A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr². Is the following statement true or false and justify your answer
Solution:
Given, a solid cylinder of radius r and height h is placed over another cylinder of same height and radius.
We have to determine if the total surface area of the shape so formed is 4πrh + 4πr².
Total surface area = curved surface area + area of the bases.
Curved surface area is defined as the area of only curved surface leaving the top and bottom bases.
Curved surface area of cylinder = 2πrh
Area of bases = 2πr²
So, the total surface area of the cylinder = 2πrh + 2πr²
Given, two cylinders of same dimensions are placed one over another.
Total surface area of cylinder = 2(Total surface of single cylinder) – 2(Area of base of cylinder)
=2(2πrh + 2πr²) - 2(πr²)
= 4πrh + 2πr²
Therefore, the total surface area of the shape so formed is 4πrh + 2πr².
✦ Try This: A solid cylinder has base radius 5 cm and height 7 cm. Find the volume of the cylinder.
Given, radius of cylinder = 5 cm
Height of cylinder = 7 cm
We have to find the volume of the cylinder.
Volume of the cylinder = πr²h
= (22/7)(5)²(7)
= (22)(25)
= 550 cm³
Therefore, the volume of the cylinder is 550 cm³.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 13
NCERT Exemplar Class 10 Maths Exercise 12.2 Problem 2
A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr². Is the following statement true or false and justify your answer
Summary:
The statement “A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr².” is false
☛ Related Questions:
- A solid cone of radius r and height h is placed over a solid cylinder having same base radius and he . . . .
- A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is (4/3)πa³. . . . .
- The volume of the frustum of a cone is πh/3[r₁² + r₂² - r₁r₂], where h is vertical height of the fru . . . .