Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
1. Find the curved surface area and total surface area of a cylinder, the diameter of
whose base is 7 cm and height is 60 cm.
Solution:
We have,
Diameter of cylinder = 7 cm
So, Radius of cylinder = 7/2 cm
Height of cylinder = 60 cm
By using the formula,
Curved surface area of cylinder = 2prh
= 2 × 22/7 × 7/2 × 60
= 1320 cm
2
Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7/2 (60 + 7/2)
= 22 (127/2)
= 1397 cm
2
2. The curved surface area of a cylindrical road is 132 cm
2
. Find its length if the
radius is 0.35 cm.
Solution:
We have,
Curved surface area of cylindrical road =132 cm
2
Radius of road = 0.35 cm
Let length of road be ‘h’ cm
By using the formula,
Curved surface area of cylindrical road = 2prh
So, 2prh = 132
2 × 22/7 × 0.35 × h = 132
h = 132×7 / 2×22×0.35
= 924 / 15.4
= 60cm
? Length of road is 60 cm.
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 2.5 cm.
Find the curved surface area of the cylinder.
Solution:
We have,
Page 2
1. Find the curved surface area and total surface area of a cylinder, the diameter of
whose base is 7 cm and height is 60 cm.
Solution:
We have,
Diameter of cylinder = 7 cm
So, Radius of cylinder = 7/2 cm
Height of cylinder = 60 cm
By using the formula,
Curved surface area of cylinder = 2prh
= 2 × 22/7 × 7/2 × 60
= 1320 cm
2
Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7/2 (60 + 7/2)
= 22 (127/2)
= 1397 cm
2
2. The curved surface area of a cylindrical road is 132 cm
2
. Find its length if the
radius is 0.35 cm.
Solution:
We have,
Curved surface area of cylindrical road =132 cm
2
Radius of road = 0.35 cm
Let length of road be ‘h’ cm
By using the formula,
Curved surface area of cylindrical road = 2prh
So, 2prh = 132
2 × 22/7 × 0.35 × h = 132
h = 132×7 / 2×22×0.35
= 924 / 15.4
= 60cm
? Length of road is 60 cm.
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 2.5 cm.
Find the curved surface area of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 2.5 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So, pr
2
= 616
22/7 r
2
= 616
r
2
= 616×7 / 22
= 196
r = v196
= 14cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 2.5
= 1540/7
= 220 cm
2
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
its curved surface area and total surface area.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88×7 / 44
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 15
= 1320 cm
2
? Total surface area area of cylinder = 2pr (h+r)
= 2 × 22/7 × 14 (15 + 14)
= 2 × 22/7 × 14 × 29
= 2552 cm
2
Page 3
1. Find the curved surface area and total surface area of a cylinder, the diameter of
whose base is 7 cm and height is 60 cm.
Solution:
We have,
Diameter of cylinder = 7 cm
So, Radius of cylinder = 7/2 cm
Height of cylinder = 60 cm
By using the formula,
Curved surface area of cylinder = 2prh
= 2 × 22/7 × 7/2 × 60
= 1320 cm
2
Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7/2 (60 + 7/2)
= 22 (127/2)
= 1397 cm
2
2. The curved surface area of a cylindrical road is 132 cm
2
. Find its length if the
radius is 0.35 cm.
Solution:
We have,
Curved surface area of cylindrical road =132 cm
2
Radius of road = 0.35 cm
Let length of road be ‘h’ cm
By using the formula,
Curved surface area of cylindrical road = 2prh
So, 2prh = 132
2 × 22/7 × 0.35 × h = 132
h = 132×7 / 2×22×0.35
= 924 / 15.4
= 60cm
? Length of road is 60 cm.
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 2.5 cm.
Find the curved surface area of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 2.5 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So, pr
2
= 616
22/7 r
2
= 616
r
2
= 616×7 / 22
= 196
r = v196
= 14cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 2.5
= 1540/7
= 220 cm
2
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
its curved surface area and total surface area.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88×7 / 44
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 15
= 1320 cm
2
? Total surface area area of cylinder = 2pr (h+r)
= 2 × 22/7 × 14 (15 + 14)
= 2 × 22/7 × 14 × 29
= 2552 cm
2
5. A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find the total
surface area of the solid thus generated.
Solution:
We have,
Dimension of rectangular strip = 25 cm × 7cm
When this strip is rotated about its longer side,
Height of cylinder becomes = 25 cm
Radius = 7 cm
? Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7 (25 + 7)
= 2 × 22/7 × 7 × 32
= 1408 cm
2
6. A rectangular sheet of paper, 44 cm× 20 cm, is rolled along its length to form a
cylinder. Find the total surface area of the cylinder thus generated.
Solution:
We have,
Dimensions of rectangular sheet of paper = 44cm × 20cm
When this sheet of paper is rolled along its length,
Circumference of base becomes = 44 cm
By using the formula,
Circumference of base = 2pr
So, 2pr = 44
2 × 22/7 × r = 44
r = 44×7 / 44
= 7cm
Radius = 7cm
Height = 20cm
? Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7 (20 + 7)
= 2 × 22/7 × 7 × 27
= 1188 cm
2
7. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio
5:3. Calculate the ratio of their curved surface areas.
Solution:
We have,
Ratio of radius of two cylinder, r
1
:r
2
= 2:3
Page 4
1. Find the curved surface area and total surface area of a cylinder, the diameter of
whose base is 7 cm and height is 60 cm.
Solution:
We have,
Diameter of cylinder = 7 cm
So, Radius of cylinder = 7/2 cm
Height of cylinder = 60 cm
By using the formula,
Curved surface area of cylinder = 2prh
= 2 × 22/7 × 7/2 × 60
= 1320 cm
2
Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7/2 (60 + 7/2)
= 22 (127/2)
= 1397 cm
2
2. The curved surface area of a cylindrical road is 132 cm
2
. Find its length if the
radius is 0.35 cm.
Solution:
We have,
Curved surface area of cylindrical road =132 cm
2
Radius of road = 0.35 cm
Let length of road be ‘h’ cm
By using the formula,
Curved surface area of cylindrical road = 2prh
So, 2prh = 132
2 × 22/7 × 0.35 × h = 132
h = 132×7 / 2×22×0.35
= 924 / 15.4
= 60cm
? Length of road is 60 cm.
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 2.5 cm.
Find the curved surface area of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 2.5 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So, pr
2
= 616
22/7 r
2
= 616
r
2
= 616×7 / 22
= 196
r = v196
= 14cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 2.5
= 1540/7
= 220 cm
2
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
its curved surface area and total surface area.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88×7 / 44
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 15
= 1320 cm
2
? Total surface area area of cylinder = 2pr (h+r)
= 2 × 22/7 × 14 (15 + 14)
= 2 × 22/7 × 14 × 29
= 2552 cm
2
5. A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find the total
surface area of the solid thus generated.
Solution:
We have,
Dimension of rectangular strip = 25 cm × 7cm
When this strip is rotated about its longer side,
Height of cylinder becomes = 25 cm
Radius = 7 cm
? Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7 (25 + 7)
= 2 × 22/7 × 7 × 32
= 1408 cm
2
6. A rectangular sheet of paper, 44 cm× 20 cm, is rolled along its length to form a
cylinder. Find the total surface area of the cylinder thus generated.
Solution:
We have,
Dimensions of rectangular sheet of paper = 44cm × 20cm
When this sheet of paper is rolled along its length,
Circumference of base becomes = 44 cm
By using the formula,
Circumference of base = 2pr
So, 2pr = 44
2 × 22/7 × r = 44
r = 44×7 / 44
= 7cm
Radius = 7cm
Height = 20cm
? Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7 (20 + 7)
= 2 × 22/7 × 7 × 27
= 1188 cm
2
7. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio
5:3. Calculate the ratio of their curved surface areas.
Solution:
We have,
Ratio of radius of two cylinder, r
1
:r
2
= 2:3
Ratio of their heights, h
1
:h
2
= 5:3
r
1
/r
2
= 2/3
h
1
/h
2
= 5/3
so,
Curved surface area of cylinder1 / curved surface area of cylinder2 = 2pr
1
h
1
/ 2pr
2
h
2
= (2 × 22/7 × 2 × 5) / (2 × 22/7 × 3 × 3)
= 10/9
? Ratio of their curved surface area is 10:9
8. The ratio between the curved surface area and the total surface area of a right
circular cylinder is 1:2. Prove that its height and radius are equal.
Solution:
We have,
Let radius of cylinder be ‘r’
Let height of cylinder be ‘h’
Curved surface area of cylinder / total surface area of cylinder = 1/2
2prh / 2pr (h+r) = 1/2
h/(h+r) = 1/2
2h = h+r
2h – h = r
h = r
Height = Radius
Hence proved.
9. The curved surface area of a cylinder is 1320 cm
2
and its base has diameter 21
cm. Find the height of the cylinder.
Solution:
We have,
Diameter of base = 21 cm
Radius of cylinder = 21/2 cm
Let height of cylinder be ‘h’ cm
Curved surface area of cylinder = 1320 cm
2
By using the formula,
Curved surface area of cylinder = 2prh
So,
2prh = 1320
2 × 22/7 × 21/2 × h = 1320
66h = 1320
h = 1320/66 = 20cm
Page 5
1. Find the curved surface area and total surface area of a cylinder, the diameter of
whose base is 7 cm and height is 60 cm.
Solution:
We have,
Diameter of cylinder = 7 cm
So, Radius of cylinder = 7/2 cm
Height of cylinder = 60 cm
By using the formula,
Curved surface area of cylinder = 2prh
= 2 × 22/7 × 7/2 × 60
= 1320 cm
2
Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7/2 (60 + 7/2)
= 22 (127/2)
= 1397 cm
2
2. The curved surface area of a cylindrical road is 132 cm
2
. Find its length if the
radius is 0.35 cm.
Solution:
We have,
Curved surface area of cylindrical road =132 cm
2
Radius of road = 0.35 cm
Let length of road be ‘h’ cm
By using the formula,
Curved surface area of cylindrical road = 2prh
So, 2prh = 132
2 × 22/7 × 0.35 × h = 132
h = 132×7 / 2×22×0.35
= 924 / 15.4
= 60cm
? Length of road is 60 cm.
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 2.5 cm.
Find the curved surface area of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 2.5 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So, pr
2
= 616
22/7 r
2
= 616
r
2
= 616×7 / 22
= 196
r = v196
= 14cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 2.5
= 1540/7
= 220 cm
2
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
its curved surface area and total surface area.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88×7 / 44
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Curved surface area of cylinder = 2prh
= 2 × 22/7 × 14 × 15
= 1320 cm
2
? Total surface area area of cylinder = 2pr (h+r)
= 2 × 22/7 × 14 (15 + 14)
= 2 × 22/7 × 14 × 29
= 2552 cm
2
5. A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find the total
surface area of the solid thus generated.
Solution:
We have,
Dimension of rectangular strip = 25 cm × 7cm
When this strip is rotated about its longer side,
Height of cylinder becomes = 25 cm
Radius = 7 cm
? Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7 (25 + 7)
= 2 × 22/7 × 7 × 32
= 1408 cm
2
6. A rectangular sheet of paper, 44 cm× 20 cm, is rolled along its length to form a
cylinder. Find the total surface area of the cylinder thus generated.
Solution:
We have,
Dimensions of rectangular sheet of paper = 44cm × 20cm
When this sheet of paper is rolled along its length,
Circumference of base becomes = 44 cm
By using the formula,
Circumference of base = 2pr
So, 2pr = 44
2 × 22/7 × r = 44
r = 44×7 / 44
= 7cm
Radius = 7cm
Height = 20cm
? Total surface area of cylinder = 2pr (h+r)
= 2 × 22/7 × 7 (20 + 7)
= 2 × 22/7 × 7 × 27
= 1188 cm
2
7. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio
5:3. Calculate the ratio of their curved surface areas.
Solution:
We have,
Ratio of radius of two cylinder, r
1
:r
2
= 2:3
Ratio of their heights, h
1
:h
2
= 5:3
r
1
/r
2
= 2/3
h
1
/h
2
= 5/3
so,
Curved surface area of cylinder1 / curved surface area of cylinder2 = 2pr
1
h
1
/ 2pr
2
h
2
= (2 × 22/7 × 2 × 5) / (2 × 22/7 × 3 × 3)
= 10/9
? Ratio of their curved surface area is 10:9
8. The ratio between the curved surface area and the total surface area of a right
circular cylinder is 1:2. Prove that its height and radius are equal.
Solution:
We have,
Let radius of cylinder be ‘r’
Let height of cylinder be ‘h’
Curved surface area of cylinder / total surface area of cylinder = 1/2
2prh / 2pr (h+r) = 1/2
h/(h+r) = 1/2
2h = h+r
2h – h = r
h = r
Height = Radius
Hence proved.
9. The curved surface area of a cylinder is 1320 cm
2
and its base has diameter 21
cm. Find the height of the cylinder.
Solution:
We have,
Diameter of base = 21 cm
Radius of cylinder = 21/2 cm
Let height of cylinder be ‘h’ cm
Curved surface area of cylinder = 1320 cm
2
By using the formula,
Curved surface area of cylinder = 2prh
So,
2prh = 1320
2 × 22/7 × 21/2 × h = 1320
66h = 1320
h = 1320/66 = 20cm
? Height of cylinder is 20cm.
10. The height of a right circular cylinder is 10.5 cm. If three times the sum of the
areas of its two circular faces is twice the area of the curved surface area. Find the
radius of its base.
Solution:
We have,
Height of cylinder = 10.5 cm
Let radius of cylinder be ‘r’ cm
So,
Area of two bases of cylinder = 2pr
2
Area of curved surface of cylinder = 2prh
Now,
3 (2pr
2
) = 2 (2prh)
6pr
2
= 4prh
pr
2
/ pr = 4/6 h
r = 2/3 h
= 2×10.5 /3
= 7cm
? Radius of base of cylinder is 7cm.
11. Find the cost of plastering the inner surface of a well at Rs 9.50 per m
2
, if it is 21
m deep and diameter of its top is 6 m.
Solution:
We have,
Height of cylinder = 21m
Diameter of cylinder = 6m
Radius of cylinder = 6/2 = 3m
Curved surface area of cylinder = 2prh
= 2 × 22/7 × 3 × 21
= 396 m
2
? Cost of plastering the inner surface at the rate of Rs 9.50 per m
2
= 396 × 9.50 = Rs
3762
12. A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find
the cost of tin-plating it on the inside at the rate of 50 paise per hundred square
centimetre.
Solution:
We have,
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.97/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 have been prepared according to the Class 8 exam syllabus. Information about Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 covers topics
like and Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1.
Introduction of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1
In this doc you can find the meaning of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 defined & explained in the simplest way possible. Besides explaining types of
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 theory, EduRev gives you an ample number of questions to practice Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Previous Year Questions with Solutions
,mock tests for examination
,past year papers
,Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 | Mathematics (Maths) Class 8
,Extra Questions
,MCQs
,Important questions
,Summary
,video lectures
,Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 | Mathematics (Maths) Class 8
,Exam
,Viva Questions
,Objective type Questions
,Free
,practice quizzes
,ppt
,Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 | Mathematics (Maths) Class 8
,shortcuts and tricks
,Semester Notes
,study material
,Sample Paper
;
Additional Information about Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 for Class 8 Preparation
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Free PDF Download
The Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 is an invaluable resource that delves deep into the core of the Class 8 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 now and kickstart your journey towards success in the Class 8 exam.
Importance of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1
The importance of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Notes
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Notes on EduRev are your ultimate resource for success.
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Class 8 Questions
The "Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 on the App
Students of Class 8 can study Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.1 is prepared as per the latest Class 8 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup