The curved surface area of a frustum of a cone is πl(r₁ + r₂), where l = √h² + (r₁ + r₂)², r₁ and r₂ are the radii of the two ends of the frustum and h is the vertical height. Is the following statement true or false and justify your answer
Solution:
We know that
Curved surface area of a frustum of a cone = πl(r₁ + r₂),
Where l is the slant height
l = √h² + (r₁ - r₂)²,
Where h is the vertical height,
r₁ and r₂ are the radii of the two ends of the frustum
Therefore, the statement is false.
✦ Try This: Find the surface area of a frustum of cone given below:
We know that
Surface area of a frustum of cone = π(r + R)s + πr² + πR²
From the figure
R = 3 mm
r = 2 mm
s = 15 mm
Substituting the values in the formula
Surface area of a frustum of cone = π(2 + 3)15 + π(2)² + π(3)²
By further calculation
= 75π + 4π + 9π
= 88π mm²
Therefore, the surface area of a frustum of cone is 88π mm².
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 13
NCERT Exemplar Class 10 Maths Exercise 12.2 Problem 7
The curved surface area of a frustum of a cone is πl(r₁ + r₂), where l = √h² + (r₁ + r₂)², r₁ and r₂ are the radii of the two ends of the frustum and h is the vertical height. Is the following statement true or false and justify your answer
Summary:
The statement “The curved surface area of a frustum of a cone is πl(r₁ + r₂), where l = √h² + (r₁ + r₂)², r₁ and r₂ are the radii of the two ends of the frustum and h is the vertical height” is false
☛ Related Questions:
- An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base . . . .
- A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone an . . . .
- A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each . . . .