- GST
- Banking
- Shares and Dividends
- Quadratic Equations in One Variable
- Factorization
- Ratio and Proportion
- Matrices
- Arithmetic and Geometric Progression
- Reflection
- Section Formula
- Equation of Straight Line
- Similarity
- Locus
- Circles
- Constructions
- Mensuration
- Trigonometric Identities
- Trigonometric Tables
- Heights and Distances
ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.1 in Chapter 17 - Mensuration
Prev
The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm.
Find its total surface area.
Solution:
The amount of space that is covered by the curved surface and flat surface of a cylinder's bases is known as the cylinder's surface area.
Given height of the metal pipe = 77 cm
Inner diameter = 4 cm
Inner radius, r = d/2 = 4/2 = 2 cm
Outer diameter = 4.4 cm
Outer radius, R = d/2 = 4.4/2 = 2.2 cm
Area of ring = (R2-r2)
= (22/7)×(2.22-22)
= (22/7)×(4.84-4)
= (22/7)×0.84
= 2.64
Total surface area = inner surface area + outer surface area + area of two rings
= 968+ 1064.8+ 2×2.64
= 2038.08 cm2
Hence the total surface area of the metal pipe is 2038.08 cm2. 
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
