Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3
Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.3 PAGE NO: 21.22
1. Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Given details are,
Length of a cuboid = 10 cm
Breadth of a cuboid = 12 cm
Height of a cuboid = 14 cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (10×12 + 12×14 + 14×10)
= 2 (120 + 168 + 140)
= 2 (428)
= 856 cm
2
(ii) Given details are,
Length of a cuboid = 6 dm
Breadth of a cuboid = 8 dm
Height of a cuboid = 10 dm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (6×8 + 8×10 + 10×6)
= 2 (48 + 80 + 60)
= 2 (188)
= 376 dm
2
(iii) Given details are,
Length of a cuboid = 2m
Breadth of a cuboid = 4m
Height of a cuboid = 5m
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (2×4 + 4×5 + 5×2)
= 2 (8 + 20 + 10)
= 2 (38)
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.3 PAGE NO: 21.22
1. Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Given details are,
Length of a cuboid = 10 cm
Breadth of a cuboid = 12 cm
Height of a cuboid = 14 cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (10×12 + 12×14 + 14×10)
= 2 (120 + 168 + 140)
= 2 (428)
= 856 cm
2
(ii) Given details are,
Length of a cuboid = 6 dm
Breadth of a cuboid = 8 dm
Height of a cuboid = 10 dm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (6×8 + 8×10 + 10×6)
= 2 (48 + 80 + 60)
= 2 (188)
= 376 dm
2
(iii) Given details are,
Length of a cuboid = 2m
Breadth of a cuboid = 4m
Height of a cuboid = 5m
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (2×4 + 4×5 + 5×2)
= 2 (8 + 20 + 10)
= 2 (38)
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
= 76 m
2
(iv) Given details are,
Length of a cuboid = 3.2 m= 32 dm
Breadth of a cuboid = 30 dm
Height of a cuboid = 250 cm= 25 dm
We know that,
surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (32×30 + 30×25 + 25×32)
= 2 (960 + 750 + 800)
= 2 (2510)
= 5020 dm
2
2. Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m
Solution:
(i) Given,
Edge of cube = 1.2 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 1.2
2
= 6 × 1.44
= 8.64 m
2
(ii) Given,
Edge of cube = 27 cm
We know that,
Surface area of cube = 6 × side
2
= 6 × 27
2
= 6 × 729
= 4374 cm
2
(iii) Given,
Edge of cube = 3 cm
We know that,
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.3 PAGE NO: 21.22
1. Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Given details are,
Length of a cuboid = 10 cm
Breadth of a cuboid = 12 cm
Height of a cuboid = 14 cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (10×12 + 12×14 + 14×10)
= 2 (120 + 168 + 140)
= 2 (428)
= 856 cm
2
(ii) Given details are,
Length of a cuboid = 6 dm
Breadth of a cuboid = 8 dm
Height of a cuboid = 10 dm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (6×8 + 8×10 + 10×6)
= 2 (48 + 80 + 60)
= 2 (188)
= 376 dm
2
(iii) Given details are,
Length of a cuboid = 2m
Breadth of a cuboid = 4m
Height of a cuboid = 5m
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (2×4 + 4×5 + 5×2)
= 2 (8 + 20 + 10)
= 2 (38)
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
= 76 m
2
(iv) Given details are,
Length of a cuboid = 3.2 m= 32 dm
Breadth of a cuboid = 30 dm
Height of a cuboid = 250 cm= 25 dm
We know that,
surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (32×30 + 30×25 + 25×32)
= 2 (960 + 750 + 800)
= 2 (2510)
= 5020 dm
2
2. Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m
Solution:
(i) Given,
Edge of cube = 1.2 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 1.2
2
= 6 × 1.44
= 8.64 m
2
(ii) Given,
Edge of cube = 27 cm
We know that,
Surface area of cube = 6 × side
2
= 6 × 27
2
= 6 × 729
= 4374 cm
2
(iii) Given,
Edge of cube = 3 cm
We know that,
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Surface area of cube = 6 × side
2
= 6 × 3
2
= 6 × 9
= 54 cm
2
(iv) Given,
Edge of cube = 6 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 6
2
= 6 × 36
= 216 m
2
(v) Given,
Edge of cube = 2.1 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 2.1
2
= 6 × 4.41
= 26.46 m
2
3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
Solution:
Given details are,
Dimensions of cuboidal box = 5cm × 5cm × 4cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (5×5 + 5×4 + 4×5)
= 2 (25 + 20 + 20)
= 2 (65)
= 130 cm
2
4. Find the surface area of a cube whose volume is
(i) 343 m
3
(ii) 216 dm
3
Solution:
(i) Given details are,
Volume of cube = 343 m
3
Side of cube, a =
3
v(343) = 7m
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.3 PAGE NO: 21.22
1. Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Given details are,
Length of a cuboid = 10 cm
Breadth of a cuboid = 12 cm
Height of a cuboid = 14 cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (10×12 + 12×14 + 14×10)
= 2 (120 + 168 + 140)
= 2 (428)
= 856 cm
2
(ii) Given details are,
Length of a cuboid = 6 dm
Breadth of a cuboid = 8 dm
Height of a cuboid = 10 dm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (6×8 + 8×10 + 10×6)
= 2 (48 + 80 + 60)
= 2 (188)
= 376 dm
2
(iii) Given details are,
Length of a cuboid = 2m
Breadth of a cuboid = 4m
Height of a cuboid = 5m
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (2×4 + 4×5 + 5×2)
= 2 (8 + 20 + 10)
= 2 (38)
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
= 76 m
2
(iv) Given details are,
Length of a cuboid = 3.2 m= 32 dm
Breadth of a cuboid = 30 dm
Height of a cuboid = 250 cm= 25 dm
We know that,
surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (32×30 + 30×25 + 25×32)
= 2 (960 + 750 + 800)
= 2 (2510)
= 5020 dm
2
2. Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m
Solution:
(i) Given,
Edge of cube = 1.2 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 1.2
2
= 6 × 1.44
= 8.64 m
2
(ii) Given,
Edge of cube = 27 cm
We know that,
Surface area of cube = 6 × side
2
= 6 × 27
2
= 6 × 729
= 4374 cm
2
(iii) Given,
Edge of cube = 3 cm
We know that,
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Surface area of cube = 6 × side
2
= 6 × 3
2
= 6 × 9
= 54 cm
2
(iv) Given,
Edge of cube = 6 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 6
2
= 6 × 36
= 216 m
2
(v) Given,
Edge of cube = 2.1 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 2.1
2
= 6 × 4.41
= 26.46 m
2
3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
Solution:
Given details are,
Dimensions of cuboidal box = 5cm × 5cm × 4cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (5×5 + 5×4 + 4×5)
= 2 (25 + 20 + 20)
= 2 (65)
= 130 cm
2
4. Find the surface area of a cube whose volume is
(i) 343 m
3
(ii) 216 dm
3
Solution:
(i) Given details are,
Volume of cube = 343 m
3
Side of cube, a =
3
v(343) = 7m
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
We know that,
Surface area of cube = 6 × side
2
= 6 × 7
2
= 6 × 49
= 294 m
2
(ii) Given details are,
Volume of cube = 216 dm
3
Side of cube a =
3
v(216) = 6dm
We know that,
Surface area of cube = 6 × side
2
= 6 × 6
2
= 6 × 36
= 216 dm
2
5. Find the volume of a cube whose surface area is
(i) 96 cm
2
(ii) 150 m
2
Solution:
(i) Given details are,
Surface area of cube = 96 cm
2
6 × side
2
= 96cm
2
Side
2
= 96/6
= 16
Side = v16 = 4cm
? Volume of a cube = 4
3
= 64cm
3
(ii) Given details are,
Surface area of cube = 150 m
2
6 × side
2
= 150cm
2
Side
2
= 150/6
= 25
Side = v25 = 5cm
? Volume of a cube = 5
3
= 125m
3
6. The dimensions of a cuboid are in the ratio 5: 3: 1 and its total surface area is 414
m
2
. Find the dimensions.
Solution:
Given details are,
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
EXERCISE 21.3 PAGE NO: 21.22
1. Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Given details are,
Length of a cuboid = 10 cm
Breadth of a cuboid = 12 cm
Height of a cuboid = 14 cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (10×12 + 12×14 + 14×10)
= 2 (120 + 168 + 140)
= 2 (428)
= 856 cm
2
(ii) Given details are,
Length of a cuboid = 6 dm
Breadth of a cuboid = 8 dm
Height of a cuboid = 10 dm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (6×8 + 8×10 + 10×6)
= 2 (48 + 80 + 60)
= 2 (188)
= 376 dm
2
(iii) Given details are,
Length of a cuboid = 2m
Breadth of a cuboid = 4m
Height of a cuboid = 5m
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (2×4 + 4×5 + 5×2)
= 2 (8 + 20 + 10)
= 2 (38)
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
= 76 m
2
(iv) Given details are,
Length of a cuboid = 3.2 m= 32 dm
Breadth of a cuboid = 30 dm
Height of a cuboid = 250 cm= 25 dm
We know that,
surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (32×30 + 30×25 + 25×32)
= 2 (960 + 750 + 800)
= 2 (2510)
= 5020 dm
2
2. Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m
Solution:
(i) Given,
Edge of cube = 1.2 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 1.2
2
= 6 × 1.44
= 8.64 m
2
(ii) Given,
Edge of cube = 27 cm
We know that,
Surface area of cube = 6 × side
2
= 6 × 27
2
= 6 × 729
= 4374 cm
2
(iii) Given,
Edge of cube = 3 cm
We know that,
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Surface area of cube = 6 × side
2
= 6 × 3
2
= 6 × 9
= 54 cm
2
(iv) Given,
Edge of cube = 6 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 6
2
= 6 × 36
= 216 m
2
(v) Given,
Edge of cube = 2.1 m
We know that,
Surface area of cube = 6 × side
2
= 6 × 2.1
2
= 6 × 4.41
= 26.46 m
2
3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
Solution:
Given details are,
Dimensions of cuboidal box = 5cm × 5cm × 4cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
= 2 (5×5 + 5×4 + 4×5)
= 2 (25 + 20 + 20)
= 2 (65)
= 130 cm
2
4. Find the surface area of a cube whose volume is
(i) 343 m
3
(ii) 216 dm
3
Solution:
(i) Given details are,
Volume of cube = 343 m
3
Side of cube, a =
3
v(343) = 7m
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
We know that,
Surface area of cube = 6 × side
2
= 6 × 7
2
= 6 × 49
= 294 m
2
(ii) Given details are,
Volume of cube = 216 dm
3
Side of cube a =
3
v(216) = 6dm
We know that,
Surface area of cube = 6 × side
2
= 6 × 6
2
= 6 × 36
= 216 dm
2
5. Find the volume of a cube whose surface area is
(i) 96 cm
2
(ii) 150 m
2
Solution:
(i) Given details are,
Surface area of cube = 96 cm
2
6 × side
2
= 96cm
2
Side
2
= 96/6
= 16
Side = v16 = 4cm
? Volume of a cube = 4
3
= 64cm
3
(ii) Given details are,
Surface area of cube = 150 m
2
6 × side
2
= 150cm
2
Side
2
= 150/6
= 25
Side = v25 = 5cm
? Volume of a cube = 5
3
= 125m
3
6. The dimensions of a cuboid are in the ratio 5: 3: 1 and its total surface area is 414
m
2
. Find the dimensions.
Solution:
Given details are,
RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration –
II (Volumes and Surface Areas of a Cuboid and a Cube)
Ratio of dimensions of a cuboid = 5:3:1
Total surface area of cuboid = 414 m
2
The dimensions are = 5x × 3x × x
Surface area of cuboid = 414 m
2
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm
2
2 (lb + bh + hl) cm
2
= 414
2 (15x
2
+ 3x
2
+ 5x
2
) = 414
2 (23x
2
) = 414
46x
2
= 414
x
2
= 414/46
= 9
x = v9
= 3
? Dimensions are,
5x = 5 (3) = 15m
3x = 3 (3) = 9m
x = 3m
7. Find the area of the cardboard required to make a closed box of length 25 cm, 0.5
m and height 15 cm.
Solution:
Given details are,
Dimensions of closed box = 25cm × 0.5m × 15cm = 25cm × 50cm × 15cm
We know that,
Area of cardboard required = 2 (lb + bh + hl) cm
2
= 2 (25×50 + 50×15 + 15×25)
= 2 (1250 + 750 + 375)
= 2 (2375)
= 4750 cm
2
8. Find the surface area of a wooden box whose shape is of a cube, and if the edge of
the box is 12 cm.
Solution:
Given details are,
Edge of a cubic wooden box = 12 cm
We know that,
Surface area of cubic wooden box = 6 × side
2
= 6 × 12
2
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.90/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 have been prepared according to the Class 8 exam syllabus. Information about Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 covers topics
like and Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3.
Introduction of Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 | 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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3
In this doc you can find the meaning of Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 defined & explained in the simplest way possible. Besides explaining types of
Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 theory, EduRev gives you an ample number of questions to practice Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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
ppt
,shortcuts and tricks
,Free
,mock tests for examination
,Exam
,Previous Year Questions with Solutions
,practice quizzes
,video lectures
,Viva Questions
,Summary
,Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 | Mathematics (Maths) Class 8
,Extra Questions
,MCQs
,Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 | Mathematics (Maths) Class 8
,Objective type Questions
,Sample Paper
,past year papers
,Important questions
,Semester Notes
,study material
,Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 | Mathematics (Maths) Class 8
;
Additional Information about Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 for Class 8 Preparation
Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 Free PDF Download
The Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 now and kickstart your journey towards success in the Class 8 exam.
Importance of Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3
The importance of Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 Notes
Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3.
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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 Notes on EduRev are your ultimate resource for success.
Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 Class 8 Questions
The "Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 on the App
Students of Class 8 can study Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3,
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 - II (Volumes and Surface Areas of a Cuboid and a Cube) - EXERCISE 21.3 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