Chapter 7 — Front Matter
Purpose and Overview
Right triangles are among the most important figures in all of mathematics. They appear in architecture, engineering, navigation, physics, computer graphics, surveying, and design. They connect algebra, measurement, coordinate geometry, and trigonometry in ways that no other triangle can.
In Chapter 6, you studied similarity and proportional reasoning. You learned that similar triangles have equal angle measures and proportional side lengths, and that this proportionality can be used to find unknown distances. Chapter 7 builds directly on that foundation. Right triangles are a special case of triangles, but their special structure — one angle measuring exactly 90 degrees — gives them remarkable properties that allow precise calculation.
This chapter will develop those properties carefully. You will prove and apply the Pythagorean Theorem, use its converse to identify right triangles, work with the special patterns of 45-45-90 and 30-60-90 triangles, find distances on the coordinate plane, simplify radical expressions that arise from triangle measurement, and explore how the altitude of a right triangle creates geometric mean relationships among the sides.
Every idea in this chapter is a bridge. The Pythagorean Theorem bridges algebra and geometry. Distance on the coordinate plane bridges measurement and coordinates. Special right triangles bridge exact ratio patterns and algebraic simplification. Geometric mean bridges proportionality and right-triangle structure. Together, these ideas prepare you for right-triangle trigonometry in Chapter 8, coordinate geometry in Chapter 13, and applications throughout physics, engineering, and design.
Opening Question
A cable needs to be attached from the top of a 30-foot vertical wall to a point on the ground that is 40 feet from the base of the wall. How long must the cable be? Before you read further, think about what you already know that might help you solve this problem. By the end of this chapter, you will be able to answer this question exactly — and many more like it.
14Sections
16Objectives
14Key Terms
15+Worked Examples
Why This Chapter Matters
Measurement and design. Any time you need to find a length that cannot be measured directly — a diagonal, a height, a distance across a gap — a right triangle often provides the path. Builders, engineers, surveyors, and designers use right-triangle relationships constantly.
Coordinates and distance. The formula for distance between two points on the coordinate plane comes directly from the Pythagorean Theorem. Every distance calculation in coordinate geometry depends on what you will learn in this chapter.
Trigonometry. Sine, cosine, and tangent — the fundamental tools of trigonometry — are defined using the ratios of sides in right triangles. Chapter 8 begins exactly where Chapter 7 ends.
Physics and science. Vector components, projectile motion, forces on inclined planes, optics, and wave behavior all rely on right-triangle relationships.
Algebra and radicals. Right triangles produce square roots naturally. Working with exact radical lengths rather than decimal approximations is an important algebraic skill that this chapter develops.
Learning Objectives
- 1Identify the parts of a right triangle: legs, hypotenuse, and right angle.
- 2State, apply, and prove the Pythagorean Theorem.
- 3Use the Converse of the Pythagorean Theorem to determine whether a triangle is right, acute, or obtuse.
- 4Recognize and apply Pythagorean triples.
- 5Find the distance between two points on the coordinate plane using the distance formula.
- 6Apply the 45-45-90 triangle relationships to find exact side lengths.
- 7Apply the 30-60-90 triangle relationships to find exact side lengths.
- 8Simplify radical expressions that arise from geometric measurement.
- 9Explain what geometric mean means and how it applies to right triangles.
- 10Apply the altitude-to-hypotenuse theorem and its related proportions.
- 11Identify and use similar right triangles created by the altitude to the hypotenuse.
- 12Solve applied measurement and design problems using right-triangle relationships.
- 13Recognize and correct common right-triangle mistakes.
- 14Connect right-triangle relationships to the upcoming study of trigonometric ratios.
Sections 7.1 – 7.4
Right Triangles, the Pythagorean Theorem, and Integer Solutions
Right Triangles · Pythagorean Theorem · Converse · Pythagorean Triples
Section 7.1
Right Triangles
A right triangle is a triangle with exactly one angle measuring 90 degrees. That 90-degree angle is called the right angle, and it is typically marked in diagrams with a small square symbol at the vertex.
Every right triangle has three sides. The two sides that meet at the right angle are called the legs. The side that is opposite the right angle is called the hypotenuse. The hypotenuse is always the longest side of a right triangle, because the largest side is always opposite the largest angle, and the right angle is always the largest angle in a right triangle.
Diagram Instruction
Draw a right triangle. Label the right angle vertex C, and place the right angle square at C. Label the other two vertices A and B. The side AC is one leg. The side BC is the other leg. The side AB, opposite the right angle at C, is the hypotenuse. Mark the hypotenuse with a small label c along the side, and label the legs a and b opposite their respective vertices.
Key Rule
A right triangle can be oriented in many positions. What matters is which angle measures 90 degrees and which side is opposite that angle. Always identify the right angle first, and then identify the hypotenuse as the side that is opposite it.
Student Note — Algebra in Right Triangles
In a right triangle, you will use algebra to find missing sides regularly. The process always involves setting up an equation, substituting the known values, and solving for the unknown. Keep track of which side is the hypotenuse, because the Pythagorean Theorem treats the hypotenuse differently from the legs.
Section 7.2
The Pythagorean Theorem
Theorem
The Pythagorean Theorem
In a right triangle with legs of length a and b and hypotenuse of length c:
a² + b² = c²
This equation holds for every right triangle without exception. If you square both leg lengths and add the results, you always get the square of the hypotenuse length.
Geometric Interpretation
Draw a right triangle with legs of length 3 and 4 and hypotenuse of length 5. On each side, draw a square. The square on the leg of length 3 has area 9. The square on the leg of length 4 has area 16. The square on the hypotenuse has area 25. Label each square with its area. Observe that 9 + 16 = 25, confirming the theorem. The theorem literally connects the areas of squares built on the triangle's sides.
Proof of the Pythagorean Theorem
Proof by area — arranging four copies of the right triangle in a large square
GivenA right triangle with legs a and b and hypotenuse c.
Provea² + b² = c²
PlanBuild a large square with side length (a + b). Arrange four congruent copies of the right triangle inside it so that their hypotenuses form a tilted inner square of side length c. Compute the total area two ways and set them equal.
Solve
Large square area: (a + b)² = a² + 2ab + b²
Four triangles total area: 4 · (½ab) = 2ab
Inner tilted square area: c²
Setting equal: c² + 2ab = a² + 2ab + b²
Subtract 2ab: c² = a² + b² ∎
ReasoningBoth expressions describe the same total area of the large square. Setting them equal and simplifying isolates c², giving the theorem.
Example 7.2.1
Finding the Hypotenuse
- ProblemA right triangle has legs of length 6 and 8. Find the hypotenuse.
- ThinkI know both legs. I need the hypotenuse. Use a² + b² = c² with a = 6, b = 8.
- Solve
6² + 8² = c²
36 + 64 = c²
100 = c²
c = √100 = 10
- Check6² + 8² = 36 + 64 = 100 = 10² ✓
Answer: The hypotenuse is 10.
Example 7.2.2
Finding a Missing Leg
- ProblemA right triangle has one leg of length 5 and a hypotenuse of length 13. Find the other leg.
- ThinkI know one leg and the hypotenuse. Use a² + b² = c² with a = 5, c = 13, and solve for b.
- Solve
5² + b² = 13²
25 + b² = 169
b² = 144
b = 12
- Check5² + 12² = 25 + 144 = 169 = 13² ✓
Answer: The missing leg is 12.
Example 7.2.3
A Result That Is Not a Whole Number
- ProblemA right triangle has legs of length 3 and 7. Find the hypotenuse in simplest radical form.
- ThinkUse a² + b² = c². The result may not be a perfect square.
- Solve
3² + 7² = c²
9 + 49 = c²
58 = c²
c = √58
- Check58 = 2 × 29. No perfect square factors, so √58 is already in simplest form. ✓
Answer: The hypotenuse is √58.
Student Note — Exact vs. Approximate
Not every right triangle produces a whole-number hypotenuse. When the result is not a perfect square, leave the answer in radical form unless a decimal approximation is specifically requested. Radical form is exact; a decimal approximation introduces rounding error.
Opening Question — Solved
The 30-foot wall and 40-foot cable
- SetupThe wall is vertical and the ground is horizontal, so they form a right angle. The cable is the hypotenuse. Legs: 30 and 40.
- Solve
30² + 40² = c²
900 + 1600 = c²
2500 = c²
c = 50
Answer: The cable must be 50 feet long.
Section 7.3
The Converse of the Pythagorean Theorem
Converse and Triangle Classification
For three side lengths a, b, c where c is the longest side:
a² + b² = c² → Right triangle
a² + b² > c² → Acute triangle
a² + b² < c² → Obtuse triangle
As the angle opposite side c increases above 90°, c gets longer and c² grows larger than a² + b². As the angle decreases below 90°, c gets shorter and c² falls below a² + b². This reasoning connects to the Law of Cosines, which you will study in Trigonometry and Precalculus.
Example 7.3.1
Identifying Triangle Types from Side Lengths
- ProblemClassify each set of sides: (a) 7, 24, 25 (b) 5, 7, 9 (c) 4, 5, 8
- ThinkIdentify the longest side as c. Compute a² + b² and c² and compare.
- Solve (a)
c = 25
a² + b² = 49 + 576 = 625
c² = 625
625 = 625 → Right triangle
- Solve (b)
c = 9
a² + b² = 25 + 49 = 74
c² = 81
74 < 81 → Obtuse triangle
- Solve (c)
c = 8 (check: 4 + 5 = 9 > 8 ✓)
a² + b² = 16 + 25 = 41
c² = 64
41 < 64 → Obtuse triangle
Answers: (a) Right triangle (b) Obtuse triangle (c) Obtuse triangle
Student Warning — Triangle Inequality First
Always verify that three given lengths actually form a triangle before applying the converse. The Triangle Inequality (Chapter 5) requires that the sum of any two sides exceeds the third. If the sides do not form a triangle, the question of acute, right, or obtuse does not apply.
Section 7.4
Pythagorean Triples
A Pythagorean triple is a set of three positive integers (a, b, c) such that a² + b² = c². Because all three values are whole numbers, Pythagorean triples produce right triangles with no radical lengths. Every whole-number multiple of a Pythagorean triple is also a Pythagorean triple.
| Triple | Verification | Common Multiples |
|---|
(3, 4, 5) | 9 + 16 = 25 | (6,8,10), (9,12,15), (15,20,25) |
(5, 12, 13) | 25 + 144 = 169 | (10,24,26), (15,36,39) |
(8, 15, 17) | 64 + 225 = 289 | (16,30,34) |
(7, 24, 25) | 49 + 576 = 625 | (14,48,50) |
(9, 40, 41) | 81 + 1600 = 1681 | (18,80,82) |
Example 7.4.1
Recognizing a Multiple of a Triple
- ProblemA right triangle has legs of length 9 and 12. Find the hypotenuse without computing.
- ThinkThe ratio 9 : 12 simplifies to 3 : 4. This matches (3, 4, 5) scaled by 3. So the hypotenuse should be 5 × 3 = 15.
- Verify
9² + 12² = 81 + 144 = 225 = 15² ✓
Answer: The hypotenuse is 15.
Find the missing side, classify each triangle, or verify/identify triples as directed. Leave answers in simplest radical form.
- Legs: 3, 4
- Legs: 5, 12
- Legs: 8, 15
- Legs: 7, 7
- Leg 6, Hyp 10
- Leg 9, Hyp 15
- Leg 5, Hyp 13
- Leg 8, Hyp 17
- Sides 5, 12, 13: classify
- Sides 6, 7, 8: classify
- Sides 3, 4, 6: classify
- Sides 7, 7, 10: classify
Sections 7.5 – 7.8
Distance, Special Triangles, and Radicals
Distance Formula · 45-45-90 · 30-60-90 · Simplifying Radical Lengths
Section 7.5
Distance on the Coordinate Plane
Any two points that are not on the same horizontal or vertical line form the endpoints of the hypotenuse of a right triangle. The legs run horizontally and vertically, and their lengths can be read directly from the coordinates.
Diagram Instruction
Draw two points A = (1, 2) and B = (5, 5) on a coordinate plane. Draw a horizontal segment from A to C = (5, 2), and a vertical segment from C to B. Triangle ACB is a right triangle with the right angle at C. Label the horizontal leg AC = 4 (because 5 − 1 = 4) and the vertical leg CB = 3 (because 5 − 2 = 3). The hypotenuse is AB. Apply the Pythagorean Theorem: AB² = 4² + 3² = 25, so AB = 5.
The Distance Formula
For any two points (x₁, y₁) and (x₂, y₂):
d = √((x₂ − x₁)² + (y₂ − y₁)²)
This is the Pythagorean Theorem applied to coordinate geometry. Squaring the differences eliminates the need for absolute value signs.
Example 7.5.1
Using the Distance Formula
- ProblemFind the distance between P = (2, −1) and Q = (−3, 4).
- Solve
d = √((−3 − 2)² + (4 − (−1))²)
= √((−5)² + (5)²)
= √(25 + 25)
= √50
= 5√2
- CheckHorizontal distance is 5, vertical distance is 5 — right isosceles triangle. Hypotenuse of legs-5 isosceles right triangle: 5√2. ✓
Answer: 5√2 ≈ 7.07
Example 7.5.2
Verifying a Right Triangle on the Coordinate Plane
- ProblemTriangle RST has vertices R = (0, 0), S = (4, 0), T = (0, 3). Is it a right triangle?
- Solve
RS = √(16 + 0) = 4
RT = √(0 + 9) = 3
ST = √(16 + 9) = √25 = 5
RS² + RT² = 16 + 9 = 25 = ST² ✓
Answer: Yes — right angle at R. (RS and RT are perpendicular along the axes.)
Section 7.6
Special Right Triangles: 45-45-90
A 45-45-90 triangle is an isosceles right triangle — both legs are equal. Since the two legs are equal, both base angles are 45°. Let each leg have length x. The Pythagorean Theorem gives the hypotenuse:
45-45-90 Side Ratio
leg : leg : hypotenuse = x : x : x√2 = 1 : 1 : √2
Given a leg → hypotenuse = leg × √2
Given the hypotenuse → leg = hypotenuse ÷ √2 = hypotenuse · √2 / 2
Diagram Instruction
Draw an isosceles right triangle. Label both legs with variable x. Label the hypotenuse x√2. Mark the right angle with a square symbol. Mark both 45° angles at the base vertices. Add matching tick marks to both legs to show they are equal.
Example 7.6.1
Finding the Hypotenuse (45-45-90)
- ProblemA 45-45-90 triangle has legs of length 7. Find the hypotenuse.
- Solve
c = 7√2
Answer: 7√2
Example 7.6.2
Finding a Leg from the Hypotenuse (45-45-90)
- ProblemA 45-45-90 triangle has a hypotenuse of length 10. Find the legs.
- Solve
leg = 10 / √2 = (10 / √2) · (√2 / √2) = 10√2 / 2 = 5√2
Answer: Each leg is 5√2.
Example 7.6.3
Diagonal of a Square (45-45-90)
- ProblemA square floor tile has a side length of 12 inches. Find the length of its diagonal.
- ThinkThe diagonal divides the square into two 45-45-90 triangles. Each leg is 12 inches.
- Solve
d = 12√2 ≈ 16.97 inches
Answer: 12√2 inches ≈ 16.97 inches.
Section 7.7
Special Right Triangles: 30-60-90
A 30-60-90 triangle has angle measures of exactly 30°, 60°, and 90°. It appears naturally when an altitude divides an equilateral triangle in half. Starting with an equilateral triangle of side 2s, the altitude creates a right triangle with hypotenuse 2s, short leg s, and long leg s√3 (from the Pythagorean Theorem: (2s)² − s² = 3s², so h = s√3).
30-60-90 Side Ratio
short leg : long leg : hypotenuse = s : s√3 : 2s = 1 : √3 : 2
Short leg (opposite 30°) = s
Long leg (opposite 60°) = s√3
Hypotenuse (opposite 90°) = 2s
The hypotenuse is always exactly twice the short leg.
Diagram Instruction
Draw a right triangle with the right angle at C. Label the angle at A as 30° and the angle at B as 60°. Label the short leg (opposite 30°) as s, the long leg (opposite 60°) as s√3, and the hypotenuse (opposite 90°) as 2s.
Example 7.7.1
Given the Short Leg (30-60-90)
- ProblemA 30-60-90 triangle has a short leg of length 4. Find the other sides.
- Solve
Long leg = 4√3
Hypotenuse = 8
Answer: Long leg = 4√3, hypotenuse = 8.
Example 7.7.2
Given the Hypotenuse (30-60-90)
- ProblemA 30-60-90 triangle has a hypotenuse of length 14. Find the legs.
- Solve
2s = 14 → s = 7
Short leg = 7
Long leg = 7√3
Answer: Short leg = 7, long leg = 7√3.
Example 7.7.3
Given the Long Leg (30-60-90)
- ProblemA 30-60-90 triangle has a long leg of length 6√3. Find the other sides.
- Solve
s√3 = 6√3 → s = 6
Short leg = 6
Hypotenuse = 12
Answer: Short leg = 6, hypotenuse = 12.
Memory Rule
The hypotenuse is twice the short leg. Start there. If you know any one side, you can find all three: identify s, then multiply by 1, √3, and 2 for the three sides.
Section 7.8
Simplifying Radical Lengths
Right triangles frequently produce side lengths that are not whole numbers. Simplest radical form means the expression under the square root has no perfect square factors other than 1. Use the product rule: √(ab) = √a · √b. Factor out any perfect square, then pull it outside the radical.
Simplification Method
√72 = √(36 × 2) = √36 · √2 = 6√2
√180 = √(36 × 5) = 6√5
√50 = √(25 × 2) = 5√2
To rationalize a denominator: multiply numerator and denominator by √r.
10/√2 · (√2/√2) = 10√2/2 = 5√2
Example 7.8.1
Simplifying a Radical from a Triangle
- ProblemA right triangle has legs of length 6 and 9. Find the hypotenuse in simplest radical form.
- Solve
c² = 36 + 81 = 117
c = √117 = √(9 × 13) = 3√13
Answer: 3√13
Student Warning — A Critical Nonrule
√(a + b) ≠ √a + √b. For example, √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7. Always simplify under the radical sign before taking the root.
- 45-45-90, legs = 5
- 45-45-90, hyp = 8
- 45-45-90, hyp = 6√2
- Square, side = 11; find diagonal
- 30-60-90, short = 3
- 30-60-90, hyp = 10
- 30-60-90, hyp = 8√3
- 30-60-90, long = 9
- Equilateral, side = 12; find altitude
- Simplify √72
- Simplify √200
- Simplify √48
- Distance: (0,0) to (3,4)
- Distance: (1,2) to (4,6)
- Distance: (−2,3) to (4,−5)
- Distance: (3,−1) to (−3,7)
Sections 7.9 – 7.11
Geometric Mean, Altitude, and Similar Right Triangles
Geometric Mean · Altitude to the Hypotenuse · Similar Right Triangles
Section 7.9
Geometric Mean as an Introduction
Definition — Geometric Mean
The geometric mean of two positive numbers a and b is the positive number x such that:
a / x = x / b → x² = ab → x = √(ab)
Example 7.9.1
Computing Geometric Means
- Find GM of 4 and 9
x = √(4 × 9) = √36 = 6
Check: 4/6 = 6/9 = 2/3 ✓
- Find GM of 3 and 12
x = √(3 × 12) = √36 = 6
- Find GM of 5 and 8
x = √40 = 2√10
Why This Matters
Geometric mean appears naturally when the altitude is drawn from the right angle to the hypotenuse. The altitude itself equals the geometric mean of the two segments it creates. This is developed in Section 7.10.
Section 7.10
Altitude to the Hypotenuse
In right triangle △ABC with the right angle at C, draw altitude CD from C perpendicular to hypotenuse AB. Let D be the foot of the altitude on AB. This creates two segments: AD = p and DB = q, with the full hypotenuse c = p + q. The altitude CD = h, leg AC = b, and leg BC = a.
Diagram Instruction
Draw right triangle △ABC with the right angle at C. Draw altitude CD to hypotenuse AB. Label AD = p, DB = q, CD = h, AC = b, BC = a, and AB = c = p + q. Mark all right angles: the original right angle at C and the two right angles at D. The altitude creates two smaller right triangles inside the original.
The Three Geometric Mean Relationships
Drawing the altitude from C to AB creates three similar triangles: △ABC, △ACD, and △CBD. Their proportional sides produce:
h = √(p · q) (altitude = GM of the two hypotenuse segments)
b = √(c · p) (each leg = GM of hypotenuse and adjacent segment)
a = √(c · q)
Example 7.10.1
Finding the Altitude
- ProblemIn △ABC, altitude CD is drawn to AB. AD = 4, DB = 9. Find CD.
- ThinkThe altitude is the geometric mean of the two hypotenuse segments.
- Solve
CD = √(4 × 9) = √36 = 6
Answer: CD = 6.
Example 7.10.2
Finding a Leg from Hypotenuse Segments
- ProblemIn △ABC, AB = 20 and AD = 5. Find leg AC.
- ThinkAC is adjacent to segment AD. So AC² = AB × AD.
- Solve
AC² = 20 × 5 = 100
AC = 10
Answer: AC = 10.
Section 7.11
Similar Right Triangles
Right triangles are similar if they share one pair of equal acute angles (AA Similarity, since both already have a 90° angle). When writing similarity statements, matching vertices must have matching angle measures. Incorrect vertex order produces wrong proportions.
Student Warning — Vertex Order Matters
If you write a similarity statement with vertices in the wrong order, every proportion you write from it will be wrong. Always identify which angles correspond — match them by measure — before writing the similarity statement.
Example 7.11.1
Finding All Three Sides from Hypotenuse Segments
- ProblemIn △ABC with right angle at C, altitude CD is drawn to AB. AD = 3, DB = 12. Find AC, BC, and CD.
- ThinkAB = 3 + 12 = 15. Apply all three geometric mean relationships.
- Solve
CD = √(3 × 12) = √36 = 6
AC = √(15 × 3) = √45 = 3√5
BC = √(15 × 12) = √180 = 6√5
- Check
AC² + BC² = 45 + 180 = 225 = 15² = AB² ✓
Answer: CD = 6, AC = 3√5, BC = 6√5.
- Find the geometric mean of 4 and 25.
- Find the geometric mean of 6 and 24.
- Find the geometric mean of 3 and 27.
- In △ABC with altitude CD to AB: AD = 5, DB = 20. Find CD.
- In △ABC with altitude CD to AB: AD = 6, DB = 24. Find CD, AC, and BC.
- In △ABC: AB = 25, AD = 9. Find AC and CD.
- In △ABC: AD = 4, DB = 9, AB = 13. Find CD, AC, and BC. Verify with the Pythagorean Theorem.
- In △PQR with altitude RS to PQ: PS = 8, SQ = 2. Find RS, PR, and QR.
Sections 7.12 – 7.13
Applications, Common Mistakes, and Preparation
Real-World Applications · Common Errors · Preparing for Trigonometry
Section 7.12
Right Triangles in Measurement and Design
The Pythagorean Theorem and the right-triangle relationships of this chapter appear in virtually every technical field. Builders use the 3-4-5 method to check right angles. Navigators find distances across water using triangle reasoning. Physicists resolve vector components into legs. Architects calculate rafter lengths, stair rise-and-run, and ramp slopes. Surveyors measure distances across inaccessible terrain by constructing right triangles with measurable legs.
Example 7.12.1
Ladder Problem
- ProblemA ladder 20 feet long leans against a vertical wall. The base is 8 feet from the wall. How high does the ladder reach?
- DiagramThe wall, ground, and ladder form a right triangle. The ladder (20 ft) is the hypotenuse. The ground distance (8 ft) is one leg.
- Solve
8² + b² = 20²
64 + b² = 400
b² = 336
b = √336 = 4√21 ≈ 18.33 ft
Answer: 4√21 ≈ 18.33 feet up the wall.
Example 7.12.2
Diagonal Bracing
- ProblemA rectangular gate is 6 feet wide and 4 feet tall. How long is the diagonal brace?
- Solve
c = √(6² + 4²) = √(36 + 16) = √52 = 2√13 ≈ 7.21 ft
Answer: 2√13 ≈ 7.21 feet.
Example 7.12.3
Surveying Across a Pond
- ProblemA surveyor sets up a right angle at C, measures AC = 120 m and BC = 160 m. Find AB.
- Solve
AB² = 120² + 160² = 14400 + 25600 = 40000
AB = 200 m
- Check(120, 160, 200) is (3, 4, 5) scaled by 40. ✓
Answer: 200 meters.
MGU Library Connection
- MGU Physics supplement: Vectors and Components
- MGU Engineering supplement: Applied Measurement and Structural Analysis
- MGU Architecture and Construction: Roof Pitch, Stairs, and Bracing
- MGU Technical Measurement Guide: Surveying and Navigation
Section 7.13
Common Right-Triangle Mistakes
Mistake 1 — Treating a leg as the hypotenuse
5² + 13² = c² (using hyp 13 as a leg)
The hypotenuse is the side opposite the right angle — always the longest side. Identify the right angle first. Then c goes opposite it.
Mistake 2 — Forgetting to take the square root
c² = 100, so c = 100
After solving for c², always take the positive square root. c² = 100 means c = √100 = 10.
Mistake 3 — Incorrect radical simplification
√50 = 5√10 or √(a+b) = √a + √b
√50 = √(25 × 2) = 5√2. The product rule is √(ab) = √a · √b. Never split a sum under the radical.
Mistake 4 — Confusing 45-45-90 and 30-60-90 ratios
Using 1 : 1 : √2 for a 30-60-90 triangle
Always identify the angles of the triangle before applying a special ratio. 45-45-90 gives 1 : 1 : √2. 30-60-90 gives 1 : √3 : 2.
Mistake 5 — Reversing short and long legs in a 30-60-90 triangle
Labeling the long leg as opposite 30°
The short leg is opposite the smallest angle (30°). The long leg is opposite 60°. Larger angle → larger side.
Mistake 6 — Wrong altitude-to-hypotenuse formula
CD = (AD + DB) / 2 = arithmetic mean instead of geometric mean
CD = √(AD × DB). The altitude is the geometric mean of the two hypotenuse segments. Each leg is the geometric mean of the whole hypotenuse and the adjacent segment.
Mistake 7 — Applying Pythagorean Theorem to non-right triangles
Using a² + b² = c² for an obtuse triangle
The Pythagorean Theorem only holds for right triangles. Confirm a right angle exists before applying it.
Mistake 8 — Distance formula errors
Forgetting to square the differences, or forgetting the square root
Write the full formula: d = √((x₂−x₁)² + (y₂−y₁)²). Squaring the differences makes order irrelevant. But you must take the final square root.
Section 7.14
Preparing for Trigonometry
Chapter 7 established that right triangles have precise, predictable relationships between their sides and angles. Chapter 8 will take one more step: it will name those side ratios and show that they depend only on the angle, not on the size of the triangle.
In Section 7.6, every 45-45-90 triangle had a leg-to-hypotenuse ratio of 1/√2 = √2/2. In Section 7.7, every 30-60-90 triangle had a short-leg-to-hypotenuse ratio of 1/2 and a long-leg-to-hypotenuse ratio of √3/2. These ratios are consistent because similar triangles have proportional sides. Any two right triangles with the same acute angle measures are similar and therefore have identical side ratios.
| Ratio | Name in Trigonometry |
|---|
| opposite / hypotenuse | Sine |
| adjacent / hypotenuse | Cosine |
| opposite / adjacent | Tangent |
Readiness Checklist for Chapter 8
Before beginning trigonometry, confirm you can: (1) identify legs and hypotenuse in any right triangle, (2) apply the Pythagorean Theorem to find missing sides, (3) simplify radical expressions, (4) work with 45-45-90 and 30-60-90 ratios, (5) use the distance formula on the coordinate plane, and (6) read proportions from similar right triangles.
Chapter 7 — Back Matter
Chapter Summary
- Sec 7.1A right triangle has exactly one 90° angle. The sides forming the right angle are the legs; the side opposite the right angle is the hypotenuse, always the longest side.
- Sec 7.2The Pythagorean Theorem: a² + b² = c². It can be proven by arranging four copies of the right triangle in a large square and computing area two ways. It finds any side when two are known.
- Sec 7.3The Converse classifies triangles from side lengths using c (longest side): a² + b² = c² (right), a² + b² > c² (acute), a² + b² < c² (obtuse). Always verify the Triangle Inequality first.
- Sec 7.4Pythagorean triples are integer solutions to a² + b² = c². The most common are (3,4,5), (5,12,13), (8,15,17), and (7,24,25). Any whole-number multiple of a triple is also a triple.
- Sec 7.5The distance formula d = √((x₂−x₁)² + (y₂−y₁)²) is the Pythagorean Theorem applied to coordinates. Any two points form the hypotenuse of a right triangle with horizontal and vertical legs.
- Sec 7.645-45-90 triangles have sides in ratio 1 : 1 : √2. Hypotenuse = leg × √2. Leg = hypotenuse × √2/2. Appears whenever a square is divided by its diagonal.
- Sec 7.730-60-90 triangles have sides in ratio 1 : √3 : 2. Hypotenuse = 2 × short leg. Long leg = s√3. Derived from the altitude of an equilateral triangle.
- Sec 7.8Simplest radical form: no perfect square factors under the radical sign. Use √(ab) = √a · √b. Rationalize denominators by multiplying by √r/√r. Never split a sum: √(a+b) ≠ √a + √b.
- Sec 7.9The geometric mean of a and b is √(ab) — the number x where a/x = x/b. It is the multiplicative "middle" between two numbers.
- Sec 7.10Altitude CD to hypotenuse AB (with AD = p, DB = q): CD = √(pq); each leg = √(hypotenuse × adjacent segment). All three triangles formed are similar.
- Sec 7.11Right triangles are similar when they share one pair of equal acute angles (AA). Similarity must be written with corresponding vertices in matching order to produce correct proportions.
- Sec 7.12Right triangles appear in construction (square corners), surveying (distance across obstacles), navigation, physics (vector components), architecture (rafters, ramps), and engineering (bracing).
- Sec 7.13Eight major errors: misidentifying the hypotenuse, forgetting the square root, incorrect radical simplification, mixing special triangle ratios, reversing 30-60-90 legs, wrong altitude formula, applying theorem to non-right triangles, distance formula mistakes.
- Sec 7.14Trigonometry names the consistent side ratios of similar right triangles as sine, cosine, and tangent. These ratios depend only on the angle measure, which follows from the proportionality of similar triangles established in Chapter 6.
Concept Review Questions
- What makes a triangle a right triangle? How is the hypotenuse identified?
- State the Pythagorean Theorem in words, then as an equation.
- What does the Converse of the Pythagorean Theorem allow you to determine? What must c represent?
- What is a Pythagorean triple? Give three examples, including one that is a multiple of a basic triple.
- How is the distance formula derived from the Pythagorean Theorem?
- What is the side ratio of a 45-45-90 triangle? Why is the hypotenuse equal to the leg times √2?
- What is the side ratio of a 30-60-90 triangle? Which angle is opposite the short leg? How is the ratio derived?
- What does it mean for a radical to be in simplest form? Give an example of simplifying √72.
- Define geometric mean. How do you find the geometric mean of two numbers?
- When the altitude is drawn from the right angle to the hypotenuse, how many similar triangles are created? What are the three geometric mean relationships?
- What does the Pythagorean Theorem have to do with trigonometry?
Applications
For each problem: draw and label a diagram, write the relevant equation or proportion, solve, and verify your answer in context.
- Ladder. A 25-foot ladder leans against a building. The base is 7 feet from the wall. How high does the ladder reach?
- Baseball diamond. A standard diamond is a square with 90 feet between bases. How far is it from home plate to second base? Leave in simplest radical form.
- Ramp. A ramp rises 5 feet over a horizontal distance of 12 feet. Find the length of the ramp surface.
- Window diagonal. A rectangular window is 36 inches wide and 48 inches tall. Find the diagonal.
- City blocks. A person walks 8 blocks east and 6 blocks north. If each block is 100 meters, how far are they from the start?
- Square park. A square park has side length 200 meters. Find the diagonal path. Leave in simplest radical form.
- Altitude. In △ABC with right angle at C, altitude CD is drawn. AD = 5, DB = 20. Find CD, AC, and BC.
- Surveying. A surveyor creates a right angle at P and measures PQ = 180 m, PR = 240 m. Find QR.
- Roof rafter. A roof has a horizontal run of 15 feet and a vertical rise of 8 feet. Find the rafter length.
- Metal plate. A rectangular metal plate is 12 inches by 16 inches. A diagonal cut is made from corner to corner. Find its length.
Error Analysis
Each problem contains a common mistake. Identify the mistake and show the correct solution.
Problem 1 — Misidentifying the hypotenuse
A right triangle has leg 9 and hypotenuse 15.
Student writes: 9² + 15² = c²
81 + 225 = c²
c = √306 = 3√34
What error did the student make? What is the correct answer?
Problem 2 — Incorrect radical simplification
Student simplifies √48:
√48 = √16 + √32 = 4 + √32
What error did the student make? What is the correct simplified form?
Problem 3 — Wrong special triangle formula
45-45-90 triangle, hypotenuse = 10.
Student writes: leg = 10√2
What error did the student make? What is the correct answer?
Problem 4 — Arithmetic mean instead of geometric mean
△ABC with altitude CD to AB: AD = 4, DB = 9.
Student writes: CD = (4 + 9) / 2 = 6.5
What error did the student make? What is the correct value of CD?
Problem 5 — Converse applied with wrong c
Classify sides 5, 8, 10:
Student uses c = 8 (middle value):
5² + 10² = 25 + 100 = 125
8² = 64; 125 ≠ 64, so not a right triangle.
(Student's conclusion is correct, but setup is wrong.)
What mistake did the student make in the setup? What is the correct procedure?
Chapter 7 Checkpoint
Chapter 7 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 7
Section 1 — Pythagorean Theorem
- Find the hypotenuse of a right triangle with legs 10 and 24.
- Find the missing leg of a right triangle with one leg of 20 and hypotenuse of 29.
- Determine whether a triangle with sides 11, 14, and 18 is acute, right, or obtuse.
Section 2 — Pythagorean Triples
- Is (20, 48, 52) a Pythagorean triple? Show your work.
- A right triangle has one leg of 16 and hypotenuse of 20. Identify the other leg by recognizing the triple.
Section 3 — Coordinate Distance
- Find the distance from A = (−2, 1) to B = (4, 9).
- Find the perimeter of a right triangle with vertices at (0, 0), (5, 0), and (5, 12).
Section 4 — Special Right Triangles
- A 45-45-90 triangle has legs of length 6√2. Find the hypotenuse.
- A 30-60-90 triangle has a hypotenuse of 16. Find both legs.
- An equilateral triangle has sides of length 10. Find the altitude.
Section 5 — Simplifying Radicals
- Simplify √200.
- Simplify √98.
- Rationalize the denominator: 12/√3.
Section 6 — Altitude to the Hypotenuse
- In △ABC, altitude CD is drawn to AB. AD = 6, DB = 24. Find CD, AC, and BC.
- AD = 8, AB = 18. Find AC.
Section 7 — Applications
- A kite string 50 meters long makes a right triangle with the ground. The horizontal distance from the person to directly below the kite is 30 meters. How high is the kite?
- A rectangular swimming pool is 15 meters long and 20 meters wide. Find the diagonal distance across the pool.
Bridge to Chapter 8
Chapter 8: Right-Triangle Trigonometry
Chapter 7 gave you a complete toolkit for right-triangle geometry: finding any side, testing for right triangles, working with exact radical lengths, using special angle patterns, finding distances on the coordinate plane, and analyzing geometric mean relationships.
Chapter 8 names the side ratios and shows they depend only on the angle — not the triangle's size. This is possible because similar right triangles are proportional, a fact established in Chapter 6 and reinforced throughout Chapter 7. In Chapter 8 you will study sine, cosine, and tangent as consistent ratios; use them to find missing sides and angles; work with angles of elevation and depression; and apply trigonometry to surveying, navigation, physics, and engineering. Everything you built in this chapter leads directly there.
MGU Library Connections
- Chapter 8: Right-Triangle Trigonometry — follows immediately
- Chapter 13: Coordinate Geometry and Algebraic Proof — distance formula revisited
- Appendix F: Right Triangle Formula Reference
- Appendix G: Trigonometry Ratio Reference
- MGU Physics supplement: Vectors, Components, and Inclined Planes
- MGU Engineering supplement: Applied Measurement and Structural Diagonal Bracing
- Algebra II, Chapter 1: Algebra II Readiness — radical expressions revisited
- Trigonometry and Precalculus: unit circle, radian measure, and the full trigonometric functions