Volume of Composite Shapes Worksheet - Calculate the volume of each shaded region.

To solve the problem of finding the volumes of the shaded regions in the composite shapes, we need to break down each shape into simpler geometric components and calculate their volumes step by step. Below is a detailed explanation for each part:

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Problem 1:

The first shape consists of a cylinder with a hemisphere removed from its top.

#### Step 1: Volume of the Cylinder
The formula for the volume of a cylinder is:
\[
V_{\text{cylinder}} = \pi r^2 h
\]
Here, the radius \( r = 3 \) cm and the height \( h = 5 \) cm.
\[
V_{\text{cylinder}} = \pi (3)^2 (5) = \pi \cdot 9 \cdot 5 = 45\pi \, \text{cm}^3
\]

#### Step 2: Volume of the Hemisphere
The formula for the volume of a sphere is:
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
Since we have a hemisphere, its volume is half of the sphere's volume:
\[
V_{\text{hemisphere}} = \frac{1}{2} \left( \frac{4}{3} \pi r^3 \right) = \frac{2}{3} \pi r^3
\]
Here, the radius \( r = 3 \) cm.
\[
V_{\text{hemisphere}} = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi \cdot 27 = 18\pi \, \text{cm}^3
\]

#### Step 3: Volume of the Shaded Region
The shaded region is the volume of the cylinder minus the volume of the hemisphere:
\[
V_{\text{shaded}} = V_{\text{cylinder}} - V_{\text{hemisphere}} = 45\pi - 18\pi = 27\pi \, \text{cm}^3
\]

Using \( \pi \approx 3.1416 \):
\[
V_{\text{shaded}} \approx 27 \cdot 3.1416 = 84.8232 \, \text{cm}^3
\]

Rounding to two decimal places:
\[
V_{\text{shaded}} \approx 84.82 \, \text{cm}^3
\]

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Problem 2:

The second shape consists of a rectangular prism with a cylindrical hole through its center.

#### Step 1: Volume of the Rectangular Prism
The formula for the volume of a rectangular prism is:
\[
V_{\text{prism}} = l \cdot w \cdot h
\]
Here, the dimensions are \( l = 10 \) cm, \( w = 6 \) cm, and \( h = 4 \) cm.
\[
V_{\text{prism}} = 10 \cdot 6 \cdot 4 = 240 \, \text{cm}^3
\]

#### Step 2: Volume of the Cylindrical Hole
The formula for the volume of a cylinder is:
\[
V_{\text{cylinder}} = \pi r^2 h
\]
Here, the radius \( r = 2 \) cm and the height \( h = 4 \) cm.
\[
V_{\text{cylinder}} = \pi (2)^2 (4) = \pi \cdot 4 \cdot 4 = 16\pi \, \text{cm}^3
\]

#### Step 3: Volume of the Shaded Region
The shaded region is the volume of the rectangular prism minus the volume of the cylindrical hole:
\[
V_{\text{shaded}} = V_{\text{prism}} - V_{\text{cylinder}} = 240 - 16\pi
\]

Using \( \pi \approx 3.1416 \):
\[
V_{\text{shaded}} \approx 240 - 16 \cdot 3.1416 = 240 - 50.2656 = 189.7344 \, \text{cm}^3
\]

Rounding to two decimal places:
\[
V_{\text{shaded}} \approx 189.73 \, \text{cm}^3
\]

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Problem 3:

The third shape consists of a triangular prism with a conical cavity.

#### Step 1: Volume of the Triangular Prism
The formula for the volume of a triangular prism is:
\[
V_{\text{prism}} = \text{Base Area} \cdot \text{Height}
\]
The base is a right triangle with legs \( 6 \) cm and \( 8 \) cm. The area of the triangle is:
\[
\text{Base Area} = \frac{1}{2} \cdot 6 \cdot 8 = 24 \, \text{cm}^2
\]
The height of the prism is \( 10 \) cm.
\[
V_{\text{prism}} = 24 \cdot 10 = 240 \, \text{cm}^3
\]

#### Step 2: Volume of the Conical Cavity
The formula for the volume of a cone is:
\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h
\]
Here, the radius \( r = 3 \) cm and the height \( h = 10 \) cm.
\[
V_{\text{cone}} = \frac{1}{3} \pi (3)^2 (10) = \frac{1}{3} \pi \cdot 9 \cdot 10 = 30\pi \, \text{cm}^3
\]

#### Step 3: Volume of the Shaded Region
The shaded region is the volume of the triangular prism minus the volume of the conical cavity:
\[
V_{\text{shaded}} = V_{\text{prism}} - V_{\text{cone}} = 240 - 30\pi
\]

Using \( \pi \approx 3.1416 \):
\[
V_{\text{shaded}} \approx 240 - 30 \cdot 3.1416 = 240 - 94.248 = 145.752 \, \text{cm}^3
\]

Rounding to two decimal places:
\[
V_{\text{shaded}} \approx 145.75 \, \text{cm}^3
\]

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Final Answers:

\[
\boxed{84.82, 189.73, 145.75}
\]