Geometry and Volume Study Guide: Rectangular Prisms, Pyramids, and Compound Solids

Given Dimensions: * The base of the net is a square with side lengths of $3\,\text{m}$ . * The net consists of four identical triangular faces that meet at a point when assembled. * The slant height (the height of each triangular face) is $5\,\text{m}$ .
Surface Area Components: * Base Area (Square): Calculated using the formula $A_{\text{base}} = s^2$ . * $3\,\text{m} \times 3\,\text{m} = 9\,\text{m}^2$ * Lateral Area (Four Triangles): Calculated using the formula for the area of a triangle $A_{\text{triangle}} = \frac{1}{2} \times b \times h$ . * Area of one triangle: $\frac{1}{2} \times 3\,\text{m} \times 5\,\text{m} = 7.5\,\text{m}^2$ * Total lateral area: $4 \times 7.5\,\text{m}^2 = 30\,\text{m}^2$
Total Surface Area: * $\text{Total Area} = \text{Base Area} + \text{Lateral Area}$ * $9\,\text{m}^2 + 30\,\text{m}^2 = 39\,\text{m}^2$

Given Dimensions: * Width: $4\,\text{in}$ * Height: $13\,\text{in}$ * Length: $10\,\text{in}$
Volume Formula: $V = l \times w \times h$
Calculation: * $10\,\text{in} \times 4\,\text{in} \times 13\,\text{in}$ * $40\,\text{in}^2 \times 13\,\text{in} = 520\,\text{in}^3$

Rectangular Container Specifications: * Height: $20\,\text{cm}$ * Base Dimensions: $4\,\text{cm} \times 8\,\text{cm}$ * Volume Calculation: $20\,\text{cm} \times 4\,\text{cm} \times 8\,\text{cm} = 640\,\text{cm}^3$
Cube Toy Specifications: * Edge length: $4\,\text{cm}$ * Volume Calculation: $V = s^3 = 4\,\text{cm} \times 4\,\text{cm} \times 4\,\text{cm} = 64\,\text{cm}^3$
Quantity of Toys within Container: * To find how many toys fit, divide the total container volume by the volume of one cube. * $\frac{640\,\text{cm}^3}{64\,\text{cm}^3} = 10\,\text{toys}$

Given Data: * The rhombus is composed of four congruent right triangles defined by diagonals intersecting at a central point. * Horizontal semi-diagonal: $6\,\text{cm}$ * Vertical semi-diagonal: $4\,\text{cm}$
Calculation Method: * Total horizontal diagonal ( $d_1$ ): $6\,\text{cm} + 6\,\text{cm} = 12\,\text{cm}$ * Total vertical diagonal ( $d_2$ ): $4\,\text{cm} + 4\,\text{cm} = 8\,\text{cm}$ * Area Formula: $A = \frac{d_1 \times d_2}{2}$ * Calculation: $\frac{12\,\text{cm} \times 8\,\text{cm}}{2} = 48\,\text{cm}^2$

Box Specifications: * The provided net indicates the box is a cube with side lengths of $5\,\text{in}$ .
Part A: Surface Area (Total area covered by paint): * A cube has $6$ identical square faces. * Area of one face: $5\,\text{in} \times 5\,\text{in} = 25\,\text{in}^2$ * Total surface area: $6 \times 25\,\text{in}^2 = 150\,\text{in}^2$
Part B: Total Edge Length (Ribbon calculation): * A cube has $12$ edges. * Total inches of ribbon: $12 \times 5\,\text{in} = 60\,\text{in}$ * Conversion to feet: Since $12\,\text{in} = 1\,\text{ft}$ , $\frac{60\,\text{in}}{12\,\text{in/ft}} = 5\,\text{ft}$ .

Structure Description: The solid is formed by two joined prisms.
Prism 1 (Left): * Dimensions: $6\,\text{ft} \times 3\,\text{ft} \times 4\frac{1}{2}\,\text{ft}$ * Fractional conversion: $4\frac{1}{2} = 4.5$ * Volume: $6\,\text{ft} \times 3\,\text{ft} \times 4.5\,\text{ft} = 81\,\text{ft}^3$
Prism 2 (Right): * Dimensions: $9\,\text{ft} \times 10\,\text{ft} \times 4\frac{1}{2}\,\text{ft}$ * Volume: $9\,\text{ft} \times 10\,\text{ft} \times 4.5\,\text{ft} = 405\,\text{ft}^3$
Total Volume: * $81\,\text{ft}^3 + 405\,\text{ft}^3 = 486\,\text{ft}^3$

Given Dimensions: * Base 1 ( $b_1$ ): $19\,\text{cm}$ * Base 2 ( $b_2$ ): $11\,\text{cm}$ * Height ( $h$ ): $15\,\text{cm}$ * Slant side length: $13\,\text{cm}$
Area Formula: $A = \frac{b_1 + b_2}{2} \times h$
Calculation: * $\frac{19\,\text{cm} + 11\,\text{cm}}{2} \times 15\,\text{cm}$ * $\frac{30\,\text{cm}}{2} \times 15\,\text{cm} = 15\,\text{cm} \times 15\,\text{cm} = 225\,\text{cm}^2$

Given Dimensions: * Length: $14\,\text{in}$ * Width: $8\,\text{in}$ * Height: $5\,\text{in}$ * Additional value provided in figure: $35\,\text{in}$ (Context suggests this might be the area of one face, but standard volume requires the three primary edges).
Calculation: * $V = 14\,\text{in} \times 8\,\text{in} \times 5\,\text{in}$ * $14\,\text{in} \times 40\,\text{in}^2 = 560\,\text{in}^3$