Honors Geometry Unit 7: Measurement & Modeling in 2D & 3D Review

Module 18 – Volume Formulas

• 18.1 – Volume of Prisms and Cylinders: This section covers the calculation of the total space occupied by three-dimensional solids with uniform cross-sections. The review worksheet identifies four specific types of solids for volume calculations:     * Right Triangular Prism: Requires calculating the area of the triangular base ($B = \frac{1}{2} \times b \times h$) and multiplying by the prism's height ($H$). The general formula used is $V = B \times H$.     * Rectangular Prism: A solid where all faces are rectangles. The volume is calculated using the formula $V = l \times w \times h$.     * Hexagonal Prism: Requires determining the area of a regular hexagonal base (often involving the apothem) and multiplying by the height of the prism.     * Cylinder: A solid with two congruent circular bases. The volume formula is given by $V = \pi \times r^2 \times h$, where $r$ is the radius of the base and $h$ is the vertical height.

• 18.2 – Volume of Pyramids: This module focuses on solids that taper from a base to a single point (the apex). The volume of a pyramid is exactly one-third the volume of a prism with the same base and height, represented by the formula $V = \frac{1}{3} \times B \times h$.     * Square Pyramid: A pyramid with a square base of side length $s$. The volume is $V = \frac{1}{3} \times s^2 \times h$.     * Hexagonal Pyramid: A pyramid with a six-sided base. The area of the hexagonal base ($B$) must be calculated first before applying the standard pyramid volume formula.

• 18.3 – Volume of Cones: Similar to pyramids, the volume of a cone is one-third the volume of a cylinder with the same radius and height.     * The universal formula for the volume of a cone is $V = \frac{1}{3} \times \pi \times r^2 \times h$.

• 18.4 – Volume of Spheres and Composite Figures:     * Standard Sphere: The volume is calculated using the radius ($r$) with the formula $V = \frac{4}{3} \times \pi \times r^3$.     * Sphere via Great Circle: Problem 12 describes a sphere with a great circle having an area $A = 9\pi\,in^2$. To find the volume, one must first solve for the radius from the area of the circle ($A = \pi \times r^2$) before using the volume formula.     * Composite Figures (Volume): Problems 9, 10, 13, and 14 involve figures made of two or more combined solids (e.g., a cone on top of a cylinder or a hemisphere on a prism). The total volume is the sum or difference of the individual volumes of the component parts.

Module 19 – Visualizing Solids

• 19.1 – Cross Sections and Solids of Rotation:     * Nets: Problems 15 and 16 require identifying three-dimensional solids based on their two-dimensional "unfolded" patterns or nets.     * Cross Sections: Problems 17 and 18 involve describing the two-dimensional shape created when a plane intersects a three-dimensional solid (e.g., a horizontal cross-section of a cylinder is a circle; a vertical cross-section of a cone through the apex is a triangle).     * Solids of Rotation: Problems 19 and 20 describe the 3D figure formed by rotating a 2D shape (like a rectangle or right triangle) around a fixed axis. For example, rotating a rectangle around one side forms a cylinder; rotating a right triangle around one leg forms a cone.

• 19.2 – Surface Area of Prisms and Cylinders:     * Square Prism: The sum of the areas of the two square bases and the four rectangular lateral faces.     * Triangular Prism: The sum of the areas of the two triangular bases and the three rectangular lateral faces.     * Cylinder: Calculated using the formula $SA = 2 \times \pi \times r^2 + 2 \times \pi \times r \times h$, representing the area of two circles plus the lateral area (circumference times height).     * Subtractive Solids: Problem 24 involves a $5\,inch$ cube with a cylinder removed. The surface area calculation must account for the exterior faces of the cube, the subtraction of the cylinder's circular tops from the cube's faces, and the addition of the cylinder's interior lateral surface area.

• 19.3 – Surface Area of Pyramids and Cones:     * Rectangular Pyramid: Includes the area of the rectangular base and the four triangular faces. This often requires calculating slant height ($l$).     * Pentagonal Pyramid: Base area of the pentagon plus five triangular lateral faces.     * Cone: The formula is $SA = \pi \times r^2 + \pi \times r \times l$, where $l$ is the slant height.     * Composite Surface Area: Problem 28 requires finding the total external surface area of a combined figure, ensuring that overlapping faces are not counted.

• 19.4 – Surface Area of Spheres:     * Sphere: The total surface area is determined by $SA = 4 \times \pi \times r^2$.     * Closed Hemisphere: Includes the curved surface area (half of a sphere) and the flat circular base area. The formula is $SA = 2 \times \pi \times r^2 + \pi \times r^2 = 3 \times \pi \times r^2$.

Module 20 – Modeling and Problem Solving

• 20.1 – Scale Factor:     * Parallelogram Scaling: In problem 31, if the dimensions of a parallelogram (base $7\,m$, height $5\,m$) are multiplied by a scale factor of $k = \frac{2}{5}$, the perimeter changes by a factor of $k$ and the area changes by a factor of $k^2 = (\frac{2}{5})^2 = \frac{4}{25}$.     * Sphere Volume Scaling: In problem 32, if the radius is multiplied by a scale factor of $k = \frac{6}{5}$, the volume changes by a factor of $k^3 = (\frac{6}{5})^3 = \frac{216}{125}$. Given an original volume of $6\pi\,mm^3$, the new volume is $6\pi \times \frac{216}{125}$.     * Circle Scaling: In problem 33, a radius is tripled ($k = 3$), resulting in an area of $72\pi$. To find the exact circumference of the original circle, find the original area ($\frac{72\pi}{3^2} = 8\pi$), solve for original radius ($r^2 = 8$), and find $C = 2 \times \pi \times r$.

• 20.2 – Modeling and Density:     * Population Density: Calculated as $\text{Density} = \frac{\text{Population}}{\text{Area}}$. For a town modeled as a rectangular area ($6\,miles \times 7.5\,miles$) with a population of $14,850$, the density is $\frac{14850}{45}\,\text{people/mile}^2$.     * Energy Density: Problem 35 describes a cylindrical tank for propane ($2550\,Btu/ft^3$). The tank must hold at least $30,000\,Btu$ and have a height of $2\,ft$. The required volume is $V = \frac{30000}{2550}$. Using $V = \pi \times r^2 \times h$, solve for the minimum radius $r$.     * Mass Density: For a solid cube of aluminum with sides of $0.5\,meters$ and density $2800\,kg/m^3$, the volume is $V = (0.5)^3 = 0.125\,m^3$. The mass is calculated as $\text{Mass} = \text{Density} \times \text{Volume} = 2800 \times 0.125$.

• 20.3 – Modeling and Problem Solving (Constraints):     * Rectangular Pool Design: Problem 37 specifies a capacity of at least $1920\,ft^3$, depth of $8\,ft$, and width of $12\,ft$. The length ($L$) must satisfy $8 \times 12 \times L \geq 1920$.     * Cylinder Constraints: Problem 38 specifies a volume of $850\,ft^3$ and a radius limit where $r \leq 6\,ft$. To find the smallest possible height, utilize the maximum possible radius in the formula $850 = \pi \times 6^2 \times h$.