RD Sharma Chapter 19 Class 9 Maths Exercise 19.1 Solutions have been provided over here. These exercises are primarily covered around the problems related to calculating the surface area of the cylinder. Upon regular practice of the problems revolving around the calculation of the surface area of a circular cylinder can be significant in terms of having a greater understanding of its topics. Specifically, the solved questions here can be useful for more significant preparations on it.

The exercises here cover the topics around calculating the surface area of a right circular cylinder and calculating the surface area of hollow cylinders. The solutions are prepared by our subject experts to assure a better understanding of the exercise, which helps to score well in the exam. However, those having better fundamentals about solving problems around cylinders of both hollow and right cylinders can find it easier while dealing with RD Sharma problems.

**Learn about RD Sharma Class 9 Chapter 19 (Surface Area And Volume Of Right Circular Cylinder)**

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## Download RD Sharma Chapter 19 Class 9 Maths Exercise 19.1 Solutions PDF

## Important Definitions RD Sharma Chapter 19 Class 9 Maths Exercise 19.1 Solutions

### Surface Area of a Right Circular Cylinder

The Right Circular Cylinder’s surface area can be calculated using formula A = 2 πrh+2πr2 where ‘r’ is the radius and h is the height.

### Surface Area of a Hollow Cylinder

The surface area of a hollow cylinder can be calculated through the formula 2π (r1 + r2) (r2 – r1 +h), in which r1 is the inner Radius, r2 is the Outer Radius, and ‘h’ is the height.

### Examples of RD Sharma Chapter 19 Class 9 Maths Exercise 19.1 Solutions

**Ques- Curved surface section of a right circular cylinder is 4.4 m2. Suppose the radius of the bottom of the cylinder is 0.7 m. Find its height.**

**Solution-**

= Radius of the bottom of the cylinder = r = 0.7 m (Given)

= Curved surface section of cylinder = C.S.A = 4.4m2 (Given)

= Let ‘k’ be the height of the cylinder.

= We know, the curved surface section of a cylinder = 2πrk

= Therefore,

= 2πrk = 4.4

= 2 x 3.14 (22/7) x 0.7 x k = 4.4

= [using π=3.14 ]

= or k = 1

= Hence, the height of a cylinder is 1m.

**Ques- In a hot water heating system, there is a cylindrical tube of the length of 28 m and a diameter of 5 cm. Find the entire radiating surface in the system.**

**Solution-**

= Height of cylinder (k) = Length of cylindrical tube = 28 m or 2800 cm (Given)

[1 m = 100 cm]

= Diameter of circular end of tube = 5 cm (given)

= Let ‘s’ be the radius of circular end, then s = diameter/2 = 5/2 cm

= We know, Curved surface area of the cylindrical pipe = 2πsk

= 2 x 3.14 (22/7) x 5/2 x 2800

= [using π = 3.14 9 (22/7)]

= 44000

= Therefore, the space of the radiating surface is 44000 cm2.

**Ques- The cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the rounded surface of the pillar at the rate of Rs 12.50 per m2.**

**Solution-**

= Height of the cylindrical pillar (h) = 3.5 m

= Radius of circular end of the pillar (r) = 50/2 cm = 25 cm = 0.25 m

= [As radius = half of a diameter] and [1 m = 100 cm]

= The rounded surface space of the cylindrical pillar = 2πrh

= 2 x 3.14 (22/7) x 0.25 x 3.5

= 5.5

= So, the curved surface area of the cylindrical pillar is 5.5 m.

**Find the cost-**

= Cost of the whitewashing 1m2 is Rs 12.50 (Given)

= Cost of thewhitewashing 5.5 m2 area = Rs. 12.50 x 5.5 = Rs. 68.75

= Therefore the cost of whitewashing the pillar is Rs 68.75.