Maha Board (MSBHSE) Class 8 Math (गणित) Surface area and Volume Practice Set 16.2 Solution (Page No. 110) in our website. Here candidates of Marathi Medium & English Medium of Maharashtra State can follow this to get Most Easy Solution of Maharashtra Board Class 8 Chapter 16: Surface area and Volume (Practice Set 16.2).
Maharashtra Board Class 8 Math Practice Set 16.2 Solution | BalBharati Class 8 Mathematics Chapter 16 Solution:
Subject | Mathematics (गणित) |
Class | 8 (८) |
Chapter | 16 |
Practice Set | 16.2 |
Practice Set 16.2 Solution (Page No. 110)
(1) In each example given below, radius of base of a cylinder and and its height are given. Then find the curved surface area and total surface area.
(1) r = 7 cm, h = 10 cm
Solution: Given r = 7 cm, h = 10 cm,
∴ Curved Surface area = 2πrh
= 2π x 7 x 10
= 2 x 3.14 x 7 x 10
= 439.6 sq cm
Total Surface area = 2πr(r + h)
= 2 x 3.14 x 7(7 + 10)
= 2 x 3.14 x 7 x 17
= 747.32 sq cm.
(2) r = 1.4 cm, h = 2.1 cm
Solution: Given, r = 1.4 cm, h = 2.1 cm.
∴ Curved Surface area = 2πrh
= 2π x 1.4 x 2.1
= 2 x 22/7 x 1.4 x 2.1
= 18.48 sq cm.
Total surface area = 2πr(r + h)
= 2 x 22/7 x 1.4 (1.4 + 2.1)
= 2 x 22/7 x 1.4 x 3.5
= 30.80 sq cm.
(3) r = 2.5 cm, h = 7 cm
Solution: Given, r = 2.5 cm, h = 7 cm.
∴ Curved surface area = 2πrh
= 2 x 22/7 x 2.5 x 7
= 110 sq cm.
Total Surface area = 2πr(r + h)
= 2 x 22/7 x 2.5 (2.5 + 7)
= 2 x 22/7 x 2.5 x 9.5
= 149.28 sq cm.
(4) r = 70 cm, h = 1.4 cm
Solution: Given, r = 70 cm, h = 1.4 cm.
∴ Curved surface area = 2πrh
= 2 x 22/7 x 70 x 1.4
= 616 sq cm.
∴ Total surface area = 2πr (r + h)
= 2 x 22/7 x 70 (70 + 1.4)
= 2 x 22/7 x 70 x 71.4
= 31416 sq cm.
(5) r = 4.2 cm, h = 14 cm
Solution: Given, r = 4.2 cm, h = 14 cm.
∴ Curved Surface area = 2πrh
= 2 x 22/7 x 4.2 x 14
= 369.60 sq cm.
Total Surface area = 2πr(r + h)
= 2 x 22/7 x 4.2 (4.2 + 14)
= 2 x 22/7 x 4.2 x 18.2
= 480.48 sq cm.
(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)
Solution: Diameter = 50 cm
=) r = 50/2
=) 25 cm
Total Surface area of the closed cylindrical drum
= 2πr( r + h)
= 2 x 22/7 x 25 (25 + 45)
= 2 x 3.14 x 25 x 70
= 10990 sq cm.
(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.
Solution: Curved Surface area = 660 sq cm.
We know that, 2πrh = 660
=) 2 x 22/7 x r x 21 = 660
=) r = 660 x 7/ 2 x 21 x 22
= 5 cm.
Area of base = πr2
= 22/7 x 5 x 5
= 78.57 sq cm.
(4) Find the area of the sheet required to make a cyclindrical 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.
Rules:
Curved surface area of cylinder = Area of rectangle
= length x breadth
= Circumference of the base of the cylinder x height of the cylinder
= 2πr x h
= 2πrh
* Total Surface area
= Curved Surface Area + 2 x area of cylinder faces
= 2πrh + 2πr
= 2πr(r + h)
Latest Edition BalBharati Math Solution for Standard 8th Chapter 16.2. Any Kinds of problem to understand, Feel free to comment us below. Thank You.