Mathematics: Exponents, Square Roots, and the Pythagorean Theorem

Definition of Power: A power consists of a base and an exponent together (e.g., $3^5$ ).
Base: The number that is being multiplied by itself.
Exponent: Also known as the index, it tells us how many times the base should be multiplied by itself.
Representations of Numerical Operations: * Expanded Form / Repeated Multiplication: For example, $3 imes 3 imes 3 imes 3 imes 3$ is the repeated multiplication form of $3^5$ . * Exponential Form: The shorthand notation where the base is written with a superscript exponent (e.g., $3^5$ ). * Standard Form / Answer: The final calculated value of the expression (e.g., $243$ ).
Examples of Conversions: * Standard to Exponential: * $2 imes 2 imes 2 imes 2 imes 2 = 2^5$ * $5 imes 5 imes 5 imes 5 = 5^4$ * Exponential to Standard: * $4^3 = 4 imes 4 imes 4 = 64$ * $7^2 = 7 imes 7 = 49$ * $8^6 = 8 imes 8 imes 8 imes 8 imes 8 imes 8 = 262,144$ * $6^3 = 6 imes 6 imes 6 = 216$

Concept of Area: Area refers to the total number of square units contained inside an object.
Calculating Area: There are two primary methods to find the area of a square or rectangle: 1. Counting: Counting every individual square unit inside the shape. 2. Multiplication: Multiplying the length by the width ( $A = l imes w$ ).
Perfect Squares: * A perfect square is formed when a number is multiplied by itself (e.g., $4 imes 4 = 16$ ). * This is visually represented by a shape where length and width are identical, forming a geometric square. * Notation: $4 imes 4$ is written as $4^2$ .
Group Task (Rectangular Dimensions for Specific Areas): * Area of 20: Possible dimensions include $1 imes 20$ , $2 imes 10$ , and $4 imes 5$ . * Area of 18: Possible dimensions include $1 imes 18$ , $2 imes 9$ , and $3 imes 6$ . * Area of 16: Possible dimensions include $1 imes 16$ , $2 imes 8$ , and $4 imes 4$ . Note that $4 imes 4$ is a perfect square.

Definitions: * Factor: A number that divides evenly into another number. * Formatting: Factors must be listed in brackets from smallest to largest.
Factor Listing Examples: * 28: Factors are $1 imes 28$ , $2 imes 14$ , $4 imes 7$ . List: $(1, 2, 4, 7, 14, 28)$ . * 45: Factors are $1 imes 45$ , $3 imes 15$ , $5 imes 9$ . List: $(1, 3, 5, 9, 15, 45)$ . * 12: List: $(1, 2, 3, 4, 6, 12)$ . * 25: List: $(1, 5, 25)$ . (Perfect Square). * 16: List: $(1, 2, 4, 8, 16)$ . (Perfect Square). * 20: List: $(1, 2, 4, 5, 10, 20)$ . * 36: List: $(1, 2, 3, 4, 6, 9, 12, 18, 36)$ . (Perfect Square).
Square Roots ( $\sqrt{\quad}$ ): * The square root is the inverse operation of squaring. * If a square has an area of $36$ , finding the side length requires taking the square root: $\sqrt{36} = 6$ . * Mathematical logic: $\sqrt{36} = \sqrt{6 \times 6} = \sqrt{6^2} = 6$ .

Problem: How to find the side length of a square when it does not align perfectly with grid lines.
Step-by-Step Procedure: 1. Divide the square into four congruent triangles and one smaller central square. 2. Calculate the area of one triangle: $A_{\text{triangle}} = \frac{\text{length} \times \text{width}}{2}$ . 3. Calculate the area of the central square: $A_{\text{square}} = s^2$ . 4. Multiply the triangle area by four ( $4 imes A_{\text{triangle}}$ ). 5. Add the combined triangle areas to the central square area to find the total area. 6. The side length of the original large square is the square root of the total area.
Numerical Example from Lesson: * Triangle area ( $\text{base 4, height 2}$ ): $\frac{4 \times 2}{2} = 4$ . * Small square area ( $\text{side 2}$ ): $2^2 = 4$ . * Total Area: $(4 \times 4) + 4 = 20$ . * Side Length: $\sqrt{20} \approx 4.4$ .

Perfect Square Reference Chart: * $\sqrt{1} = 1$ * $\sqrt{4} = 2$ * $\sqrt{9} = 3$ * $\sqrt{16} = 4$ * $\sqrt{25} = 5$ * $\sqrt{36} = 6$ * $\sqrt{49} = 7$ * $\sqrt{64} = 8$ * $\sqrt{81} = 9$ * $\sqrt{100} = 10$ * $\sqrt{121} = 11$ * $\sqrt{144} = 12$ * $\sqrt{169} = 13$ * $\sqrt{196} = 14$ * $\sqrt{225} = 15$
Estimation Strategy: When a number is not a perfect square, estimate by finding the two perfect squares it falls between. * Case Study: $\sqrt{18}$ * 18 falls between perfect squares 16 and 25. * Therefore, $\sqrt{18}$ falls between $\sqrt{16}$ ( $4$ ) and $\sqrt{25}$ ( $5$ ). * Halfway between 16 and 25 is 20.5. Since 18 is less than 20.5, $\sqrt{18}$ must be less than 4.5. * Case Study: $\sqrt{58}$ * Surrounding perfect squares: $\sqrt{49}$ (7) and $\sqrt{64}$ (8). * Practice Tasks: Estimate $\sqrt{12}$ , $\sqrt{7}$ , $\sqrt{21}$ , and $\sqrt{46}$ .

Triangle Classifications: * Right Triangle: Contains one $90^{\circ}$ angle. * Acute Triangle: All angles are less than $90^{\circ}$ . * Obtuse Triangle: Contains one angle greater than $90^{\circ}$ .
Theorem Principles: * Applicable only to right-angled triangles. * Hypotenuse: The longest side, always opposite the right angle (denoted as side $c$ ). * Legs: The two sides that form the right angle (denoted as sides $a$ and $b$ ). * Formula: The area of the square of the hypotenuse is equal to the sum of the areas of the squares of the other two sides: $a^2 + b^2 = c^2$ .
Verification Example: * Sides: 6, 8, 10. * Areas: $6^2 = 36$ , $8^2 = 64$ , $10^2 = 100$ . * Validation: $36 + 64 = 100$ . This is a right triangle.
Proving Non-Right Triangles: * If $a^2 + b^2 \neq c^2$ , the triangle is not a right-angle triangle. * Test sets: $(10, 26, 24)$ , $(9, 9, 15)$ , $(30, 9, 25)$ .
Pythagorean Triple: A set of three positive integers that perfectly satisfy the theorem (e.g., $7, 24, 25$ ). * Proof: $7^2 + 24^2 = 49 + 576 = 625$ . * Since $25^2 = 625$ , the theorem holds.
Algebraic Application: * To find hypotenuse: $c = \sqrt{a^2 + b^2}$ * To find a missing leg: $a = \sqrt{c^2 - b^2}$

Ladder Problem: A 7-foot ladder ( $c = 7$ ) leans against a house. Its base is 2 feet from the house ( $b = 2$ ). To find height ( $a$ ): * $a^2 = 7^2 - 2^2$ * $a^2 = 49 - 4 = 45$ * $a = \sqrt{45} \approx 6.71\text{ feet}$
Walking Shortcut: Suzie and Jackie walk to a playground. They can walk along the sidewalk (legs of a triangle, lengths 5 and 8) or take a shortcut (hypotenuse). To find the difference in distance: 1. Distance around sidewalk: $5 + 8 = 13$ . 2. Shortcut distance: $\sqrt{5^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89} \approx 9.43$ . 3. Difference: $13 - 9.43 = 3.57$ .