# Class 8 RD Sharma Solutions- Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.4 | Set 2

**Question 11. A field is 150m long and 100m wide. A plot (outside the field) 50m long and 30m wide is dug to a depth of 8m and the earth taken out from the plot is spread evenly in the field. By how much is the level of field raised?**

**Solution:**

Given, the length of field = 150 m

The width of the field = 100 m

So, the area of field = Length × breadth

= 150 × 100 = 15000 m

^{2}Also, the length of the plot = 50 m

The width of the plot = 30 m

The depth of the plot = 8 m

We can now find the volume of the plot,

The volume of plot = l × b × h

= 50 × 30 × 8 = 12000 m

^{3}Let

hbe the height of earth spread on the fieldWe know, volume = l × b × h

So, h = (volume)/l × b

h = 12000/(150 × 100)

So, h = 0.8 m = 80 cm

Hence, the level of field raised by 80 cm

**Question 12. Two cubes, each of volume 512 cm**^{3} are joined end to end. Find the surface area of the resulting cuboid.

^{3}are joined end to end. Find the surface area of the resulting cuboid.

**Solution:**

Given, the volume of each cube is 512 cm

^{3}Let

abe the edge of each cubeAs we know, the Volume of cube = (edge)

^{3}So, 512 = a

^{3}a = 8 cm

When these two cubes are joined end to end a cuboid is formed,

So, the length of cuboid = 8 + 8 = 16 cm

The breadth of cuboid = 8 cm

The height of cuboid = 8 cm

So, the surface area of the resulting cuboid will be 2 (l × b + b × h + h × l)

= 2 (16 × 8 + 8 × 8 + 8 × 16) = 640 cm

^{2}

Hence, the surface area of the resulting cuboid is 640 cm^{2}

**Question 13. Three cubes whose edges measure 3 cm, 4 cm, and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.**

**Solution:**

Given, the edge of three cubes are 3 cm, 4 cm, and 5 cm respectively.

As we know, the Volume of cube = (edge)

^{3}The volume of the first cube = 3

^{3}= 27 cm^{3}The volume of the second cube = 4

^{3}= 64 cm^{3}The volume of the third cube = 5

^{3}= 125 cm^{3}Now, the sum of volumes of given three cubes = 27 + 64 + 125 cm

^{3}= 216 cm

^{3}When these three cubes are melted, a new cube is formed,

Let

abe the edge of the new cubeAs we know, the Volume of cube = (edge)

^{3}So, 216 = a

^{3}a = 6 cm

So, the surface area of the new cube so formed = 6 x (edge)

^{2}= 6 × 6

^{2}= 216 cm^{2}

Hence, the surface area of the new cube formed is 216 cm^{2}

**Question 14. The cost of preparing the walls of a room 12m long at the rate of Rs 1.35 per square meter is Rs 340.20 and the cost of matting the floor at 85 paise per square meter is Rs 91.80. Find the height of the room**

**Solution:**

Given, the length of room is 12 m

Let

bbe the breadth of roomLet

hbe the breadth of roomAs we know, the area of floor = l × b

= 12b m

^{2}Given that cost of matting floor at 85 paise per square meter is Rs 91.80

So, we can write 12b × 0.85 = 91.80

b = 9 m

The breadth of room is 9 m

We know area of 4 walls = 2(l × h + b × h)

= 2(12 × h + 9 × h) = 42h m

^{2}Also, the cost of preparing the walls at the rate of Rs 1.35 per square meter is Rs 340.20

So, we can write 42h × 1.35 = 340.20

h = 6 m

Hence, the height of room is 6 m and breadth is 9 m

**Question 15. The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall.**

**Solution:**

Given, the length of hall is 18 m

The width of hall is 12 m

Let

hbe the height of hallWe know, sum of the areas of the floor and the flat roof = (l × b + l × b)

= (18 × 12 + 18 × 12) = 432 m

^{2}Also, the areas of the four walls = 2(l × h + b × h)

= 2(18 × h + 12 × h) = 60h m

^{2}It is given that sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls,

So, 60h = 432

h = 7.2 m

Hence, the height of hall is 7.2 m

**Question 16. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.**

**Solution:**

Given, the edge of metal cube is 12 cm

The edge of two smaller cubes are 6 cm and 8 cm

Let

abe the edge of third smaller cubeWe know that volume of metal cube will be equal to sum of volume of three smaller cubes

So, 12

^{3}= 6^{3}+ 8^{3}+ a^{3}a = 10 cm

Hence, the edge of third smaller cube is 10 cm

**Question 17. The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person required 150 m**^{3} of air?

^{3}of air?

**Solution:**

Given, the length of hall = 100 m

The breadth of hall = 50 m

The height of hall = 18 m

So, the volume of cinema hall = l × b × h

= 100 × 50 × 18 = 90000 m

^{3}It is given that each person require 150 m

^{3}of airThe number of person that can sit in hall = Volume of hall/Volume of each person

= 90000/150 = 600

Hence, 600 canpersons can sit in the hall

**Question 18. The external dimensions of a closed wooden box are 48 cm, 36 cm and 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm× 3 cm× 0.75 cm can be put in this box?**

**Solution:**

Given, the length of box = 48 cm

The breadth of box = 36 cm

The height of box = 30 cm

The dimensions of bricks are 6 cm× 3 cm× 0.75 cm

So, the volume of 1 brick = l × b × h

= 6 × 3 × 0.75 = 13.5 cm

^{3}Also, the thickness of wood is 1.5 cm

So, the internal dimensions of box can be given as (48 – 2 × 1.5) × (36 – 2 × 1.5) × (30 – 2 × 1.5)

= (45 × 33 × 27) cm

The internal volume of box = l × b × h

= 45 × 33 × 27 = 40095 cm

^{3}So, the number of bricks that can be put inside the box = Internal volume of box/Volume of 1 brick

= 40095/13.5 = 2970 bricks

Hence, 2970 bricks can be put inside the given wooden box

**Question 19. The dimensions of a rectangular box are in the ratio of 2: 3: 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m**^{2} is Rs 1248. Find the dimensions of the box.

^{2}is Rs 1248. Find the dimensions of the box.

**Solution:**

Given, the dimensions of a rectangular box are in the ratio of 2: 3: 4

So, the length of box be

2aThe breadth will be

3aAnd, the height be

4aThe area of sheet that is required to cover the box will be the total surface area of cuboid,

= 2 (l × b + b × h + h × l)

= 2 (2a × 3a + 3a × 4a + 4a × 2a) = 52a

^{2}m^{2}Also, the cost of covering it with sheet of paper at the rates of Rs 8 is 52a

^{2}× 8= Rs 416a

^{2}And, the cost of covering it with sheet of paper at the rates of Rs 9.50 is 52a

^{2}× 9.50= Rs 494a

^{2}We know that difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m

^{2}is Rs 1248So, 494a

^{2}– 416a^{2}= 1248a = 4

So, the length is 2a = 8 m

The breadth = 3a = 15 m

The height = 4a = 16 m

Hence, the dimensions of box are 8 m × 12 m × 16 m

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