# Maharashtra Board Class 8 Math Surface area and Volume Chapter Practice Set 2 Solution

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

= 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.

Updated: March 17, 2020 — 3:54 pm