Find The Volume Of The Composite Figure

Find the volume of the composite figure is a common task in geometry that combines the volumes of simpler solids to determine the total space occupied by a more complex shape. Whether you are calculating the capacity of a water tank built from a cylinder and a hemisphere, or determining how much material is needed for a sculpture made of a prism and a pyramid, mastering this skill is essential for both academic success and real‑world problem solving. In this guide we will walk through the concept, the step‑by‑step procedure, and several worked examples that illustrate how to handle addition and subtraction of volumes in composite figures.

What Is a Composite Figure?

A composite figure (also called a composite solid) is a three‑dimensional shape formed by joining two or more basic solids—such as prisms, cylinders, pyramids, cones, and spheres—along their faces. The resulting object may look irregular, but its volume can be found by breaking it down into the known parts, calculating each part’s volume, and then combining those volumes appropriately.

Key points to remember

• Addition is used when solids are attached to each other (e.g., a cylinder sitting on top of a prism).
• Subtraction is used when a portion of a solid is removed (e.g., a cylindrical hole drilled through a rectangular block). - The units of volume are always cubic units (cm³, m³, in³, etc.), so keep all measurements in the same unit before computing.

Step‑by‑Step Procedure to Find the Volume of a Composite Figure

Follow these logical steps to ensure accuracy and avoid common pitfalls.

1. Identify the basic solids that make up the composite figure.
   Look for familiar shapes: rectangular prisms, triangular prisms, cylinders, cones, pyramids, spheres, or hemispheres.

2. Separate the figure mentally (or on paper) into those basic solids.
   Sketch a quick diagram if it helps; label each part with its dimensions.

3. Write down the volume formula for each basic solid.

   • Rectangular prism: (V = l \times w \times h)
   • Triangular prism: (V = \frac{1}{2} b h_{\text{tri}} \times l) - Cylinder: (V = \pi r^{2} h)
   • Cone: (V = \frac{1}{3} \pi r^{2} h)
   • Pyramid: (V = \frac{1}{3} B h) (where (B) is the area of the base)
   • Sphere: (V = \frac{4}{3} \pi r^{3})
   • Hemisphere: (V = \frac{2}{3} \pi r^{3})

4. Calculate each volume using the given dimensions.
   Keep π as π (or use 3.1416 if a decimal approximation is required) until the final step to minimize rounding error.

5. Combine the volumes according to how the solids are joined: - Add volumes for solids that are attached externally.

   • Subtract the volume of any removed solid (hole, cavity) from the volume of the larger solid.

6. State the final answer with the correct cubic units.
   Double‑check that all length measurements were in the same unit; if not, convert before computing.

Worked Examples### Example 1: Rectangular Prism with a Half‑Cylinder on Top

Problem: A storage container consists of a rectangular prism (length = 10 cm, width = 6 cm, height = 8 cm) topped by a half‑cylinder whose diameter equals the width of the prism (6 cm) and whose length equals the prism’s length (10 cm). Find the total volume.

Solution:

1. Identify solids: rectangular prism + half‑cylinder.
2. Prism volume:
   [ V_{\text{prism}} = lwh = 10 \times 6 \times 8 = 480 \text{ cm}^3 ]
3. Half‑cylinder:
   • Radius (r = \frac{6}{2}=3) cm
   • Length (height of cylinder) (h = 10) cm - Full cylinder volume: (V_{\text{cyl}} = \pi r^{2} h = \pi \times 3^{2} \times 10 = 90\pi) cm³
   • Half of that: (V_{\text{half‑cyl}} = \frac{1}{2} \times 90\pi = 45\pi) cm³ ≈ 141.37 cm³
4. Add volumes:
   [ V_{\text{total}} = 480 + 45\pi \approx 480 + 141.37 = 621.37 \text{ cm}^3 ]
5. Answer: The container holds approximately 621.4 cm³ of material.

Example 2: Cone on Top of a Cylinder (a “Silos” Shape)

Problem: A grain silo is made of a cylinder (radius = 4 m, height = 10 m) with a cone (same radius, height = 3 m) on its roof. Find the volume of the silo.

Solution:

1. Cylinder volume: [ V_{\text{cyl}} = \pi r^{2} h = \pi \times 4^{2} \times 10 = 160\pi \text{ m}^3 ] 2. Cone volume:
   [ V_{\text{cone}} = \frac{1}{3} \pi r^{2} h = \frac{1}{3} \pi \times 4^{2} \times 3 = \frac{1}{3} \pi \times 16 \times 3 = 16\pi \text{ m}^3 ]
2. Add:
   [ V_{\text{total}} = 160\pi + 16\pi = 176\pi \text{ m}^3 \approx 552.92 \text{ m}^3 ]
3. Answer: The silo can hold about 553 m³ of grain.

Example 3: Rectangular Block with a Cylindrical Hole (Subtraction)

Problem: A solid rectangular block measures 12 cm × 8 cm × 5 cm. A cylindrical hole of radius 2 cm is drilled completely through the block along its