Pltw Ied Activity 5.1 Calculating Properties Of Shapes Answer Key

The ability to accurately calculate the properties of shapes, such as area, perimeter, volume, and surface area, is a foundational skill in various fields, ranging from engineering and architecture to computer graphics and even everyday problem-solving. Within the context of Project Lead The Way (PLTW) Introduction to Engineering Design (IED), Activity 5.1 focuses specifically on mastering these calculations for common geometric shapes. Understanding the correct formulas and applying them with precision is crucial for success in this activity and for building a solid foundation for future engineering studies.

This article will serve as a comprehensive guide to understanding and calculating the properties of shapes as required in PLTW IED Activity 5.1, providing a detailed breakdown of the necessary formulas, practical examples, and insights into common pitfalls to avoid. While providing general instruction, the goal is to equip you with the knowledge and skills to confidently tackle any shape-related calculations.

Understanding the Fundamentals: Key Properties of Shapes

Before diving into specific calculations, it's essential to define the key properties we'll be working with:

• Area: The measure of the two-dimensional space enclosed by a shape. It is typically expressed in square units (e.g., square inches, square meters).
• Perimeter: The total distance around the outside of a two-dimensional shape. It is expressed in linear units (e.g., inches, meters).
• Volume: The measure of the three-dimensional space occupied by a solid object. It is expressed in cubic units (e.g., cubic inches, cubic meters).
• Surface Area: The total area of all the surfaces of a three-dimensional object. It is expressed in square units (e.g., square inches, square meters).

Two-Dimensional Shapes: Formulas and Examples

Let's explore the formulas for calculating the area and perimeter of common two-dimensional shapes.

1. Square

• Area: Area = side * side = s^2 where s is the length of a side.
• Perimeter: Perimeter = 4 * side = 4s

Example: A square has a side length of 5 inches.

• Area = 5 inches * 5 inches = 25 square inches
• Perimeter = 4 * 5 inches = 20 inches

2. Rectangle

• Area: Area = length * width = lw where l is the length and w is the width.
• Perimeter: Perimeter = 2 * (length + width) = 2(l + w)

Example: A rectangle has a length of 8 cm and a width of 3 cm.

• Area = 8 cm * 3 cm = 24 square cm
• Perimeter = 2 * (8 cm + 3 cm) = 22 cm

3. Triangle

• Area: Area = 1/2 * base * height = 1/2 bh where b is the base and h is the height (perpendicular distance from the base to the opposite vertex).
• Perimeter: Perimeter = side1 + side2 + side3 (sum of all three sides)

Example: A triangle has a base of 6 inches, a height of 4 inches, and sides of length 5 inches, 5 inches, and 6 inches.

• Area = 1/2 * 6 inches * 4 inches = 12 square inches
• Perimeter = 5 inches + 5 inches + 6 inches = 16 inches

4. Circle

• Area: Area = π * radius^2 = πr^2 where r is the radius and π (pi) is approximately 3.14159.
• Circumference (Perimeter): Circumference = 2 * π * radius = 2πr or Circumference = π * diameter = πd where d is the diameter.

Example: A circle has a radius of 7 meters.

• Area = π * (7 meters)^2 = π * 49 square meters ≈ 153.94 square meters
• Circumference = 2 * π * 7 meters = 14π meters ≈ 43.98 meters

5. Parallelogram

• Area: Area = base * height = bh where b is the base and h is the height (perpendicular distance between the base and the opposite side).
• Perimeter: Perimeter = 2 * (side1 + side2) where side1 and side2 are the lengths of the adjacent sides.

Example: A parallelogram has a base of 10 cm, a height of 5 cm, and adjacent sides of length 6 cm and 10 cm.

• Area = 10 cm * 5 cm = 50 square cm
• Perimeter = 2 * (6 cm + 10 cm) = 32 cm

6. Trapezoid

• Area: Area = 1/2 * (base1 + base2) * height = 1/2 (b1 + b2)h where b1 and b2 are the lengths of the two parallel bases and h is the height (perpendicular distance between the bases).
• Perimeter: Perimeter = side1 + side2 + base1 + base2 (sum of all four sides)

Example: A trapezoid has bases of 8 inches and 12 inches, a height of 5 inches, and sides of length 6 inches and 7 inches.

• Area = 1/2 * (8 inches + 12 inches) * 5 inches = 50 square inches
• Perimeter = 6 inches + 7 inches + 8 inches + 12 inches = 33 inches

Three-Dimensional Shapes: Formulas and Examples

Now, let's explore the formulas for calculating the volume and surface area of common three-dimensional shapes.

1. Cube

• Volume: Volume = side * side * side = s^3 where s is the length of a side.
• Surface Area: Surface Area = 6 * side^2 = 6s^2 (since a cube has 6 equal square faces)

Example: A cube has a side length of 4 cm.

• Volume = 4 cm * 4 cm * 4 cm = 64 cubic cm
• Surface Area = 6 * (4 cm)^2 = 6 * 16 square cm = 96 square cm

2. Rectangular Prism (Cuboid)

• Volume: Volume = length * width * height = lwh
• Surface Area: Surface Area = 2 * (length * width + length * height + width * height) = 2(lw + lh + wh)

Example: A rectangular prism has a length of 7 inches, a width of 3 inches, and a height of 5 inches.

• Volume = 7 inches * 3 inches * 5 inches = 105 cubic inches
• Surface Area = 2 * (7 inches * 3 inches + 7 inches * 5 inches + 3 inches * 5 inches) = 2 * (21 + 35 + 15) square inches = 2 * 71 square inches = 142 square inches

3. Cylinder

• Volume: Volume = π * radius^2 * height = πr^2h
• Surface Area: Surface Area = 2 * π * radius * height + 2 * π * radius^2 = 2πrh + 2πr^2 (curved surface area + area of the two circular bases)

Example: A cylinder has a radius of 2 meters and a height of 6 meters.

• Volume = π * (2 meters)^2 * 6 meters = π * 4 square meters * 6 meters = 24π cubic meters ≈ 75.40 cubic meters
• Surface Area = 2 * π * 2 meters * 6 meters + 2 * π * (2 meters)^2 = 24π square meters + 8π square meters = 32π square meters ≈ 100.53 square meters

4. Cone

• Volume: Volume = 1/3 * π * radius^2 * height = 1/3 πr^2h
• Surface Area: Surface Area = π * radius * slant height + π * radius^2 = πrs + πr^2 where s is the slant height (the distance from the tip of the cone to a point on the edge of the circular base). If the slant height is not given, it can be calculated using the Pythagorean theorem: s = √(r^2 + h^2)

Example: A cone has a radius of 3 cm and a height of 4 cm.

• First, calculate the slant height: s = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 cm
• Volume = 1/3 * π * (3 cm)^2 * 4 cm = 1/3 * π * 9 square cm * 4 cm = 12π cubic cm ≈ 37.70 cubic cm
• Surface Area = π * 3 cm * 5 cm + π * (3 cm)^2 = 15π square cm + 9π square cm = 24π square cm ≈ 75.40 square cm

5. Sphere

• Volume: Volume = 4/3 * π * radius^3 = 4/3 πr^3
• Surface Area: Surface Area = 4 * π * radius^2 = 4πr^2

Example: A sphere has a radius of 5 inches.

• Volume = 4/3 * π * (5 inches)^3 = 4/3 * π * 125 cubic inches = 500π/3 cubic inches ≈ 523.60 cubic inches
• Surface Area = 4 * π * (5 inches)^2 = 4 * π * 25 square inches = 100π square inches ≈ 314.16 square inches

6. Pyramid (Square Base)

• Volume: Volume = 1/3 * base area * height = 1/3 * s^2 * h where s is the side length of the square base and h is the height (perpendicular distance from the apex to the base).
• Surface Area: Surface Area = base area + 4 * (1/2 * base * slant height) = s^2 + 2 * s * s_h where s_h is the slant height (the distance from the apex to the midpoint of a side of the base). The slant height can be found using the Pythagorean theorem if needed.

Example: A square pyramid has a base side length of 6 meters and a height of 4 meters.

• To find the slant height, imagine a right triangle formed by the height of the pyramid (4m), half the base side (3m), and the slant height. So, s_h = √(4^2 + 3^2) = √25 = 5 meters
• Volume = 1/3 * (6 meters)^2 * 4 meters = 1/3 * 36 square meters * 4 meters = 48 cubic meters
• Surface Area = (6 meters)^2 + 2 * 6 meters * 5 meters = 36 square meters + 60 square meters = 96 square meters

Composite Shapes: Breaking Down the Problem

Many real-world objects are composed of multiple basic shapes. To calculate their properties, you need to:

1. Decompose the shape: Identify the individual basic shapes that make up the composite shape.
2. Calculate individual properties: Calculate the area, perimeter, volume, or surface area of each individual shape.
3. Combine the results: Add or subtract the individual properties as needed to find the overall property of the composite shape. For example, if a shape consists of a rectangle with a triangle on top, you would add the area of the rectangle and the area of the triangle to find the total area. If a shape has a hole in it (like a washer), you would subtract the area of the hole from the area of the larger shape.

Example: A shape consists of a rectangle (length = 10 cm, width = 5 cm) with a semicircle on one of its shorter sides (radius = 2.5 cm).

1. Area of the rectangle = 10 cm * 5 cm = 50 square cm
2. Area of the semicircle = 1/2 * π * (2.5 cm)^2 = 1/2 * π * 6.25 square cm ≈ 9.82 square cm
3. Total area = 50 square cm + 9.82 square cm = 59.82 square cm

Tips for Success in PLTW IED Activity 5.1

• Understand the formulas: Memorize the formulas for common shapes. Create flashcards or a reference sheet.
• Draw diagrams: Visualizing the shape can help you identify the correct dimensions and apply the formulas correctly.
• Pay attention to units: Use consistent units throughout your calculations and include the units in your final answer.
• Double-check your work: Review your calculations carefully to avoid errors.
• Practice, practice, practice: The more you practice, the more comfortable you will become with calculating the properties of shapes.
• Use a calculator: A scientific calculator can help you with complex calculations, especially those involving π.
• Break down complex problems: Decompose complex shapes into simpler shapes that you can easily calculate.
• Understand the difference between perimeter/circumference, area, and volume: This seems obvious, but it's a common source of errors.
• Know when to use the Pythagorean theorem: This is often needed to find slant heights or other missing dimensions.

Common Mistakes to Avoid

• Using the wrong formula: Ensure you are using the correct formula for the specific shape.
• Incorrectly identifying dimensions: Double-check that you are using the correct values for length, width, height, radius, etc.
• Forgetting units: Always include units in your answers.
• Mixing up radius and diameter: Remember that the diameter is twice the radius.
• Rounding errors: Avoid rounding intermediate calculations too early, as this can lead to inaccuracies in the final answer. Keep as many decimal places as possible until the final step.
• Misunderstanding the concept of height: The height must be perpendicular to the base.
• Not accounting for all surfaces: When calculating surface area, make sure you include all faces of the object.

Example Problems and Solutions

Let's work through a few example problems to illustrate the application of these formulas.

Problem 1:

A rectangular garden is 12 feet long and 8 feet wide. What is the area of the garden, and how much fencing is needed to enclose it?

Solution:

• Area = length * width = 12 feet * 8 feet = 96 square feet
• Perimeter = 2 * (length + width) = 2 * (12 feet + 8 feet) = 2 * 20 feet = 40 feet

Therefore, the area of the garden is 96 square feet, and 40 feet of fencing is needed.

Problem 2:

A cylindrical tank has a radius of 4 meters and a height of 10 meters. What is the volume of the tank?

Solution:

• Volume = π * radius^2 * height = π * (4 meters)^2 * 10 meters = π * 16 square meters * 10 meters = 160π cubic meters ≈ 502.65 cubic meters

Therefore, the volume of the tank is approximately 502.65 cubic meters.

Problem 3:

A square pyramid has a base side length of 8 cm and a height of 6 cm. What is the volume of the pyramid?

Solution:

• Volume = 1/3 * base area * height = 1/3 * (8 cm)^2 * 6 cm = 1/3 * 64 square cm * 6 cm = 128 cubic cm

Therefore, the volume of the pyramid is 128 cubic cm.

Problem 4:

A composite shape consists of a square with a side length of 6 inches, with a circle cut out of the center. The circle has a diameter of 4 inches. What is the area of the composite shape?

Solution:

1. Area of the square = side * side = 6 inches * 6 inches = 36 square inches
2. Radius of the circle = diameter / 2 = 4 inches / 2 = 2 inches
3. Area of the circle = π * radius^2 = π * (2 inches)^2 = 4π square inches ≈ 12.57 square inches
4. Area of the composite shape = Area of square - Area of circle = 36 square inches - 12.57 square inches = 23.43 square inches

Therefore, the area of the composite shape is approximately 23.43 square inches.

Conclusion

Mastering the calculation of properties of shapes is a crucial skill for success in PLTW IED Activity 5.1 and in many other areas of engineering and design. By understanding the formulas, practicing regularly, and avoiding common mistakes, you can confidently tackle any shape-related calculation. Remember to break down complex problems into simpler steps, double-check your work, and always pay attention to units. With dedication and practice, you will develop a strong foundation in this essential skill. Good luck!