Maharashtra State BoardSSC (English Medium) 8th Standard
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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. - Mathematics

Sum

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.
(1) r = 7 cm, h = 10 cm
(2) r = 1.4 cm, h = 2.1 cm
(3) r = 2.5 cm, h = 7 cm
(4) r = 70 cm, h = 1.4 cm
(5) r = 4.2 cm, h = 14 cm

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Solution

We know that
Curved surface area of a cylinder = 2 π rh
Total surface area = 2 πr (h + r)

(1) r = 7 cm, h = 10 cm
Curved surface area of a cylinder = 2 π rh = 2 π × 7 × 10 = 140 π = 140 × `22/7` = 440 sq cm

Total surface area = 2 π r (h + r) = 2 × `22/7` (10 + 7) = 748 sq cm 

(2) r = 1.4 cm, h = 2.1 cm
Curved surface area of a cylinder = 2 π rh = 2 × `22/7` × 1.4 × 2.1 = 18.48 sq cm

Total surface area = 2 π r (h + r) = 2 × `22/7` ×1.4 (2.1 + 1.4) = 30.8 sq cm 

(3) r = 2.5 cm, h = 7 cm
Curved surface area of a cylinder = 2 π rh = 2 × `22/7` × 2.5× 7 = 110 sq cm

Total surface area = 2 π r (h + r) = 2 × `22/7` × 2.5 (2.5 + 7) = 149.28 sq cm 

(4) r = 70 cm, h = 1.4 cm
Curved surface area of a cylinder = 2 π rh = 2 × `22/7` × 70 × 1.4 = 616 sq cm

Total surface area = 2 π r (h + r) = 2 × `22/7` × 70  (70  + 1.4) = 31416 sq cm 

(5) r = 4.2 cm, h = 14 cm
Curved surface area of a cylinder = 2 π rh = 2 × `22/7` × 4.2 × 14 = 369.6 sq cm

Total surface area = 2 π r (h + r) = 2 × `22/7` × 4.2  (4.2  + 1.4) = 480.48 sq cm 

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APPEARS IN

Balbharati Mathematics 8th Standard Maharashtra State Board
Chapter 16 Surface area and Volume
Practice Set 16.2 | Q 1 | Page 110
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