Pythagorean Theorem - Tenth Grade Geometry
1. Pythagorean Theorem
Pythagorean Theorem: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides
Named After: Greek mathematician Pythagoras (c. 570-495 BC)
Applies to: ONLY right triangles (triangles with one 90° angle)
Key Terms:
• Hypotenuse (c): The longest side, opposite the right angle
• Legs (a, b): The two sides that form the right angle
Named After: Greek mathematician Pythagoras (c. 570-495 BC)
Applies to: ONLY right triangles (triangles with one 90° angle)
Key Terms:
• Hypotenuse (c): The longest side, opposite the right angle
• Legs (a, b): The two sides that form the right angle
Pythagorean Theorem Formula:
$$a^2 + b^2 = c^2$$
Where:
• $a$ and $b$ = lengths of the two legs (sides forming the right angle)
• $c$ = length of the hypotenuse (side opposite the right angle)
Alternative Forms:
To find hypotenuse:
$$c = \sqrt{a^2 + b^2}$$
To find a leg (if you know hypotenuse and other leg):
$$a = \sqrt{c^2 - b^2}$$
$$b = \sqrt{c^2 - a^2}$$
$$a^2 + b^2 = c^2$$
Where:
• $a$ and $b$ = lengths of the two legs (sides forming the right angle)
• $c$ = length of the hypotenuse (side opposite the right angle)
Alternative Forms:
To find hypotenuse:
$$c = \sqrt{a^2 + b^2}$$
To find a leg (if you know hypotenuse and other leg):
$$a = \sqrt{c^2 - b^2}$$
$$b = \sqrt{c^2 - a^2}$$
Example 1: Find the hypotenuse
A right triangle has legs of length 3 and 4. Find the hypotenuse.
Using $a^2 + b^2 = c^2$:
$$3^2 + 4^2 = c^2$$
$$9 + 16 = c^2$$
$$25 = c^2$$
$$c = \sqrt{25} = 5$$
Answer: Hypotenuse = 5 units
(This is the famous 3-4-5 Pythagorean triple!)
A right triangle has legs of length 3 and 4. Find the hypotenuse.
Using $a^2 + b^2 = c^2$:
$$3^2 + 4^2 = c^2$$
$$9 + 16 = c^2$$
$$25 = c^2$$
$$c = \sqrt{25} = 5$$
Answer: Hypotenuse = 5 units
(This is the famous 3-4-5 Pythagorean triple!)
Example 2: Find a leg
A right triangle has hypotenuse 13 and one leg 5. Find the other leg.
Using $a^2 + b^2 = c^2$:
$$5^2 + b^2 = 13^2$$
$$25 + b^2 = 169$$
$$b^2 = 144$$
$$b = \sqrt{144} = 12$$
Answer: Other leg = 12 units
(This is the 5-12-13 Pythagorean triple!)
A right triangle has hypotenuse 13 and one leg 5. Find the other leg.
Using $a^2 + b^2 = c^2$:
$$5^2 + b^2 = 13^2$$
$$25 + b^2 = 169$$
$$b^2 = 144$$
$$b = \sqrt{144} = 12$$
Answer: Other leg = 12 units
(This is the 5-12-13 Pythagorean triple!)
Example 3: With decimals
Find the hypotenuse if legs are 6 and 8.
$$6^2 + 8^2 = c^2$$
$$36 + 64 = c^2$$
$$100 = c^2$$
$$c = 10$$
Answer: c = 10 units
(This is the 6-8-10 triple, which is 2 times the 3-4-5 triple)
Find the hypotenuse if legs are 6 and 8.
$$6^2 + 8^2 = c^2$$
$$36 + 64 = c^2$$
$$100 = c^2$$
$$c = 10$$
Answer: c = 10 units
(This is the 6-8-10 triple, which is 2 times the 3-4-5 triple)
Pythagorean Triples
Pythagorean Triple: A set of three positive integers (a, b, c) that satisfy $a^2 + b^2 = c^2$
Special Property: These are exact whole numbers (no decimals or radicals)
Useful: Recognizing these can speed up calculations
Special Property: These are exact whole numbers (no decimals or radicals)
Useful: Recognizing these can speed up calculations
Common Pythagorean Triples:
Basic Triples:
• 3, 4, 5
• 5, 12, 13
• 8, 15, 17
• 7, 24, 25
• 9, 40, 41
• 11, 60, 61
• 13, 84, 85
Multiples of Basic Triples:
• If (a, b, c) is a Pythagorean triple, then (ka, kb, kc) is also a triple
• Examples:
- 2(3, 4, 5) = (6, 8, 10)
- 3(3, 4, 5) = (9, 12, 15)
- 2(5, 12, 13) = (10, 24, 26)
Basic Triples:
• 3, 4, 5
• 5, 12, 13
• 8, 15, 17
• 7, 24, 25
• 9, 40, 41
• 11, 60, 61
• 13, 84, 85
Multiples of Basic Triples:
• If (a, b, c) is a Pythagorean triple, then (ka, kb, kc) is also a triple
• Examples:
- 2(3, 4, 5) = (6, 8, 10)
- 3(3, 4, 5) = (9, 12, 15)
- 2(5, 12, 13) = (10, 24, 26)
Applications of Pythagorean Theorem
Real-World Applications:
• Construction: Ensuring corners are square (90°)
• Navigation: Finding shortest distances
• Architecture: Designing roofs and ramps
• Sports: Calculating distances on fields
• Technology: Screen sizes (diagonal measurement)
• Surveying: Measuring land
• Physics: Vector components and resultants
• Construction: Ensuring corners are square (90°)
• Navigation: Finding shortest distances
• Architecture: Designing roofs and ramps
• Sports: Calculating distances on fields
• Technology: Screen sizes (diagonal measurement)
• Surveying: Measuring land
• Physics: Vector components and resultants
2. Prove the Pythagorean Theorem
Why Prove? Understanding the proof deepens comprehension and shows why the theorem works
Number of Proofs: Over 370 known proofs exist!
Methods: Geometric, algebraic, and similar triangles approaches
Number of Proofs: Over 370 known proofs exist!
Methods: Geometric, algebraic, and similar triangles approaches
Proof Method 1: Algebraic Method (Rearrangement)
Algebraic Proof Steps:
Step 1: Take 4 identical right triangles with legs $a$ and $b$, and hypotenuse $c$
Step 2: Arrange them to form a large square with side length $(a + b)$
Step 3: The inner space forms a tilted square with side length $c$
Step 4: Calculate areas two ways:
Method 1 - Outer square:
$$\text{Area} = (a + b)^2 = a^2 + 2ab + b^2$$
Method 2 - Inner square + 4 triangles:
$$\text{Area} = c^2 + 4 \times \left(\frac{1}{2}ab\right) = c^2 + 2ab$$
Step 5: Set equal:
$$a^2 + 2ab + b^2 = c^2 + 2ab$$
Step 6: Subtract $2ab$ from both sides:
$$a^2 + b^2 = c^2$$
Proof Complete! ✓
Step 1: Take 4 identical right triangles with legs $a$ and $b$, and hypotenuse $c$
Step 2: Arrange them to form a large square with side length $(a + b)$
Step 3: The inner space forms a tilted square with side length $c$
Step 4: Calculate areas two ways:
Method 1 - Outer square:
$$\text{Area} = (a + b)^2 = a^2 + 2ab + b^2$$
Method 2 - Inner square + 4 triangles:
$$\text{Area} = c^2 + 4 \times \left(\frac{1}{2}ab\right) = c^2 + 2ab$$
Step 5: Set equal:
$$a^2 + 2ab + b^2 = c^2 + 2ab$$
Step 6: Subtract $2ab$ from both sides:
$$a^2 + b^2 = c^2$$
Proof Complete! ✓
Proof Method 2: Similar Triangles
Similar Triangles Proof:
Step 1: Start with right triangle ABC with right angle at B
Step 2: Draw altitude BD from B perpendicular to hypotenuse AC
Step 3: This creates three similar triangles:
• △ABC (original)
• △ABD (left smaller triangle)
• △BDC (right smaller triangle)
Step 4: From similarity of △ABC ~ △ABD:
$$\frac{AB}{AC} = \frac{AD}{AB}$$
$$AB^2 = AC \times AD$$
Step 5: From similarity of △ABC ~ △BDC:
$$\frac{BC}{AC} = \frac{DC}{BC}$$
$$BC^2 = AC \times DC$$
Step 6: Add the two equations:
$$AB^2 + BC^2 = AC \times AD + AC \times DC$$
$$AB^2 + BC^2 = AC(AD + DC)$$
Step 7: Since $AD + DC = AC$:
$$AB^2 + BC^2 = AC \times AC = AC^2$$
Therefore: $a^2 + b^2 = c^2$ ✓
Step 1: Start with right triangle ABC with right angle at B
Step 2: Draw altitude BD from B perpendicular to hypotenuse AC
Step 3: This creates three similar triangles:
• △ABC (original)
• △ABD (left smaller triangle)
• △BDC (right smaller triangle)
Step 4: From similarity of △ABC ~ △ABD:
$$\frac{AB}{AC} = \frac{AD}{AB}$$
$$AB^2 = AC \times AD$$
Step 5: From similarity of △ABC ~ △BDC:
$$\frac{BC}{AC} = \frac{DC}{BC}$$
$$BC^2 = AC \times DC$$
Step 6: Add the two equations:
$$AB^2 + BC^2 = AC \times AD + AC \times DC$$
$$AB^2 + BC^2 = AC(AD + DC)$$
Step 7: Since $AD + DC = AC$:
$$AB^2 + BC^2 = AC \times AC = AC^2$$
Therefore: $a^2 + b^2 = c^2$ ✓
Proof Method 3: Area Method (Geometric)
Visual Geometric Proof:
Concept: Compare areas of squares built on each side
Setup:
• Build a square on each leg (sides $a$ and $b$)
• Build a square on the hypotenuse (side $c$)
Result:
Area of square on hypotenuse = Sum of areas of squares on two legs
$$c^2 = a^2 + b^2$$
This can be shown by dissecting and rearranging the areas
Concept: Compare areas of squares built on each side
Setup:
• Build a square on each leg (sides $a$ and $b$)
• Build a square on the hypotenuse (side $c$)
Result:
Area of square on hypotenuse = Sum of areas of squares on two legs
$$c^2 = a^2 + b^2$$
This can be shown by dissecting and rearranging the areas
3. Converse of the Pythagorean Theorem
Converse: The reverse of a statement
Original Theorem: If triangle is right → then $a^2 + b^2 = c^2$
Converse: If $a^2 + b^2 = c^2$ → then triangle is right
Use: Determine if a triangle is a RIGHT triangle given side lengths
Original Theorem: If triangle is right → then $a^2 + b^2 = c^2$
Converse: If $a^2 + b^2 = c^2$ → then triangle is right
Use: Determine if a triangle is a RIGHT triangle given side lengths
Converse of Pythagorean Theorem:
If the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is a RIGHT triangle.
$$\text{If } c^2 = a^2 + b^2, \text{ then the triangle has a right angle}$$
Where:
• $c$ = longest side
• $a$ and $b$ = other two sides
• The right angle is opposite the longest side (c)
If the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is a RIGHT triangle.
$$\text{If } c^2 = a^2 + b^2, \text{ then the triangle has a right angle}$$
Where:
• $c$ = longest side
• $a$ and $b$ = other two sides
• The right angle is opposite the longest side (c)
Steps to Use the Converse:
Step 1: Identify the longest side (this would be the hypotenuse if it's a right triangle)
Step 2: Square all three sides
Step 3: Check if: (longest side)² = (side 1)² + (side 2)²
Step 4: Make conclusion:
• If equal → RIGHT triangle
• If not equal → NOT a right triangle
Step 1: Identify the longest side (this would be the hypotenuse if it's a right triangle)
Step 2: Square all three sides
Step 3: Check if: (longest side)² = (side 1)² + (side 2)²
Step 4: Make conclusion:
• If equal → RIGHT triangle
• If not equal → NOT a right triangle
Example 1: Is it a right triangle?
Triangle has sides 5, 12, and 13. Is it a right triangle?
Step 1: Identify longest side: 13
Step 2: Check if $13^2 = 5^2 + 12^2$:
$$13^2 = 169$$
$$5^2 + 12^2 = 25 + 144 = 169$$
Step 3: $169 = 169$ ✓
Answer: YES, it is a right triangle!
Triangle has sides 5, 12, and 13. Is it a right triangle?
Step 1: Identify longest side: 13
Step 2: Check if $13^2 = 5^2 + 12^2$:
$$13^2 = 169$$
$$5^2 + 12^2 = 25 + 144 = 169$$
Step 3: $169 = 169$ ✓
Answer: YES, it is a right triangle!
Example 2: Not a right triangle
Triangle has sides 3, 4, and 6. Is it a right triangle?
Step 1: Longest side: 6
Step 2: Check if $6^2 = 3^2 + 4^2$:
$$6^2 = 36$$
$$3^2 + 4^2 = 9 + 16 = 25$$
Step 3: $36 \neq 25$ ✗
Answer: NO, it is NOT a right triangle
Triangle has sides 3, 4, and 6. Is it a right triangle?
Step 1: Longest side: 6
Step 2: Check if $6^2 = 3^2 + 4^2$:
$$6^2 = 36$$
$$3^2 + 4^2 = 9 + 16 = 25$$
Step 3: $36 \neq 25$ ✗
Answer: NO, it is NOT a right triangle
Example 3: Practical application
A contractor measures sides of a triangular plot: 20 ft, 21 ft, and 29 ft. Is it a right triangle?
Longest side: 29
$$29^2 = 841$$
$$20^2 + 21^2 = 400 + 441 = 841$$
$841 = 841$ ✓
Answer: YES, the plot has a right angle (this is a 20-21-29 Pythagorean triple)
A contractor measures sides of a triangular plot: 20 ft, 21 ft, and 29 ft. Is it a right triangle?
Longest side: 29
$$29^2 = 841$$
$$20^2 + 21^2 = 400 + 441 = 841$$
$841 = 841$ ✓
Answer: YES, the plot has a right angle (this is a 20-21-29 Pythagorean triple)
4. Pythagorean Inequality Theorems
Pythagorean Inequalities: Extended versions that determine if a triangle is acute, right, or obtuse
Purpose: Classify triangles by their largest angle
Method: Compare $c^2$ with $a^2 + b^2$
Key Idea: The relationship between sides determines the type of triangle
Purpose: Classify triangles by their largest angle
Method: Compare $c^2$ with $a^2 + b^2$
Key Idea: The relationship between sides determines the type of triangle
Pythagorean Inequality Theorems:
Let $c$ be the longest side of a triangle with sides $a$, $b$, and $c$.
1. RIGHT Triangle:
$$c^2 = a^2 + b^2$$
The angle opposite side $c$ is exactly 90°
2. ACUTE Triangle:
$$c^2 < a^2 + b^2$$
The angle opposite side $c$ is less than 90°
(All angles in the triangle are acute)
3. OBTUSE Triangle:
$$c^2 > a^2 + b^2$$
The angle opposite side $c$ is greater than 90°
(The triangle has one obtuse angle opposite the longest side)
Let $c$ be the longest side of a triangle with sides $a$, $b$, and $c$.
1. RIGHT Triangle:
$$c^2 = a^2 + b^2$$
The angle opposite side $c$ is exactly 90°
2. ACUTE Triangle:
$$c^2 < a^2 + b^2$$
The angle opposite side $c$ is less than 90°
(All angles in the triangle are acute)
3. OBTUSE Triangle:
$$c^2 > a^2 + b^2$$
The angle opposite side $c$ is greater than 90°
(The triangle has one obtuse angle opposite the longest side)
How to Remember:
Compare $c^2$ with $a^2 + b^2$:
• $c^2 = a^2 + b^2$ → Right triangle (equal sign = right angle)
• $c^2 < a^2 + b^2$ → Acute triangle (less than = smaller angle)
• $c^2 > a^2 + b^2$ → Obtuse triangle (greater than = bigger angle)
Think: If $c^2$ is bigger, the angle is bigger (obtuse)
If $c^2$ is smaller, the angle is smaller (acute)
Compare $c^2$ with $a^2 + b^2$:
• $c^2 = a^2 + b^2$ → Right triangle (equal sign = right angle)
• $c^2 < a^2 + b^2$ → Acute triangle (less than = smaller angle)
• $c^2 > a^2 + b^2$ → Obtuse triangle (greater than = bigger angle)
Think: If $c^2$ is bigger, the angle is bigger (obtuse)
If $c^2$ is smaller, the angle is smaller (acute)
Steps to Classify a Triangle:
Step 1: Identify the longest side ($c$)
Step 2: Calculate $c^2$
Step 3: Calculate $a^2 + b^2$ (sum of squares of other two sides)
Step 4: Compare:
• If $c^2 = a^2 + b^2$ → RIGHT
• If $c^2 < a^2 + b^2$ → ACUTE
• If $c^2 > a^2 + b^2$ → OBTUSE
Step 1: Identify the longest side ($c$)
Step 2: Calculate $c^2$
Step 3: Calculate $a^2 + b^2$ (sum of squares of other two sides)
Step 4: Compare:
• If $c^2 = a^2 + b^2$ → RIGHT
• If $c^2 < a^2 + b^2$ → ACUTE
• If $c^2 > a^2 + b^2$ → OBTUSE
Example 1: Acute triangle
Triangle has sides 5, 7, and 8. Classify the triangle.
Step 1: Longest side: $c = 8$
Step 2: Calculate:
$c^2 = 8^2 = 64$
$a^2 + b^2 = 5^2 + 7^2 = 25 + 49 = 74$
Step 3: Compare:
$64 < 74$
$c^2 < a^2 + b^2$
Answer: ACUTE triangle (all angles less than 90°)
Triangle has sides 5, 7, and 8. Classify the triangle.
Step 1: Longest side: $c = 8$
Step 2: Calculate:
$c^2 = 8^2 = 64$
$a^2 + b^2 = 5^2 + 7^2 = 25 + 49 = 74$
Step 3: Compare:
$64 < 74$
$c^2 < a^2 + b^2$
Answer: ACUTE triangle (all angles less than 90°)
Example 2: Obtuse triangle
Triangle has sides 3, 5, and 7. Classify the triangle.
Step 1: Longest side: $c = 7$
Step 2: Calculate:
$c^2 = 7^2 = 49$
$a^2 + b^2 = 3^2 + 5^2 = 9 + 25 = 34$
Step 3: Compare:
$49 > 34$
$c^2 > a^2 + b^2$
Answer: OBTUSE triangle (angle opposite longest side is > 90°)
Triangle has sides 3, 5, and 7. Classify the triangle.
Step 1: Longest side: $c = 7$
Step 2: Calculate:
$c^2 = 7^2 = 49$
$a^2 + b^2 = 3^2 + 5^2 = 9 + 25 = 34$
Step 3: Compare:
$49 > 34$
$c^2 > a^2 + b^2$
Answer: OBTUSE triangle (angle opposite longest side is > 90°)
Example 3: Right triangle
Triangle has sides 6, 8, and 10. Classify the triangle.
Step 1: Longest side: $c = 10$
Step 2: Calculate:
$c^2 = 10^2 = 100$
$a^2 + b^2 = 6^2 + 8^2 = 36 + 64 = 100$
Step 3: Compare:
$100 = 100$
$c^2 = a^2 + b^2$
Answer: RIGHT triangle (has a 90° angle)
Triangle has sides 6, 8, and 10. Classify the triangle.
Step 1: Longest side: $c = 10$
Step 2: Calculate:
$c^2 = 10^2 = 100$
$a^2 + b^2 = 6^2 + 8^2 = 36 + 64 = 100$
Step 3: Compare:
$100 = 100$
$c^2 = a^2 + b^2$
Answer: RIGHT triangle (has a 90° angle)
Example 4: Mixed practice
Classify each triangle:
a) Sides: 4, 5, 6
b) Sides: 9, 12, 15
c) Sides: 2, 3, 4
a) 4, 5, 6:
$c^2 = 36$, $a^2 + b^2 = 16 + 25 = 41$
$36 < 41$ → ACUTE
b) 9, 12, 15:
$c^2 = 225$, $a^2 + b^2 = 81 + 144 = 225$
$225 = 225$ → RIGHT
c) 2, 3, 4:
$c^2 = 16$, $a^2 + b^2 = 4 + 9 = 13$
$16 > 13$ → OBTUSE
Classify each triangle:
a) Sides: 4, 5, 6
b) Sides: 9, 12, 15
c) Sides: 2, 3, 4
a) 4, 5, 6:
$c^2 = 36$, $a^2 + b^2 = 16 + 25 = 41$
$36 < 41$ → ACUTE
b) 9, 12, 15:
$c^2 = 225$, $a^2 + b^2 = 81 + 144 = 225$
$225 = 225$ → RIGHT
c) 2, 3, 4:
$c^2 = 16$, $a^2 + b^2 = 4 + 9 = 13$
$16 > 13$ → OBTUSE
Pythagorean Theorem Summary
| Theorem | Formula | Use |
|---|---|---|
| Pythagorean Theorem | $a^2 + b^2 = c^2$ | Find missing side in RIGHT triangle |
| Converse | If $a^2 + b^2 = c^2$, then right triangle | Determine if triangle IS right |
| Acute Inequality | $c^2 < a^2 + b^2$ | Triangle is ACUTE |
| Obtuse Inequality | $c^2 > a^2 + b^2$ | Triangle is OBTUSE |
Common Pythagorean Triples
| Triple (a, b, c) | Type | Verification |
|---|---|---|
| 3, 4, 5 | Basic | $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ✓ |
| 5, 12, 13 | Basic | $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓ |
| 8, 15, 17 | Basic | $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓ |
| 7, 24, 25 | Basic | $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ ✓ |
| 6, 8, 10 | Multiple of 3-4-5 | $2 \times (3, 4, 5)$ |
| 9, 12, 15 | Multiple of 3-4-5 | $3 \times (3, 4, 5)$ |
| 10, 24, 26 | Multiple of 5-12-13 | $2 \times (5, 12, 13)$ |
| 20, 21, 29 | Basic | $20^2 + 21^2 = 400 + 441 = 841 = 29^2$ ✓ |
Triangle Classification Quick Reference
| Comparison | Type | Largest Angle | Example |
|---|---|---|---|
| $c^2 = a^2 + b^2$ | RIGHT | Exactly 90° | 3, 4, 5: $25 = 9 + 16$ |
| $c^2 < a^2 + b^2$ | ACUTE | Less than 90° | 5, 7, 8: $64 < 74$ |
| $c^2 > a^2 + b^2$ | OBTUSE | Greater than 90° | 3, 5, 7: $49 > 34$ |
Formulas for Finding Sides
| To Find | Given | Formula | Example |
|---|---|---|---|
| Hypotenuse (c) | Both legs | $c = \sqrt{a^2 + b^2}$ | If a=3, b=4: $c = \sqrt{9+16} = 5$ |
| Leg (a) | Hypotenuse & other leg | $a = \sqrt{c^2 - b^2}$ | If c=13, b=12: $a = \sqrt{169-144} = 5$ |
| Leg (b) | Hypotenuse & other leg | $b = \sqrt{c^2 - a^2}$ | If c=13, a=5: $b = \sqrt{169-25} = 12$ |
Problem-Solving Steps
| Problem Type | Steps | Key Formula |
|---|---|---|
| Find missing side | 1. Identify right angle 2. Label sides 3. Use $a^2 + b^2 = c^2$ 4. Solve | $c = \sqrt{a^2 + b^2}$ or $a = \sqrt{c^2 - b^2}$ |
| Check if right triangle | 1. Find longest side (c) 2. Calculate $c^2$ and $a^2 + b^2$ 3. Compare | If $c^2 = a^2 + b^2$ → Right |
| Classify triangle | 1. Find longest side (c) 2. Calculate both values 3. Compare to classify | Compare $c^2$ with $a^2 + b^2$ |
Success Tips for Pythagorean Theorem:
✓ Only works for RIGHT triangles (one 90° angle)
✓ Hypotenuse (c) is always the longest side, opposite the right angle
✓ Formula: $a^2 + b^2 = c^2$ (sum of squares of legs = square of hypotenuse)
✓ Converse: If $a^2 + b^2 = c^2$, then triangle IS right
✓ Memorize common triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25
✓ Acute: $c^2 < a^2 + b^2$ (smaller angle opposite longest side)
✓ Obtuse: $c^2 > a^2 + b^2$ (larger angle opposite longest side)
✓ Always identify the longest side FIRST when classifying
✓ Double-check your arithmetic - common errors come from calculation mistakes!
✓ Only works for RIGHT triangles (one 90° angle)
✓ Hypotenuse (c) is always the longest side, opposite the right angle
✓ Formula: $a^2 + b^2 = c^2$ (sum of squares of legs = square of hypotenuse)
✓ Converse: If $a^2 + b^2 = c^2$, then triangle IS right
✓ Memorize common triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25
✓ Acute: $c^2 < a^2 + b^2$ (smaller angle opposite longest side)
✓ Obtuse: $c^2 > a^2 + b^2$ (larger angle opposite longest side)
✓ Always identify the longest side FIRST when classifying
✓ Double-check your arithmetic - common errors come from calculation mistakes!