The radii of the ends of a frustum of a cone of height h cm are r₁ cm and r₂ cm. The volume in cm³ of the frustum of the cone is
a. πh/3[r₁² + r₂² + r₁r₂]
b. πh/3[r₁² + r₂² - r₁r₂]
c. πh/3[r₁² - r₂² + r₁r₂]
d. πh/3[r₁² - r₂² - r₁r₂]
Solution:
Given, the radii of frustum of a cone are r₁ cm and r₂ cm.
The height of the frustum cone is h cm.
We have to find the volume of the frustum cone in cm³
Consider a frustum of a cone with slant height ‘l’ and height ‘h’
We know that cross sectional area of the cone = (1/3)πh[r₁² + r₂² + r₁r₂]
Extend sides AD and BC to meet at O.
Frustum of a cone is the difference of two right circular cones OAB and OCD.
Let h₁ and l₁ be the eight and slant height of cone OAB
Let h₂ and l₂ be the height and slant height of cone OCD
Considering triangles OPA and OQD,
As the cones are right circular, ∠OPA = ∠OQD = 90°
∠POA = ∠QOD = common
By AAA criteria, the triangles OPA and OQD are similar.
By the property of similar triangles,
The corresponding sides are proportional
PA/QD = OA/DO = PO/OQ
From the figure,
r₁/r₂ = h₁/h₂ = l₁/l₂
So, r₁/r₂ = h₁/h₂ (or) r₂/r₁ = h₂/h₁
Subtracting 1 from both sides,
r₁/r₂ - 1= h₁/h₂ - 1
(r₁ - r₂)/r₂ = (h₁ - h₂)/h₂
From the figure,
h₁ - h₂ = h
(r₁ - r₂)/r₂ = h/h₂
h₂ = r₂h/(r₁ - r₂) ------------------------- (1)
Considering r₂/r₁ = h₂/h₁
Subtracting 1 from both sides,
r₂/r₁ - 1= h₂/h₁ - 1
(r₂ - r₁)/r₁ = (h₂ - h₁)/h₁
(r₁ - r₂)/r₁ = (h₁ - h₂)/h₁
(r₁ - r₂)/r₁ = h/h₁
h₁ = r₁h/(r₁ - r₂) --------------------------- (2)
Volume of frustum cone = volume of cone OAB - volume of cone OCD
Volume of cone OAB = (1/3)r₁²h₁
Volume of cone OCD = (1/3)r₂²h₂
Volume of frustum cone = (1/3)r₁²h₁ - (1/3)r₂²h₂
Substituting (1) and (2),
= π/3[r₁²(r₁h/(r₁ - r₂)) - r₂²(r₂h/(r₁ - r₂))]
= πh/3 [(r₁³ - r₂³)/(r₁ - r₂)]
By using algebraic identity,
(a³ - b³) = (a - b)(a² + ab + b²)
= πh/3[(r₁ - r₂)(r₁² + r₁r₂ + r₂²)/(r₁ - r₂)]
= πh/3[r₁² + r₁r₂ + r₂²]
Therefore, the volume of frustum cone is πh/3[r₁² + r₁r₂ + r₂²]
✦ Try This: The radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Find its slant height.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 13
NCERT Exemplar Class 10 Maths Exercise 12.1 Sample Problem 4
The radii of the ends of a frustum of a cone of height h cm are r₁ cm and r₂ cm. The volume in cm³ of the frustum of the cone is a. πh/3[r₁² + r₂² + r₁r₂], b. πh/3[r₁² + r₂² - r₁r₂], c. πh/3[r₁² - r₂² + r₁r₂], d. πh/3[r₁² - r₂² - r₁r₂]
Summary:
The radii of the ends of a frustum of a cone of height h cm are r₁ cm and r₂ cm. The volume in cm³ of the frustum of the cone is πh/3[r₁² + r₁r₂ + r₂²]
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