Volume of a Pyramid Textbook Exercise - Corbettmaths - Free Printable

To solve the problem of finding the volume of each pyramid, we will use the formula for the volume of a pyramid:

\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]

We will calculate the volume for each pyramid step by step.

Step 1: Understand the Base Shapes

- For pyramids with rectangular bases, the base area is calculated as:
\[
\text{Base Area} = \text{Length} \times \text{Width}
\]
- For pyramids with triangular bases, the base area is calculated using the formula for the area of a triangle:
\[
\text{Base Area} = \frac{1}{2} \times \text{Base} \times \text{Height of Triangle}
\]

Step 2: Calculate the Volume for Each Pyramid

#### (a) Pyramid with a Rectangular Base
- Base Dimensions: \(9 \, \text{cm} \times 6 \, \text{cm}\)
- Height: \(7 \, \text{cm}\)
- Base Area:
\[
\text{Base Area} = 9 \times 6 = 54 \, \text{cm}^2
\]
- Volume:
\[
V = \frac{1}{3} \times 54 \times 7 = \frac{1}{3} \times 378 = 126 \, \text{cm}^3
\]

#### (b) Pyramid with a Triangular Base
- Base Dimensions: \(6 \, \text{cm} \times 8 \, \text{cm}\) (right triangle)
- Height: \(6 \, \text{cm}\)
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times 6 \times 8 = \frac{1}{2} \times 48 = 24 \, \text{cm}^2
\]
- Volume:
\[
V = \frac{1}{3} \times 24 \times 6 = \frac{1}{3} \times 144 = 48 \, \text{cm}^3
\]

#### (c) Pyramid with a Rectangular Base
- Base Dimensions: \(4 \, \text{cm} \times 3 \, \text{cm}\)
- Height: \(3 \, \text{m} = 300 \, \text{cm}\) (convert meters to centimeters)
- Base Area:
\[
\text{Base Area} = 4 \times 3 = 12 \, \text{cm}^2
\]
- Volume:
\[
V = \frac{1}{3} \times 12 \times 300 = \frac{1}{3} \times 3600 = 1200 \, \text{cm}^3
\]

#### (d) Pyramid with a Triangular Base
- Base Dimensions: \(7 \, \text{cm} \times 5 \, \text{cm}\) (isosceles triangle)
- Height: \(12 \, \text{cm}\)
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times 7 \times 5 = \frac{1}{2} \times 35 = 17.5 \, \text{cm}^2
\]
- Volume:
\[
V = \frac{1}{3} \times 17.5 \times 12 = \frac{1}{3} \times 210 = 70 \, \text{cm}^3
\]

#### (e) Pyramid with a Triangular Base
- Base Dimensions: \(15 \, \text{cm} \times 10 \, \text{cm}\) (right triangle)
- Height: \(20.6 \, \text{cm}\)
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times 15 \times 10 = \frac{1}{2} \times 150 = 75 \, \text{cm}^2
\]
- Volume:
\[
V = \frac{1}{3} \times 75 \times 20.6 = \frac{1}{3} \times 1545 = 515 \, \text{cm}^3
\]

#### (f) Pyramid with a Rectangular Base
- Base Dimensions: \(2 \, \text{m} \times 1.5 \, \text{m}\)
- Height: \(90 \, \text{cm} = 0.9 \, \text{m}\) (convert centimeters to meters)
- Base Area:
\[
\text{Base Area} = 2 \times 1.5 = 3 \, \text{m}^2
\]
- Volume:
\[
V = \frac{1}{3} \times 3 \times 0.9 = \frac{1}{3} \times 2.7 = 0.9 \, \text{m}^3
\]

Final Answers

\[
\boxed{
\begin{array}{ll}
(a) & 126 \, \text{cm}^3 \\
(b) & 48 \, \text{cm}^3 \\
(c) & 1200 \, \text{cm}^3 \\
(d) & 70 \, \text{cm}^3 \\
(e) & 515 \, \text{cm}^3 \\
(f) & 0.9 \, \text{m}^3 \\
\end{array}
}
\]