Volume of Composite Figures Worksheet - Practice Problems for Calculating Volume of Combined Geometric Shapes

Let's solve each of the four composite figure volume problems step by step. We'll use standard volume formulas:

- Rectangular prism: $ V = l \times w \times h $
- Cylinder: $ V = \pi r^2 h $
- Cone: $ V = \frac{1}{3} \pi r^2 h $
- Pyramid: $ V = \frac{1}{3} \times \text{Base Area} \times h $

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Problem 1: Two Rectangular Prisms

Description: A smaller rectangular prism (7 m high, 4 m wide, 5 m deep) sits on top of a larger one (12 m long, 9 m wide, 6 m high).

We need to find the total volume.

#### Step 1: Volume of the bottom prism
- Length = 12 m
- Width = 9 m
- Height = 6 m
$ V_1 = 12 \times 9 \times 6 = 648 \, \text{m}^3 $

#### Step 2: Volume of the top prism
- Length = 5 m (same as depth)
- Width = 4 m
- Height = 7 m
$ V_2 = 5 \times 4 \times 7 = 140 \, \text{m}^3 $

#### Total Volume:
$ V = V_1 + V_2 = 648 + 140 = \boxed{788} \, \text{m}^3 $

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Problem 2: Cylinder on Top of a Larger Cylinder

Given:
- Bottom cylinder: diameter = 12 cm → radius $ r = 6 $ cm, height = 10 cm
- Top cylinder: diameter = 8 cm → radius $ r = 4 $ cm, height = 6 cm

#### Volume of bottom cylinder:
$ V_1 = \pi r^2 h = \pi \times 6^2 \times 10 = \pi \times 36 \times 10 = 360\pi \, \text{cm}^3 $

#### Volume of top cylinder:
$ V_2 = \pi \times 4^2 \times 6 = \pi \times 16 \times 6 = 96\pi \, \text{cm}^3 $

#### Total Volume:
$ V = 360\pi + 96\pi = 456\pi \, \text{cm}^3 $

Using $ \pi \approx 3.14 $:
$ V \approx 456 \times 3.14 = \boxed{1428.24} \, \text{cm}^3 $

(Alternatively, leave as $ \boxed{456\pi} \, \text{cm}^3 $ if exact form is preferred.)

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Problem 3: Cone on Top of a Cylinder

Given:
- Cone: radius = 11 in, height = 10 in
- Cylinder: radius = 11 in, height = 20 in

#### Volume of cone:
$ V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 11^2 \times 10 = \frac{1}{3} \pi \times 121 \times 10 = \frac{1210}{3} \pi \approx 403.33\pi \, \text{in}^3 $

#### Volume of cylinder:
$ V_{\text{cyl}} = \pi r^2 h = \pi \times 11^2 \times 20 = \pi \times 121 \times 20 = 2420\pi \, \text{in}^3 $

#### Total Volume:
$ V = 403.33\pi + 2420\pi = 2823.33\pi \, \text{in}^3 $

Approximate value:
$ V \approx 2823.33 \times 3.14 \approx \boxed{8867.5} \, \text{in}^3 $

(Or exactly: $ \boxed{\frac{8470}{3}\pi} \, \text{in}^3 $)

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Problem 4: Pyramid on Top of a Cube

Given:
- Base: cube with side = 10 mm
- Pyramid: square base (10 mm × 10 mm), height = 13 mm

#### Volume of cube:
$ V_{\text{cube}} = 10 \times 10 \times 10 = 1000 \, \text{mm}^3 $

#### Volume of pyramid:
$ V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times h = \frac{1}{3} \times (10 \times 10) \times 13 = \frac{1}{3} \times 100 \times 13 = \frac{1300}{3} \approx 433.33 \, \text{mm}^3 $

#### Total Volume:
$ V = 1000 + 433.33 = \boxed{1433.33} \, \text{mm}^3 $

(Or exactly: $ \boxed{\frac{4300}{3}} \, \text{mm}^3 $)

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✔ Final Answers:

1. $ \boxed{788} \, \text{m}^3 $
2. $ \boxed{456\pi} \, \text{cm}^3 $ or $ \boxed{1428.24} \, \text{cm}^3 $
3. $ \boxed{2823.33\pi} \, \text{in}^3 $ or $ \boxed{8867.5} \, \text{in}^3 $
4. $ \boxed{1433.33} \, \text{mm}^3 $ or $ \boxed{\frac{4300}{3}} \, \text{mm}^3 $

Let me know if you'd like these rounded differently or formatted for a worksheet!