Comprehensive Study Guide for 45-45-90 Special Right Triangles

Characteristics of Isosceles Right Triangles (45-45-90)

• Definition of Isosceles: An isosceles triangle is defined as a triangle where at least two sides are congruent (equal in length).
• 45-45-90 Triangle Context:     * This specific triangle is an isosceles right triangle.     * It can be visualized by taking a square and drawing a diagonal from one corner to the opposite corner. This creates two identical 45-45-90 triangles.     * The angles are always $45^{\circ}$, $45^{\circ}$, and $90^{\circ}$.

Fundamental Formulas and Properties

• The Legs: The two legs (the sides adjacent to the right angle) are always equal in length.     * Formula: $\text{leg}_1 = \text{leg}_2$
• The Hypotenuse: The hypotenuse (the side opposite the $90^{\circ}$ angle) is calculated by multiplying the length of a leg by the square root of two.     * Formula: $\text{hypotenuse} = \text{leg} \times \sqrt{2}$     * Alternative Formula: $\text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}}$
• Cognitive Requirement for Problem Solving: It is mandatory to draw the triangle for every single problem. Doing so supports the cognitive process and provides the brain with the necessary visual information to complete the calculations correctly. Failure to draw the triangle results in a loss of points and hinders the learning process.

Column Practice: Leg and Hypotenuse Calculations

• Problem 3:     * Given: Leg $a = 6$.     * Leg Equality: Because the legs are equal, $b = 6$.     * Hypotenuse Calculation: $c = 6 \times \sqrt{2}$.     * Final Answer: $a = 6$, $b = 6$, $c = 6\sqrt{2}$.
• Problem 4:     * Given: Leg $b = \frac{4}{5}$.     * Leg Equality: $a = b = \frac{4}{5}$.     * Hypotenuse Calculation: $c = \frac{4}{5} \times \sqrt{2}$.     * Formatting Note: When written, the $\sqrt{2}$ goes to the top of the fraction, resulting in $\frac{4\sqrt{2}}{5}$.
• Problem 5:     * Given: Leg $a = \sqrt{5}$.     * Leg Equality: $b = \sqrt{5}$.     * Hypotenuse Calculation: $c = \sqrt{5} \times \sqrt{2}$.     * Simplification: Both numbers are under the radical, so they can be multiplied. $c = \sqrt{10}$.
• Problem 6:     * Given: Hypotenuse $c = 8\sqrt{2}$.     * Method: Use the formula $\text{hypotenuse} = \text{leg} \times \sqrt{2}$.     * Equation: $8\sqrt{2} = a \times \sqrt{2}$.     * Calculation: Divide both sides by $\sqrt{2}$. The $\sqrt{2}$ terms cancel out.     * Final Answer: $a = 8$, and because legs are equal, $b = 8$.
• Problem 8:     * Given: Hypotenuse $c = \sqrt{22}$.     * Equation: $\sqrt{22} = a \times \sqrt{2}$.     * Calculation: Divide both sides by $\sqrt{2}$.     * Result: $a = \frac{\sqrt{22}}{\sqrt{2}} = \sqrt{11}$.     * Final Answer: $a = \sqrt{11}$, $b = \sqrt{11}$.

The Rationalization Process

• Definition of Rationalizing: Rationalize is the process of removing a radical (square root) from the denominator of a fraction. In geometry and high-level mathematics, answers are typically not left with a square root in the denominator, especially on multiple-choice tests like the ACT.
• Mathematical Concept: Rationalizing is similar to reducing a fraction like $\frac{4}{8}$ to $\frac{1}{2}$. It simplifies the expression to a standard form without changing its numerical value.
• Example Problem (Problem 7):     * Given: Hypotenuse $c = 6$.     * Equation: $6 = a \times \sqrt{2}$.     * Isolation: $a = \frac{6}{\sqrt{2}}$.     * Step 1: Multiply the numerator and the denominator by $\sqrt{2}$. (Note: $\frac{\sqrt{2}}{\sqrt{2}}$ is equal to $1$, so the value remains the same).     * Step 2: $a = \frac{6 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{6\sqrt{2}}{\sqrt{4}}$.     * Step 3: Simplify the denominator: $\sqrt{4} = 2$. Now the expression is $\frac{6\sqrt{2}}{2}$.     * Step 4: Reduce the outside numbers. $6$ divided by $2$ is $3$.     * Final Answer: $a = 3\sqrt{2}$.

Worksheet Practice and Categorization

• Problem 16 (Finding Hypotenuse from a Complex Leg):     * Given: Leg $b = 9\sqrt{2}$.     * Leg Equality: $a = 9\sqrt{2}$.     * Hypotenuse Calculation: $u = (9\sqrt{2}) \times \sqrt{2}$.     * Step: Multiply the square roots. $\sqrt{2} \times \sqrt{2} = 2$.     * Calculation: $u = 9 \times 2 = 18$.     * Common Mistake: Students often think they don't need to multiply by $\sqrt{2}$ because the leg already contains a $\sqrt{2}$. This is incorrect; you must always multiply the leg by $\sqrt{2}$.
• Problem 8 (Worksheet Version):     * Given: Hypotenuse is $15\sqrt{2}$.     * Calculation: Divide by $\sqrt{2}$ to get legs.     * Final Answer: $x = 15$, $y = 15$.
• Problem 9 (Rationalization Example):     * Given: Hypotenuse is $2$.     * Equation: $2 = x \times \sqrt{2}$.     * Isolation: $x = \frac{2}{\sqrt{2}}$.     * Rationalize: $\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2}$.     * Simplify: The outside twos cancel out.     * Final Answer: $x = \sqrt{2}$, $y = \sqrt{2}$.
• Problem 19 (Odd Number Rationalization):     * Given: Hypotenuse is $11$.     * Equation: $11 = x \times \sqrt{2}$.     * Isolation: $x = \frac{11}{\sqrt{2}}$.     * Rationalize: $\frac{11}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{11\sqrt{2}}{2}$.     * Note: Since $11$ and $2$ do not divide evenly, the answer is left as a fraction. While it could be written as $5.5\sqrt{2}$, the fractional form $\frac{11\sqrt{2}}{2}$ is the most standard for tests.
• Categorization of Worksheet Problems:     * Identical to Problem 8 (Easy Division): 6, 7, 12, 13.     * Identical to Problem 16 (Multiplication): 2, 5, 10, 11.     * Easy/Standard: 17.     * Require Rationalization: 9, 13, 14, 15, 18, 19.

Geometry Curriculum Overview

• Big Ideas in Geometry:     * Parallel Lines: Corresponding angles and alternate interior angles (the "zigzag" concept) where angles are equal.     * Congruent Triangles: Used primarily for geometric proofs.     * Similar Triangles: Setting up ratios and cross-multiplying to solve for unknown sides.     * Chapter 9 Key Topics:         * Pythagorean Theorem.         * 45-45-90 Special Right Triangles.         * 30-60-90 Special Right Triangles (upcoming).         * Trigonometric Functions (Sine, Cosine, Tangent) scheduled for next Monday and Tuesday.     * Circles: A future topic that will be covered to ensure exposure, though it may not be on the formal test.

Questions & Discussion

• Question (Rainy): What is happening?
• Response (Instructor): Provides guidance on rationalizing radicals and clarifies the upcoming curriculum (trig functions and special triangles).
• Discussion on Personal Plans: The teacher discusses gas prices reaching $9.09$ (possibly hyperbole regarding price trends) versus a current crazy price of $4.79$. There is a brief exchange about whether the teacher will return next year; the teacher mentions potentially needing time off for themselves but wanting to stay for this specific school.
• Classroom Management: The teacher jokes with Joaquin about his phone use, warning that a call to the office might be made if he continues to be on it rather than focusing on the "gift" of a good grade.