14.2 The Pythagorean Theorem, Distance Formula and...

§ 14.2 The Pythagorean Theorem, DistanceFormula and Equation of a Circle

Pythagoras of Samos

c. 569 BC - (500-475) BC

Settled in Crotona (Greek colony in southern Italy) where hefounded a philosophical and religious school

All things are numbers. Mathematics is the basis for everything,and geometry is the highest form of mathematical studies. Thephysical world can be understood through mathematics.Numbers have personalities, characteristics, strengths andweaknesses.Certain symbols have a mystical significanceNot sure how he died

Pythagoras of Samos

c. 569 BC - (500-475) BCSettled in Crotona (Greek colony in southern Italy) where hefounded a philosophical and religious school


Pythagoras of Samos


All things are numbers. Mathematics is the basis for everything,and geometry is the highest form of mathematical studies. Thephysical world can be understood through mathematics.

Numbers have personalities, characteristics, strengths andweaknesses.Certain symbols have a mystical significanceNot sure how he died

Pythagoras of Samos


All things are numbers. Mathematics is the basis for everything,and geometry is the highest form of mathematical studies. Thephysical world can be understood through mathematics.Numbers have personalities, characteristics, strengths andweaknesses.

Certain symbols have a mystical significanceNot sure how he died

Pythagoras of Samos


All things are numbers. Mathematics is the basis for everything,and geometry is the highest form of mathematical studies. Thephysical world can be understood through mathematics.Numbers have personalities, characteristics, strengths andweaknesses.Certain symbols have a mystical significance

Not sure how he died

Pythagoras of Samos



Attributed to Pythagoras

The sum of the angles of a triangle is equal to two right angles.

The theorem of Pythagoras - for a right-angled triangle thesquare on the hypotenuse is equal to the sum of the squares onthe other two sides.

Constructing figures of a given area and geometrical algebra.

The discovery of irrational numbers is attributed to thePythagoreans, but seems unlikely to have been the idea ofPythagoras.

The Pythagorean Theorem

TheoremIf a right triangle has legs of lengths a and b and hypotenuse of lengthc, then c2 = a2 + b2.

There are 98 different known proofs, including one by PresidentJames Garfield.

Proof

The sides of the outside square are all of length c, so the area of thewhole thing is c2.Each triangle is a right triangle with legs a and b, so each of the fourhave area 1

2 ab and since there are 4 of them, we have 4(12 ab) = 2ab.

The square in the middle has each side of length b− a, so the area ofthat square is (b− a)2.When we put it all together, we get

c2 = 2ab + (b− a)2 = 2ab + b2 − 2ab + a2 = a2 + b2

Proof

The sides of the outside square are all of length c, so the area of thewhole thing is c2.

Each triangle is a right triangle with legs a and b, so each of the fourhave area 1



c2 = 2ab + (b− a)2 = 2ab + b2 − 2ab + a2 = a2 + b2

Proof




c2 = 2ab + (b− a)2 = 2ab + b2 − 2ab + a2 = a2 + b2

Proof



The square in the middle has each side of length b− a, so the area ofthat square is (b− a)2.

When we put it all together, we get

c2 = 2ab + (b− a)2 = 2ab + b2 − 2ab + a2 = a2 + b2

Proof




c2 = 2ab + (b− a)2 = 2ab + b2 − 2ab + a2 = a2 + b2

Example

Application of the Pythagorean TheoremSuppose you leave your house and run 4 miles east. Then, you turnand run 2 miles north. You turn again and run 4 miles east. When youfinish the 10 miles, how far from home are you?

• 4 // •

Home•4

// •2

OO

Solution

• 4 // •

Home•

/7

4// •

2

OO

• 4 // •

Home•

/7

4// •

2

OO

So, we really have a right triangle with legs of lengths 2 and 8, so thedistance from home is√

22 + 82 =√

4 + 64 =√

68 = 2√

17

And, of course, we need a proper label, so the distance is 2√

17 miles.

Example

Application of the Pythagorean TheoremWhile decorating for an outdoor birthday party, you put up a bannerthat is 20 feet long. You tie one end to the top of a gazebo that is 12feet high and the other end to the edge of a picnic table at a height of3 feet. What is the horizontal distance from the gazebo to the table?

x2 + 92 = 202 ⇒ x2 + 81 = 400⇒ x2 = 319

So, the distance between the two is√

319 feet, which is roughly 17.86feet.

Application of the Pythagorean Theorem

ExampleHow far is it from home to second?

√902 + 902 = 90

√2 ≈ 127.45 feet.

Special Right Triangles

45◦ − 45◦ − 90◦ triangleThis triangle has the property that if the legs are of length x then thehypotenuse is of length x

√2.

30◦ − 60◦ − 90◦

Missing Lengths

ExampleSolve for x.

��

4

31

x

What do we do first?Once we know we have similar triangles, we can set up ratios.

13

=x

hypotenuse

Missing Lengths

ExampleSolve for x.

��

4

31

x

What do we do first?

Once we know we have similar triangles, we can set up ratios.

13

=x

hypotenuse

Missing Lengths

ExampleSolve for x.

��

4

31

x

What do we do first?Once we know we have similar triangles, we can set up ratios.

13

=x

hypotenuse

Solution (cont.)

��

4

31

x

So, we need the hypotenuse of the larger triangle.

32 + 42 = hyp2 ⇒ hyp = 5

Now we ...

3x = 5⇒ x =53

Missing Lengths

ExampleSolve for x and y if ∠ABD and ∠ADC are both right angles.

��

��

��@

@@@@@

x

4√

3y 4

AB

C

D

Solution

��

��

��@@@@@@

x

4√

3y 4

AB

C

D

What do we do first?

I’d find x ...(4√

3)2

+ 42 = x2 ⇒ x2 = 64⇒ x = 8

Now, why are ∆ADC and ∆CDB similar?Now, we set up ratios.

84

=4y

So, y = 2

Solution

��

��

��@@@@@@

x

4√

3y 4

AB

C

D

What do we do first? I’d find x ...(4√

3)2

+ 42 = x2 ⇒ x2 = 64⇒ x = 8


84

=4y

So, y = 2

Solution

��

��

��@@@@@@

x

4√

3y 4

AB

C

D


3)2

+ 42 = x2 ⇒ x2 = 64⇒ x = 8

Now, why are ∆ADC and ∆CDB similar?

Now, we set up ratios.84

=4y

So, y = 2

Solution

��

��

��@@@@@@

x

4√

3y 4

AB

C

D


3)2

+ 42 = x2 ⇒ x2 = 64⇒ x = 8


84

=4y

So, y = 2

Rhombus

ExampleFind the area of the given rhombus.

JJJJJ

JJJJJ

5

2

Rhombus Solution

What do we know about the diagonals of a rhombus?

The areperpendicular bisectors of each other. Why does this help?

Area =d1d2

2Now ...

52 = 22 + d2 ⇒ d =√

21

This means that the vertical diagonal is of length 2√

21. Therefore thearea is

d1d2

2=

2√

21 · 42

= 4√

21 square units

Rhombus Solution

What do we know about the diagonals of a rhombus? The areperpendicular bisectors of each other. Why does this help?

Area =d1d2

2Now ...

52 = 22 + d2 ⇒ d =√

21

This means that the vertical diagonal is of length 2√

21. Therefore thearea is

d1d2

2=

2√

21 · 42

= 4√

21 square units

Triangles

Example

What kind of triangle has vertices at (5, 6), (−1, 2) and (9, 0)?

To solve, we need the distance formula.

D =

√(x2 − x1)2 + (y2 − y1)2

Solution

So let’s find the lengths.

√(5− (−1))2 + (6− 2)2 =

√36 + 16 =

√52 = 2

√13√

(5− 9)2 + (6− 0)2 =√

16 + 36 =√

52 = 2√

13√(−1− 9)2 + (2− 0)2 =

√100 + 4 =

√104 = 2

√26

What type of triangle is this?Isosceles triangle

Solution


√(5− (−1))2 + (6− 2)2 =

√36 + 16 =

√52 = 2

√13√

(5− 9)2 + (6− 0)2 =√

16 + 36 =√

52 = 2√

13√(−1− 9)2 + (2− 0)2 =

√100 + 4 =

√104 = 2

√26

What type of triangle is this?

Isosceles triangle

Solution


√(5− (−1))2 + (6− 2)2 =

√36 + 16 =

√52 = 2

√13√

(5− 9)2 + (6− 0)2 =√

16 + 36 =√

52 = 2√

13√(−1− 9)2 + (2− 0)2 =

√100 + 4 =

√104 = 2

√26

What type of triangle is this?Isosceles triangle

Same Triangle

ExampleSo what is the perimeter of the triangle in the last example?

2√

13 + 2√

13 + 2√

26 = 4√

13 + 2√

26 units

Same Triangle

ExampleCould we find the area?

(2√

13)2

=(√

26)2

+ h2

52 = 26 + h2

h =√

26

So, each of the two triangles has area

base× height2

=

√26√

262

=262

= 13

And, since there are two of them, the area is 26 square units.

Circles

Does anyone remember the Pythagorean Identity?

Circles

Now, if we look at this in terms of coordinates, let cos(θ) = x andsin(θ) = y and let the center be at the point (h, k).

Circles

Then the distance between the center and the point on the radius is

r =√

(x− h)2 + (y− k)2

And finally, if we square both sides, we get

FormulaThe equation of a circle centered at the point (h, k) with radius r isgiven by

(x− h)2 + (y− k)2 = r2

Circles

Example

If a circle has the formula (x− 4)2 + y2 = 4, what is the center? theradius?

The center is (4, 0) and the radius is 2.

Circles

ExampleSuppose we have two circles, given by the formulas

(x− 2)2 + (y + 3)2 = 16 (1)

(x− 2)2 + (y + 3)2 = 4 (2)

What is the area between the circles?

The radius of circle (1) is 4, so the area of the circle is 16π squareunits.The radius of circle (2) is 2, so the area of the circle is 4π square units.So, the area between them is 16π − 4π = 12π square units.

Circles


(x− 2)2 + (y + 3)2 = 16 (1)

(x− 2)2 + (y + 3)2 = 4 (2)


The radius of circle (1) is 4, so the area of the circle is 16π squareunits.

The radius of circle (2) is 2, so the area of the circle is 4π square units.So, the area between them is 16π − 4π = 12π square units.

Circles


(x− 2)2 + (y + 3)2 = 16 (1)

(x− 2)2 + (y + 3)2 = 4 (2)


The radius of circle (1) is 4, so the area of the circle is 16π squareunits.The radius of circle (2) is 2, so the area of the circle is 4π square units.

So, the area between them is 16π − 4π = 12π square units.

Circles


(x− 2)2 + (y + 3)2 = 16 (1)

(x− 2)2 + (y + 3)2 = 4 (2)


The radius of circle (1) is 4, so the area of the circle is 16π squareunits.The radius of circle (2) is 2, so the area of the circle is 4π square units.So, the area between them is 16π − 4π = 12π square units.

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