Pythagorean Theorem and Special Right Triangles Notes

THE PYTHAGOREAN THEOREM

  • The Pythagorean Theorem applies to right triangles and relates the lengths of the three sides.

  • Notation:

    • Legs: the two sides that form the right angle, commonly labeled as a and b.

    • Hypotenuse: the side opposite the right angle, labeled as c.

  • Core formula:

    • a2+b2=c2a^2 + b^2 = c^2

  • Significance: gives the exact relation between the sides; used to determine unknown side lengths in right triangles and to check whether a triangle is right (via the converse).

  • Converse (from Page 13):

    • If the square of the longest side equals the sum of squares of the other two sides, the triangle is a right triangle.

    • In symbols: If c2=a2+b2c^2 = a^2 + b^2, then the triangle is right.

RIGHT TRIANGLE BASICS

  • Right triangle definition: a triangle with one angle equal to 90°.

  • Hypotenuse: side opposite the right angle.

  • Legs: the other two sides that form the right angle.

  • Key relationships:

    • In a right triangle, the hypotenuse is the longest side.

    • The Pythagorean Theorem relates the squares of the side lengths.

WHO DISCOVERED THIS THEOREM

  • Pythagoras (c. 570 BCE) was an ancient Greek philosopher and mathematician.

  • Born on Samos; founded the Pythagorean school in Croton (now Croto, Italy).

  • Contributions span geometry, arithmetic, and number theory.

APPLYING THE PYTHAGOREAN THEOREM: EXAMPLES

  • Example 1: Find the missing value when a = 5, b = 12.

    • Use a2+b2=c2a^2 + b^2 = c^2

    • Compute: 52+122=c2 25+144=c2 c2=1695^2 + 12^2 = c^2 \ 25 + 144 = c^2 \ c^2 = 169

    • Solve: c=169=13c = \sqrt{169} = 13

    • Answer: c = 13

  • Example 2: Find missing value when a = 9, c = 15.

    • Setup: a2+b2=c2a^2 + b^2 = c^2 → 92+b2=1529^2 + b^2 = 15^2

    • Compute: 81+b2=225 b2=225−81 b2=14481 + b^2 = 225 \ b^2 = 225 - 81 \ b^2 = 144

    • Solve: b=144=12b = \sqrt{144} = 12

    • Answer: b = 12

  • Example 3: Find missing value when a = ?, b = 6, c = 10.

    • Setup: a2+62=102a^2 + 6^2 = 10^2

    • Compute: a2+36=100 a2=64a^2 + 36 = 100 \ a^2 = 64

    • Solve: a=64=8a = \sqrt{64} = 8

    • Answer: a = 8

  • Example 4: A 15-ft tree and its 8-ft shadow form a right triangle with legs 15 and 8.

    • Compute: 152+82=c2 225+64=c2 c2=28915^2 + 8^2 = c^2 \ 225 + 64 = c^2 \ c^2 = 289

    • Hypotenuse: c=289=17c = \sqrt{289} = 17

    • Answer: c = 17 ft

  • Example 5: Tent front view with base 3 ft and hypotenuse 5 ft; find height (other leg).

    • Setup: 32+b2=523^2 + b^2 = 5^2

    • Compute: 9+b2=25 b2=169 + b^2 = 25 \ b^2 = 16

    • Solve: b=16=4b = \sqrt{16} = 4

    • Answer: b = 4 ft

CONVERSE OF THE PYTHAGOREAN THEOREM

  • Statement (Page 13):

    • If the square of the longest side equals the sum of the squares of the other two sides, the triangle is right.

  • Process:

    • Determine the longest side (put it as the potential hypotenuse).

    • Check if c2=a2+b2c^2 = a^2 + b^2. If true → triangle is right.

LET'S TRY (APPLYING THEOREM TO IDENTIFY TRIANGLES)

  • Given different side-lengths, determine whether the triangle is right, obtuse, or acute using the comparison of the squares:

    • If c2=a2+b2c^2 = a^2 + b^2 → right triangle.

    • If c^2 > a^2 + b^2 → obtuse triangle.

    • If c^2 < a^2 + b^2 → acute triangle.

CHECKING PYTHAGOREAN TRIPLES

  • A Pythagorean triple is a set of three positive integers (a, b, c) satisfying a2+b2=c2a^2 + b^2 = c^2.

  • Common triples (from the material):

    • (3, 4, 5)

    • (5, 12, 13)

    • (6, 8, 10) (a multiple of 3,4,5)

    • (7, 24, 25)

    • (8, 15, 17)

    • (9, 40, 41)

  • These can be used to quickly verify right triangles without a calculator.

ACTIVITY AND PRACTICE STRUCTURE SUMMARY

  • Activity 1: Given sets of side measures in cm, determine the kind of triangle (right/obtuse/acute).

  • Activity 2: Use the Pythagorean Theorem to find an unknown side when the other two sides are given in right triangles.

  • Typical right-triangle problem patterns include 3-4-5, 5-12-13, 6-8-10, 8-15-17, 9-40-41 templates.

SPECIAL RIGHT TRIANGLES

  • There are two main families:

    • 45°-45°-90° triangle (isosceles right triangle)

    • 30°-60°-90° triangle

45°-45°-90° TRIANGLE

  • Properties:

    • The two legs are congruent (equal in length): if each leg = x, then the hypotenuse = x√2.

    • Hypotenuse to leg ratio: c = x√2, and x = c/√2 = c√2/2.

  • Formulas:

    • c=x2c = x\sqrt{2}

    • x=c2=c22x = \frac{c}{\sqrt{2}} = \frac{c\sqrt{2}}{2}

  • Examples from the material:

    • If a leg = 5 cm, hypotenuse = $5\sqrt{2}$ cm.

    • If hypotenuse = 15 cm, each leg = $\frac{15}{\sqrt{2}} = \frac{15\sqrt{2}}{2}$ cm.

  • Notes on rationalizing denominators:

    • a2=a22\frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}

45°-45°-90° Example Workflows (Selected from transcript)

  • Example: Given a leg = 5 cm, find the hypotenuse: hyp = $5\sqrt{2}$ cm.

  • Example: Given hypotenuse = 15 cm, find legs: legs = $15/\sqrt{2} = 15\sqrt{2}/2$ cm each.

30°-60°-90° TRIANGLE

  • Core ratios (shorter leg opposite 30°, longer leg opposite 60°, hypotenuse):

    • Shorter leg = x

    • Longer leg = x√3

    • Hypotenuse = 2x

  • Formulas (in terms of hypotenuse and legs):

    • If hypotenuse is given: shorter leg x = \frac{\text{hypotenuse}}{2}

    • Longer leg = x√3 = \frac{\text{hypotenuse}}{2}√3

  • Worked examples from transcript:

    • If hypotenuse = 20 cm:

    • Shorter leg x = 10 cm

    • Longer leg y = 10√3 cm

    • If a 30°-60°-90° triangle has shorter leg x and longer leg y, then y = x√3 and hypotenuse = 2x.

  • Another type of application example:

    • Given hypotenuse = 160 m in a kite problem (60° angle):

    • Shorter leg (opposite 30°) = 160/2 = 80 m

    • Longer leg (opposite 60°) = 80√3 m

    • Vertical distance uses the longer leg in the 60° angle setup.

ADDITIONAL PRACTICAL PROBLEMS AND CHECKS

  • Practical pattern: Use the 3-4-5 family and its multiples for quick checks and constructions.

  • Example check: If a triangle has sides 9, 40, 41, then 92+402=81+1600=1681=4129^2 + 40^2 = 81 + 1600 = 1681 = 41^2, so it is a right triangle.

  • Application strategy for 30°-60°-90° problems:

    • Identify the 30° and 60° angles to label the shorter and longer legs.

    • Use the ratios: shorter leg : longer leg : hypotenuse = 1 : \sqrt{3} : 2 (in length units).

PRACTICE QUESTIONS (BACK POCKET-TYPE IDEAS)

  • 1) What makes a triangle a "special right triangle"?

  • 2) Why do we only need one side to find the others in these triangles?

  • 3) What are the angle measures in a 45-45-90 triangle? How do we know it's isosceles?

  • 4) Where does the \sqrt{2} come from in a 45-45-90 triangle?

  • 5) Why does a 30-60-90 triangle have the ratios it does? Can you explain using an equilateral triangle?

QUICK REFERENCE FORMULAS

  • Pythagorean Theorem: a2+b2=c2a^2 + b^2 = c^2

  • Converse: If c2=a2+b2c^2 = a^2 + b^2, triangle is right.

  • 45°-45°-90° triangle:

    • Hypotenuse: c=a2=b2c = a\sqrt{2} = b\sqrt{2}

    • Legs: a=c2=c22a = \frac{c}{\sqrt{2}} = \frac{c\sqrt{2}}{2}

  • 30°-60°-90° triangle:

    • Shorter leg (opposite 30°): xx

    • Longer leg (opposite 60°): x3x\sqrt{3}

    • Hypotenuse: 2x2x

    • If given hypotenuse HH, then shorter leg x=H2x = \frac{H}{2} and longer leg x3=H32x\sqrt{3} = \frac{H\sqrt{3}}{2}

  • Common Pythagorean triples (verification):

    • 32+42=523^2 + 4^2 = 5^2

    • 52+122=1325^2 + 12^2 = 13^2

    • 62+82=1026^2 + 8^2 = 10^2

    • 72+242=2527^2 + 24^2 = 25^2

    • 82+152=1728^2 + 15^2 = 17^2

    • 92+402=4129^2 + 40^2 = 41^2

Note: Use these relationships to quickly identify right triangles and solve for missing sides in a variety of problems.