An open storage container is designed in the shape of a frustum of a right circular cone. … — JAMB UTME

An open storage container is designed in the shape of a frustum of a right circular cone. The top radius of the container is $4\text{ m}$ , the bottom radius is $1\text{ m}$ , and its vertical height is $7\text{ m}$ . Taking $\pi = \frac{22}{7}$ , what is the volume of the container in cubic metres ( $\text{m}^3$ )?

Answer: 154 m³

Answer

The volume of the container is

154\text{ m}^3

Using the volume formula for a frustum of a cone

V = \frac{1}{3}\pi h (R^2 + Rr + r^2)

with

R = 4\text{ m}

r = 1\text{ m}

h = 7\text{ m}

, and

\pi = \frac{22}{7}

yields

V = \frac{1}{3} \times \frac{22}{7} \times 7 \times (16 + 4 + 1) = 22 \times 7 = 154\text{ m}^3

Step-by-Step Solution

Identify the given dimensions of the frustum of the cone.

Top radius

R = 4\text{ m}

, bottom radius

r = 1\text{ m}

, height

h = 7\text{ m}

, and

\pi = \frac{22}{7}

Establishing known parameter values for the frustum volume formula.

Apply the standard formula for the volume of a frustum of a circular cone.

V = \frac{1}{3}\pi h (R^2 + Rr + r^2)

This formula accounts for the non-uniform cross-sectional area of a truncated cone.

Calculate the sum of the squares and the product of the radii.

R^2 + Rr + r^2 = 4^2 + (4)(1) + 1^2 = 16 + 4 + 1 = 21

Evaluating the quadratic radius factor in the frustum equation.

Substitute all numeric values and simplify.

V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 21 = 22 \times 7 = 154\text{ m}^3

Simplifying by cancelling

7

in the numerator and denominator, then dividing

21

3

yields

22 \times 7 = 154

Key Concept

Volume of a Frustum of a Cone

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