NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes
Get here NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes.
More Detail
Latest News
Finished viewing? Save it for later —
Download NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes (PDF · 38 pages)Downloaded 108 times
About NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes
NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes is available here for free download. Published by NCERT for Class 9, this solution can be viewed online or downloaded as a PDF (38 pages). Candidates preparing for Class 9 can use NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes to understand the exam pattern, the type of questions asked, and the overall difficulty level.
Frequently Asked Questions
How can I download NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes?
Open this page and click the Download button to save NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes as a PDF. It is completely free on AglaSem Docs.
Is NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes free to download?
Yes. NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes can be viewed online and downloaded as a PDF free of cost on AglaSem Docs.
How many pages does NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes have?
NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes contains 38 pages, which you can read online or download together as a single PDF.
Where can I find more Class 9 study material?
You can find more Class 9 question papers, sample papers, syllabus, and answer keys on AglaSem Docs.
NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes – Text
Read the full text of this solution below — useful to quickly search, copy and reference the content online without downloading the PDF.
📄 View text version (38 pages)
Page 1
NCERT
SOLUTIONS
CLASS - 9th
aglase .co
SOLUTIONS
CLASS - 9th
aglase .co
Page 2
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Class : 9th
Subject : Maths
Chapter : 13
Chapter Name : SURFACE AREAS AND VOLUMES
Exercise 13.1
Q1 A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top.
Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m2 costs ` 20.
Answer. It is given that, length (l) of box =1.5m
Breadth (b) of box 1.25 m
Depth (h) of box = 0.65 m
(i) Box is to be open at top.
Area of sheet required
=2lh+2bh+lb
=[2 x 1.5 x 0.65 +2 x 1.25 x 0.65 + 1.5 x 1.25] m
2
=(1.95 + 1.625 + 1.875) m2 =5.45 m 2
(ii) Cost of sheet per m area =Rs 20
2
Cost of sheet of 5.45 m area =Rs (5.45x20)
2
=Rs 109
Page : 213 , Block Name : Exercise 13.1
Q2 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 ` 7.50 per m2 .
Answer. It is given that
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
Page 1 of 37 Aglasem Schools
Class : 9th
Subject : Maths
Chapter : 13
Chapter Name : SURFACE AREAS AND VOLUMES
Exercise 13.1
Q1 A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top.
Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m2 costs ` 20.
Answer. It is given that, length (l) of box =1.5m
Breadth (b) of box 1.25 m
Depth (h) of box = 0.65 m
(i) Box is to be open at top.
Area of sheet required
=2lh+2bh+lb
=[2 x 1.5 x 0.65 +2 x 1.25 x 0.65 + 1.5 x 1.25] m
2
=(1.95 + 1.625 + 1.875) m2 =5.45 m 2
(ii) Cost of sheet per m area =Rs 20
2
Cost of sheet of 5.45 m area =Rs (5.45x20)
2
=Rs 109
Page : 213 , Block Name : Exercise 13.1
Q2 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 ` 7.50 per m2 .
Answer. It is given that
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
Page 1 of 37 Aglasem Schools
Page 3
Book : Mathematics Ncert Solutions | Chapter-13 Maths
It can be observed that four walls and the ceiling of the room are to be white- washed. The oor of
the room is not to be white-washed.
Area to be white-washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
=[2x5x3+2x4x3+5x4] m 2
(30 + 24 + 20) m 2
= 74 m 2
Cost of white-washing per m area = Rs 7.50
2
Cost of white-washlng 74 m area =Rs (74x7.50)
2
= RS 555
Page : 213 , Block Name : Exercise 13.1
Q3 The oor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at
the rate of 10perm2is 15000, nd the height of the hall. [Hint : Area of the four walls = Lateral
surface area.]
Answer. Let length, breadth, and height of the rectangular hall be I m, b m, and h mrespectively.
Area of four walls = 21h + 2bh
=2( l + b )h
Perimeter of the oor of hall =2(l + b)
=250 m
Area of four walls =2( l+b )h =250 m 2
Cost of painting per m area = Rs 10
2
Cost of painting 250h m area = Rs (250h x 10) = Rs 2500h
2
However, it is given that the cost of painting the walls is Rs 15000.
15000 = 2500h
h=6
Therefore, the height of the hall is 6 m.
Page : 213 , Block Name : Exercise 13.1
Q4 The paint in a certain container is suf cient to paint an area equal to 9.375 m2 . How many
bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Answer. Total surface area of one brick = 2(lb + bh + /h)
= [2(22.5 X10 + 10 x 7.5 + 22.5 x 7.5)] cm 2
= 2(225 + 75 + 168.75) cm 2
= (2 x 468.75) cm 2
= 937.5 cm 2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n x937.5) cm = 937.5n cm
2 2
Area that can be painted by the paint of the container = 9.375 m 2
= 93750 cm 2
0=93750 - 937.5n
Page 2 of 37 Aglasem Schools
It can be observed that four walls and the ceiling of the room are to be white- washed. The oor of
the room is not to be white-washed.
Area to be white-washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
=[2x5x3+2x4x3+5x4] m 2
(30 + 24 + 20) m 2
= 74 m 2
Cost of white-washing per m area = Rs 7.50
2
Cost of white-washlng 74 m area =Rs (74x7.50)
2
= RS 555
Page : 213 , Block Name : Exercise 13.1
Q3 The oor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at
the rate of 10perm2is 15000, nd the height of the hall. [Hint : Area of the four walls = Lateral
surface area.]
Answer. Let length, breadth, and height of the rectangular hall be I m, b m, and h mrespectively.
Area of four walls = 21h + 2bh
=2( l + b )h
Perimeter of the oor of hall =2(l + b)
=250 m
Area of four walls =2( l+b )h =250 m 2
Cost of painting per m area = Rs 10
2
Cost of painting 250h m area = Rs (250h x 10) = Rs 2500h
2
However, it is given that the cost of painting the walls is Rs 15000.
15000 = 2500h
h=6
Therefore, the height of the hall is 6 m.
Page : 213 , Block Name : Exercise 13.1
Q4 The paint in a certain container is suf cient to paint an area equal to 9.375 m2 . How many
bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Answer. Total surface area of one brick = 2(lb + bh + /h)
= [2(22.5 X10 + 10 x 7.5 + 22.5 x 7.5)] cm 2
= 2(225 + 75 + 168.75) cm 2
= (2 x 468.75) cm 2
= 937.5 cm 2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n x937.5) cm = 937.5n cm
2 2
Area that can be painted by the paint of the container = 9.375 m 2
= 93750 cm 2
0=93750 - 937.5n
Page 2 of 37 Aglasem Schools
Page 4
Book : Mathematics Ncert Solutions | Chapter-13 Maths
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
Page : 213 , Block Name : Exercise 13.1
Q5 A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8
cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has
the smaller total surface area and by how much?
Answer. (i) Edge of cube = 10 cm
Length (j) of box = 12.5 cm
Breadth (b) of box = 10 cm
Height (h) of box = 8 cm
Lateral surface area of cubical box = 4( edge ) 2
=400 cm 2
Lateral surface area of cuboidal box =2[lh+bh]
= [2(12.5 x 10 x 8)] cm 2
= (2 x 180) cm 2
= 360 cm 2
Clearly, the lateral surface area of the cubical box is greater than the lateral surface
area of the cuboidal box.
Lateral surface area of cubical box - Lateral surface area of cuboidal box = 400 cm2
- 360 cm2 = 40 cm 2
Therefore, the lateral surface area of the cubical box is greater than the lateral
surface area of the cuboidal box by 40cm 2
(ii) Total surface area of cubical box =6( edge ) =6(10cm) = 600 cm
2 2 2
Total surface area of cuboidal box
= 2(lh + bh + lb]
=[2(12.5 x 8 + 10 x 8+ 12.5 x 100]cm2
610 cm 2
Clearly, the total surface area of the cubical box is smaller than that of the cuboidal
box.
Total surface area of cuboidal box - Total surface area of cubical box = 610 cm2
600cm = 10 cm
2 2
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10
cm .
2
Page : 213 , Block Name : Exercise 13.1
Q6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held
together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Page 3 of 37 Aglasem Schools
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
Page : 213 , Block Name : Exercise 13.1
Q5 A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8
cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has
the smaller total surface area and by how much?
Answer. (i) Edge of cube = 10 cm
Length (j) of box = 12.5 cm
Breadth (b) of box = 10 cm
Height (h) of box = 8 cm
Lateral surface area of cubical box = 4( edge ) 2
=400 cm 2
Lateral surface area of cuboidal box =2[lh+bh]
= [2(12.5 x 10 x 8)] cm 2
= (2 x 180) cm 2
= 360 cm 2
Clearly, the lateral surface area of the cubical box is greater than the lateral surface
area of the cuboidal box.
Lateral surface area of cubical box - Lateral surface area of cuboidal box = 400 cm2
- 360 cm2 = 40 cm 2
Therefore, the lateral surface area of the cubical box is greater than the lateral
surface area of the cuboidal box by 40cm 2
(ii) Total surface area of cubical box =6( edge ) =6(10cm) = 600 cm
2 2 2
Total surface area of cuboidal box
= 2(lh + bh + lb]
=[2(12.5 x 8 + 10 x 8+ 12.5 x 100]cm2
610 cm 2
Clearly, the total surface area of the cubical box is smaller than that of the cuboidal
box.
Total surface area of cuboidal box - Total surface area of cubical box = 610 cm2
600cm = 10 cm
2 2
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10
cm .
2
Page : 213 , Block Name : Exercise 13.1
Q6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held
together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Page 3 of 37 Aglasem Schools
Page 5
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer. (i) Length (l) of greenhouse = 30 cm
Breadth (b) of greenhouse =25 cm
Height (h) of greenhouse = 25 cm
Total surface area of greenhouse
=2[lb+lh+bh]
= [2(30 x 25 + 30 x 25 + 25 x 25)] cm2
= [2(750 + 750 + 625)) cm 2
= (2 x 2125) cm 2
=4250 cm 2
Therefore, the area of glass is 4250 cm2.
It can be observed that tape is required alongside AD, OC, CD, DA, EF, FG, GH, HE,
AH, BE, DG, and CF.
Total length of tape =4(l+ b + h)
= [4(30 + 25 + 25)) cm
= 320 cm
Therefore, 320 cm tape is required for all the 12 edges.
Page : 213 , Block Name : Exercise 13.1
Q7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets.
Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller
of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required
extra. If the cost of the cardboard is ` 4 for 1000 cm2 , nd the cost of cardboard required for
supplying 250 boxes of each kind.
Answer. Length (h) of bigger box = 25 cm
Breadth (bl) of bigger box = 20 cm
Height (m) of bigger box = 5 cm
Total surface area of bigger box = 2(lb + lh + bh)
[2(25 x 20 + 25 x s +20 x 5)] cm2
= [2(500 + 125 + 100)]cm 2
= 1450cm 2
Extra area required for overlapping = ( 1450×5
100
) cm
2
=72.5 cm 2
While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) cm =1522.5cm
2 2
Area of cardboard sheet required for 250 such bigger boxes
Page 4 of 37 Aglasem Schools
Answer. (i) Length (l) of greenhouse = 30 cm
Breadth (b) of greenhouse =25 cm
Height (h) of greenhouse = 25 cm
Total surface area of greenhouse
=2[lb+lh+bh]
= [2(30 x 25 + 30 x 25 + 25 x 25)] cm2
= [2(750 + 750 + 625)) cm 2
= (2 x 2125) cm 2
=4250 cm 2
Therefore, the area of glass is 4250 cm2.
It can be observed that tape is required alongside AD, OC, CD, DA, EF, FG, GH, HE,
AH, BE, DG, and CF.
Total length of tape =4(l+ b + h)
= [4(30 + 25 + 25)) cm
= 320 cm
Therefore, 320 cm tape is required for all the 12 edges.
Page : 213 , Block Name : Exercise 13.1
Q7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets.
Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller
of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required
extra. If the cost of the cardboard is ` 4 for 1000 cm2 , nd the cost of cardboard required for
supplying 250 boxes of each kind.
Answer. Length (h) of bigger box = 25 cm
Breadth (bl) of bigger box = 20 cm
Height (m) of bigger box = 5 cm
Total surface area of bigger box = 2(lb + lh + bh)
[2(25 x 20 + 25 x s +20 x 5)] cm2
= [2(500 + 125 + 100)]cm 2
= 1450cm 2
Extra area required for overlapping = ( 1450×5
100
) cm
2
=72.5 cm 2
While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) cm =1522.5cm
2 2
Area of cardboard sheet required for 250 such bigger boxes
Page 4 of 37 Aglasem Schools
Page 6
Book : Mathematics Ncert Solutions | Chapter-13 Maths
= (1522.5 x 250) cm2 380625cm 2
Similarly, total surface area of smaller box = [2(15 + 15 x 5 + 12 x 5)] cm 2
= [2(180 + 75 + 60)] cm 2
= (2 x 315) cm 2
= 630 cm 2
Therefore, extra area required for overlapping = ( 630×5
100
) cm
2
= 31.5
Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) cm = 661.5 cm2 2
Area of cardboard sheet required for 250 smaller boxes = (250 x 661.5) cm 2
= 165375 cm 2
Total cardboard sheet required = (380625 + 165375) cm 2
= 546000 cm 2
Cost of 1000 cm cardboard sheet =Rs 4
2
Cost of 546000 cm2 cardboard sheet = Rs(
546000×4
) = Rs 2184
1000
Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will
be Rs 2184.
Page : 213 , Block Name : Exercise 13.1
Q8 Parveen wanted to make a temporary shelter for her car, by making a box-like structure with
tarpaulin that covers all the four sides and the top of the car (with the front face as a ap which
can be rolled up). Assuming that the stitching margins are very small, and therefore negligible,
how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4
m × 3 m?
Answer. Length (l) of shelter = 4 m
Breadth (b) of shelter = 3 m
Height (h) of shelter = 2.5 m
Tarpaulin will be required for the top and four wall sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4 x 2.5+3 x 2.5) +4 x 3] m 2
= [2(10 + 7.5) + 12] m 2
=47m 2
Therefore, 47 m tarpaulin will be required.
2
Page : 213 , Block Name : Exercise 13.1
Q1 The curved surface area of a right circular cylinder of height 14 cm is 88 cm2 . Find the
diameter of the base of the cylinder.
Answer. Height (h) of cylinder = 14 cm
Let the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
Page 5 of 37 Aglasem Schools
= (1522.5 x 250) cm2 380625cm 2
Similarly, total surface area of smaller box = [2(15 + 15 x 5 + 12 x 5)] cm 2
= [2(180 + 75 + 60)] cm 2
= (2 x 315) cm 2
= 630 cm 2
Therefore, extra area required for overlapping = ( 630×5
100
) cm
2
= 31.5
Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) cm = 661.5 cm2 2
Area of cardboard sheet required for 250 smaller boxes = (250 x 661.5) cm 2
= 165375 cm 2
Total cardboard sheet required = (380625 + 165375) cm 2
= 546000 cm 2
Cost of 1000 cm cardboard sheet =Rs 4
2
Cost of 546000 cm2 cardboard sheet = Rs(
546000×4
) = Rs 2184
1000
Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will
be Rs 2184.
Page : 213 , Block Name : Exercise 13.1
Q8 Parveen wanted to make a temporary shelter for her car, by making a box-like structure with
tarpaulin that covers all the four sides and the top of the car (with the front face as a ap which
can be rolled up). Assuming that the stitching margins are very small, and therefore negligible,
how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4
m × 3 m?
Answer. Length (l) of shelter = 4 m
Breadth (b) of shelter = 3 m
Height (h) of shelter = 2.5 m
Tarpaulin will be required for the top and four wall sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4 x 2.5+3 x 2.5) +4 x 3] m 2
= [2(10 + 7.5) + 12] m 2
=47m 2
Therefore, 47 m tarpaulin will be required.
2
Page : 213 , Block Name : Exercise 13.1
Q1 The curved surface area of a right circular cylinder of height 14 cm is 88 cm2 . Find the
diameter of the base of the cylinder.
Answer. Height (h) of cylinder = 14 cm
Let the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
Page 5 of 37 Aglasem Schools
Page 7
Book : Mathematics Ncert Solutions | Chapter-13 Maths
2nrh SS cm2 (r is the radius of the base of the cylinder)
ndh = 88 crn2 (d = 2r)
d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
Page : 216 , Block Name : Exercise 13.2
Q2 It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a
metal sheet. How many square metres of the sheet are required for the same?
Answer. height(h) of the cylindrical tank =1m
Base radius(r) of cylindrical tank= = ( 140
2
) cm = 70cm = 0.7m
Area of sheet required = total surface of tank =2πr(r + h)m
22 2
= [2 × × 0.7(0.7 + 1)] m
7
=(4.4 × 1.7)m 2
=7.48m 2
Therefore , it will required 7.48m area of sheet.
2
Page : 216 , Block Name : Exercise 13.2
Q3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter
being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
Answer. Inner radius r1 of cylindrical pipe = = ( 4
) cm = 2cm
2
outer radius r2 of cylindrical pipe == ( 4.4
2
) cm = 2.2cm
Height (h) Of cylindrical pipe = Length Of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe=2πr h 1
=(2 × 22
7
× 2 × 77) cm
2
=968cm 2
(ii) CSA of inner surface of pipe=2πr h 2
Page 6 of 37 Aglasem Schools
2nrh SS cm2 (r is the radius of the base of the cylinder)
ndh = 88 crn2 (d = 2r)
d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
Page : 216 , Block Name : Exercise 13.2
Q2 It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a
metal sheet. How many square metres of the sheet are required for the same?
Answer. height(h) of the cylindrical tank =1m
Base radius(r) of cylindrical tank= = ( 140
2
) cm = 70cm = 0.7m
Area of sheet required = total surface of tank =2πr(r + h)m
22 2
= [2 × × 0.7(0.7 + 1)] m
7
=(4.4 × 1.7)m 2
=7.48m 2
Therefore , it will required 7.48m area of sheet.
2
Page : 216 , Block Name : Exercise 13.2
Q3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter
being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
Answer. Inner radius r1 of cylindrical pipe = = ( 4
) cm = 2cm
2
outer radius r2 of cylindrical pipe == ( 4.4
2
) cm = 2.2cm
Height (h) Of cylindrical pipe = Length Of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe=2πr h 1
=(2 × 22
7
× 2 × 77) cm
2
=968cm 2
(ii) CSA of inner surface of pipe=2πr h 2
Page 6 of 37 Aglasem Schools
Page 8
Book : Mathematics Ncert Solutions | Chapter-13 Maths
=(2 ×
22 2
× 2.2 × 77) cm
7
=(22 × 22 × 2.2)cm 2
2
1064.8cm
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area
of both circular ends of pipe
2 2
= 2πr1 h + 2πr2 h + 2π (r − r )
2 1
2 2 2
= [968 + 1064.8 + 2π {(2.2) − (2) }] cm
22 2
= (2032.8 + 2 × × 0.84) cm
7
2
= (2032.8 + 5.28)cm
2
= 2038.08cm
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm 2
Page : 216 , Block Name : Exercise 13.2
Q4 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to
move once over to level a playground. Find the area of the playground in m2 .
Answer. It can be observed that a roller is cylindrical.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end roller = (
84
) cm = 42cm
2
CSA of rollar = 2nrh
22 2
= (2 × × 42 × 120) cm
7
2
= 31680cm
Area of eld = 500x CSA of roller
2
= (500 × 31680)cm
2
= 1584m
Page : 217 , Block Name : Exercise 13.2
Q5 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the
curved surface of the pillar at the rate of ` 12.50 per ( \mathrm{m}^{2}\)
Answer. Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar =
50
= 25cm
2
=0.25m
=(2 × 22
7
× 0.25 × 3.5) m
2
=(44 × 0.125)m 2
=5.5m 2
Cost of painting 1 m area = Rs. 12.50
2
Cost of painting 5.5 m area Rs (5.5 x 12.50)
2
=Rs 68.75
Page 7 of 37 Aglasem Schools
=(2 ×
22 2
× 2.2 × 77) cm
7
=(22 × 22 × 2.2)cm 2
2
1064.8cm
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area
of both circular ends of pipe
2 2
= 2πr1 h + 2πr2 h + 2π (r − r )
2 1
2 2 2
= [968 + 1064.8 + 2π {(2.2) − (2) }] cm
22 2
= (2032.8 + 2 × × 0.84) cm
7
2
= (2032.8 + 5.28)cm
2
= 2038.08cm
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm 2
Page : 216 , Block Name : Exercise 13.2
Q4 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to
move once over to level a playground. Find the area of the playground in m2 .
Answer. It can be observed that a roller is cylindrical.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end roller = (
84
) cm = 42cm
2
CSA of rollar = 2nrh
22 2
= (2 × × 42 × 120) cm
7
2
= 31680cm
Area of eld = 500x CSA of roller
2
= (500 × 31680)cm
2
= 1584m
Page : 217 , Block Name : Exercise 13.2
Q5 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the
curved surface of the pillar at the rate of ` 12.50 per ( \mathrm{m}^{2}\)
Answer. Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar =
50
= 25cm
2
=0.25m
=(2 × 22
7
× 0.25 × 3.5) m
2
=(44 × 0.125)m 2
=5.5m 2
Cost of painting 1 m area = Rs. 12.50
2
Cost of painting 5.5 m area Rs (5.5 x 12.50)
2
=Rs 68.75
Page 7 of 37 Aglasem Schools
Page 9
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Therefore, the cost of painting the CSA of the pillar is Rs 68.75 .
Page : 217 , Block Name : Exercise 13.2
Q6 Curved surface area of a right circular cylinder is 4.4 m2 . If the radius of the base of the
cylinder is 0.7 m, nd its height.
Answer. Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA Of cylinder = 4.4 m 2
2
2πrh = 4.4m
22 2
(2 × × 0.7 × h) m = 4.4m
7
h=1m
Therefore, the height of the cylinder is 1 m.
Page : 217 , Block Name : Exercise 13.2
Q7 The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ` 40 per m
2
Answer. Inner radius (r) Of circular well = (
3.5
) m = 1.75m
2
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
22 2
= (2 × × 1.75 × 10) m
7
2
= (44 × 0.25 × 10)m
2
= 110m
Therefore, the inner curved surface area of the circular well is 110 m 2
Cost of plastering 1 m area Rs 40
2
Cost of plastering 100 m area = Rs (110 x 40)
2
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 4400.
Page : 217 , Block Name : Exercise 13.2
Q8 In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm.
Find the total radiating surface in the system.
Answer. Height (h) of cylindrical pipe Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe = = 2.5cm =0.025m
5
2
CSA of cylindrical pipe = 2nrh
Page 8 of 37 Aglasem Schools
Therefore, the cost of painting the CSA of the pillar is Rs 68.75 .
Page : 217 , Block Name : Exercise 13.2
Q6 Curved surface area of a right circular cylinder is 4.4 m2 . If the radius of the base of the
cylinder is 0.7 m, nd its height.
Answer. Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA Of cylinder = 4.4 m 2
2
2πrh = 4.4m
22 2
(2 × × 0.7 × h) m = 4.4m
7
h=1m
Therefore, the height of the cylinder is 1 m.
Page : 217 , Block Name : Exercise 13.2
Q7 The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ` 40 per m
2
Answer. Inner radius (r) Of circular well = (
3.5
) m = 1.75m
2
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
22 2
= (2 × × 1.75 × 10) m
7
2
= (44 × 0.25 × 10)m
2
= 110m
Therefore, the inner curved surface area of the circular well is 110 m 2
Cost of plastering 1 m area Rs 40
2
Cost of plastering 100 m area = Rs (110 x 40)
2
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 4400.
Page : 217 , Block Name : Exercise 13.2
Q8 In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm.
Find the total radiating surface in the system.
Answer. Height (h) of cylindrical pipe Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe = = 2.5cm =0.025m
5
2
CSA of cylindrical pipe = 2nrh
Page 8 of 37 Aglasem Schools
Page 10
Book : Mathematics Ncert Solutions | Chapter-13 Maths
22 2
= (2 × × 0.025 × 28) m
7
2
= 4.4m
The area of the radiating surface of the system is = 4.4m 2
Page : 217 , Block Name : Exercise 13.2
Q9 Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in
diameter and 4.5 m high.
(ii) how much steel was actually used, if 1 12 of the steel actually used was wasted in making the
tank.
Answer. Height (h) of the cylindrical tank = 4.5m
radius(r) of the circular end of cylindrical tank = ( 4.2
2
) m = 2.1m
(i) lateral or curved surface area of tank = 2nrh
22 2
= (2 × × 2.1 × 4.5) m
7
2
= (44 × 0.3 × 4.5)m
2
= 59.4m
Therefore , CSA of tank is 59.4 m 2
22 2
= [2 × × 2.1 × (2.1 + 4.5)] m
7
2
= (44 × 0.3 × 6.6)m
2
= 87.12m
Let A steel m sheet be actually used in making the tank.
2
1 2
A (1 − ) = 87.12m
12
12 2
A) = ( × 87.12) m
11
2
A = 95.04m
Therefore , 95.04 A steel was used in actual while making such a tank.
Page : 217 , Block Name : Exercise 13.2
Q10 In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The
frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for
folding it over the top and bottom of the frame. Find how much cloth is required for covering the
lampshade.
Page 9 of 37 Aglasem Schools
22 2
= (2 × × 0.025 × 28) m
7
2
= 4.4m
The area of the radiating surface of the system is = 4.4m 2
Page : 217 , Block Name : Exercise 13.2
Q9 Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in
diameter and 4.5 m high.
(ii) how much steel was actually used, if 1 12 of the steel actually used was wasted in making the
tank.
Answer. Height (h) of the cylindrical tank = 4.5m
radius(r) of the circular end of cylindrical tank = ( 4.2
2
) m = 2.1m
(i) lateral or curved surface area of tank = 2nrh
22 2
= (2 × × 2.1 × 4.5) m
7
2
= (44 × 0.3 × 4.5)m
2
= 59.4m
Therefore , CSA of tank is 59.4 m 2
22 2
= [2 × × 2.1 × (2.1 + 4.5)] m
7
2
= (44 × 0.3 × 6.6)m
2
= 87.12m
Let A steel m sheet be actually used in making the tank.
2
1 2
A (1 − ) = 87.12m
12
12 2
A) = ( × 87.12) m
11
2
A = 95.04m
Therefore , 95.04 A steel was used in actual while making such a tank.
Page : 217 , Block Name : Exercise 13.2
Q10 In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The
frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for
folding it over the top and bottom of the frame. Find how much cloth is required for covering the
lampshade.
Page 9 of 37 Aglasem Schools
Page 11
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer.
Height (h) of the frame of lampshade =(2.5 +30 +2.5) cm = 35cm
Radius (r) of the circular end of the frame of lampshade = (
20
) cm = 10cm
2
Cloth required for covering the lampshade = 2πrh
22 2
= (2 × × 10 × 35) cm
7
2
= 2200cm
Hence, for covering the lampshade,2200 cm cloth will be required. 2
Page : 217 , Block Name : Exercise 13.2
Q11 The students of a Vidyalaya were asked to participate in a competition for making and
decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was
to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with
cardboard. If there were 35 competitors, how much cardboard was required to be bought for the
competition?
Answer. Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) Of penholder =10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
2
= 2πrh + πr
22 22 2 2
= [2 × × 3 × 10.5 + × (3) ] cm
7 7
198 2
= (132 × 1.5 + ) cm
7
198 2
= (198 + ) cm
7
1584 2
= cm
7
Area Of cardboard sheet used by 1 competitor = 1584
7
cm
2
Area of cardboard sheet used by 35 competitors == (
1584 2
× 35) cm
7
Therefore ,7920cm cardboard sheet will be bought.
2
Page : 217 , Block Name : Exercise 13.2
Q1 Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface
area.
Page 10 of 37 Aglasem Schools
Answer.
Height (h) of the frame of lampshade =(2.5 +30 +2.5) cm = 35cm
Radius (r) of the circular end of the frame of lampshade = (
20
) cm = 10cm
2
Cloth required for covering the lampshade = 2πrh
22 2
= (2 × × 10 × 35) cm
7
2
= 2200cm
Hence, for covering the lampshade,2200 cm cloth will be required. 2
Page : 217 , Block Name : Exercise 13.2
Q11 The students of a Vidyalaya were asked to participate in a competition for making and
decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was
to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with
cardboard. If there were 35 competitors, how much cardboard was required to be bought for the
competition?
Answer. Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) Of penholder =10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
2
= 2πrh + πr
22 22 2 2
= [2 × × 3 × 10.5 + × (3) ] cm
7 7
198 2
= (132 × 1.5 + ) cm
7
198 2
= (198 + ) cm
7
1584 2
= cm
7
Area Of cardboard sheet used by 1 competitor = 1584
7
cm
2
Area of cardboard sheet used by 35 competitors == (
1584 2
× 35) cm
7
Therefore ,7920cm cardboard sheet will be bought.
2
Page : 217 , Block Name : Exercise 13.2
Q1 Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface
area.
Page 10 of 37 Aglasem Schools
Page 12
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer. Radius (r) of the base of cone = ( 10.5
2
) cm = 5.25cm
Slant height (l) of cone = 10 cm
CSA of cone = Πrl
22 2 2 2
= ( × 5.25 × 10) cm = (22 × 0.75 × 10)cm = 165cm
7
Therefore, the curved surface area of the cone is 165 cm 2
Page : 221 , Block Name : Exercise 13.3
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer. Radius (r) of the base of cone = ( 24
2
) m=12m
Slant height (l) of cone = 21 m
Total surface area of cone = pi r(r + l)
22 2
= [ × 12 × (12 + 21)] m
7
22 2
= ( × 12 × 33) m
7
2
= 1244.57m
Page : 221 , Block Name : Exercise 13.3
Q3 Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Answer. (i) Slant height (l) of cone = 14 cm
Let the radius Of the circular end Of the cone be r.
We know, CSA of cone = Πrl
2 22
(308)cm = ( × r × 14) cm
7
308
r = ( ) cm = 7cm
44
Therefore, the radius Of the circular end Of the cone is = 7
(ii) Total surface area of cone = CSA of cone + Area of base
nrl + nr v
2
22 2 2
= [308 + × (7) ] cm
7
2
= (308 + 154)cm
2
= 462cm
Therefore, the total surface area of the cone is 462 cm 2
Page : 221 , Block Name : Exercise 13.3
Page 11 of 37 Aglasem Schools
Answer. Radius (r) of the base of cone = ( 10.5
2
) cm = 5.25cm
Slant height (l) of cone = 10 cm
CSA of cone = Πrl
22 2 2 2
= ( × 5.25 × 10) cm = (22 × 0.75 × 10)cm = 165cm
7
Therefore, the curved surface area of the cone is 165 cm 2
Page : 221 , Block Name : Exercise 13.3
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer. Radius (r) of the base of cone = ( 24
2
) m=12m
Slant height (l) of cone = 21 m
Total surface area of cone = pi r(r + l)
22 2
= [ × 12 × (12 + 21)] m
7
22 2
= ( × 12 × 33) m
7
2
= 1244.57m
Page : 221 , Block Name : Exercise 13.3
Q3 Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Answer. (i) Slant height (l) of cone = 14 cm
Let the radius Of the circular end Of the cone be r.
We know, CSA of cone = Πrl
2 22
(308)cm = ( × r × 14) cm
7
308
r = ( ) cm = 7cm
44
Therefore, the radius Of the circular end Of the cone is = 7
(ii) Total surface area of cone = CSA of cone + Area of base
nrl + nr v
2
22 2 2
= [308 + × (7) ] cm
7
2
= (308 + 154)cm
2
= 462cm
Therefore, the total surface area of the cone is 462 cm 2
Page : 221 , Block Name : Exercise 13.3
Page 11 of 37 Aglasem Schools
Page 13
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Q4 A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is 70.
Answer.
(i) Let ABC be a conical tent.
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l.
In △ABO
2 2 2
AB = AO + BO
2 2 2
l = h + r
== (10m) 2
+ (24m)
2
2
= 676m
l=26m
Therefore, the slant height Of the tent is 26 m.
(ii) CSA of tent = nrl
22 2
= ( × 24 × 26) m
7
13728 2
= m
7
Cost of 1m canvas = Rs 70 2
Cost of m canvas =
13728
7
2
13728
Rs( × 70)
7
=Rs 137280
Therefore, the cost of the canvas required to make such a tent is RS 137280.
Page : 221 , Block Name : Exercise 13.3
Q5 What length of tarpaulin 3m wide will be required to make conical tent of height 8 m and base
radius 6 m? Assume that the extra length of material that will be required for stitching margins
and wastage in cutting is approximately 20 cm (Use π = 3.14).
Answer. Height (h) Of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent = √r + h 2 2
2 2
= (√6 + 8 ) m = (√100)m = 10m
CSA of conical tent = πrl
2
= (3.14 × 6 × 10)m
Page 12 of 37 Aglasem Schools
Q4 A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is 70.
Answer.
(i) Let ABC be a conical tent.
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l.
In △ABO
2 2 2
AB = AO + BO
2 2 2
l = h + r
== (10m) 2
+ (24m)
2
2
= 676m
l=26m
Therefore, the slant height Of the tent is 26 m.
(ii) CSA of tent = nrl
22 2
= ( × 24 × 26) m
7
13728 2
= m
7
Cost of 1m canvas = Rs 70 2
Cost of m canvas =
13728
7
2
13728
Rs( × 70)
7
=Rs 137280
Therefore, the cost of the canvas required to make such a tent is RS 137280.
Page : 221 , Block Name : Exercise 13.3
Q5 What length of tarpaulin 3m wide will be required to make conical tent of height 8 m and base
radius 6 m? Assume that the extra length of material that will be required for stitching margins
and wastage in cutting is approximately 20 cm (Use π = 3.14).
Answer. Height (h) Of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent = √r + h 2 2
2 2
= (√6 + 8 ) m = (√100)m = 10m
CSA of conical tent = πrl
2
= (3.14 × 6 × 10)m
Page 12 of 37 Aglasem Schools
Page 14
Book : Mathematics Ncert Solutions | Chapter-13 Maths
2
= 188.4m
Let the length of tarpaulin sheet required be l.
As 20 cm will be wasted, therefore, the effective length will be ( l- 0.2 m).
Breadth of tarpaulin = 3m
Area Of sheet = CSA Of tent
2
[(l − 0.2m) × 3]m = 188.4m
l - 0.2m = 62.8m
l=63m
Therefore, the length of the required tarpaulin sheet will be 63 m.
Page : 221 , Block Name : Exercise 13.3
Q6 The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the
cost of white-washing its curved surface at the rate of ` 210 per 100 m2.
Answer. Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb
14
= 7m
2
CSA of conical tomb = πrl
22 2
= ( × 7 × 25) m
7
2
= 550m
Cost of white-washing 100 m area = R 210 2
210×550
Cost of white-washing 550 m area = =
Rs( )
2 100
=Rs 1155
Therefore ,it will cost of 1155 while white-washing such a conical tomb.
Page : 221 , Block Name : Exercise 13.3
Q7 A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find
the area of the sheet required to make 10 such caps.
Answer. Radius (r) of conical cap = 7 cm
Height (h) of conical cap 24 cm
Slant height (l) of the conical cap = √r + h 2 2
2 2
= [√(7) + (24) ] cm = (√625)cm = 25cm
CSA of 1 conical cap = πrl
22 2 2
= ( × 7 × 25) cm = 550cm
7
CSA of 10 such caps =(10x550)cm =5500 cm 2 2
Therefore , 5500 cm sheet will be required.
2
Page : 221 , Block Name : Exercise 13.3
Q8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of
Page 13 of 37 Aglasem Schools
2
= 188.4m
Let the length of tarpaulin sheet required be l.
As 20 cm will be wasted, therefore, the effective length will be ( l- 0.2 m).
Breadth of tarpaulin = 3m
Area Of sheet = CSA Of tent
2
[(l − 0.2m) × 3]m = 188.4m
l - 0.2m = 62.8m
l=63m
Therefore, the length of the required tarpaulin sheet will be 63 m.
Page : 221 , Block Name : Exercise 13.3
Q6 The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the
cost of white-washing its curved surface at the rate of ` 210 per 100 m2.
Answer. Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb
14
= 7m
2
CSA of conical tomb = πrl
22 2
= ( × 7 × 25) m
7
2
= 550m
Cost of white-washing 100 m area = R 210 2
210×550
Cost of white-washing 550 m area = =
Rs( )
2 100
=Rs 1155
Therefore ,it will cost of 1155 while white-washing such a conical tomb.
Page : 221 , Block Name : Exercise 13.3
Q7 A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find
the area of the sheet required to make 10 such caps.
Answer. Radius (r) of conical cap = 7 cm
Height (h) of conical cap 24 cm
Slant height (l) of the conical cap = √r + h 2 2
2 2
= [√(7) + (24) ] cm = (√625)cm = 25cm
CSA of 1 conical cap = πrl
22 2 2
= ( × 7 × 25) cm = 550cm
7
CSA of 10 such caps =(10x550)cm =5500 cm 2 2
Therefore , 5500 cm sheet will be required.
2
Page : 221 , Block Name : Exercise 13.3
Q8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of
Page 13 of 37 Aglasem Schools
Page 15
Book : Mathematics Ncert Solutions | Chapter-13 Maths
recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of
each of the cones is to be painted and the cost of painting is ` 12 per m2 , what will be the cost of
painting all these cones? (Use π = 3.14 and take 1.04 = 1.02)
Answer. Radius (r) Of cone = 40
= 20cm =0.2cm
2
Height (h) Of cone =1m
Slant height of cone = √h 2
+ r
2
2 2
= [√(1) + (0.2) ] m = (√1.04)m = 1.02m
CSA of each cone= πrl
2 2
= (3.14 × 0.2 × 1.02)m = 0.64056m
CSA of 50 such cones =(50x0.64056) ( \mathrm{m}^{2}\)
2
= 32.028m
Cost of painting 1 ( \mathrm{m}^{2}\)area Rs 12
Cost of painting 32.02S ( \mathrm{m}^{2}\) area Rs (32.028 x 12)
=Rs 384.336
= Rs 384.34 (approximately)
Therefore, it Will cost Rs 384.34 in painting 50 such hollow cones.
Page : 221 , Block Name : Exercise 13.3
Q1 Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Answer. Ardius (r) of sphere=10.5 cm
Surface area of sphere = 4πr 2
22 2 2
= [4 × × (10.5) ] cm
7
22 2
= (4 × × 10.5 × 10.5) cm
7
2
= (88 × 1.5 × 10.5)cm
2
= 1386cm
Therefore, the surface area Of a sphere having radius 10.5cm is 1386 cm 2
(ii) Radius(r) of sphere =5.6 cm
Surface area of sphere = 4πr 2
22 2 2
= [4 × × (5.6) ] cm
7
2
= (88 × 0.8 × 5.6)cm
2
= 394.24cm
Therefore, the surface area Of a sphere having radius 5.6cm is 394.24 cm . 2
Page 14 of 37 Aglasem Schools
recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of
each of the cones is to be painted and the cost of painting is ` 12 per m2 , what will be the cost of
painting all these cones? (Use π = 3.14 and take 1.04 = 1.02)
Answer. Radius (r) Of cone = 40
= 20cm =0.2cm
2
Height (h) Of cone =1m
Slant height of cone = √h 2
+ r
2
2 2
= [√(1) + (0.2) ] m = (√1.04)m = 1.02m
CSA of each cone= πrl
2 2
= (3.14 × 0.2 × 1.02)m = 0.64056m
CSA of 50 such cones =(50x0.64056) ( \mathrm{m}^{2}\)
2
= 32.028m
Cost of painting 1 ( \mathrm{m}^{2}\)area Rs 12
Cost of painting 32.02S ( \mathrm{m}^{2}\) area Rs (32.028 x 12)
=Rs 384.336
= Rs 384.34 (approximately)
Therefore, it Will cost Rs 384.34 in painting 50 such hollow cones.
Page : 221 , Block Name : Exercise 13.3
Q1 Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Answer. Ardius (r) of sphere=10.5 cm
Surface area of sphere = 4πr 2
22 2 2
= [4 × × (10.5) ] cm
7
22 2
= (4 × × 10.5 × 10.5) cm
7
2
= (88 × 1.5 × 10.5)cm
2
= 1386cm
Therefore, the surface area Of a sphere having radius 10.5cm is 1386 cm 2
(ii) Radius(r) of sphere =5.6 cm
Surface area of sphere = 4πr 2
22 2 2
= [4 × × (5.6) ] cm
7
2
= (88 × 0.8 × 5.6)cm
2
= 394.24cm
Therefore, the surface area Of a sphere having radius 5.6cm is 394.24 cm . 2
Page 14 of 37 Aglasem Schools
Page 16
Book : Mathematics Ncert Solutions | Chapter-13 Maths
(iii) Radius(r) of sphere =14 cm
Surface area of sphere = 4πr 2
22 2 2
= [4 × × (14) ] cm
7
2
= (4 × 44 × 14)cm
2
= 2464cm
Therefore, the surface area Of a sphere having radius 14 cm is 2464 cm . 2
Page : 225 , Block Name : Exercise 13.4
Q2 Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Answer. (i) radius (r) of sphere =
Diameter 14
= ( ) cm = 7cm
2 2
Surface area of sphere = 4πr 2
22 2 2
= (4 × × (7) ) cm
7
2
= (88 × 7)cm
2
= 616cm
Therefore, the surface area Of a sphere having diameter 14 cm is 616 \mathrm{cm}^{2}\).
(ii) radius (r) of sphere =
Diameter 21
= ( ) cm = 10.5cm
2 2
Surface area of sphere = 4πr 2
22 2 2
= (4 × × (10.5) ) cm
7
2
= (88 × 10.5)cm
= 1386cm v 2
Therefore, the surface area Of a sphere having diameter 21 cm is 1386 \mathrm{cm}^{2}\).
(iii) radius (r) of sphere =
Diameter 3.5
= ( ) cm = 1.75cm
2 2
Surface area of sphere = 4πr 2
22 2 2
= (4 × × (1.75) ) cm
7
2
= (88 × 1.75)cm
2
= 38.5cm
Therefore, the surface area Of a sphere having diameter 3.5 cm is 38.5 (\mathrm{cm}^{2}\).
Page : 225 , Block Name : Exercise 13.4
Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Answer.
Page 15 of 37 Aglasem Schools
(iii) Radius(r) of sphere =14 cm
Surface area of sphere = 4πr 2
22 2 2
= [4 × × (14) ] cm
7
2
= (4 × 44 × 14)cm
2
= 2464cm
Therefore, the surface area Of a sphere having radius 14 cm is 2464 cm . 2
Page : 225 , Block Name : Exercise 13.4
Q2 Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Answer. (i) radius (r) of sphere =
Diameter 14
= ( ) cm = 7cm
2 2
Surface area of sphere = 4πr 2
22 2 2
= (4 × × (7) ) cm
7
2
= (88 × 7)cm
2
= 616cm
Therefore, the surface area Of a sphere having diameter 14 cm is 616 \mathrm{cm}^{2}\).
(ii) radius (r) of sphere =
Diameter 21
= ( ) cm = 10.5cm
2 2
Surface area of sphere = 4πr 2
22 2 2
= (4 × × (10.5) ) cm
7
2
= (88 × 10.5)cm
= 1386cm v 2
Therefore, the surface area Of a sphere having diameter 21 cm is 1386 \mathrm{cm}^{2}\).
(iii) radius (r) of sphere =
Diameter 3.5
= ( ) cm = 1.75cm
2 2
Surface area of sphere = 4πr 2
22 2 2
= (4 × × (1.75) ) cm
7
2
= (88 × 1.75)cm
2
= 38.5cm
Therefore, the surface area Of a sphere having diameter 3.5 cm is 38.5 (\mathrm{cm}^{2}\).
Page : 225 , Block Name : Exercise 13.4
Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Answer.
Page 15 of 37 Aglasem Schools
Page 17
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Total surface area Of hemisphere =CSA Of hemisphere + Area Of circular end Of
hemisphere
2 2
= 2πr + πr
2
= 3πr
2 2
= [3 × 3.14 × (10) ] cm
2
= 942cm
Therefore, the total surface area of such a hemisphere is 942 ( \mathrm{cm}^{2}\)
Page : 225 , Block Name : Exercise 13.4
Q4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it.
Find the ratio of surface areas of the balloon in the two cases.
Answer. Radius (rt) of spherical balloon = 7 cm
Radius (r2) of spherical balloon, when air is pumped into it = 14 cm
Initial surface area
=
Surface area after pumping air into balloon
2 2
4πr r1
1
= = ( )
2 r2
4πr
2
2
7 1
= ( ) =
14 4
Therefore, the ratio between the surface areas in these two cases is 1:4.
Page : 225 , Block Name : Exercise 13.4
Q5 A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it
on the inside at the rate of ` 16 per 100 cm2 .
Answer. Inner radius (r) of hemispherical bowl = (
10.5
) cm = 5.25cm
2
Surface area Of hemispherical bowl = 2πr 2
22 2 2
= [2 × × (5.25) ] cm
7
2
= 173.25cm
Cost of tin-plating 100 (\mathrm{cm}^{2}\) area = Rs 16
Cost of tin-plating 173.25 (\mathrm{cm}^{2}\) area = Rs ( 16×173.25
100
)
=Rs27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs
27.72.
Page : 225 , Block Name : Exercise 13.4
Q6 Find the radius of a sphere whose surface area is 154 cm2 .
Page 16 of 37 Aglasem Schools
Total surface area Of hemisphere =CSA Of hemisphere + Area Of circular end Of
hemisphere
2 2
= 2πr + πr
2
= 3πr
2 2
= [3 × 3.14 × (10) ] cm
2
= 942cm
Therefore, the total surface area of such a hemisphere is 942 ( \mathrm{cm}^{2}\)
Page : 225 , Block Name : Exercise 13.4
Q4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it.
Find the ratio of surface areas of the balloon in the two cases.
Answer. Radius (rt) of spherical balloon = 7 cm
Radius (r2) of spherical balloon, when air is pumped into it = 14 cm
Initial surface area
=
Surface area after pumping air into balloon
2 2
4πr r1
1
= = ( )
2 r2
4πr
2
2
7 1
= ( ) =
14 4
Therefore, the ratio between the surface areas in these two cases is 1:4.
Page : 225 , Block Name : Exercise 13.4
Q5 A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it
on the inside at the rate of ` 16 per 100 cm2 .
Answer. Inner radius (r) of hemispherical bowl = (
10.5
) cm = 5.25cm
2
Surface area Of hemispherical bowl = 2πr 2
22 2 2
= [2 × × (5.25) ] cm
7
2
= 173.25cm
Cost of tin-plating 100 (\mathrm{cm}^{2}\) area = Rs 16
Cost of tin-plating 173.25 (\mathrm{cm}^{2}\) area = Rs ( 16×173.25
100
)
=Rs27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs
27.72.
Page : 225 , Block Name : Exercise 13.4
Q6 Find the radius of a sphere whose surface area is 154 cm2 .
Page 16 of 37 Aglasem Schools
Page 18
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer. Let the radius of the sphere be r.
Surface area of sphere = 154
2 2
4πr = 154cm
2 154×7 2 7×7 2
r = ( ) cm = ( ) cm
4×22 2×2
7
r = ( ) cm = 3.5cm
2
Therefore, the radius of the sphere whose surface area is 154cm is 3.5 cm. 2
Page : 225 , Block Name : Exercise 13.4
Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the
ratio of their surface areas.
Answer. Let the diameter of earth be d. Therefore, the diameter of mcn)n will be
d
4
Radius of earth= d
2
Radius of moon= 1 d d
× =
2 4 8
2
Surface area of moon= 4π(
d
)
8
2
d
4π( )
Surface area of earth==
8
2
d
4π( )
2
Required ratio=
4 1
=
64 16
Therefore, the ratio between their surface areas will be 1:16.
Page : 225 , Block Name : Exercise 13.4
Q8 A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find
the outer curved surface area of the bowl.
Answer. Inner radius Of hemispherical bowl =5cm
Thickness of the bowl = 0.25 cm
Outer radius (r) Of hemispherical bowl = (5 + 0.25) cm
=5.25 cm
Outer CSA of hemispherical bowl = 2πr 2
22 2 2
= 2 × × (5.25cm) = 173.25cm
7
Therefore, the outer curved surface area Of the bowl is 173.25 ( \mathrm{cm}^{2}\)
Page : 225 , Block Name : Exercise 13.4
Q9 A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Page 17 of 37 Aglasem Schools
Answer. Let the radius of the sphere be r.
Surface area of sphere = 154
2 2
4πr = 154cm
2 154×7 2 7×7 2
r = ( ) cm = ( ) cm
4×22 2×2
7
r = ( ) cm = 3.5cm
2
Therefore, the radius of the sphere whose surface area is 154cm is 3.5 cm. 2
Page : 225 , Block Name : Exercise 13.4
Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the
ratio of their surface areas.
Answer. Let the diameter of earth be d. Therefore, the diameter of mcn)n will be
d
4
Radius of earth= d
2
Radius of moon= 1 d d
× =
2 4 8
2
Surface area of moon= 4π(
d
)
8
2
d
4π( )
Surface area of earth==
8
2
d
4π( )
2
Required ratio=
4 1
=
64 16
Therefore, the ratio between their surface areas will be 1:16.
Page : 225 , Block Name : Exercise 13.4
Q8 A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find
the outer curved surface area of the bowl.
Answer. Inner radius Of hemispherical bowl =5cm
Thickness of the bowl = 0.25 cm
Outer radius (r) Of hemispherical bowl = (5 + 0.25) cm
=5.25 cm
Outer CSA of hemispherical bowl = 2πr 2
22 2 2
= 2 × × (5.25cm) = 173.25cm
7
Therefore, the outer curved surface area Of the bowl is 173.25 ( \mathrm{cm}^{2}\)
Page : 225 , Block Name : Exercise 13.4
Q9 A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Page 17 of 37 Aglasem Schools
Page 19
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer.
Answer. (i) Surface area of sphere \(4 \mathrm{pi\r^{2}\)
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder = 2nrh
=2nr (2r)
=4 \mathrm{nr}^{2}
(iii) =
Surface area of sphere
CSA of cylinder
2
4πr
= 2
4πr
1
=
1
Therefore, the ratio between these two surface areas is 1:1.
Page : 225 , Block Name : Exercise 13.4
Q1 A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing
12 such boxes?
Answer. Matchbox is a cuboid having its length (j), breadth (b), height (h) as 4 cm, 2.5cm,and 1.5
cm.
Volume of 1 match box = / x b x h
= (4 x 2.5 x 1.5) cm = 15 cm 3 3
=Volume of 1 match box =(15x12) cm 3
=180 cm 3
Therefore, the volume Of 12 match boxes is 180 cm 3
Page : 228 , Block Name : Exercise 13.5
Q2 A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it
hold? (1 m3 = 1000 l)
Answer. The given cuboidal water tank has its length (l) as 6 m, breadth (b) as 5 m, and
height (h) as 4.5 m.
Volume of tank = / x b x h
= (6 x 5 x 4.5) m = 135 m v
3 3
Page 18 of 37 Aglasem Schools
Answer.
Answer. (i) Surface area of sphere \(4 \mathrm{pi\r^{2}\)
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder = 2nrh
=2nr (2r)
=4 \mathrm{nr}^{2}
(iii) =
Surface area of sphere
CSA of cylinder
2
4πr
= 2
4πr
1
=
1
Therefore, the ratio between these two surface areas is 1:1.
Page : 225 , Block Name : Exercise 13.4
Q1 A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing
12 such boxes?
Answer. Matchbox is a cuboid having its length (j), breadth (b), height (h) as 4 cm, 2.5cm,and 1.5
cm.
Volume of 1 match box = / x b x h
= (4 x 2.5 x 1.5) cm = 15 cm 3 3
=Volume of 1 match box =(15x12) cm 3
=180 cm 3
Therefore, the volume Of 12 match boxes is 180 cm 3
Page : 228 , Block Name : Exercise 13.5
Q2 A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it
hold? (1 m3 = 1000 l)
Answer. The given cuboidal water tank has its length (l) as 6 m, breadth (b) as 5 m, and
height (h) as 4.5 m.
Volume of tank = / x b x h
= (6 x 5 x 4.5) m = 135 m v
3 3
Page 18 of 37 Aglasem Schools
Page 20
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Amount of water that 1 m volume can hold = 1000 litres
3
Amount Of water that 135 m volume can hold = (135 x 1000) litres
3
=135000 litres
Therefore, such tank can hold up to 135000 litres of Water.
Page : 228 , Block Name : Exercise 13.5
Q3 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic
metres of a liquid?
Answer. Length (l) of vessel = 10 m
Width (b) Of vessel = 8 m
Volume of vessel =380 m 3
L x b x h = 380
[(10) (8)h] m = 380 m v
2 3
h=4.75 mv
Therefore, the height Of the vessel should be 4.75 m.
Page : 228 , Block Name : Exercise 13.5
Q4 Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ` 30 per
m3 .
Answer. The given cuboidal pit has its length (l) as 8 m, width (b) as 6 m, and depth (h)as 3 m.
Volume Of pit =(8 x 6 x 3) m = 144 m
3 3
Cost of digging per m volume = Rs 30
3
Cost of digging 144 m volume = Rs (144 x 30) = Rs 4320
3
Page : 228 , Block Name : Exercise 13.5
5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length
and depth are respectively 2.5 m and 10 m.
Answer. Let the breadth Of the tank be b m.
Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.
Volume of tank / x b x h
= (2.5 x bx 10) m3
= 25b m 3
Capacity of tank = 25b m = 25000 b litres
3
25000 b = 50000
Therefore, the breadth of the tank is 2 m.
Page : 228 , Block Name : Exercise 13.5
Page 19 of 37 Aglasem Schools
Amount of water that 1 m volume can hold = 1000 litres
3
Amount Of water that 135 m volume can hold = (135 x 1000) litres
3
=135000 litres
Therefore, such tank can hold up to 135000 litres of Water.
Page : 228 , Block Name : Exercise 13.5
Q3 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic
metres of a liquid?
Answer. Length (l) of vessel = 10 m
Width (b) Of vessel = 8 m
Volume of vessel =380 m 3
L x b x h = 380
[(10) (8)h] m = 380 m v
2 3
h=4.75 mv
Therefore, the height Of the vessel should be 4.75 m.
Page : 228 , Block Name : Exercise 13.5
Q4 Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ` 30 per
m3 .
Answer. The given cuboidal pit has its length (l) as 8 m, width (b) as 6 m, and depth (h)as 3 m.
Volume Of pit =(8 x 6 x 3) m = 144 m
3 3
Cost of digging per m volume = Rs 30
3
Cost of digging 144 m volume = Rs (144 x 30) = Rs 4320
3
Page : 228 , Block Name : Exercise 13.5
5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length
and depth are respectively 2.5 m and 10 m.
Answer. Let the breadth Of the tank be b m.
Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.
Volume of tank / x b x h
= (2.5 x bx 10) m3
= 25b m 3
Capacity of tank = 25b m = 25000 b litres
3
25000 b = 50000
Therefore, the breadth of the tank is 2 m.
Page : 228 , Block Name : Exercise 13.5
Page 19 of 37 Aglasem Schools
Page 21
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Q6 A village, having a population of 4000, requires 150 litres of water per head per day. It has a
tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Answer. The given tank is cuboidal in shape having its length (l) as 20 m, breadth (b) as 15 m, and
height (h) as 6 m.
Capacity of tank I x bx h
= (20 x 15 x 6) = 1800 m = 1800000 litres
3
Water consumed by the people Of the village in 1 day = (4000 x 150) litres
= 600000 litres
Let water in this tank last for n days.
Water consumed by all people of village in n days = Capacity of tank
n x 600000 = 1800000
n=3
Therefore, the water of this tank will last for 3 days.
Page : 228 , Block Name : Exercise 13.5
Q7 A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each
measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Answer. The godown has its length\left(I_{1}\right) as 40 m, breadth as 25 m, height
\left(h_{1}\right) as 10 m,
while the wooden crate has its length (\left(I_{2}\right) as 1.5 m, breadth (\left(b_{2}\right)) as 1.25
m, and height (\left(h_{2}\right)) as 0.5 m.
Therefore, volume of godown =I × b × h1 1 1
= (40 x 25 x 10) m3
= 10000 m 3
Volume of 1 wooden crate = I × b × h
2 2 2
= (1.5 x 1.25 x 0.5) m 3
= 0.9375 m 3
Let n wooden crates can be stored in the godown.
Therefore, volume of n vvcn)den crates Volume of godown
0.9375 x n = 10000
= 10666.66
Therefore, 10666 Wooden crates can be stored in the godown.
Page : 228 , Block Name : Exercise 13.5
Q8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the
new cube? Also, nd the ratio between their surface areas.
Answer. Side (a) of cube = 12 cm
Volume of cube (a) =(12cm) 1728 cm\)=1728cm
3 3 3
Let the side of the smaller cube be aa . 1
Page 20 of 37 Aglasem Schools
Q6 A village, having a population of 4000, requires 150 litres of water per head per day. It has a
tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Answer. The given tank is cuboidal in shape having its length (l) as 20 m, breadth (b) as 15 m, and
height (h) as 6 m.
Capacity of tank I x bx h
= (20 x 15 x 6) = 1800 m = 1800000 litres
3
Water consumed by the people Of the village in 1 day = (4000 x 150) litres
= 600000 litres
Let water in this tank last for n days.
Water consumed by all people of village in n days = Capacity of tank
n x 600000 = 1800000
n=3
Therefore, the water of this tank will last for 3 days.
Page : 228 , Block Name : Exercise 13.5
Q7 A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each
measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Answer. The godown has its length\left(I_{1}\right) as 40 m, breadth as 25 m, height
\left(h_{1}\right) as 10 m,
while the wooden crate has its length (\left(I_{2}\right) as 1.5 m, breadth (\left(b_{2}\right)) as 1.25
m, and height (\left(h_{2}\right)) as 0.5 m.
Therefore, volume of godown =I × b × h1 1 1
= (40 x 25 x 10) m3
= 10000 m 3
Volume of 1 wooden crate = I × b × h
2 2 2
= (1.5 x 1.25 x 0.5) m 3
= 0.9375 m 3
Let n wooden crates can be stored in the godown.
Therefore, volume of n vvcn)den crates Volume of godown
0.9375 x n = 10000
= 10666.66
Therefore, 10666 Wooden crates can be stored in the godown.
Page : 228 , Block Name : Exercise 13.5
Q8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the
new cube? Also, nd the ratio between their surface areas.
Answer. Side (a) of cube = 12 cm
Volume of cube (a) =(12cm) 1728 cm\)=1728cm
3 3 3
Let the side of the smaller cube be aa . 1
Page 20 of 37 Aglasem Schools
Page 22
Book : Mathematics Ncert Solutions | Chapter-13 Maths
volume of 1 smaller cube=(
1728 3 3
) cm = 216cm
8
3 3
(a1 ) = 216cm
a1 = 6cm
Therefore, the side of the smaller cubes will be 6 cm.
Ratio between surface areas of cube=
Surface area of bigger cube
Surface area of smaller cube
2
2 (12)
6a
= =
2 2
6a (6)
1
4
=
1
Therefore, the ratio between the surface areas of these cubes is 4:1.
Page : 228 , Block Name : Exercise 13.5
Q9 A river 3 m deep and 40 m wide is owing at the rate of 2 km per hour. How much water will fall
into the sea in a minute?
Answer. Rate of water ow = 2 km per hour
2000
= ( ) m/min
60
100
= ( ) m/min
3
Depth (h) of river = 3 m
Width (b) Of =40 m
Volume of water owed in 1 min =( 100
3
× 40 × 3) m
3
= 4000m
3
Therefore, in 1 minute, 4000 m water will fall in the sea.
3
Page : 228 , Block Name : Exercise 13.5
Q1 The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How
many litres of water can it hold? (1000 cm3 = 1l)
Answer. Let the radius Of the cylindrical vessel be r.
Height (h) of vessel = 25 cm
Circumference of vessel = 132 cm
2πr= 132 cm
132×7
r = ( ) cm = 21cm
2×22
Volume of cylindrical vessel=pir h
2
22 2 3
= [ × (21) × 25] cm
7
=34650cm 3
liter
34650
= ( )
1000
=34.65 liter
Therefore, such vessel can hold 34.65 litres Of water.
Page 21 of 37 Aglasem Schools
volume of 1 smaller cube=(
1728 3 3
) cm = 216cm
8
3 3
(a1 ) = 216cm
a1 = 6cm
Therefore, the side of the smaller cubes will be 6 cm.
Ratio between surface areas of cube=
Surface area of bigger cube
Surface area of smaller cube
2
2 (12)
6a
= =
2 2
6a (6)
1
4
=
1
Therefore, the ratio between the surface areas of these cubes is 4:1.
Page : 228 , Block Name : Exercise 13.5
Q9 A river 3 m deep and 40 m wide is owing at the rate of 2 km per hour. How much water will fall
into the sea in a minute?
Answer. Rate of water ow = 2 km per hour
2000
= ( ) m/min
60
100
= ( ) m/min
3
Depth (h) of river = 3 m
Width (b) Of =40 m
Volume of water owed in 1 min =( 100
3
× 40 × 3) m
3
= 4000m
3
Therefore, in 1 minute, 4000 m water will fall in the sea.
3
Page : 228 , Block Name : Exercise 13.5
Q1 The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How
many litres of water can it hold? (1000 cm3 = 1l)
Answer. Let the radius Of the cylindrical vessel be r.
Height (h) of vessel = 25 cm
Circumference of vessel = 132 cm
2πr= 132 cm
132×7
r = ( ) cm = 21cm
2×22
Volume of cylindrical vessel=pir h
2
22 2 3
= [ × (21) × 25] cm
7
=34650cm 3
liter
34650
= ( )
1000
=34.65 liter
Therefore, such vessel can hold 34.65 litres Of water.
Page 21 of 37 Aglasem Schools
Page 23
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Page : 230 , Block Name : Exercise 13.6
Q2 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The
length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Answer. Inner radius of cylindrical pipe =(
28
) cm = 14cm
2
Outer radius of cylindrical pipe =(
28
) cm = 14cm
2
Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe =π (r − r ) h 2
2
2
1
22 2 2 3
= [ × (14 − 12 ) × 35] cm
7
=110x52 cm 3
=5720 cm 3
Mass of 1 cm wood= 0.6 g
3
Mass of 5720 cm wood = (5720 x0.6) g
3
=3432 g
=3.432 kg
Page : 230 , Block Name : Exercise 13.6
Q3 A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and
width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm
and height 10 cm. Which container has greater capacity and by how much?
Answer. The tin can will be cuboidal in shape while the plastic cylinder Hill be cylindrical in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) Of tin can = 15 cm
Capacity of tin can =l x b x h
Page 22 of 37 Aglasem Schools
Page : 230 , Block Name : Exercise 13.6
Q2 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The
length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Answer. Inner radius of cylindrical pipe =(
28
) cm = 14cm
2
Outer radius of cylindrical pipe =(
28
) cm = 14cm
2
Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe =π (r − r ) h 2
2
2
1
22 2 2 3
= [ × (14 − 12 ) × 35] cm
7
=110x52 cm 3
=5720 cm 3
Mass of 1 cm wood= 0.6 g
3
Mass of 5720 cm wood = (5720 x0.6) g
3
=3432 g
=3.432 kg
Page : 230 , Block Name : Exercise 13.6
Q3 A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and
width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm
and height 10 cm. Which container has greater capacity and by how much?
Answer. The tin can will be cuboidal in shape while the plastic cylinder Hill be cylindrical in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) Of tin can = 15 cm
Capacity of tin can =l x b x h
Page 22 of 37 Aglasem Schools
Page 24
Book : Mathematics Ncert Solutions | Chapter-13 Maths
(5 x 4 x 15) cm 3
=300 cm 3
Radius (r) of circular end of plastic cylinder = (
7
) cm = 3.5cm
2
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder=πr H 2
22 2 3
[ × (3.5) × 10] cm
7
=(11x35) cm 3
Therefore, plastic cylinder has the greater capacity.
Difference in capacity = (385 — 300) cm 3
= 85 cm 3
Page : 230 , Block Name : Exercise 13.6
Q4 If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then nd (i) radius of its
base (ii) its volume. (Use π = 3.14)
Answer. (i) Height (h) of cylinder =5 cm
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm 2
2πrh= 94.2 cm
2
(2 x 3.14 x r x 5) cm =94.2 cm 2
r = 3 cm
(ii) Volume Of cylinder =πr H 2
== (3.14 × (3) × 5) cm 2 3
= 141.3 cm 3
Page : 230 , Block Name : Exercise 13.6
Q5 It costs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of
painting is at the rate of ` 20 per m2 , nd
(i) inner curved surface area of the vessel,
Page 23 of 37 Aglasem Schools
(5 x 4 x 15) cm 3
=300 cm 3
Radius (r) of circular end of plastic cylinder = (
7
) cm = 3.5cm
2
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder=πr H 2
22 2 3
[ × (3.5) × 10] cm
7
=(11x35) cm 3
Therefore, plastic cylinder has the greater capacity.
Difference in capacity = (385 — 300) cm 3
= 85 cm 3
Page : 230 , Block Name : Exercise 13.6
Q4 If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then nd (i) radius of its
base (ii) its volume. (Use π = 3.14)
Answer. (i) Height (h) of cylinder =5 cm
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm 2
2πrh= 94.2 cm
2
(2 x 3.14 x r x 5) cm =94.2 cm 2
r = 3 cm
(ii) Volume Of cylinder =πr H 2
== (3.14 × (3) × 5) cm 2 3
= 141.3 cm 3
Page : 230 , Block Name : Exercise 13.6
Q5 It costs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of
painting is at the rate of ` 20 per m2 , nd
(i) inner curved surface area of the vessel,
Page 23 of 37 Aglasem Schools
Page 25
Book : Mathematics Ncert Solutions | Chapter-13 Maths
(ii) radius of the base,
(iii) capacity of the vessel.
Answer. (i) Rs 20 is the cost Of painting 1m : area. 2
RS 2200 is the cost Of painting =(
1 2
× 2200) m
20
= 110m area 2
Therefore, the inner surface area of the vessel is 110m 2
(ii) Let the radius of the base of the vessel be r.
Height (h) Of vessel = 10 m
Surface area 2πrh = 110 m 2
22 2
Rightarrow (2 × × r × 10) m = 110m
7
7
⇒ r = ( ) m = 1.75m
4
(iii) volume of vessel =πr H 2
22 2 3
= [ × (1.75) × 10] m
7
3
= 96.25m
Therefore, the capacity of the vessel is 96.25 m 3
or 96250 litres
Page : 231 , Block Name : Exercise 13.6
Q6 The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres
of metal sheet would be needed to make it?
Answer. Let the radius of the circular end be r.
Height (h) Of cylindrical vessel =1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m 3
2 3
πr h = 0.0154m
r=0.07m
Total surface area of vessel = 2πr(r + h)
22 2
= [2 × × 0.07(0.07 + 1)] m
7
2
= 0.44 × 1.07m
2
= 0.4708m
Therefore, 0.4708 m of the metal sheet would be required to make the cylindrical vessel.
2
Page : 231 , Block Name : Exercise 13.6
Q7 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite lled in the
interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length
of the pencil is 14 cm, nd the volume of the wood and that of the graphite.
Page 24 of 37 Aglasem Schools
(ii) radius of the base,
(iii) capacity of the vessel.
Answer. (i) Rs 20 is the cost Of painting 1m : area. 2
RS 2200 is the cost Of painting =(
1 2
× 2200) m
20
= 110m area 2
Therefore, the inner surface area of the vessel is 110m 2
(ii) Let the radius of the base of the vessel be r.
Height (h) Of vessel = 10 m
Surface area 2πrh = 110 m 2
22 2
Rightarrow (2 × × r × 10) m = 110m
7
7
⇒ r = ( ) m = 1.75m
4
(iii) volume of vessel =πr H 2
22 2 3
= [ × (1.75) × 10] m
7
3
= 96.25m
Therefore, the capacity of the vessel is 96.25 m 3
or 96250 litres
Page : 231 , Block Name : Exercise 13.6
Q6 The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres
of metal sheet would be needed to make it?
Answer. Let the radius of the circular end be r.
Height (h) Of cylindrical vessel =1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m 3
2 3
πr h = 0.0154m
r=0.07m
Total surface area of vessel = 2πr(r + h)
22 2
= [2 × × 0.07(0.07 + 1)] m
7
2
= 0.44 × 1.07m
2
= 0.4708m
Therefore, 0.4708 m of the metal sheet would be required to make the cylindrical vessel.
2
Page : 231 , Block Name : Exercise 13.6
Q7 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite lled in the
interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length
of the pencil is 14 cm, nd the volume of the wood and that of the graphite.
Page 24 of 37 Aglasem Schools
Page 26
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer.
Radius (r) of pencil =( =0.35cm
7 0.7
) mm = ( ) cm
2 2
Radius of graphite = ( =0.05
1 0.1
) mm = ( ) cm
2 2
Height (h) of pencil = 14cm
Volume of wood in pencil =π (r 2
1
2
− r )h
2
v
22 2 2 3
= [ {(0.35) − (0.05) × 14}] cm
7
22 3
= [ (0.1225 − 0.0025) × 14] cm
7
3
= (44 × 0.12)cm
3
= 5.28cm
2 22 2 3
= πr h = [ × (0.05) × 14] cm
2 7
3
= (44 × 0.0025)cm
3
= 0.11cm
Page : 231 , Block Name : Exercise 13.6
Q8 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is
lled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250
patients?
Answer.
Page 25 of 37 Aglasem Schools
Answer.
Radius (r) of pencil =( =0.35cm
7 0.7
) mm = ( ) cm
2 2
Radius of graphite = ( =0.05
1 0.1
) mm = ( ) cm
2 2
Height (h) of pencil = 14cm
Volume of wood in pencil =π (r 2
1
2
− r )h
2
v
22 2 2 3
= [ {(0.35) − (0.05) × 14}] cm
7
22 3
= [ (0.1225 − 0.0025) × 14] cm
7
3
= (44 × 0.12)cm
3
= 5.28cm
2 22 2 3
= πr h = [ × (0.05) × 14] cm
2 7
3
= (44 × 0.0025)cm
3
= 0.11cm
Page : 231 , Block Name : Exercise 13.6
Q8 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is
lled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250
patients?
Answer.
Page 25 of 37 Aglasem Schools
Page 27
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Radius (r) of cylindrical bowl =(
7
) cm = 3.5cm
2
Height (h) of bowl, up to which bowl is lled with soup = 4 cm
Volume Of soup in I bowl = πr H 2
22 2 3
= ( × (3.5) × 4) cm
7
3
= (11 × 3.5 × 4)cm
3
= 154cm
Volume of soup given to 250 patients =(250x154)cm 3
3
= 38500cm
= 38.5
Page : 231 , Block Name : Exercise 13.6
Q1 Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm .
Answer. (i) Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone== πr h
1 2
3
1 22 2 3
= [ × × (6) × 7] cm
3 7
3
= (12 × 22)cm
3
= 264cm
Therefore, the volume of the cone is 264 cm 3
Page : 233 , Block Name : Exercise 13.7
Q2 Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm .
Answer. (i) Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height of cone = √l − r 2 2
2 2
= (√25 − 7 ) cm
3
= (154 × 8)cm
3
= 1232cm
Therefore, capacity of the conical vessel
1232 3
= ( ) litres (1 litre = 1000cm )
1000
=1.232 litres
(ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Page 26 of 37 Aglasem Schools
Radius (r) of cylindrical bowl =(
7
) cm = 3.5cm
2
Height (h) of bowl, up to which bowl is lled with soup = 4 cm
Volume Of soup in I bowl = πr H 2
22 2 3
= ( × (3.5) × 4) cm
7
3
= (11 × 3.5 × 4)cm
3
= 154cm
Volume of soup given to 250 patients =(250x154)cm 3
3
= 38500cm
= 38.5
Page : 231 , Block Name : Exercise 13.6
Q1 Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm .
Answer. (i) Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone== πr h
1 2
3
1 22 2 3
= [ × × (6) × 7] cm
3 7
3
= (12 × 22)cm
3
= 264cm
Therefore, the volume of the cone is 264 cm 3
Page : 233 , Block Name : Exercise 13.7
Q2 Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm .
Answer. (i) Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height of cone = √l − r 2 2
2 2
= (√25 − 7 ) cm
3
= (154 × 8)cm
3
= 1232cm
Therefore, capacity of the conical vessel
1232 3
= ( ) litres (1 litre = 1000cm )
1000
=1.232 litres
(ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Page 26 of 37 Aglasem Schools
Page 28
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Radius (r) of cone= √l 2
− h
2
2 2
= (√13 − 12 ) cm
=5
Volume of cone==
1 2
πr h
3
1 22 2 3
= [ × × (5) × 12] cm
3 7
22 3
= (4 × × 25) cm
7
2200 3
= ( ) cm
7
Therefore, capacity of the conical vessel
2200 3
( ) litres (1 litre = 1000cm )
7000
11
=
35 litres
Page : 233 , Block Name : Exercise 13.7
Q3 The height of a cone is 15 cm. If its volume is 1570 cm3 , nd the radius of the base. (Use π =
3.14)
Answer. Height (h) of cone = 15 cm
Let the radius of the cone be r.
Volume of cone = 1570 cm 3
1 2 3
πr h = 1570cm
3
1 2 3
Rightarrow ( × 3.14 × r × 15) cm = 1570cm
3
2 2
⇒ r = 100cm
r = 10 cm
Therefore, the radius of the base of cone is 10 cm.
Q4 If the volume of a right circular cone of height 9 cm is 48 π cm3 , nd the diameter of its base.
Answer. Height (h) of cone = 9 cm
Let the radius of the cone be r.
Volume of cone =48πcm 3
1 2 3
⇒ πr h = 48πcm
3
1 2 3
Rightarrow ( πr × 9) cm = 48πcm
3
2 2
⇒ r = 16cm
r =4 cm
Diameter of base = 2r = 8 cm
Page : 233 , Block Name : Exercise 13.7
Q5 A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Page 27 of 37 Aglasem Schools
Radius (r) of cone= √l 2
− h
2
2 2
= (√13 − 12 ) cm
=5
Volume of cone==
1 2
πr h
3
1 22 2 3
= [ × × (5) × 12] cm
3 7
22 3
= (4 × × 25) cm
7
2200 3
= ( ) cm
7
Therefore, capacity of the conical vessel
2200 3
( ) litres (1 litre = 1000cm )
7000
11
=
35 litres
Page : 233 , Block Name : Exercise 13.7
Q3 The height of a cone is 15 cm. If its volume is 1570 cm3 , nd the radius of the base. (Use π =
3.14)
Answer. Height (h) of cone = 15 cm
Let the radius of the cone be r.
Volume of cone = 1570 cm 3
1 2 3
πr h = 1570cm
3
1 2 3
Rightarrow ( × 3.14 × r × 15) cm = 1570cm
3
2 2
⇒ r = 100cm
r = 10 cm
Therefore, the radius of the base of cone is 10 cm.
Q4 If the volume of a right circular cone of height 9 cm is 48 π cm3 , nd the diameter of its base.
Answer. Height (h) of cone = 9 cm
Let the radius of the cone be r.
Volume of cone =48πcm 3
1 2 3
⇒ πr h = 48πcm
3
1 2 3
Rightarrow ( πr × 9) cm = 48πcm
3
2 2
⇒ r = 16cm
r =4 cm
Diameter of base = 2r = 8 cm
Page : 233 , Block Name : Exercise 13.7
Q5 A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Page 27 of 37 Aglasem Schools
Page 29
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer. Radius (r) of pit = (
3.5
) m = 1.75m
2
Height (h) of pit = depth of pit = 12
Volume of pit = πr h
1 2
3
1 22 2 3
= [ × × (1.75) × 12] cm
3 7
3
= 38.5m
Thus, capacity of the pit = (38.5 x 1) kilolitres =38.5 kilolitres
Page : 233 , Block Name : Exercise 13.7
Q6 The volume of a right circular cone is 9856 cm3 . If the diameter of the base is 28 cm, nd
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Answer. (i) radius of cone =(
28
) cm = 14cm
2
Let the height of the cone be h.
Volume of cone = 9856 cm 3
1 2 3
⇒ πr h = 9856cm
3
1 22 2 2 3
Rightarrow [ × × (14) × h] cm = 9856cm
3 7
h=48 cm
Therefore, the height of the cone is 48 cm.
Page : 233 , Block Name : Exercise 13.7
Q7 A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find
the volume of the solid so obtained.
Answer.
When right-angled △ABC is revolved about its side 12 cm, a cone With height (h) as 12 cm,
radius (r) as 5 cm, and slant height (l) 13 cm will be formed.
Page 28 of 37 Aglasem Schools
Answer. Radius (r) of pit = (
3.5
) m = 1.75m
2
Height (h) of pit = depth of pit = 12
Volume of pit = πr h
1 2
3
1 22 2 3
= [ × × (1.75) × 12] cm
3 7
3
= 38.5m
Thus, capacity of the pit = (38.5 x 1) kilolitres =38.5 kilolitres
Page : 233 , Block Name : Exercise 13.7
Q6 The volume of a right circular cone is 9856 cm3 . If the diameter of the base is 28 cm, nd
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Answer. (i) radius of cone =(
28
) cm = 14cm
2
Let the height of the cone be h.
Volume of cone = 9856 cm 3
1 2 3
⇒ πr h = 9856cm
3
1 22 2 2 3
Rightarrow [ × × (14) × h] cm = 9856cm
3 7
h=48 cm
Therefore, the height of the cone is 48 cm.
Page : 233 , Block Name : Exercise 13.7
Q7 A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find
the volume of the solid so obtained.
Answer.
When right-angled △ABC is revolved about its side 12 cm, a cone With height (h) as 12 cm,
radius (r) as 5 cm, and slant height (l) 13 cm will be formed.
Page 28 of 37 Aglasem Schools
Page 30
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Volume of cone =
1 2
πr h
3
1 2 3
= [ × π × (5) × 12] cm
3
== 100ncm 3
=Therefore, the volume of the cone so formed is loon ( \mathrm{cm}^{3}
Page : 233 , Block Name : Exercise 13.7
Q8 If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then nd the
volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in
Questions 7 and 8.
Answer.
When right-angled △ABC is revolved about its side 5 cm, a cone will be formed having radius (r)
as 12 cm, height (h) as S cm, and slant height (l) as 13 cm.
Volume of cone = πr h 1
3
2
1 2 3
= [ × π × (12) × 5] cm
3
3
= 240πcm
Therefore, the volume of the cone so formed is 240π ( \mathrm{cm}^{3}\)
5
= = 5 : 12
12
Page : 233 , Block Name : Exercise 13.7
Q9 A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its
volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas
required.
Answer. Radius (r) of heap == (
10.5
) m = 5.25m
2
Height (h) of heap = 3m
Volume of heap = πr h
1 2
3
1 22 2 3
= ( × × (5.25) × 3) m
3 7
3
= 86.625m
Therefore, the volume of the heap of wheat is 86.625 m . 3
Area of canvas required = CSA of cone
Page 29 of 37 Aglasem Schools
Volume of cone =
1 2
πr h
3
1 2 3
= [ × π × (5) × 12] cm
3
== 100ncm 3
=Therefore, the volume of the cone so formed is loon ( \mathrm{cm}^{3}
Page : 233 , Block Name : Exercise 13.7
Q8 If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then nd the
volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in
Questions 7 and 8.
Answer.
When right-angled △ABC is revolved about its side 5 cm, a cone will be formed having radius (r)
as 12 cm, height (h) as S cm, and slant height (l) as 13 cm.
Volume of cone = πr h 1
3
2
1 2 3
= [ × π × (12) × 5] cm
3
3
= 240πcm
Therefore, the volume of the cone so formed is 240π ( \mathrm{cm}^{3}\)
5
= = 5 : 12
12
Page : 233 , Block Name : Exercise 13.7
Q9 A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its
volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas
required.
Answer. Radius (r) of heap == (
10.5
) m = 5.25m
2
Height (h) of heap = 3m
Volume of heap = πr h
1 2
3
1 22 2 3
= ( × × (5.25) × 3) m
3 7
3
= 86.625m
Therefore, the volume of the heap of wheat is 86.625 m . 3
Area of canvas required = CSA of cone
Page 29 of 37 Aglasem Schools
Page 31
Book : Mathematics Ncert Solutions | Chapter-13 Maths
2 2
= πrl = πr√r + h
22 2 2 2
= [ × 5.25 × √(5.25) + (3) ] m
7
22 2
= ( × 5.25 × 6.05) m
7
2
= 99.825m
Therefore, 99.825 m canvas will be required to protect the heap from rain.
2
Page : 233 , Block Name : Exercise 13.7
Q1 Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Answer. (i) Radius of sphere = 7 cm
Volume of sphere = πr
4 3
3
4 22 3 3
= [ × × (7) ] cm
3 7
4312 3
= ( ) cm
3
1 3
= 1437 cm
3
Therefore , the volume of the sphere is = 1437
1 3
cm
3
(ii) Radius of sphere = 0.63 cm
Volume of sphere = πr 4
3
3
4 22 3 3
= [ × × (0.63) ] cm
3 7
3
= 1.0478m
Therefore , the volume of the sphere is= 1.0478m 3
Page : 236 , Block Name : Exercise 13.8
Q2 Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m .
Answer. (i) Radius (r) of ball = ( 28
2
) cm = 14cm
Volume of ball =
4 3
πr
3
=[
4 22 3 3
× × (14) ] cm
3 7
2 3
= 11498 cm
3
Therefore ,the volume of the sphere = 11498
2 3
cm
3
(ii) radius of ball = ( 0.21
2
) m = 0.105m
Volume of the ball =
4 3
πr
3
Page 30 of 37 Aglasem Schools
2 2
= πrl = πr√r + h
22 2 2 2
= [ × 5.25 × √(5.25) + (3) ] m
7
22 2
= ( × 5.25 × 6.05) m
7
2
= 99.825m
Therefore, 99.825 m canvas will be required to protect the heap from rain.
2
Page : 233 , Block Name : Exercise 13.7
Q1 Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Answer. (i) Radius of sphere = 7 cm
Volume of sphere = πr
4 3
3
4 22 3 3
= [ × × (7) ] cm
3 7
4312 3
= ( ) cm
3
1 3
= 1437 cm
3
Therefore , the volume of the sphere is = 1437
1 3
cm
3
(ii) Radius of sphere = 0.63 cm
Volume of sphere = πr 4
3
3
4 22 3 3
= [ × × (0.63) ] cm
3 7
3
= 1.0478m
Therefore , the volume of the sphere is= 1.0478m 3
Page : 236 , Block Name : Exercise 13.8
Q2 Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m .
Answer. (i) Radius (r) of ball = ( 28
2
) cm = 14cm
Volume of ball =
4 3
πr
3
=[
4 22 3 3
× × (14) ] cm
3 7
2 3
= 11498 cm
3
Therefore ,the volume of the sphere = 11498
2 3
cm
3
(ii) radius of ball = ( 0.21
2
) m = 0.105m
Volume of the ball =
4 3
πr
3
Page 30 of 37 Aglasem Schools
Page 32
Book : Mathematics Ncert Solutions | Chapter-13 Maths
4 22 3 3
= [ × × (0.105) ] m
3 7
3
= 0.004851m
Therefore ,the volume of the sphere = 0.004851m 3
Page : 236 , Block Name : Exercise 13.8
Q3 The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the
metal is 8.9 g per cm3 ?
Answer. Radius (r)of metallic ball = (
4.2
) cm = 2.1cm
2
Volume of the metallic ball =
4 3
πr
3
=[
4 22 3 3
× × (2.1) ] cm
3 7
3
= 38.808cm
density==
Mass
Volume
Mass = density x volume
= (8.9 × 38.808)g
= 345.3912g
Hence, the mass of the ball is 345.39 g (approximately).
Page : 236 , Block Name : Exercise 13.8
Q4 The diameter of the moon is approximately one-fourth of the diameter of the earth. What
fraction of the volume of the earth is the volume of the moon?
Answer. Let the diameter of earth be d. Therefore, the radius of earth will be
d
2
Diameter of moon will be and the radius of moon will be
d d
4 8
3
Volume of moon =
4 3 4 d 1 4 3
πr = π( ) = × πd
3 3 8 512 3
3
Volume of earth =
4 3 4 d 1 4 3
πr = π( ) = × πd
3 3 2 8 3
1 4 3
× πd
Volume of moon 512 3
=
Volume of earth 1 4 3
× πd
8 3
1
=
64
Therefore, the volume = of moon is of the volume of earth.
1
64
Page : 236 , Block Name : Exercise 13.8
Q5 How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer. Radius (r) of hemisphere bowl = ( =
10.5
) cm 5.25cm
2
Volume of hemisphere =
2 3
πr
3
Page 31 of 37 Aglasem Schools
4 22 3 3
= [ × × (0.105) ] m
3 7
3
= 0.004851m
Therefore ,the volume of the sphere = 0.004851m 3
Page : 236 , Block Name : Exercise 13.8
Q3 The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the
metal is 8.9 g per cm3 ?
Answer. Radius (r)of metallic ball = (
4.2
) cm = 2.1cm
2
Volume of the metallic ball =
4 3
πr
3
=[
4 22 3 3
× × (2.1) ] cm
3 7
3
= 38.808cm
density==
Mass
Volume
Mass = density x volume
= (8.9 × 38.808)g
= 345.3912g
Hence, the mass of the ball is 345.39 g (approximately).
Page : 236 , Block Name : Exercise 13.8
Q4 The diameter of the moon is approximately one-fourth of the diameter of the earth. What
fraction of the volume of the earth is the volume of the moon?
Answer. Let the diameter of earth be d. Therefore, the radius of earth will be
d
2
Diameter of moon will be and the radius of moon will be
d d
4 8
3
Volume of moon =
4 3 4 d 1 4 3
πr = π( ) = × πd
3 3 8 512 3
3
Volume of earth =
4 3 4 d 1 4 3
πr = π( ) = × πd
3 3 2 8 3
1 4 3
× πd
Volume of moon 512 3
=
Volume of earth 1 4 3
× πd
8 3
1
=
64
Therefore, the volume = of moon is of the volume of earth.
1
64
Page : 236 , Block Name : Exercise 13.8
Q5 How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer. Radius (r) of hemisphere bowl = ( =
10.5
) cm 5.25cm
2
Volume of hemisphere =
2 3
πr
3
Page 31 of 37 Aglasem Schools
Page 33
Book : Mathematics Ncert Solutions | Chapter-13 Maths
2 22 3 3
= [ × × (5.25) ] cm
3 7
Capacity of bowl = ( letre
303.1875
)
1000
= 0.3031875 litre litre (approximately)
= 0.303
Therefore ,the volume hemisphere bowl is 0.303 litres
Page : 236 , Block Name : Exercise 13.8
Q6 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then
nd the volume of the iron used to make the tank.
Answer. Inner radius of hemispherical tank = 1 m
Thickness of hemispherical tank 1 cm = 0.01 m
Outer radius of hemispherical tank (1 + 0.01) m = 1.01 m
Volume of iron is used to make such a tank = (r − r )
2 3 3
3 2 1
2 22 3 3 3
= [ × × {(1.01) − (1) }] m
3 7
44 3
= [ × (1.030301 − 1)] m
21
= 0.06348m
3
(approximately)
Page : 236 , Block Name : Exercise 13.8
Q7 Find the volume of a sphere whose surface area is 154 cm2 .
Answer. Let radius of sphere be r.
Surface area of sphere = 154cm 2
2 154×7 2
⇒ r = ( ) cm
4×22
7
⇒ r = ( ) cm = 3.5cm
2
Volume of sphere =
4 3
πr
3
4 22 3 3
= [ × × (3.5) ] cm
3 7
2 3
= 179 cm
3
Therefore , the volume of the sphere is =179
2 3
cm
3
Page : 236 , Block Name : Exercise 13.8
Q8 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the
cost of ` 498.96 If the cost of white-washing is 2.00 per square metre, nd the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Answer. (i) Cost of white washing the dome from inside = Rs 498.96
Page 32 of 37 Aglasem Schools
2 22 3 3
= [ × × (5.25) ] cm
3 7
Capacity of bowl = ( letre
303.1875
)
1000
= 0.3031875 litre litre (approximately)
= 0.303
Therefore ,the volume hemisphere bowl is 0.303 litres
Page : 236 , Block Name : Exercise 13.8
Q6 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then
nd the volume of the iron used to make the tank.
Answer. Inner radius of hemispherical tank = 1 m
Thickness of hemispherical tank 1 cm = 0.01 m
Outer radius of hemispherical tank (1 + 0.01) m = 1.01 m
Volume of iron is used to make such a tank = (r − r )
2 3 3
3 2 1
2 22 3 3 3
= [ × × {(1.01) − (1) }] m
3 7
44 3
= [ × (1.030301 − 1)] m
21
= 0.06348m
3
(approximately)
Page : 236 , Block Name : Exercise 13.8
Q7 Find the volume of a sphere whose surface area is 154 cm2 .
Answer. Let radius of sphere be r.
Surface area of sphere = 154cm 2
2 154×7 2
⇒ r = ( ) cm
4×22
7
⇒ r = ( ) cm = 3.5cm
2
Volume of sphere =
4 3
πr
3
4 22 3 3
= [ × × (3.5) ] cm
3 7
2 3
= 179 cm
3
Therefore , the volume of the sphere is =179
2 3
cm
3
Page : 236 , Block Name : Exercise 13.8
Q8 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the
cost of ` 498.96 If the cost of white-washing is 2.00 per square metre, nd the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Answer. (i) Cost of white washing the dome from inside = Rs 498.96
Page 32 of 37 Aglasem Schools
Page 34
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Cost of white washing 1 m area = Rs 2.00 2
Therefore .CSA of the inner side of dome (
498.96 2
)m
2
2
= 249.48m
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome =249.48 m 2
2 2
2πr = 249.48m
22 2 2
⇒ 2 × × r = 249.48m
7
2 249.48×7 2 2
⇒ r = ( )m = 39.69m
2×22
r=6.3m
Volume of air inside the dome = Volume of hemispherical dome
2 3
= πr
3
2 22 3 3
= [ × × (6.3) ] m
3 7
= 523.908m
3
(approximately)
Page : 236 , Block Name : Exercise 13.8
Q9 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere
with surface area S′.
Find the
(i) radius r′ of the new sphere,
(ii) ratio of S and S.
Answer. (i)Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere = πr
4 3
3
Volume of 27 solid iron spheres = 27 × 4
3
πr
3
27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this
iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this
new sphere be r'.
Volume of new solid iron sphere = πr
4 3
3
4 3 4 3
πr = 27 × πr
3 3
′3 3
r = 27r
′
r = 3r
(ii) surface area of 1 solid iron sphere of radius = 4πr 2
Surface area of iron sphere of radius r' = 4n(r ) ′ 2
2 2
= 4π(3r) = 36πr
2
S 4πr 1
′
= 2
= = 1 : 9
S 36πr 9
Page : 236 , Block Name : Exercise 13.8
Q10 A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in
Page 33 of 37 Aglasem Schools
Cost of white washing 1 m area = Rs 2.00 2
Therefore .CSA of the inner side of dome (
498.96 2
)m
2
2
= 249.48m
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome =249.48 m 2
2 2
2πr = 249.48m
22 2 2
⇒ 2 × × r = 249.48m
7
2 249.48×7 2 2
⇒ r = ( )m = 39.69m
2×22
r=6.3m
Volume of air inside the dome = Volume of hemispherical dome
2 3
= πr
3
2 22 3 3
= [ × × (6.3) ] m
3 7
= 523.908m
3
(approximately)
Page : 236 , Block Name : Exercise 13.8
Q9 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere
with surface area S′.
Find the
(i) radius r′ of the new sphere,
(ii) ratio of S and S.
Answer. (i)Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere = πr
4 3
3
Volume of 27 solid iron spheres = 27 × 4
3
πr
3
27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this
iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this
new sphere be r'.
Volume of new solid iron sphere = πr
4 3
3
4 3 4 3
πr = 27 × πr
3 3
′3 3
r = 27r
′
r = 3r
(ii) surface area of 1 solid iron sphere of radius = 4πr 2
Surface area of iron sphere of radius r' = 4n(r ) ′ 2
2 2
= 4π(3r) = 36πr
2
S 4πr 1
′
= 2
= = 1 : 9
S 36πr 9
Page : 236 , Block Name : Exercise 13.8
Q10 A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in
Page 33 of 37 Aglasem Schools
Page 35
Book : Mathematics Ncert Solutions | Chapter-13 Maths
mm3 ) is needed to ll this capsule?
Answer. Radius (r) of capsule = (
3.5
) mm = 1.75mm
2
Volume of spherical capsule =
4 3
πr
3
4 22 3 3
= [ × × (1.75) ] mm
3 7
= 22.458mm 3
=22.46mm (approximately)
3
Therefore ,the volume of the spherical capsule is 22.46 mm 3
Page : 236 , Block Name : Exercise 13.8
Q1 A wooden bookshelf has external dimensions as follows: Height = 110 cm,
Depth = 25 cm, Breadth = 85 cm (see Fig. 13.31). The thickness of the plank is 5 cm everywhere.
The external faces are to be polished and the inner faces are to be painted. If the rate of polishing
is 20 paise per cm2 and the rate of painting is 10 paise per cm2 , nd the total expenses required
for polishing and painting the surface of the bookshelf.
Answer.
External height (l) of book self = 85 cm
External breadth (b) of book self = 25 cm
External height (h) of book self = 110 cm
External surface area of shelf while leaving out the front face of the shelf
=lh + 2 ( lb + bh )
= [85 x 110 + 2 (85 x 25 + 25 x 110)) cm 2
Page 34 of 37 Aglasem Schools
mm3 ) is needed to ll this capsule?
Answer. Radius (r) of capsule = (
3.5
) mm = 1.75mm
2
Volume of spherical capsule =
4 3
πr
3
4 22 3 3
= [ × × (1.75) ] mm
3 7
= 22.458mm 3
=22.46mm (approximately)
3
Therefore ,the volume of the spherical capsule is 22.46 mm 3
Page : 236 , Block Name : Exercise 13.8
Q1 A wooden bookshelf has external dimensions as follows: Height = 110 cm,
Depth = 25 cm, Breadth = 85 cm (see Fig. 13.31). The thickness of the plank is 5 cm everywhere.
The external faces are to be polished and the inner faces are to be painted. If the rate of polishing
is 20 paise per cm2 and the rate of painting is 10 paise per cm2 , nd the total expenses required
for polishing and painting the surface of the bookshelf.
Answer.
External height (l) of book self = 85 cm
External breadth (b) of book self = 25 cm
External height (h) of book self = 110 cm
External surface area of shelf while leaving out the front face of the shelf
=lh + 2 ( lb + bh )
= [85 x 110 + 2 (85 x 25 + 25 x 110)) cm 2
Page 34 of 37 Aglasem Schools
Page 36
Book : Mathematics Ncert Solutions | Chapter-13 Maths
= (9350 + 9750)cm 2
= 19100 cm 2
Area of front face = [85 x 110 - 75 x 100 + 2 (75 x 5)] cm 2
= 1850 + 750 cm 2
= 2600 cm 2
Area to be polished = (19100 + 2600)cm = 21700 cm2 2
Cost of polishing 1cm area = Rs 0.20
2
Cost of polishing 21700cm area Rs (21700 x 0.20) = Rs 4340
2
It can be observed that length (l), breadth (b), and height (h) of each row of the
book shelf is 75 cm, 20 cm, and 30 cm respectively.
Area to be painted in 1 row = 2 (l + h) b + lh
= [2 (75 + 30) x 20 + 75 x 30) cm 2
- (4200 + 2250) cm 2
= 6450 cm 2
Area to be painted in 3 rows = (3 x 6450) cm = 19350 cm
2 2
Cost of painting 1cm area = Rs 0.10
2
Cost of painting 19350cm area = Rs (19350 x 0.1)
2
= RS 1935
Total expense required for polishing and painting = Rs (4340 + 1935)
= Rs 6275
Therefore, it will cost Rs 6275 for polishing and painting the surface of the bookshelf.
Page : 236 , Block Name : Exercise 13.8 (optional)
Q2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed
on small supports as shown in Fig 13.32. Eight such spheres are used for this purpose, and are to
be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted
black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5
paise per cm2 .
Page 35 of 37 Aglasem Schools
= (9350 + 9750)cm 2
= 19100 cm 2
Area of front face = [85 x 110 - 75 x 100 + 2 (75 x 5)] cm 2
= 1850 + 750 cm 2
= 2600 cm 2
Area to be polished = (19100 + 2600)cm = 21700 cm2 2
Cost of polishing 1cm area = Rs 0.20
2
Cost of polishing 21700cm area Rs (21700 x 0.20) = Rs 4340
2
It can be observed that length (l), breadth (b), and height (h) of each row of the
book shelf is 75 cm, 20 cm, and 30 cm respectively.
Area to be painted in 1 row = 2 (l + h) b + lh
= [2 (75 + 30) x 20 + 75 x 30) cm 2
- (4200 + 2250) cm 2
= 6450 cm 2
Area to be painted in 3 rows = (3 x 6450) cm = 19350 cm
2 2
Cost of painting 1cm area = Rs 0.10
2
Cost of painting 19350cm area = Rs (19350 x 0.1)
2
= RS 1935
Total expense required for polishing and painting = Rs (4340 + 1935)
= Rs 6275
Therefore, it will cost Rs 6275 for polishing and painting the surface of the bookshelf.
Page : 236 , Block Name : Exercise 13.8 (optional)
Q2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed
on small supports as shown in Fig 13.32. Eight such spheres are used for this purpose, and are to
be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted
black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5
paise per cm2 .
Page 35 of 37 Aglasem Schools
Page 37
Book : Mathematics Ncert Solutions | Chapter-13 Maths
Answer. Radius (r) Of wooden sphere = (
21
) cm = 10.5cm
2
Surface area of wooden sphere = 4πr 2
22 2 2 2
= [4 × × (10.5) ] cm = 1386cm
7
Radius of the circular end of cylindrical support =1.5 cm
Height (h) of cylindrical support = 7 cm
CSA of cylindrical support = 2πrh
22 2 2
= [2 × × (1.5) × 7] cm = 66cm
7
Area of the circular end of cylindrical support = [ 22
7
2
× (1.5) ] cm
2
=7.07 cm 2
= x (1386 7.07)] cm2
Area to be painted silver =[ 8 x (1386 - 7.07 ) ] cm 2
(8 x 1378.93) cm 2
= 11031.44 cm 2
cost for painting With silver colour = RS (11031.44 x 0.25) = RS 2757.86
Area to be painted black = (8 x 66) cm v 2
= 528 cm 2
Cost for painting with black colour = Rs (528 x 0.05) = Rs 26.40
Total cost in painting = Rs (2757.86 + 26.40)
= Rs 2784.26
Therefore, it will cost Rs 2784.26 in painting in such a way.
Page : 237 , Block Name : Exercise 13.8 (optional)
Q3 The diameter of a sphere is decreased by 25%. By what percent does its curved surface area
decrease?
Answer. Radius (n) of sphere =
d
2
New radius of sphere =
d 25 3
(1 − ) = d
2 100 8
CSA S of sphere = 4πr
1
2
1
2
d 2
= 4π( ) = πd
2
CSA S of sphere when the radius is decreased = 4πr
2
2
2
=πd 2 9 2
− πd
16
Page 36 of 37 Aglasem Schools
Answer. Radius (r) Of wooden sphere = (
21
) cm = 10.5cm
2
Surface area of wooden sphere = 4πr 2
22 2 2 2
= [4 × × (10.5) ] cm = 1386cm
7
Radius of the circular end of cylindrical support =1.5 cm
Height (h) of cylindrical support = 7 cm
CSA of cylindrical support = 2πrh
22 2 2
= [2 × × (1.5) × 7] cm = 66cm
7
Area of the circular end of cylindrical support = [ 22
7
2
× (1.5) ] cm
2
=7.07 cm 2
= x (1386 7.07)] cm2
Area to be painted silver =[ 8 x (1386 - 7.07 ) ] cm 2
(8 x 1378.93) cm 2
= 11031.44 cm 2
cost for painting With silver colour = RS (11031.44 x 0.25) = RS 2757.86
Area to be painted black = (8 x 66) cm v 2
= 528 cm 2
Cost for painting with black colour = Rs (528 x 0.05) = Rs 26.40
Total cost in painting = Rs (2757.86 + 26.40)
= Rs 2784.26
Therefore, it will cost Rs 2784.26 in painting in such a way.
Page : 237 , Block Name : Exercise 13.8 (optional)
Q3 The diameter of a sphere is decreased by 25%. By what percent does its curved surface area
decrease?
Answer. Radius (n) of sphere =
d
2
New radius of sphere =
d 25 3
(1 − ) = d
2 100 8
CSA S of sphere = 4πr
1
2
1
2
d 2
= 4π( ) = πd
2
CSA S of sphere when the radius is decreased = 4πr
2
2
2
=πd 2 9 2
− πd
16
Page 36 of 37 Aglasem Schools
Page 38
Book : Mathematics Ncert Solutions | Chapter-13 Maths
7 2
πd
16
Percentage decrease in surface area of sphere =
S1 −S2
× 100
S1
2
7πd 700
= × 100 = = 43.75%
16πd 2 16
Page : 237 , Block Name : Exercise 13.8 (optional)
Page 37 of 37 Aglasem Schools
7 2
πd
16
Percentage decrease in surface area of sphere =
S1 −S2
× 100
S1
2
7πd 700
= × 100 = = 43.75%
16πd 2 16
Page : 237 , Block Name : Exercise 13.8 (optional)
Page 37 of 37 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 9 |
| Type | Solution |
| Pages | 38 |
| Updated | 30 Apr 2026 |