14.2 the pythagorean theorem, distance formula and...
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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
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
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
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
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
c. 569 BC - (500-475) BCSettled 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
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.
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.
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.
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
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
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
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
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
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.
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.
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.
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.
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.
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.
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◦
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
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
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
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
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
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
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
Now, why are ∆ADC and ∆CDB similar?Now, we set up ratios.
84
=4y
So, y = 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
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
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
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
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
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
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.
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.
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.
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
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
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
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
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
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.