RD Sharma Solutions Class 9 Chapter 18 consists of various topics which are important from exam perspective. Below are some topics you will be seeing here:
Surface area: The surface area of a solid object is a measure of the total area that the surface of the object occupies.
Volume: Volume is the quantity of three-dimensional space enclosed by closed surface i.e, the space that a substance or shape occupies or contains. The SI unit of volume is ‘cubic meter’.
The volume of a cuboid: A cuboid is a three-dimensional shape, so to find the volume of cuboid we need to know 3 measurements. i.e. Length, Width and Height.
The volume is found using the formula : Volume = Length * Height * Width
Surface Area of a Cuboid: The total surface area of a cuboid is the sum of the areas of its six faces.
Formula : 2 (length.width + width.length + height.length)
Surface Area of a Cube: The surface area of a cube is the area of the six squares that cover it. Formula: 6(Side)2
The formula for Volume of a cube: (Side)3
Our RD Sharma Solutions for Class 9 are designed according to the syllabus prescribed by the Central Board of Secondary Education and in accordance with the guidelines given by CCE. Here students will find three exercises of Class 9 Maths Chapter 18 – Surface Area and Volume of Cuboid and Cube. Click on the below link to access all chapter 18 exercises.
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Exercise 18.1 Page No: 18.14
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.
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
Therefore, Total Surface Area = 11200 cm2
Lateral Surface Area = 2[l + b]h
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.
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
= 600 cm2
Question 3: Find the ratio of the total surface area and lateral surface area of a cube.
Total Surface Area of the Cube (TSA) = 6 Side2
Lateral surface area of the Cube (LSA) = 4 Side2
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?
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]
Surface Area of the wooden box is 11200 cm2
The Area of each sheet of the paper = 40×40 cm2 = 1600 cm2
The total number of sheets required = (Surface area of the box )/(Area of one sheet of paper)
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.
Formula: Total Area to be washed = lb + 2(l + b)h ….(1)
Where, l = length , b = breadth and h = height.
Length = l = 5 m
Breadth = b = 4 m
Height = h = 3 m
Total area to be white washed = (5×4) + 2(5 + 4)3
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.
Let breadth of the cuboid = a
Then, length of the new cuboid = 3a and
Height of the new cuboid = a
Total surface area of the new cuboid (TSA) = 2(lb+bh+hl)
= 2(3a x a + a x a + a x 3a)
Total Surface area of three cubes = 3 x (6 side2 )
= 3 x 6a2
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.
Edge of the cube = 4 cm (Given)
Volume of the cube = Side3 = 43 = 64
Volume of the cube is 64cm3
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.
Dimensions of the hall are:
Length = 18m
Width = 12m
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.
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)
Number of tiles required = (Surface Area of Tank ) / (Area of each Tile)
= (5×150×150) / 25×25
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.
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
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
Percentage increase in the surface area = ( (15/2)a2 / 6a2 ) x 100
= 15/12 x 100
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?
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
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?
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.
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
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
Dimensions of a cube are:
Length = l = a
Breadth = b = b
Height = h = c
We know, Volume of the cube (V) = lbh
Or V = abc ….(1)
Surface area of the cube (S) = 2 (lb+bh+hl)
or S = 2 (ab+bc+ca) ……(2)
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.
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
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.
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.
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)
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?
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?
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
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.
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
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.
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.
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.
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
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.
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.
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.
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, 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 which meet the base of a cube are called its lateral faces.
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.
Lateral surface area of cube = 4a2 = 4×82 = 256 cm2.
RD Sharma Solutions for Class 9 Maths Chapter 18 Surface Area and Volume of Cuboid and Cube
In the 18h chapter of class 9 RD Sharma Solutions students will study important concepts listed below:
- Volume of a Cuboid
- Volume of a Cube