8th Grade Math Worksheets | Free Downloads Available - Free Printable

To solve the problem of calculating the volumes of pyramids, cones, and spheres, we will use the following formulas:

1. Volume of a Pyramid:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]

2. Volume of a Cone:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the base and \( h \) is the height.

3. Volume of a Sphere:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.

Let's solve each problem step by step.

---

Problem 1: Pyramid

- Base: Square with side length \( 4 \, \text{cm} \)
- Height: \( 9 \, \text{cm} \)

Step 1: Calculate the area of the base.
\[
\text{Base Area} = 4 \times 4 = 16 \, \text{cm}^2
\]

Step 2: Use the volume formula for a pyramid.
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 16 \times 9 = \frac{1}{3} \times 144 = 48 \, \text{cm}^3
\]

Answer:
\[
\boxed{48.00}
\]

---

Problem 2: Pyramid

- Base: Rectangle with dimensions \( 12 \, \text{cm} \times 6 \, \text{cm} \)
- Height: \( 15 \, \text{cm} \)

Step 1: Calculate the area of the base.
\[
\text{Base Area} = 12 \times 6 = 72 \, \text{cm}^2
\]

Step 2: Use the volume formula for a pyramid.
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 72 \times 15 = \frac{1}{3} \times 1080 = 360 \, \text{cm}^3
\]

Answer:
\[
\boxed{360.00}
\]

---

Problem 3: Pyramid

- Base: Triangle with base \( 7 \, \text{cm} \) and height \( 5 \, \text{cm} \)
- Height: \( 20 \, \text{cm} \)

Step 1: Calculate the area of the triangular base.
\[
\text{Base Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 5 = \frac{1}{2} \times 35 = 17.5 \, \text{cm}^2
\]

Step 2: Use the volume formula for a pyramid.
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 17.5 \times 20 = \frac{1}{3} \times 350 = 116.67 \, \text{cm}^3
\]

Answer:
\[
\boxed{116.67}
\]

---

Problem 4: Pyramid

- Base: Trapezoid with bases \( 3 \, \text{cm} \) and \( 2 \, \text{cm} \), and height \( 3 \, \text{cm} \)
- Height: \( 7 \, \text{cm} \)

Step 1: Calculate the area of the trapezoidal base.
\[
\text{Base Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} = \frac{1}{2} \times (3 + 2) \times 3 = \frac{1}{2} \times 5 \times 3 = \frac{1}{2} \times 15 = 7.5 \, \text{cm}^2
\]

Step 2: Use the volume formula for a pyramid.
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 7.5 \times 7 = \frac{1}{3} \times 52.5 = 17.50 \, \text{cm}^3
\]

Answer:
\[
\boxed{17.50}
\]

---

Problem 5: Cone

- Radius: \( 3 \, \text{cm} \)
- Height: \( 9 \, \text{cm} \)

Step 1: Use the volume formula for a cone.
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (9) = \frac{1}{3} \pi \times 9 \times 9 = \frac{1}{3} \pi \times 81 = 27\pi \, \text{cm}^3
\]

Step 2: Approximate the value.
\[
V \approx 27 \times 3.1416 = 84.82 \, \text{cm}^3
\]

Answer:
\[
\boxed{84.82}
\]

---

Problem 6: Cone

- Radius: \( 1.5 \, \text{cm} \)
- Height: \( 7 \, \text{cm} \)

Step 1: Use the volume formula for a cone.
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.5)^2 (7) = \frac{1}{3} \pi \times 2.25 \times 7 = \frac{1}{3} \pi \times 15.75 = 5.25\pi \, \text{cm}^3
\]

Step 2: Approximate the value.
\[
V \approx 5.25 \times 3.1416 = 16.49 \, \text{cm}^3
\]

Answer:
\[
\boxed{16.49}
\]

---

Problem 7: Cone

- Radius: \( 20 \, \text{cm} \)
- Height: \( 24 \, \text{cm} \)

Step 1: Use the volume formula for a cone.
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (20)^2 (24) = \frac{1}{3} \pi \times 400 \times 24 = \frac{1}{3} \pi \times 9600 = 3200\pi \, \text{cm}^3
\]

Step 2: Approximate the value.
\[
V \approx 3200 \times 3.1416 = 10053.09 \, \text{cm}^3
\]

Answer:
\[
\boxed{10053.09}
\]

---

Problem 8: Cone

- Diameter: \( 7.2 \, \text{mm} \)
- Radius: \( \frac{7.2}{2} = 3.6 \, \text{mm} \)
- Height: \( 7.7 \, \text{mm} \)

Step 1: Use the volume formula for a cone.
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3.6)^2 (7.7) = \frac{1}{3} \pi \times 12.96 \times 7.7 = \frac{1}{3} \pi \times 100.032 = 33.344\pi \, \text{mm}^3
\]

Step 2: Approximate the value.
\[
V \approx 33.344 \times 3.1416 = 104.72 \, \text{mm}^3
\]

Answer:
\[
\boxed{104.72}
\]

---

Final Answers

\[
\boxed{48.00, 360.00, 116.67, 17.50, 84.82, 16.49, 10053.09, 104.72}
\]