Kerala Syllabus Class 9 Maths Chapter 12 Solutions Prisms

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 12 Prisms Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 12 Solutions Prisms

Prisms Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 12 Prisms Solutions Questions and Answers

Class 9 Maths Chapter 12 Kerala Syllabus – Volume

Textual Questions And Answers

Question 1.
The base of a prism is an equilateral triangle of perimeter 18 centimetres and its height is 5 centimetres. Calculate its volume.
Answer:
Given that,
The base perimeter of an equilateral triangle = 18 cm
Therefore the base edge = \(\frac{18}{2}\) = 6 cm
Height = 5 cm
Base area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × (6)²
= 9√3 cm²
Volume = Base area × Height
= 9√3 × 5
= 45√3 cm³

Question 2.
The base of a prism is a triangle of sides 13 centimetres, 14 centimetres and 15 centimetres, and its height is 20 centimetres. Calculate its volume.
Answer:
Considering the base edges of a triangular prism as a, b and c where,
a = 13 cm,
b = 14 cm,
c = 15 cm
Height = 20 cm

Considering the semi-perimeter of a triangular prism as S,
S = \(\frac{\text { Sum of the base edges }}{2}=\frac{a+b+c}{2}\)
S = \(\frac{13+14+15}{2}\)
= 21 cm

Base area = \(\sqrt{S(S-a)(S-b)(S-c)}=\sqrt{21(21-13)(21-14)(21-15)}\)
= \(\sqrt{21 \times 8 \times 7 \times 6}\)
= \(\sqrt{7056}\)
= 84 cm²

Volume = Base area × Height
= 84 × 20
= 1680 cm³

Question 3.
There is a hexagonal pit in the school ground to collect rain water. Each side of the hexagon is 2 metres and the depth of the pit is 3 metres. It now contains water one meter deep. How much litres is this? How much more litres can it contain?
Answer:
One side of the hexagon = 2 m
It is given that the depth of the pit is 3 meter and the water level is up to 1 meter.

Base area of a hexagon = 6 × \(\frac{\sqrt{3}}{4}\) × (side)²
= 6 × \(\frac{\sqrt{3}}{4}\) × 2²
= 6√3 m²

Volume of water = Volume of the hexagonal prism
= Base area × height
= 6√3 × 1
= 6 × 1.73 × 1000
= 10380 cubic litre

Depth of the pit = Height of the prism = 3m
Total volume of the hexagonal prism = Base area × height
= 6√3 × 3
= 18√3
= 18 × 1.73 × 1000
= 31140 cubic litre

More liters of water does the prism can hold = 31140 – 10380
= 20760 cubic litre

Question 4.
A hollow prism with base a square of sides 16 centimetres contains water 10 centimetres high. If a cube of edges 8 centimetres is immersed in it, by how much would the water level rise?
Answer:
Base of the hollow prism is a square,
Side = 16 cm
Height =10 cm
Base area = (side)² = 16² = 256 cm²
Volume of the water = Base area × height
= 256 × 10
= 2560 cm³
Edges of the solid cube = 8 cm

Base area = (side)² = 8²
= 64 cm²
Volume of the solid cube = Base area × height
= 64 × 8
= 512 cm³

Let as consider the level of water increased be h,
Volume of the increased water level = Base area × height
=16 × 16 × h
= 256 h cm³

Volume of the increased water level = Volume of the water + Volume of the solid cube
256 h = 2560 + 512
256h = 3072
h = \(\frac{3072}{256}\) = 12
Increased water level = 12 – 10 = 2 cm

Question 5.
A rectangular block of metal has edges of lengths 6 centimetres, 9 centimetres and 15 centimetres. It is melted and recast into identical cubes of sides 3 centimetres. How many cubes would be got?
Answer:
Edges of the rectangular block = 6 cm, 9 cm, 15 cm
Base area of the rectangular block = length × breadth
= 9 × 6
= 54 cm²

Volume of the rectangular block = Base area × height
= 54 × 15
= 810 cm³

Base area of the cube = (side)² = 3² = 9 cm²
Volume of the cube = Base area × height = 9 × 3 = 27 cm³
Number of solid cubes = \(\frac{\text { Volume of the rectangular block }}{\text { Volume of the cube }}=\frac{810}{27}\) = 30

Question 6.
The base of a prism is a square of sides 6 centimetres and its height is 10 centimetres. What is its volume? What is the maximum volume of a triangular prism cut from it?
Answer:
Side of the square = 6 cm
Height = 10 cm
Base area = (side)²
= 6²
= 36 cm²

Volume of the square prism = Base area × height
= 36 × 10
= 360 cm³

To obtain a triangular prism with the maximum volume, the rectangular prism should split through its diagonal.
Base edges of the triangular prism = 6 cm, 6 cm, 10 cm
Base area = \(\frac{1}{2}\)(6 × 6)
= \(\frac{1}{2}\)
= 18 cm²
Volume of the triangular prism = Base area × Third side
= 18 × 10
= 180 cm³

Class 9 Maths Kerala Syllabus Chapter 12 Solutions – Area

Textual Questions And Answers

Question 1.
The base of a prism is an equilateral triangle of perimeter 12 centimetres and its height is 5 centimetres. What is its total surface area?
Answer:
Base perimeter of an equilateral triangle = 12 cm
One side = \(\frac{12}{3}\) = 4 cm
Height = 5 cm
Lateral surface area of the equilateral triangle = Base perimeter × height
= 12 × 5
= 60 cm2
Base area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × 4²
= 6.92 cm²
Total surface area = Lateral surface area + 2 (Base area)
= 60 + 2 (6.92)
= 60 + 13.84
= 73.84 cm²

Question 2.
Two identical prisms with right triangles as bases are joined to form a rectangular prism, as in the picture below:

What is the total surface area of this rectangular prism?
Answer:
The sides of the rectangular prism are:
Length = 12 cm
Breadth = 5 cm
Height = 15 cm

Base area = Length × Breadth
= 12 × 5
= 60 cm

Lateral surface area = base perimeter × height
= 2 (length + breadth) × height
= 2(12 + 5) × 15
= 2 × 17 × 15
= 510 cm²

Total surface area = Lateral surface area + 2 ( Base area)
= 510 + 2 (60)
= 510 + 120
= 630 cm²

Question 3.
The base of a prism is a triangle with sides 4 centimetres, 13 centimetres and 15 centimetres and its height is 25 centimetres. Calculate its lateral surface area and total surface area.
Answer:
Considering the sides of the triangular prism be a, b and c
where, a = 4 cm, b = 13 cm, c = 15 cm
Height = 25 cm

Base perimeter = a + b + c
= 4 +13 + 15
= 32 cm

Lateral surface area = base perimeter × height
= 32 × 25
= 800 cm²

Considering the semi-perimeter of a triangular prism as S,
S = \(\frac{a+b+c}{2}=\frac{4+13+15}{2}=\frac{32}{2}\)
= 16 cm

Base area = \(\sqrt{S(S-a)(S-b)(S-c)}\)
= \(\sqrt{16(16-4)(16-13)(16-15)}\)
= \(\sqrt{16 \times 12 \times 3 \times 1}\)
= \(\sqrt{576}\)
= 24 cm²

Total surface area = Lateral surface area + 2 (Base area)
= 800 + 48
= 848 cm²

Question 4.
The lateral surface area of a prism, with base an equilateral triangle, is 120 square centimetres
i) What is the lateral surface area of a prism, with base a rhombus, made by joining two such triangular prisms?
ii) What is the lateral surface area of a prism, with base an isosceles trapezium, made by joining three such triangular prisms?
iii) What is the lateral surface area of a prism, with base a regular hexagon, made by joining six such triangular prisms?
Answer:
Considering the side of a triangular prism be ‘a’ and the height be ‘h’
Lateral surface area =120 cm²
That means, 3 ah = 120
ah = \(\frac{120}{3}\) = 40

i) Number of sides of a rhombus = 4

Lateral surface area = 4 ah = 4 × 40
= 160 cm²

ii) Number of sides of an isosceles trapezium = 5

Lateral surface area = 5 a h
= 5 × 40
= 200 cm²

iii) Number of sides of a regular hexagon = 6

Lateral surface area = 6 a h
= 6 × 40
= 240 cm²

Question 5.
Six sheets of metal, each a square of sides 10 centimetres are joined to make a cube
i) What is its total surface area?
ii) How much water can it contain?
Answer:
Side of the metal sheet and the side of the cube are same
Therefore, side of the cube = 10 cm

i) Total surface area of the cube = 6 × (side)²
= 6 × (10)²
= 600 cm²

ii) Volume of the cube = (side)³
= (10)³
= 1000 cm³

Amount of water does the cube can contain = \(\frac{\text { Volume of the cube }}{1000}=\frac{1000}{1000}\)
= 1 litre

SCERT Class 9 Maths Chapter 12 Solutions – Cylinders

Textual Questions And Answers

Question 1.
The base diameter of a cylindrical tank is 1 meter and its height is 2 meters. How many litres of water can it contain?
Answer:
Base diameter ‘d’ = 1 m
Height = 2 m
Radius r = \(\frac{d}{2}=\frac{1}{2}\) = 0.5 m
Base area = πr² = (0.5)²π – 0.25 π m²

Volume = Base area × Height
= 0.25 π × 2
= 0.5 π m³
1 m³ is equivalent to 1000 liters of water.
That means, the amount of water that the cylindrical tank can contains = volume 1000
= 0.5 π 1000 ×
= 500 π × liters
= 500 × 3.1416
= 1570.8 liters

Question 2.
The base radius of an iron cylinder is 15 centimetres and its height is 32 centimetres. It is melted and recast into a cylinder of base radius 20 centimetres. What is the height of this cylinder?
Answer:
Base radius of an iron cylinder = 15 cm
Height = 32 cm
Base area = πr²
= 225 π cm²

Volume V₁ = Base area × Height
= 225 π × 32
= 7200 π cm³

Iron cylinder is melted and recast into a new cylinder, whose Radius = 20 cm
Considering the height of the new cylinder be ‘h’
Base area = πr² = (20)² π = 400 π cm²

Volume V₂ = Base area × Height
= 400 π h cm³

Volume of the cylinder and the new cylinder are same.
That means, V₁ = V₂
7200 π = 400 π h
h = \(\frac{7200 \pi}{400 \pi}\)
= 18 cm

Question 3.
The base radii of two cylinders of the same height are in the ratio 3 : 4. What is the ratio of their volumes?
Answer:
Lets consider the base radius of the two cylinders be r₁ and r₂
Therefore r₁ : r₂ = 3 : 4
That means, \(\frac{r_1}{r_2}=\frac{3}{4}\)
Considering the height of the cylinder be ‘h’ and
Volume of the cylinder be, v = πr² h

The ratio of their volumes be,
\(\frac{v_1}{v_2}=\frac{\pi r_1^2 h}{\pi r_2^2 h}=\frac{r_1^2}{r_2^2}=\frac{(3)^2}{(4)^2}=\frac{9}{16}\)
That means, v₁ : v₂ = 9 : 16

Question 4.
The base radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4
i) What is the ratio of their volumes?
ii) The volume of the smaller cylinder is 720 cubic centimeters. What is the volume of the larger one?
Answer:
i) Ratio of the base radius of the two cylinders = 2:3
Ratio of their heights = 5:4

For first cylinder consider,
Radius = 2r and Height = 5h

For the second cylinder consider,
Radius = 3r and Height = 4h

Volume of the first cylinder = Base area × Height
= π (2r)² × 5h
= 4πr² × 5 h
= 20 πr²h cm³

Volume of the second cylinder = Base area × Height
= π (3r)² × 4h
= 9 π r² × 4h
= 36 π r²h cm³

Ratio between their volumes = 20 π r²h : 36 π r²h
= 20 : 36 = 5 : 9

ii) Volume of the smaller cylinder = 720 cm3
Considering the volume of the larger cylinder = x
Therefore the ratio between the volume of the two cylinders = 5 : 9
That means, 5:9 = 720 : x
\(\frac{5}{9}=\frac{720}{x}\)
5x = 720 × 9
5 x = 6480
x = \(\frac{6480}{5}\) = 1296
Volume of the larger cylinder = 1296 cm³

Prisms Class 9 Kerala Syllabus – Curved Surface

Intext Questions And Answers

Question 1.
Rectangular sheets of thick paper, of sides 24 centimetres and 18 centimetres are bent along the longer side to make hollow prisms with bases an equilateral triangle, a square, a regular hexagon, a regular octagon. One of them is rolled into a cylinder. All the prisms have the same lateral surface area and it is also equal to the curved surface area of the cylinder. What about their volumes? Do you note anything special?

What if these shapes are made by bending or rolling along the shorter side of the rectangle? Find out the similarities and differences between these and the first ones, in lateral surface area, curved surface area and volume.
Answer:
Length of the rectangular sheet = 24 cm
Breadth = 18 cm

Bent along the longer side
Base perimeter = 24 cm
Height = 18cm

Lateral surface area
= Base perimeter × Height
= 24 × 18
= 432 cm²

Curved surface area of the cylinder
= Base circumference × Height
= 24 × 18
= 432 cm²

Shorter Side:

Bent along the shorter side
Base perimeter = 18 cm
Height = 24cm

Lateral surface area
= Base perimeter × Height
= 18 × 24
= 432 cm²

Curved surface area of the cylinder
= Base circumference × Height
= 18 × 24
= 432 cm²

Here the lateral surface area and the curved surface areas of the two prisms are equal. But their volumes are different.

1^st Prism:

Equilateral triangle
One side = \(\frac{\text { Base perimeter }}{3}=\frac{24}{3}\) = 8cm
Base area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × (8)²
= 16√3 cm²

Volume = Base area × Height
= 16√3 × 18
= 288√3
= 288 × 1.73
= 498.24 cm³

Square prism
One side = \(\frac{\text { Base perimeter }}{4}=\frac{24}{4}\)
= 6cm

Base area = (side)²
= (6)²
= 36 cm²

Volume = Base area × Height
= 36 × 18
= 648 cm³

2^nd Prism:

Equilateral triangle
One side = \(\frac{\text { Base perimeter }}{3}=\frac{18}{3}\) = 6cm
Base area = \(\frac{\sqrt{3}}{4}\) × (side)
= \(\frac{\sqrt{3}}{4}\) × (6)²
= 9√3 cm²

Volume = Base area × Height
= 9√3 × 24 = 216√3
= 216 × 1.73
= 373.68 cm³

Square prism
One side = \(\frac{\text { Base perimeter }}{4}=\frac{18}{4}\) = 4.5 cm
Base area = (side)² = (4.5)²
= 20.25 cm²

Volume = Base area × Height
= 20.25 × 24
=486 cm³

Similarly, we can find the volume of the remaining prisms.
From this we can conclude that, if the base area changes its volume also changes.

Textual Questions And Answers

Question 1.
In a school building there are 12 cylindrical pillars, each of base diameter \(\frac{1}{2}\) metre and height 4 metres.
i) What is the curved surface area of a pillar?
ii) What is the total cost of painting all the pillars at 80 rupees per square metre?
Answer:
i) Number of cylindrical pillars = 12
Base diameter = \(\frac{1}{2}\)m
Height = 4 m
Radius = \(\frac{\text { Base diameter }}{2}=\frac{1 / 2}{2}=\frac{1}{4}\) = 0.25 m
Base circumference = 2 π r
= 2 π × 0.25
= 0.5 π

Curved surface area of the pillar = Base circumference × Height
= 0.5 π × 4
= 2 π cm²

ii) Curved surface area of 12 pillars = 12 × 2π
= 24 π

Cost of painting per square meter = Rs. 80

Total cost for painting the 12 pillars = Curved surface area of 12 pillars × 80
= 24 π × 80
= 1920 π
= 1920 × 3.1416
= 6031.872 Rs

Question 2.
The drum of a road roller has base diameter 80 centimetres and length 1.2 metres:

What is the area of the road levelled when it rolls once?
Answer:
Length of the roller = 1.2 m = 120 cm
Volume = 80 cm
Radius = \(\frac{80}{2}\) = 40 cm

Base circumference = 2 × π × 40
= 80π cm
The area of the road levelled when it rolls once is equivalent to its curved surface area Therefore,
Curved surface area = Base perimeter × Height
= 80 π × 120
= 9600 π
= 9600 × 3.14
= 30,144 cm²
= 3.0144 m²

Question 3.
The curved surface area of a cylinder is equal to its base area. What is the relation between ifs base radius and height?
Answer:
Consider, the radius of the cylinder = r
Height = h
Base circumference = 2π × radius = r

Therefore,
Curved surface area = Base circumference × Height
= 2πr × h
= 2 π r h

Base area = π × (radius)2 = πr²
Given that, the curved surface area of a cylinder is equal to its base area.
That means, 2 π r h = πr²
2 h = r
That means, the radius is twice of its height.

Question 4.
A rectangular sheet of metal with sides 48 centimetres and 25 centimetres is rolled into a cylinder of height 25 centimetres and its ends are closed with exactly fitting circles. What is total surface area of this cylinder?
Answer:
Side of the rectangular sheet = 48 cm
Breadth = 25 cm
Height of the cylinder = 25 cm
Considering radius as ‘r’

The length of the rectangular sheet is equal to its circumference of the base circle.
Circumference of the base circle = 48 cm
2 π r = 48
r = \(\frac{48}{2 \pi}=\frac{48}{2 \times 3.14}\)
= 7.64 cm

From this, radius of the cylinder r = 7.64 cm
Curved surface area = Base circumference × Height
= 2πr × h
=2 × 3.1416 × 7.64 × 25
= 1200 cm²

Base area of a circle = πr²
= 3.14 × (7.64)²
= 183.37 cm²

Base area of two circles = 2 × 183.37
= 366.7 cm²

Total surface area of the cylinder = Curved surface area + Base area of two circles
= 1200 + 366.7
= 1566.7 cm²

Prisms Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
The base area of a prism is 50 square centimetres and its height is 20 centimetres. What is the volume of the prism?,
Answer:
Base area = 50 cm³
Height = 20 cm
Volume = Base area × height
= 50 × 20
= 1000 cm³

Question 2.
6 square pieces of side 10 centimetres are cut out and joined as shown in the figure. It is then folded to make a prism.

a) Find the surface area of the prism.
b) How many litres of water can it hold?
Answer:
a) one side = 10 cm
Base area of a square = (side)² = (10)²
= 100 cm²

b) Volume = Base area × height = 100 × 10
= 1000 cubic cm
= 1 litre

Question 3.
a) The lateral surface area of an equilateral triangular prism is 90 square centimetres. What is the area of one lateral face?
b) Two such prisms are put together to form a new prism. What is the lateral surface area of the new prism?
c) Six such prisms are put together to form a regular hexagonal prism. What is the lateral surface area of this prism?
Answer:
a) lateral surface area = 90 square cm
Area of one lateral face = \(\frac{90}{3}\) = 30 cm

b) Lateral surface area = 4 × 30 = 120 cm²
[If the lateral faces are joined we get the above answer and if the base faces are joined, Lateral surface area = 6 × 30 = 180sq. cm]

c) Lateral surface area = 6 × 30 = 180 cm²

Question 4
Dimensions of a rectangular box are 50 centimetres, 30 centimetres and 40 centimetres. Find area of the cardboard required for making this box. .
Answer:
Given the dimensions of a rectangular box as,
Length = 50 cm
Breadth = 30 cm
Height = 40 cm
Base perimeter = 2 (50 + 30) = 160 cm

Lateral surface area = Base perimeter × Height
= 160 × 40
= 6400 cm²

Base area = Length × Breadth
= 50 × 30
= 1500 cm²

Total surface area = Lateral surface area + 2 (Base area)
= 6400 + 2 (1500)
= 9400 cm²

Area of the cardboard required = 9400 cm²

Question 5.
If a cylinder having radius 4 centimetre and height 10 centimetre. Then what is the volume of the cylinder. Their is an another cylinder with half the radius and double the height of the first cylinder. What is the relation between the volume of the first cylinder and the second cylinder.
Answer:
Base radius = 4 cm
Height = 10 cm
Base area = πr² = π 4²
= 16 π

Volume of the cylinder = Base area × Height
= 16 π × 10
= 160 π cm³

If the cylinder having half the radius and double the height then,
Base radius = 2 cm

Height = 20 cm
Base area = πr² = π × 2² = 4 π

Volume of the second cylinder = Base area × Height
= 4π × 20
= 80π cm³

That means, volume of the second cylinder is half the volume of the first cylinder.

Question 6.
There are 10 cylindrical pillars of diameter 30 centimetres and height 5 metres in a school varandha. What is the total cost of painting the pillars at the rate of rupees 80 per square metre? (π = 3.14)
Answer:
Number of cylindrical pillars = 10
Base diameter = 30 cm = 0.3 m
Height = 5m
Base radius = \(\frac{\text { Base diameter }}{2}=\frac{0.3}{2}\) = 0.15 m
Base perimeter = 2πr = 2 × π × 0.15 = 0.3 π m

Curved surface area of a pillar = Base perimeter × height
= 0.3 π × 5
= 1.5 π cm²

Total curved surface area of 10 pillar = 10 × 1.5π
= 15π cm²

Total cost = 15π × 80
= 15 × 3.14 × 80
= 3768 Rs

Question 7.
A cylindrical water tank has a radius of 1 metre and a height of 2 metres. How many liters of water can it hold? (1 cubic metre = 1000 liters )
Ans:
Radius of the cylindrical water tank = 1 m
Height = 2 m
Base area = πr² – π 1²
= 1 π m

Volume = Base area × Height
= 1π × 2
= 2 π
— 2 × 3.14
= 6.28 cm³
= 6.28 × 1000 = 6280 liter