Volume of Compound Figures worksheet with six problems requiring calculation of shaded regions' volumes.

Let's solve each of these compound figure volume problems step by step. We are to find the volume of the shaded region in each figure, rounding to two decimal places using π = 3.14.

---

Problem 1:

Figure: A rectangular prism with a triangular prism on top.

- Rectangular prism:
- Length = 20 ft
- Width = 18 ft
- Height = 6 ft

- Triangular prism on top:
- Base triangle: base = 20 ft (same as rectangle), height = 7 ft
- The triangular prism has the same length as the base (20 ft)

#### Volume of rectangular prism:
$$
V_{\text{rect}} = \text{length} \times \text{width} \times \text{height} = 20 \times 18 \times 6 = 2160 \text{ ft}^3
$$

#### Volume of triangular prism:
$$
V_{\text{tri}} = \frac{1}{2} \times \text{base} \times \text{height} \times \text{length} = \frac{1}{2} \times 20 \times 7 \times 20 = 1400 \text{ ft}^3
$$

#### Total volume:
$$
V = 2160 + 1400 = 3560.00 \text{ ft}^3
$$

✔ Answer: 3560.00 ft³

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Problem 2:

Figure: A triangular prism with a cylinder removed from the middle.

- Triangular prism:
- Base triangle: base = 26 in, height = 15 in
- Length (depth) = 26 in

- Cylinder removed:
- Diameter = 13 in → radius = 6.5 in
- Height = 26 in (same as depth)

#### Volume of triangular prism:
$$
V_{\text{tri}} = \frac{1}{2} \times 26 \times 15 \times 26 = \frac{1}{2} \times 26 \times 15 \times 26
$$
First compute:
$$
\frac{1}{2} \times 26 = 13,\quad 13 \times 15 = 195,\quad 195 \times 26 = 5070 \text{ in}^3
$$

#### Volume of cylinder:
$$
V_{\text{cyl}} = \pi r^2 h = 3.14 \times (6.5)^2 \times 26
$$
$$
6.5^2 = 42.25,\quad 3.14 \times 42.25 = 132.615,\quad 132.615 \times 26 = 3447.99 \text{ in}^3
$$

#### Shaded volume (prism minus cylinder):
$$
V = 5070 - 3447.99 = 1622.01 \text{ in}^3
$$

✔ Answer: 1622.01 in³

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Problem 3:

Figure: A rectangular prism with a cylindrical hole through it.

- Rectangular prism:
- Length = 20 yd
- Width = 15 yd
- Height = 13 yd

- Cylindrical hole:
- Diameter = 10 yd → radius = 5 yd
- Height = 13 yd (same as prism height)

#### Volume of rectangular prism:
$$
V_{\text{rect}} = 20 \times 15 \times 13 = 3900 \text{ yd}^3
$$

#### Volume of cylinder:
$$
V_{\text{cyl}} = \pi r^2 h = 3.14 \times (5)^2 \times 13 = 3.14 \times 25 \times 13
$$
$$
= 3.14 \times 325 = 1020.5 \text{ yd}^3
$$

#### Shaded volume:
$$
V = 3900 - 1020.5 = 2879.50 \text{ yd}^3
$$

✔ Answer: 2879.50 yd³

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Problem 4:

Figure: A cylinder with a hemisphere on top.

- Cylinder:
- Diameter = 16 ft → radius = 8 ft
- Height = 10 ft

- Hemisphere:
- Radius = 8 ft

#### Volume of cylinder:
$$
V_{\text{cyl}} = \pi r^2 h = 3.14 \times (8)^2 \times 10 = 3.14 \times 64 \times 10 = 2009.6 \text{ ft}^3
$$

#### Volume of hemisphere:
$$
V_{\text{hemi}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times (8)^3 = \frac{2}{3} \times 3.14 \times 512
$$
$$
= \frac{2}{3} \times 1607.68 = 1071.79 \text{ ft}^3
$$

#### Total volume:
$$
V = 2009.6 + 1071.79 = 3081.39 \text{ ft}^3
$$

✔ Answer: 3081.39 ft³

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Problem 5:

Figure: Three stacked hemispheres (like a stack of half-spheres).

Each hemisphere has:
- Radius = 6 in

Note: Two hemispheres make a full sphere.

But here we have three hemispheres, so:
- Equivalent to 1.5 spheres

#### Volume of one sphere:
$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (6)^3 = \frac{4}{3} \times 3.14 \times 216
$$
$$
= \frac{4}{3} \times 678.24 = 904.32 \text{ in}^3
$$

#### Volume of 1.5 spheres:
$$
V = 1.5 \times 904.32 = 1356.48 \text{ in}^3
$$

Alternatively, since each hemisphere is:
$$
V_{\text{hemi}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times 216 = \frac{2}{3} \times 678.24 = 452.16 \text{ in}^3
$$
Then total for 3 hemispheres:
$$
3 \times 452.16 = 1356.48 \text{ in}^3
$$

✔ Answer: 1356.48 in³

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Problem 6:

Figure: A rectangular prism with a triangular pyramid (tetrahedron) removed from inside.

- Rectangular prism:
- Length = 28 yd
- Width = 12 yd
- Height = 12 yd

- Triangular pyramid (tetrahedron):
- Base: right triangle with legs 12 yd and 12 yd
- Height of pyramid = 12 yd (from apex to base)

Wait — the diagram shows a pyramid inscribed in the cube-like box. It appears that the pyramid has:
- Base: triangle with base = 28 yd, height = 12 yd?
But looking closely: the apex is at the center-top, and the base is a triangle formed by three corners.

Actually, this is a triangular pyramid (tetrahedron) with:
- Base: right triangle with legs 28 yd and 12 yd? But wait — the base seems to be on the bottom face.

Wait — better interpretation:

The figure shows a rectangular prism with a triangular pyramid cut out from it.

From the diagram:
- The pyramid has:
- Base: triangle with base = 28 yd, height = 12 yd (on the base of the prism)
- Height of pyramid = 12 yd (from base up to apex)

But actually, the pyramid is drawn with its apex at the top center, and base forming a triangle on the bottom face.

So:
- The base of the pyramid is a right triangle with legs 28 yd and 12 yd
- The height of the pyramid is 12 yd (vertical height)

But wait — the prism is 28 yd × 12 yd × 12 yd.

And the pyramid has:
- Base: triangle with base = 28 yd, height = 12 yd → area = $ \frac{1}{2} \times 28 \times 12 = 168 $
- Height of pyramid = 12 yd

Volume of pyramid:
$$
V_{\text{pyr}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 168 \times 12 = 672 \text{ yd}^3
$$

Volume of rectangular prism:
$$
V_{\text{rect}} = 28 \times 12 \times 12 = 4032 \text{ yd}^3
$$

Shaded volume = Prism − Pyramid:
$$
V = 4032 - 672 = 3360.00 \text{ yd}^3
$$

✔ Answer: 3360.00 yd³

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✔ Final Answers:

1) 3560.00 ft³
2) 1622.01 in³
3) 2879.50 yd³
4) 3081.39 ft³
5) 1356.48 in³
6) 3360.00 yd³

Let me know if you'd like diagrams or further explanation!