Chapter 4: Space, Shape and Measurements Study Guide

Chapter 4: Space, Shape and Measurements Overview

Learning Outcome 2: The student must be able to measure using appropriate instruments, estimate and calculate physical quantities, and interpret, describe, and represent properties of and relationships of two-dimensional objects in a variety of orientations and positions.
Assessment Standards: - Units of Measurements - Perimeter - Circumference - Area of the following shapes: - Triangle - Rectangle - Parallelogram - Square - Circle

4.1 Units of Measurement and Conversions

Specific standard identities for unit conversion include:

Volume and Capacity (Standard): - $1\,\text{cm}^3 = 1\,\text{ml}$ - $1000\,\text{cm}^3 = 1\,\text{dm}^3 = 1\,\text{l}$ - $1000\,\text{dm}^3 = 1\,\text{m}^3 = 1\,\text{kl}$
Area: - $1\,\text{ha} = 10\,000\,\text{m}^2$
Mass: - $1\,\text{Mg} = 1000\,\text{kg} = 1\,\text{ton}$

4.2 Conversions and Measurement Exercises

Calculation Prompts:

a. Determine how many ml are in a $\text{dm}^3$ .
b. Express $\text{kg/m}^3$ in terms of $2\,\text{g/cm}^3$ .
c. Perform the following conversions: - i. $1.58\,\text{l}$ into ml - ii. $0.058\,\text{cm}$ into m - iii. $12.66\,\text{cm}^3$ into ml - iv. $100\,\text{mm}$ into km - v. $4081\,\text{l}$ into ml - vi. $40^\circ\text{C}$ into K - vii. 25 acres into $\text{m}^2$ - viii. 5 ha into acre - ix. $15\,\text{km/h}$ in m/s - x. 1 litre in kg - xi. $7\,\text{cm}^2$ in $\text{m}^2$ - xii. If $1\,\text{mile} = 1.609\,\text{km}$ , express 14 miles in terms of km.
d. If $1\,\text{kg} = 2.2\,\text{lbs}$ and $1\,\text{acre} = 0.405\,\text{ha}$ , establish how many $\text{lbs/acre}$ make $1\,\text{kg/ha}$ .
e. Determine how many ha are in a $\text{km}^2$ .
f. Bridge Restriction Case Study: At a bridge, a board displayed the following restriction: "Maximum mass 5 tons".
g. What load can a lorry carry, which has a mass of $2600\,\text{g}$ , and still cross the bridge?
h. How many pockets of oranges, each having a mass of $15\,\text{kg}$ , can be carried on the lorry across the bridge?
i. Currency Exchange: On a certain day, the exchange rate between US Dollar and EURO was $1\,\text{USD} = 0.78\,\text{EUR}$ . How much was $125\,\text{EUROS}$ in US Dollars?

4.2 Perimeter

Definition: The perimeter of a polygon is the sum of the lengths of all its sides.
Example 1: Perimeter of a Rectangle - Problem: What is the perimeter of a rectangle having side-lengths of $3.4\,\text{cm}$ and $8.2\,\text{cm}$ ? - Solution: A rectangle has 4 sides. Opposite sides have the same length. Therefore, it has 2 sides of $3.4\,\text{cm}$ and 2 sides of $8.2\,\text{cm}$ . - Sum: $3.4\,\text{cm} + 3.4\,\text{cm} + 8.2\,\text{cm} + 8.2\,\text{cm} = 23.2\,\text{cm}$ .
Example 2: Perimeter of a Square - Problem: What is the perimeter of a square having side-length $74\,\text{cm}$ ? - Solution: A square has 4 sides of equal length. - Calculation: $74\,\text{cm} + 74\,\text{cm} + 74\,\text{cm} + 74\,\text{cm} = 4 \times 74\,\text{cm} = 296\,\text{cm}$ .
Example 3: Perimeter of a Regular Hexagon - Problem: What is the perimeter of a regular hexagon having side-length $2.5\,\text{m}$ ? - Solution: A hexagon is a figure with 6 sides. In a regular hexagon, each side has the same length. - Calculation: $6 \times 2.5\,\text{m} = 15\,\text{m}$ .
Example 4: Perimeter of a Trapezoid - Problem: What is the perimeter of a trapezoid having side-lengths $10\,\text{cm}$ , $7\,\text{cm}$ , $6\,\text{cm}$ , and $7\,\text{cm}$ ? - Solution: The perimeter is the sum $10\,\text{cm} + 7\,\text{cm} + 6\,\text{cm} + 7\,\text{cm} = 30\,\text{cm}$ .

4.3 Circumference

Definition: Circumference is the distance around a circle.
Formula: The circumference is equal to Pi ( $\pi$ ) times the diameter of the circle ( $d$ ).
The Constant Pi (\pi): A number that is approximately $3.14159$ .
Example: Circumference Calculation - Problem: What is the circumference of a circle having a diameter of $7.9\,\text{cm}$ , to the nearest tenth of a cm? - Solution: Using an approximation of $3.14159$ for $\pi$ . - Calculation: $\pi \times 7.9 = 3.14159 \times 7.9 = 24.81...\,\text{cm}$ . - Final Answer: $24.8\,\text{cm}$ (rounded to the nearest tenth).

4.4 Area

Definition: The area of a figure measures the size of the region enclosed by the figure, usually expressed in square units. It represents the amount of material needed to "cover" a surface completely.
Units: Common units include square meters, square centimeters, square inches, or square kilometers.

4.4.1 Area of a Triangle

Formula: For a triangle with base length $b$ and height $h$ , the area is $\frac{1}{2} \times b \times h$ .
Derivation Logic: If you take a second identical triangle, rotate it, and "paste" it to the first, it forms a parallelogram with the same base $b$ and height $h$ . The area of the resulting parallelogram is $b \times h$ . Because the parallelogram's area is twice that of the triangle, the triangle's area must be $\frac{1}{2} \times b \times h$ .
Example: - Problem: Calculate the area of a triangle with a base of $5.2\,\text{m}$ and a height of $4.2\,\text{m}$ . - Calculation: $\frac{1}{2} \times 5.2\,\text{m} \times 4.2\,\text{m} = 2.6\,\text{m} \times 4.2\,\text{m} = 10.92\,\text{m}^2$ .

4.4.2 Area of a Rectangle

Formula: The area is the product of width ( $b$ ) and length ( $l$ ).
Equation: $A = l \times b$
Example: - Problem: What is the area of a rectangle with a length of 6 and a width of 2.2? - Calculation: $6\,\text{cm} \times 2.2\,\text{cm} = 13.2\,\text{cm}^2$ .

4.4.3 Area of a Parallelogram

Formula: $A = b \times h$ , where $b$ is the base length and $h$ is the corresponding perpendicular height.
Visualization: One can "cut off" a triangle from one side of the parallelogram and "paste" it onto the other side to form a rectangle with side-lengths $b$ and $h$ . This rectangle possesses the same area as the original parallelogram ( $b \times h$ ).
Example: - Problem: Area of a parallelogram with base $20\,\text{cm}$ and height $7\,\text{cm}$ . - Calculation: $20\,\text{cm} \times 7\,\text{cm} = 140\,\text{cm}^2$ .

4.4.4 Area of a Square

Formula: If $l$ is the side-length, the area is $l^2$ or $l \times l$ .
Example: - Problem: Area of a square with side-length $3.4\,\text{m}$ . - Calculation: $3.4\,\text{m} \times 3.4\,\text{m} = 11.56\,\text{m}^2$ .

4.4.5 Area of a Trapezoid

Formula: If $a$ and $b$ are the lengths of the two parallel bases and $h$ is the height, the area is $\frac{1}{2} \times h \times (a + b)$ .
Example: - Problem: Area of a trapezoid with bases $12\,\text{cm}$ and $8\,\text{cm}$ and height $5\,\text{cm}$ . - Calculation: $\frac{1}{2} \times 5\,\text{cm} \times (12\,\text{cm} + 8\,\text{cm}) = \frac{1}{2} \times 5\,\text{cm} \times 20\,\text{cm} = \frac{1}{2} \times 100\,\text{cm}^2 = 50\,\text{cm}^2$ .

4.4.6 Area of a Circle

Formula: Area is $\pi \times r^2$ or $\text{Pi} \times r \times r$ , where $r$ is the radius.
Example: - Problem: Area of a circle with radius $4.2\,\text{cm}$ , to the nearest tenth. - Calculation: $\pi \times (4.2\,\text{cm})^2 = 3.14159 \times 17.64\,\text{cm}^2 = 55.41...\,\text{square cm}$ . - Final Answer: $55.4\,\text{square cm}$ .

4.5 Comprehensive Practice Exercises and Solutions

1. Calculate Area and Perimeter

Exercise A (Quadrilateral): - Dimensions: sides of $170\,\text{m}$ , $100\,\text{m}$ , $160\,\text{m}$ , and height/segment of $80\,\text{m}$ . - Perimeter Solution: $(170 + 80 + 100 + 160)\,\text{m} = 510\,\text{m}$ . - Area Solution (per transcript): $2 \times 80\,\text{m} \times (160\,\text{m} + 170\,\text{m}) =$
Exercise B (Complex Shape): - Side lengths: $350\,\text{m}$ , $150\,\text{m}$ , then a semi-circle with radius related to height. - Perimeter Solution: $350\,\text{m} + 150\,\text{m} + 350\,\text{m} + \pi \times 75\,\text{m} =$ - Area Solution: $350\,\text{m} \times 150\,\text{m} + \frac{\pi \times 75^2}{2} =$

2. Physical Quantity Calculations

Volume of Cylinder: - Base radius: $5\,\text{m}$ , Height: $30\,\text{m}$ . - Solution: $\text{Volume} = \text{Area of the base} \times \text{Height}$ . - Calculation: $V = \pi \times (5\,\text{m})^2 \times 30\,\text{m} =$
Volume of Rectangular Prism: - Dimensions: $60\,\text{m} \times 50\,\text{m} \times 70\,\text{m}$ . - Solution: $\text{Volume} = (60\,\text{m} \times 50\,\text{m}) \times 70\,\text{m} =$
Dimension Recovery: - If Area of a rectangle is $12\,\text{m}^2$ and breadth is $4\,\text{m}$ , calculate length. - Solution: $\text{Area} = L \times B \rightarrow L \times 4\,\text{m} = 12\,\text{m}^2$ . - Result: $\text{Length} = \frac{12\,\text{m}^2}{4\,\text{m}} = 3\,\text{m}$ .

3. Application Problems

Tiling Problem: Calculate the number of tiles ( $160\,\text{mm} \times 160\,\text{mm}$ ) to cover a square floor ( $4\,\text{m} \times 4\,\text{m}$ ). - Conversion: $160\,\text{mm} = 0.16\,\text{m}$ . - Floor Area: $4\,\text{m} \times 4\,\text{m} = 16\,\text{m}^2$ . - Tile Area: $0.16\,\text{m} \times 0.16\,\text{m} = 0.0256\,\text{m}^2$ . - Number of tiles: $\frac{16}{0.0256} =$
Fencing Problem: A farmer fences land of $300\,\text{m} \times 300\,\text{m}$ . Length of wire needed? - Solution: $\text{Perimeter} = 4 \times \text{sides} = 4 \times 300\,\text{m} = 1200\,\text{m}$ .
Thermal Expansion Problem: A rectangular metal sheet ( $8\,\text{m} \times 6\,\text{m}$ ) increases length and breadth by $6\%$ after heating. Calculate new area and percentage increase. - New length: $8\,\text{m} + (0.06 \times 8\,\text{m}) = 8.48\,\text{m}$ . - New Breadth: $6\,\text{m} + (0.06 \times 6\,\text{m}) = 6.36\,\text{m}$ . - New area: $8.48\,\text{m} \times 6.36\,\text{m} = 53.93\,\text{m}^2$ . - Original area: $8\,\text{m} \times 6\,\text{m} = 48\,\text{m}^2$ . - Percentage increase: $\frac{\text{New area} - \text{original area}}{\text{original area}} \times 100 = \frac{53.93 - 48}{48} \times 100 = 12.35\%$ .
Spherical Pot Problem: A spherical clay pot has a volume of $800\,\text{m}^3$ and a radius of $3\,\text{m}$ . Calculate its height. - Given formula: $\text{Volume of a sphere} = \pi r^2 h$ . - Calculation: $800 = \pi \times 3^2 \times h \rightarrow 800 = (3 \times 3 \times h)$ . - Result: $h = \frac{800}{12}$ .

4. Reservoir and Field Case Studies

Cylindrical Reservoir: Goal capacity is $1000\,\text{m}^3$ . - a. Height with fixed diameter: If diameter = $15\,\text{m}$ , calculate height. - $V = \pi r^2 h \rightarrow 1000 = \pi \times 7.5^2 \times h$ . - $h = \frac{1000}{\pi \times 7.5^2}$ . - b. Radius with fixed height: If height = $4\,\text{m}$ , what should the radius be? - $\frac{V}{\pi h} = r^2 \rightarrow r = \sqrt{\frac{V}{\pi h}} = \sqrt{\frac{1000}{\pi \times 4}}$ . - c. Water Withdrawal: If $5\%$ of water is withdrawn, how much remains? - Calculation: $\text{Water} = 1000\,\text{m}^3 - 53\% \times 1000\,\text{m}^3$ (Note: transcript contains "53%").
Trapezoidal Cropping Field: Sides of $10\,500\,\text{m}$ , $14\,500\,\text{m}$ , $14\,000\,\text{m}$ , and $27\,500\,\text{m}$ . The last two are parallel with a height distance of $10\,000\,\text{m}$ . - a. Fencing Poles: Poles spaced at $3.5\,\text{m}$ . - $\text{Number of poles} = \frac{10500 + 14500 + 14000 + 27500}{3.5}$ . - b. Weed Control Cost: At $R\,1000.00/\text{ha}$ . - Area: $\frac{1}{2} \times (14000 + 27500) \times 10000 = 0.5 \times 41500 \times 10000 = 207\,500\,000\,\text{m}^2$ . - Conversion to ha ( $1\,\text{ha} = 10\,000\,\text{m}^2$ ): $\frac{207\,500\,000}{10\,000} = 20\,750\,\text{ha}$ . - Cost: $20\,750\,\text{ha} \times R\,1000.00/\text{ha}$ . - c. Tillage Cost: At $R\,750.00/\text{ha}$ . - Cost: $20\,750\,\text{ha} \times R\,750.00/\text{ha}$ . - d. Longest Side Poles: Poles needed on the longest side ( $27\,500\,\text{m}$ ) with spacing of $4\,\text{m}$ . - Formula provided: $\text{Number of poles} = \frac{27000}{4} + 1$ .