RD Sharma Solutions Class 9 Maths Chapter 18 – Surface Area And Volume of A Cuboid And Cube (Updated for 2024)

RD Sharma Solutions Class 9 Maths Chapter 18: These solutions cover all the questions mentioned in the RD Sharma textbook as per the CBSE Board. The exercise-wise solutions assist the students to revise each topic included in this chapter. To know more about the RD Sharma Solutions Class 9 Maths Chapter 18, read the whole blog. 

Download RD Sharma Solutions Class 9 Maths Chapter 18 PDF

RD Sharma Class 9 Solutions Chapter 18

Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 18 PDF

RD Sharma class 9 chapter 18 exercise 18a

RD Sharma class 9 chapter 18 exercise 18b

Access answers of RD Sharma Solutions Class 9 Maths Chapter 18

Question 1: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.

Solution:

Given, Dimensions of a cuboid:

Length (l) = 80 cm

Breadth (b) = 40 cm

Height (h) = 20 cm

We know that, Total Surface Area = 2[lb + bh + hl]

By substituting the values, we get

= 2[(80)(40)+(40)(20)+(20)(80)]

= 2[3200+800+1600]

= 2[5600]

= 11200

Therefore, Total Surface Area = 11200 cm²

Now,

Lateral Surface Area = 2[l + b]h

= 2[80+40]20

= 40[120]

= 4800

Thus, Lateral Surface Area is 4800 cm².

Question 2: Find the lateral surface area and total surface area of a cube of edge 10 cm.

Solution:

Side of a Cube = 10 cm (Given)

Formula for Cube Lateral Surface Area = 4 side²

Cube Lateral Surface Area = 4(10 x 10)

= 400 cm²

Total Surface Area = 6 Side²

= 6(10²)

= 600 cm²

Question 3: Find the ratio of the total surface area and lateral surface area of a cube.

Solution:

Total Surface Area of the Cube (TSA) = 6 Side²

Lateral surface area of the Cube (LSA) = 4 Side²

Now,

Ratio of TSA and LSA = (6 Side²)/ (4 Side²) = 3/2 or 3:2.

Question 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?

Solution:

The dimensions of the wooden block are:

Length (l) = 80cm

Breadth (b) = 40cm

Height (h) = 20cm

Surface Area of the wooden box = 2[lb + bh + hl]

= 2[(80×40)+(40×20)+(20×80)]

= 2[5600]

= 11200

Surface Area of the wooden box is 11200 cm²

The Area of each sheet of the paper = 40×40 cm² = 1600 cm²

Now,

The total number of sheets required = (Surface area of the box )/(Area of one sheet of paper)

= 11200/1600

= 7

Therefore, Marry would require 7 sheets.

Question 5: The length, breadth, and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 m².

Solution:

Formula: Total Area to be washed = lb + 2(l + b)h ….(1)

Where, l = length , b = breadth and h = height.

From given:

Length = l = 5 m

Breadth = b = 4 m

Height = h = 3 m

Total area to be white washed = (5×4) + 2(5 + 4)3

(using (1))

= 74

Total area to be white washed is 74 m²

Now, cost of white washing 1 m² is Rs. 7.50 (Given)

Therefore, the cost of white washing 74 m² = (74 x 7.50)

= Rs. 555

Question 6: Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

Solution:

Let breadth of the cuboid = a

Then, length of the new cuboid = 3a and

Height of the new cuboid = a

Now,

Total surface area of the new cuboid (TSA) = 2(lb+bh+hl)

= 2(3a x a + a x a + a x 3a)

= 14a²

Again,

Total Surface area of three cubes = 3 x (6 side² )

= 3 x 6a²

= 18a²

Therefore, ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a²/18a² = 7/9 or 7:9

Therefore, required ratio is 7:9. Answer.

Question 7: A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.

Solution:

Edge of the cube = 4 cm (Given)

Volume of the cube = Side³ = 4³ = 64

Volume of the cube is 64cm³

Again,

Edge of the cube = 1 cm³

So, Total number of small cubes = 64cm³/1cm³ = 64

And, total surface area of all the cubes = 64 x 6 x 1 = 384 cm²

Question 8: 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 hall.

Solution:

Dimensions of the hall are:

Length = 18m

Width = 12m

From statement:

Area of the floor and the flat roof = Sum of the areas of four walls …(1)

Using respective formulas and given values, we have

Area of the floor and the flat roof = 2lb = 2 x 18 x 12 = 432 sq/ft …(2)

Sum of the areas of four walls = (2 x 18h + 2 x 12h)sq/ft …(3)

Using equation (2) and (3) in (1), we get

432 = 2 x 18h + 2x12h

18h + 12h = 216

or h = 7.2

Therefore, height of the hall is 7.2 m.

Question 9: Hameed has built a cubical water tank with lid for his house, with each other edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles if the cost of tiles is Rs 360 per dozen.

Solution:

Edge of the cubical tank = 1.5m or 150 cm

Surface area of the cubical tank (5 faces) = 5 x Area of one Face

= 5 x (150 x 150) cm² …….(1)

Find area of each square tile:

Side of tile = 25 cm (given)

Area of one tile = 25 x 25 cm² …….(2)

Now,

Number of tiles required = (Surface Area of Tank ) / (Area of each Tile)

= (5×150×150) / 25×25

= 180

Find cost of tiles:

Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs. 360

Therefore, cost of one tile = Rs.360/12 = Rs.30

So, the cost of 180 tiles = 180 x 30 = Rs. 5400

Question 10: Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

Solution:

Let ‘a’ be the edge of a cube.

Surface area of the cube having edge ‘a’ = 6a² …..(1)

As given, a new edge after increasing existing edge by 50%, we get

The new edge = a + 50 a /100

= 3a/2

Surface area of the cube having edge ‘3a/2’ = 6 x (3a/2)² = (27/2) a² ……(2)

Subtract equation (1) from (2) to find the increase in the Surface Area:

Increase in the Surface Area = (27/2) a² – 6a²

= (15/2)a²

Now,

Percentage increase in the surface area = ( (15/2)a² / 6a² ) x 100

= 15/12 x 100

= 125%

Therefore, percentage increase in the surface area of a cube is 125.

Exercise 18.2 Page No: 18.29

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:

Dimensions of a cuboidal water tank:

Length = l = 6m

Breadth = b = 5m

Height = h = 4.5m

We know, Volume of the cuboidal water tank = lbh

By substituting the values, we get

Volume = 6×5×4.5

= 135

Therefore, Volume of the cuboidal water tank is 135 m³

Convert into liters:

We know; 1 m³ = 1000 liters

So, 135m³ = (135×1000)liters

= 135000 liters

Hence, 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:

Dimensions of a cuboidal vessel:

Length = l = 10 m

Breadth = b = 8 m

Volume of the vessel = 380 m³ (given)

Let ‘h’ be the height of the cuboidal vessel.

We know, Volume of cuboidal vessel = lbh

lbh = 380 m³

or 10×8×h = 380

or h = (380)/(10×8)

or h = 4.75

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:

Dimensions of a cuboidal pit:

Length = l = 8 m

Breadth = b = 6 m

Depth or height = h = 3 m

We know, Volume of the Cuboidal pit = lbh

= 8×6×3

= 144

Volume of the Cuboidal pit is 144 m³

Now, find the cost:

Cost of digging 1 m³ = Rs. 30 (Given)

Cost of digging 144 m³ = 144 x 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

Solution:

Dimensions of a cube are:

Length = l = a

Breadth = b = b

Height = h = c

We know, Volume of the cube (V) = lbh

= a×b×c

Or V = abc ….(1)

Again,

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

or S = 2 (ab+bc+ca) ……(2)

Now,

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 c be the length, breadth, and height of the cuboid.

Then, x = ab, y = bc and z = ca

[Since areas of three adjacent faces of a cuboid are x, y and z (Given)]

And xyz = ab × bc × ca = (abc)² ……(1)

We know, Volume of a cuboid ( V ) = abc …..(2)

From equation (1) and (2), we have

V² = xyz

Hence proved.

Question 6: If the areas of three adjacent face of a cuboid are 8 cm² , 18 cm² and 25 cm². Find the volume of the cuboid.

Solution:

Let x, y, z denote the areas of three adjacent faces of a cuboid, then,

x = l×b = 8 cm²

y = b×h = 18 cm²

z = l×h = 25 cm²

Where l = length of a cuboid, b = breadth of a cuboid and h = height of a cuboid

xyz = 8 × 18 × 25 = 3600 ….(1)

Volume of cuboid (V) = lbh

From above results, we can write,

xyz = lb×bh×lh = (lbh)² = V² . …..(2)

Form equation (1) and (2), We get

V² = 3600

or V = 60

Thus, Volume of the cuboid is 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:

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

As per given statement,

b = 2h and b = l/2

⇒ l/2 = 2h

or l = 4h

Now, we have l = 4h and b = 2h

We know, Volume of the room = lbh

Volume of the room = 512dm³ (given)

So, 4h×2h×h=512

or h³=64

or h=4

Therefore, 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 3 m 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:

Water flow of a river = 2 km per hour = (2000/60) m/min or (100/3 )m/min

[we know: 1 km = 1000 m and 1 hour = 60 mins]

Depth of the river (h) = 3m

Width of the river (b) = 40m

Volume of the water flowing in 1 min = 100/3 × 40 × 3 = 4000 m³

Or 4000 m³ = 4000000 litres

Therefore, in 1 minute 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:

Water in the canal forms a cuboid of Width (b) and Height (h).

b = 30dm = 3m and h = 12dm = 1.2m

Here, Cuboid length = distance travelled in 30 min with a speed of 100 km per hour.

Therefore, Length of the cuboid (l) = 100 × 30/60 = 50000 metres

Volume of water used for irrigation = lbh = 50000×3×1.2 m³

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

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

Area of field = (50000 × 3 × 1.2) × 100/8

= 2250000

Thus, area of field is 2250000 m² .

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 us consider, ‘x’ be the length of each edge of the new cube.

Volume of cube = x³

⇒ x³ = (6³ + 8³ + 10³)cm³

or x³ = 1728

or x=12

Volume of the new cube = x³ = 1728 cm³

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

And, diagonal of the newly formed cube = √3a = 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:

Let ‘a’ be the side of a cube.

Volume of the cube = 512cm³ (Given)

We know volume cube = (side)³

⇒ a^3 = 512

or a = 8

Each side of a cube is 8 cm.

Now,

Dimensions of the new cuboid formed are:

Length (l) = 8+8 = 16 cm,

Breadth (b) = 8 cm and

Height (h) = 8 cm

Surface area = 2(lb+bh+hl)

= 2 (16×8+8×8+16×8)

= 640 cm²

Therefore, Surface area of a cube 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:

Volume of gold-sheet = 1/2 m³ or 0.5 m³ (Given)

Area of the gold-sheet = 1 hectare i.e. 10000 m²

Thickness of gold sheet = (Volume of solid)/(Area of gold sheet)

= 0.5 m³/10000 m²

= m/20000

Or Thickness of gold sheet = 1/200 cm

[1 m = 100 cm]

Therefore, thickness of the silver sheet is 1/200 cm.

Question 13: 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:

From the given statement, we have

Volume of the large cube = v1 + v2 + v3

Let the edge of the third cube be ‘x’ cm

12³ = 6³ + 8³ + x³

[Using formula, Volume of cube = (side)³]

1728 = 216 + 512 + x³

or 1000 = x³

or x = 10

Therefore, length of the third side is 10 cm.

---

Exercise VSAQs Page No: 18.35

Question 1: If two cubes each of side 6 cm are joined face to face, then find the volume of the resulting cuboid.

Solution:

Side of two equal cubes = 6 cm (Given)

When we join, two cubes face to face formed a cuboid.

Dimensions of a cuboid are:

Length = 6 cm + 6 cm = 12 cm

Breadth = 6 cm

Height = 6 cm

Therefore, volume of cuboid = lbh = 12 × 6 × 6 = 432 cm³

Question 2: Three cubes of metal whose edges are in the ratio 3 : 4 : 5 are melted down into a single cube whose diagonal is 12√3 cm. Find the edges of three cubes.

Solution:

Given:

Ratio of edge of 3 cubes = 3 : 4 : 5

Let edges are = 3x, 4x and 5x

Diagonal of new cube formed = 12√3 cm (given)

Volume of new cube = Volume of figure obtained after combining three cubes = (3x)³ + (4x) ³ + (5x) ³

= 216 x³ …(1)

New diagonal of a cube = √3a = 12√3

or a = 12

So, side of new cube is 12 cm.

Volume of cube with side 12 cm = (12)^3 …(2)

From equation (1) and (2), we have

(12)^3 = 216 x³

Therefore, a measure of edges are :

3x = 3×2 = 6 cm

4x = 4×2 = 8cm

5x = 5×2 = 10 cm

Question 3: If the perimeter of each face of a cube is 32 cm, find its lateral surface area. Note that four faces that meet the base of a cube are called its lateral faces.

Solution:

Perimeter of each face of a cube = 32 cm (given)

Let ‘a’ be the edge of a cube.

We know, Perimeter of each face of a cube = 4a

⇒ 4a = 32

or a = 8

Side of a cube is 8 cm.

Now,

Lateral surface area of cube = 4a² = 4×8² = 256 cm².

Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise

RD Sharma class 9 chapter 18 exercise 18a

This exercise includes topics based on the surface area of a cuboid and cube. These solutions are solved by maths expert faculty in an interactive & engaging way so that the students can build interest in Maths.

The students must practice RD Sharma class 9 chapter 18 exercise 18a regularly to prepare more efficiently for the final exams. Solving as many times as possible enables the students to strengthen their time management skills and also enhance their confidence level to acquire better marks.

RD Sharma class 9 chapter 18 exercise 18b:

These exercise solutions are given here to assist the students in understanding the concept of the surface area of a cuboid and cube easily. The students will study the topic of the volume of a cuboid and cube.

These solutions enable the students to do quick revision while preparing for mathematics final exams and also in their assignments. The students can develop problem-solving abilities by practicing RD Sharma textbook questions. This also enables them to answer any type of questions easily. 

Important Topics Class 9 Maths Chapter 18 

RD Sharma Solutions Class 9 Maths Chapter 18 include some important concepts that are listed below:

• The volume of a Cuboid
• The volume of a Cube

RD Sharma  CBSE class 9 solutions maths chapter 18 surface area and volume of a cuboid and cube are solved by the top maths experts elaborately so that the students can understand each concept with ease. For more queries, ask in the comments.

FAQs of RD Sharma Solutions Class 9 Maths Chapter 18