Frustum of a Regular Pyramid Formula Sample Problems

Question 1: The area of base b₁ of Frustum is 80 m² and the area of base b₂ of Frustum is 20 m². If the height of the frustum is 3 m, what will be the volume of the Frustum.

Solution: 

We know, Volume of Frustum = (b₁ + b₂ + (b₁ × b₂)^1/2) × h/3

Given b₁ = 80 m²

b₂ = 20 m²

h = 30 m

Putting values in the given Volume of Frustum formula,

(b₁ + b₂ + (b₁ × b₂)^1/2) × h/3

= (80 +  20 + (80 × 20)^1/2) × 30/3

= (100 + (1600)^1/2) × 1

= 100 + 40

= 140 m³

Question 2: The perimeter of the base₁ of the frustum is 50 m and the perimeter of the base₂ of the frustum is 10 m. If the slant height of the frustum is 6 m, what will be the lateral surface area of the Frustum.

Solution:  

We know, Lateral Surface Area of Frustum = (p₁ + p₂) × s/2 

Given p₁ = 50 m

p₂ = 10 m

s = 6 m

Putting values in the given Lateral Surface Area of Frustum 

= (p₁ + p₂) × s/2

= (50 + 10) × 6/2

= 60 × 6 / 2

= 180 m

Question 3: The area of base b₁ of Frustum is 25 m² and the area of base b₂ of Frustum is 4 m². If the height of frustum is 0.3 m, what will be the volume of Frustum.

Solution: 

We know, Volume of Frustum = (b₁ + b₂ + (b₁ × b₂)^1/2) × h/3

Given b₁ = 25 m²

b₂ = 4 m²

h = 0.3 m

Putting values in the given Volume of Frustum formula

(b₁ + b₂ + (b₁ x b₂)^1/2) × h/3

(25 +  4 + (25 x 4)^1/2) × 0.3/3

= (29 + 10)/ 10

= 39/10

= 3.9 m³

Question 4: The perimeter of the base₁ of the frustum is 45 m and the perimeter of the base₂ of the frustum is 15 m. If the slant height of the frustum is 5 m, what will be the lateral surface area of the Frustum.

Solution:  

We know, Lateral Surface Area of Frustum = (p₁ + p₂) × s/2

Given p₁ = 45 m

p₂ = 15 m

s = 5 m

Putting values in the given Lateral Surface Area of Frustum

= (p₁ + p₂) × s/2

= (45 + 15) × 5/2

= 60 × 5 / 2

= 30 × 5

= 150 m

Question 5: The area of base b₁ of Frustum is 10 m² and the area of base b₂ of Frustum is 40 m². If the height of the frustum is 6 m, what will be the volume of the Frustum.

Solution: 

We know, Volume of Frustum = (b₁ + b₂ + (b₁ × b₂)^1/2) × h/3

Given b₁ = 10 m²

b₂ = 40 m²

h = 6 m

Putting values in the given Volume of Frustum formula

(b₁ + b₂ + (b₁ × b₂)^1/2) × h/3

(10 +  40 + (10 × 40)^1/2) × 6/3

= (50 + 20) × 2

= 140 m³

Question 6: The perimeter of the base₁ of the frustum is 75 m and the perimeter of the base₂ of the frustum is 25 m. If the slant height of the frustum is 7 m, what will be the lateral surface area of the Frustum.

Solution:  

We know, Lateral Surface Area of Frustum = (p₁ + p₂) × s/2

Given p₁ = 75 m

p₂ = 25 m

s = 7 m

Putting values in the given Lateral Surface Area of Frustum

= (p₁ + p₂) × s/2

= (75 + 25) × 7/2

= 100 × 7 / 2

= 100 × 3.5

= 350 m

Question 7: The area of base b₁ of Frustum is 80 m² and the area of base b₂ of Frustum is 20 m². If the height of the frustum is 3 m, what will be the volume of the Frustum.

Solution: 

We know, Volume of Frustum = (b₁ + b₂ + (b₁ × b₂)^1/2) × h/3

Given b₁ = 80 m²

b₂ = 20 m²

h = 30 m

Putting values in the given Volume of Frustum formula

(b₁ + b₂ + (b₁ x b₂)^1/2) × h/3

= (80 +  20 + (80 x 20)^1/2) × 30/3

= (100 + (1600)^1/2) × 1

= 100 + 40

= 140 m³

A Pyramid is a Mathematical figure having three or four triangular faces as sides and a flat polygonal base which can be triangular, square or rectangular, etc. The side triangular faces are called Lateral faces. The common meeting point of all the triangular faces is called the apex. For a given pyramid having a base with ‘b’ sides has ‘2b’ edges and ‘b + 1’ faces and vertices.