Maharashtra Board Solutions Class 8-Maths (Practice Set 16.2): Chapter 16- Surface Area and Volume

Class 8: Maths Chapter 16 solutions. Complete Class 8 Maths Chapter 16 Notes.

Maharashtra Board Solutions Class 8-Maths (Practice Set 16.2): Chapter 16- Surface Area and Volume

Maharashtra Board 8th Maths Chapter 16, Class 8 Maths Chapter 16 solutions

Question 1.
In each example given below, radius of base of a cylinder and its height are given. Then find the curved surface area and total surface area.
i. r = 7 cm, h = 10 cm
ii. r = 1.4 cm, h = 2.1 cm
iii. r = 2.5 cm, h = 7 cm
iv. r = 70 cm, h = 1.4 cm
v. r = 4.2 cm, h = 14 cm
Solution:
i. Given: r = 7 cm and h = 10 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x 227 x 7 x 10
= 2 x 22 x 10
= 440 sq.cm
Total surface area of the cylinder:
= 2πr(h + r)
= 2 x 227 x 7(10 + 7)
= 2 x 227 x 7 x 17
= 2 x 22 x 17
= 748 sq.cm
The curved surface area of the cylinder is 440 sq.cm and its total surface area is 748 sq.cm.

ii. Given: r = 1.4 cm and h = 2.1 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x 227 x 1.4 x 2.1
= 2 x 22 x 0.2 x 2.1
= 18.48 sq.cm
Total surface area of the cylinder = 2πr (h + r)
= 2 x 227 x 1.4 (2.1 + 1.4)
= 2 x 227 x 1.4 x 3.5
= 2 x 22 x 0.2 x 3.5
= 30.80 sq.cm
∴ The curved surface area of the cylinder is 18.48 sq.cm and its total surface area is 30.80 sq.cm.

iii. Given: r = 2.5 cm and h = 7 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x 227 x 2.5 x 7
= 2 x 22 x 2.5
= 110 sq.cm
Total surface area of the cylinder = 2πr(h + r)
= 2 x 227 x 2.5 (7+ 2.5)
= 2 x 227 x 2.5 x 9.5
= 10457
= 149.29 sq.cm
∴ The curved surface area of the cylinder is 110 sq.cm and its total surface area is 149.29 sq.cm.

iv. Given: r = 70 cm and h = 1.4 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x 227 x 70 x 1.4
= 2 x 22 x 10 x 1.4
= 616 sq.cm
Total surface area of the cylinder = 2πr(h + r)
= 2 x 227 x 70(1.4 + 70)
= 2 x 227 x 70 x 71.4
= 2 x 22 x 10 x 71.4
= 2 x 22 x 714
= 31416 sq.cm
∴ The curved surface area of the cylinder is 616 sq.cm and its total surface area is 31416 sq.cm.

v. Given: r = 4.2 cm and h = 14 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x 227 x 4.2 x 14 = 2 x 22 x 4.2 x 2
= 369.60 sq.cm
Total surface area of the cylinder = 2πr (h + r)
= 2 x 227 x 4.2 (14+ 4.2)
= 2 x 227 x 4.2 x 18.2
= 2 x 22 x 0.6 x 18.2
= 480.48 sq.cm
∴ The curved surface area of the cylinder is 369.60 sq.cm and its total surface area is 480.48 sq.cm.

Question 2.
Find the total surface area of a closed cylindrical drum if its diameter is 50 cm and height is 45 cm. (π = 3.14)
Given: For cylindrical drum:
Diameter (d) = 50 cm
and height (h) = 45 cm
To find: Total surface area of the cylindrical drum
Solution:
Diameter (d) = 50 cm
∴ radius (r) = d2=502 = 25 cm
Total surface area of the cylindrical drum = 2πr (h + r)
= 2 x 3.14 x 25 (45 + 25)
= 2 x 3.14 x 25 x 70
= 10,990 sq.cm
∴ The total surface area of the cylindrical drum is 10,990 sq.cm.

Question 3.
Find the area of base and radius of a cylinder if its curved surface area is 660 sq.cm and height is 21 cm.
Given: Curved surface area = 660 sq.cm, and height = 21 cm
To find: area of base and radius of a cylinder
Solution:
i. Curved surface area of cylinder = 2πrh
∴ 660 = 2 x 227 x r x 21
∴ 660 = 2 x 22 x r x 3
∴ 6602×22×3=r
∴ 6602×66=r
∴ 5 = r
i.e., r = 5 cm

ii. Area of a base of the cylinder = πr²
= 227 x 5 x 5
= 5507
= 78.57 sq.cm
∴The radius of the cylinder is 5 cm and the area of its base is 78.57 sq.cm.

Question 4.
Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.
Given: For cylindrical container:
diameter (d) = 28 cm, height (h₁) = 20 cm
For cylindrical lid: height (h₂) = 2 cm
To find: i. Surface area of the cylinder with one side open
ii. Area of sheet required to made a lid
Solution:
diameter (d) = 28 cm
∴ radius (r) = d2=282 = 14 cm
i. Surface area of the cylinder with one side open = Curved surface area + Area of a base
= 2πrh₁ + πr²
= πr (2h₁ + r)
= 227 x 14 x (2 x 20 + 14)
= 22 x 2 x (40 + 14)
= 22 x 2 x 54
= 2376 sq.cm

ii. Area of sheet required to made a lid = Curved surface area of lid + Area of upper surface
= 2πrh₂ + πr²
= πr (2h₂ + r)
= 227 x 14 x (2 x 2 + 14)
= 22 x 2 x (4 + 14)
= 22 x 2 x 18
= 792 sq cm
∴ The area of the sheet required to make the cylindrical container is 2376 sq. cm and the approximate area of a sheet required to make the lid is 792 sq. cm.

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Maharashtra Board Solutions Class 8-Maths (Practice Set 16.2): Chapter 16- Surface Area and Volume

Download PDF: Maharashtra Board Solutions Class 8-Maths (Practice Set 16.2): Chapter 16- Surface Area and Volume PDF

Chapterwise Maharashtra Board Solutions Class 8 Maths :

• Chapter 1- Rational and Irrational Numbers (Practice Set 1.1)
• Chapter 1- Rational and Irrational Numbers (Practice Set 1.2)
• Chapter 1- Rational and Irrational Numbers (Practice Set 1.3)
• Chapter 1- Rational and Irrational Numbers (Practice Set 1.4)
• Chapter 2- Parallel Lines and Transversals (Practice Set 2.1)
• Chapter 2- Parallel Lines and Transversals (Practice Set 2.2)
• Chapter 2- Parallel Lines and Transversals (Practice Set 2.3)
• Chapter 3- Indices and Cube Root (Practice Set 3.1)
• Chapter 3- Indices and Cube Root (Practice Set 3.2)
• Chapter 3- Indices and Cube Root (Practice Set 3.3)
• Chapter 4- Altitudes and Medians of a Triangle (Practice Set 4.1)
• Chapter 5- Expansion Formulae (Practice Set 5.1)
• Chapter 5- Expansion Formulae (Practice Set 5.2)
• Chapter 5- Expansion Formulae (Practice Set 5.3)
• Chapter 5- Expansion Formulae (Practice Set 5.4)
• Chapter 6- Factorisation of Algebraic Expressions (Practice Set 6.1)
• Chapter 6- Factorisation of Algebraic Expressions (Practice Set 6.2)
• Chapter 6- Factorisation of Algebraic Expressions (Practice Set 6.3)
• Chapter 6- Factorisation of Algebraic Expressions (Practice Set 6.4)
• Chapter 7- Variation (Practice Set 7.1)
• Chapter 7- Variation (Practice Set 7.2)
• Chapter 7- Variation (Practice Set 7.3)
• Chapter 8- Quadrilateral: Constructions and Types (Practice Set 8.1)
• Chapter 8- Quadrilateral: Constructions and Types (Practice Set 8.2)
• Chapter 8- Quadrilateral: Constructions and Types (Practice Set 8.3)
• Chapter 9- Discount and Commission (Practice Set 9.1)
• Chapter 9- Discount and Commission (Practice Set 9.2)
• Chapter 10- Division of Polynomials (Practice Set 10.1)
• Chapter 10- Discount and Commission (Practice Set 10.2)
• Chapter 11- Statistics (Practice Set 11.1)
• Chapter 11- Statistics (Practice Set 11.2)
• Chapter 11- Statistics (Practice Set 11.3)
• Chapter 12- Equations in One Variable (Practice Set 12.1)
• Chapter 12- Equations in One Variable (Practice Set 12.2)
• Chapter 13- Congruence of Triangles (Practice Set 13.1)
• Chapter 13- Congruence of Triangles (Practice Set 13.2)
• Chapter 14- Compound Interest (Practice Set 14.1)
• Chapter 14- Compound Interest (Practice Set 14.2)
• Chapter 15- Area (Practice Set 15.1)
• Chapter 15- Area (Practice Set 15.2)
• Chapter 15- Area (Practice Set 15.3)
• Chapter 15- Area (Practice Set 15.4)
• Chapter 15- Area (Practice Set 15.5)
• Chapter 15- Area (Practice Set 15.6)
• Chapter 16- Surface Area and Volume (Practice Set 16.1)
• Chapter 16- Surface Area and Volume (Practice Set 16.2)
• Chapter 16- Surface Area and Volume (Practice Set 16.3)
• Chapter 17- Factorisation of Algebraic Expressions (Practice Set 17.1)
• Chapter 17- Factorisation of Algebraic Expressions (Practice Set 17.2)