can be made from wooden log =

Volume of wooden block / volume of cubical block

= (300 × 75 × 50) / (25 × 25 × 25)

= 72 blocks

**12. A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it, beads of volume 1.5 cm ^{3} each are to be made. Find the number of beads that can be made from the block.**

**Solution:**

Given details are,

Length of a cuboidal block of silver = 9cm

Breadth = 4cm

Height = 3.5cm

Volume of a cuboid = l × b × h

= 9 × 4 × 3.5 = 126cm^{3}

So, Number of beads of volume 1.5cm^{3} that can be made from the block =

Volume of silver block/volume of one bead

= 126cm^{3}/1.5cm^{3}

= 84 beads

**13. Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensions are 40 cm, 36 cm, and 24 cm.**

**Solution:**

Given details are,

Dimensions of cuboidal boxes is = 2cm × 3cm × 10 cm

Dimensions of carton is = 40cm × 36cm × 24cm

So,

Number of boxes that can be stored in carton = volume of carton / volume of one box

= (40 × 36 × 24) / (2 × 3 × 10)

= 576 cuboidal boxes

**14. A cuboidal block of solid iron has dimensions 50 cm, 45 cm, and 34 cm, how many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from this block? Assume cutting causes no wastage.**

**Solution:**

Given details are,

Dimensions of cuboidal block of iron is = 50cm × 45cm × 34cm

Size of small cuboids cutting from it is = 5cm × 3cm × 2cm

So,

Number of small cuboids that can be cut =

Volume of large iron cuboid/ volume of small cuboid

= (50 × 45 × 34) / (5 × 3 × 2)

= 2550 cuboidal blocks

**15. A cube A has side thrice as long as that of cube B. What is the ratio of the volume of cube A to that of cube B?**

**Solution:**

Given details are,

Let side of cube B be ‘x’ cm

Then, side of cube A = 3x cm

So now,

Ratio = volume of cube A / volume of cube B

= (3x)^{3} / (x)^{3}

= 27x^{3}/ x^{3} = 27/1 = 27:1

**16. An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm?**

**Solution:**

Given details are,

Dimensions of ice cream brick = 20 cm × 10cm × 7cm

Dimensions of fridge is = 100 cm × 50cm × 42 cm

So,

Number of bricks that can be put in fridge = volume of fridge / volume of one ice brick

= (100 × 50 × 42) / (20 × 10 × 7)

= 150 ice cream bricks

**17. Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volumes V _{1} and V_{2} of the cubes and compare them.**

**Solution:**

Given details are,

Edge of one cube a_{1} = 2 cm

Edge of second cube a_{2} = 4 cm

So, volume v_{1} = 2^{3} = 8cm^{3}

Volume v_{2} = 4^{3} = 64cm^{3}

v_{2} = 8v_{1}

**18. A tea-packet measures 10 cm × 6 cm × 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm × 30cm × 0.2 m?**

**Solution:**

Given details are,

Dimensions of tea packet = 10 cm × 6 cm × 4cm

Dimension of cardboard box = 50cm × 30cm × 0.2 m

So,

Number of tea packets can be put in cardboard box =

Volume of cardboard box / volume of tea packet

= (50 × 30 × 20) / (10 × 6 × 4)

= 125 tea packets

**19. The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm.**

**Solution:**

Given details are,

Dimensions of metal block = 5cm × 4cm × 3cm

Weight of block = 1 kg

Volume of box = 5×4×3 = 60 cm^{3}

Dimension of new block = 15cm × 8cm × 3cm

Volume of new box = 15 × 8 × 3 = 360 cm^{3}

We know that,

60cm^{3} = 1kg

360 cm^{3} = 6 × 60 cm^{3}

= 6 × 1

= 6 kg

**20. How many soap cakes can be placed in a box of size 56 cm × 0.4 m × 0.25 m, if the size of a soap cake is 7 cm × 5cm × 2.5 cm?**

**Solution:**

Given details are,

Dimensions of box = 56cm × 0.4m × 0.25m

Dimensions of soap cake = 7cm × 5cm × 2.5cm

So,

Number of soap cakes that can be placed in box = volume of box / volume of soap cake

= (56 × 40 × 25) / (7 × 5 × 2.5)

= 640 soap cakes

**21. The volume of a cuboidal box is 48 cm ^{3}. If its height and length are 3 cm and 4 cm respectively, find its breadth.**

**Solution:**

Given details are,

Volume of a cuboidal box = 48 cm^{3}

Length of a cuboidal box = 4 cm

Height of a cuboidal box = 3 cm

Let breadth of wooden block be ‘b’ cm

We know that, l × b × h = 48 cm^{3}

b = 48/( l × h)

= 48/(4×3)

= 4cm

**All Chapter RD Sharma Solutions For Class 8 Maths**

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