triangles, right traingles

1. About Triangles
Plane-shapes are regions that are enclosed by lines
on a flat surface.

2. About Triangles
Plane-shapes are regions that are enclosed by lines
on a flat surface.
The ones with straight line segments as boundaries:
are called polygons.

3. About Triangles
Plane-shapes are regions that are enclosed by lines
on a flat surface.
The ones with straight line segments as boundaries:
In many applications, shapes with curved boundaries
are approximated with polygons.
are called polygons.

4. About Triangles
Plane-shapes are regions that are enclosed by lines
on a flat surface.
The ones with straight line segments as boundaries:
In many applications, shapes with curved boundaries
are approximated with polygons.
To study polygons, we dissect them into triangles;
hence “trigonometry”, the study of measurements
of triangles, is pivotal.
are called polygons.

5. About Triangles
Here is a standard labeled triangle with
each side and its opposite angle labeled
with the same letter.

6. About Triangles
Here is a standard labeled triangle with
each side and its opposite angle labeled
with the same letter. Below are basic facts
about triangles that we will highlight.

7. About Triangles
* Triangles are rigid
* The Triangle Inequality
* Larger angle is opposite to the longer side
* The Sum of Angles is always 180o
* Similar triangles have the same angles and
the corresponding sides are proportional.
* The area of a triangle is A=½ Bh, B = base, h = ht
* Pythagorean Theorem (see slide 51, sect. 1)
For right triangles: a2 + b2 = c2
Here is a standard labeled triangle with
each side and its opposite angle labeled
with the same letter. Below are basic facts
about triangles that we will highlight.

8. About Triangles
* Triangles are Rigid

9. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved.

10. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved. Hence the lengths of three
sides determine a unique triangle.

11. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved. Hence the lengths of three
sides determine a unique triangle.
Shapes with 4 or more sides
may be folded into a triangle

12. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved. Hence the lengths of three
sides determine a unique triangle.
Shapes with 4 or more sides
may be folded into a triangle
or to nothing.
folding polygons

13. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved. Hence the lengths of three
sides determine a unique triangle.
Shapes with 4 or more sides
may be folded into a triangle
or to nothing. So the lengths
of the sides alone are not
sufficient to determine a
polygon with 4 or more sides. folding polygons

14. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved. Hence the lengths of three
sides determine a unique triangle.
To specify a polygon with 4 or more sides we can
dissect it into triangles and track the triangles,
Shapes with 4 or more sides
may be folded into a triangle
or to nothing. So the lengths
of the sides alone are not
sufficient to determine a
polygon with 4 or more sides. folding polygons

15. About Triangles
* Triangles are Rigid
Each side of a triangle "ties" down the other two sides
so they can't be moved. Hence the lengths of three
sides determine a unique triangle.
To specify a polygon with 4 or more sides we can
dissect it into triangles and track the triangles,
e.g.
Shapes with 4 or more sides
may be folded into a triangle
or to nothing. So the lengths
of the sides alone are not
sufficient to determine a
polygon with 4 or more sides. folding polygons
Recorded using
3 triangles

16. About Triangles
Two arms of a triangle must be
long enough to reach other.

17. About Triangles
* The Triangle Inequality
The sum of the lengths of any two sides of a triangle
must be more than the third side, or that
a + b > c, b + c > a and a + c > b.
Two arms of a triangle must be
long enough to reach other.

18. About Triangles
* The Triangle Inequality
The sum of the lengths of any two sides of a triangle
must be more than the third side, or that
a + b > c, b + c > a and a + c > b.
a = 15
b = 23
c = ?
Two arms of a triangle must be
long enough to reach other.
Example A.
Two sides of a triangle are a = 15 and b = 23.
What’s the possible length of the third side c?

19. About Triangles
* The Triangle Inequality
The sum of the lengths of any two sides of a triangle
must be more than the third side, or that
a + b > c, b + c > a and a + c > b.
a = 15
b = 23
c = ?
Two arms of a triangle must be
long enough to reach other.
Example A.
Two sides of a triangle are a = 15 and b = 23.
What’s the possible length of the third side c?
The length of c must be less than 15 + 23 = 38,

20. About Triangles
* The Triangle Inequality
The sum of the lengths of any two sides of a triangle
must be more than the third side, or that
a + b > c, b + c > a and a + c > b.
a = 15
b = 23
c = ?
Two arms of a triangle must be
long enough to reach other.
Example A.
Two sides of a triangle are a = 15 and b = 23.
What’s the possible length of the third side c?
The length of c must be less than 15 + 23 = 38,
and it must be more than 23 – 15 = 8. So 8 < c < 38.

21. About Triangles
* The Triangle Inequality
The sum of the lengths of any two sides of a triangle
must be more than the third side, or that
a + b > c, b + c > a and a + c > b.
a = 15
b = 23
c = ?
Two arms of a triangle must be
long enough to reach other.
Example A.
Two sides of a triangle are a = 15 and b = 23.
What’s the possible length of the third side c?
The length of c must be less than 15 + 23 = 38,
and it must be more than 23 – 15 = 8. So 8 < c < 38.
For general n-sided polygons, the sum of any
n – 1 sides must be more than the remaining side.

22. About Triangles
* The Triangle Inequality
The sum of the lengths of any two sides of a triangle
must be more than the third side, or that
a + b > c, b + c > a and a + c > b.
a = 15
b = 23
c = ?
Two arms of a triangle must be
long enough to reach other.
Example A.
Two sides of a triangle are a = 15 and b = 23.
What’s the possible length of the third side c?
The length of c must be less than 15 + 23 = 38,
and it must be more than 23 – 15 = 8. So 8 < c < 38.
For general n-sided polygons, the sum of any
n – 1 sides must be more than the remaining side.
So there’s no 4-sided polygons with sides 1, 1, 1 and 4.

23. About Triangles
* Larger angle faces longer side

24. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle.

25. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
A

26. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
ACB A> >

27. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
ACB A> >
then b > c > a and vice versa.
B

28. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
ACB A> >
So a triangle will have two same angles
iff* their opposites sides are the same
then b > c > a and vice versa.
*short for “if and only if”, meaning one implies the other

29. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
ACB A> >
So a triangle will have two same angles
iff* their opposites sides are the same.
These are called isosceles triangles. An isosceles triangle
then b > c > a and vice versa.
*short for “if and only if”, meaning one implies the other

30. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
ACB A> >
So a triangle will have two same angles
iff* their opposites sides are the same.
These are called isosceles triangles.
If a triangle has all angles the same
then all its sides are the same,
and vice versa.
An isosceles triangle
*short for “if and only if”, meaning one implies the other
then b > c > a and vice versa.

31. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
ACB A> >
So a triangle will have two same angles
iff* their opposites sides are the same.
These are called isosceles triangles.
If a triangle has all angles the same
then all its sides are the same,
and vice versa. These are called
equilateral triangles.
An isosceles triangle
An equilateral triangle
*short for “if and only if”, meaning one implies the other
then b > c > a and vice versa.

32. About Triangles
* Larger angle faces longer side
Given any triangle, the longest side is opposite to
the largest angle, the medium side is opposite to
the medium angle, the shortest side is opposite to
the smallest angle. So for triangle if
a
b
c
C
B
ACB A> >
So a triangle will have two same angles
iff* their opposites sides are the same.
These are called isosceles triangles.
If a triangle has all angles the same
then all its sides are the same,
and vice versa. These are called
equilateral triangles.
An isosceles triangle
An equilateral triangle
*short for “if and only if”, meaning one implies the other
then b > c > a and vice versa.

33. * The sum of all three angles is 180o
About Triangles

34. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC

35. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC BC A

36. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC
Hence the exterior angle of A is the sum of
the two interior angles B and C as shown.
BC A

37. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC
Hence the exterior angle of A is the sum of
the two interior angles B and C as shown.
the exterior
angle of A
= B + C
A
BC
B
C
BC A

38. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC
Example B.
Given the following triangles, find the angle x.
Hence the exterior angle of A is the sum of
the two interior angles B and C as shown.
the exterior
angle of A
= B + C
A
BC
B
C
22°
105°135°
x
y
BC A

39. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC
Example B.
Given the following triangles, find the angle x.
22°
105°135°
x
Let y be the angle as shown
so y = 180 – 135 = 45.
y
Hence the exterior angle of A is the sum of
the two interior angles B and C as shown.
the exterior
angle of A
= B + C
A
BC
B
C
BC A

40. * The sum of all three angles is 180o
The three angles of a triangle when joint together
form a straight line or 180o.
About Triangles
A
BC
Example B.
Given the following triangles, find the angle x.
22°
105°135°
x
Let y be the angle as shown
so y = 180 – 135 = 45.
Since 105 = (x + 22) + y
= (x + 22) + 45 so x = 38o.
y
Hence the exterior angle of A is the sum of
the two interior angles B and C as shown.
the exterior
angle of A
= B + C
A
BC
B
C
BC A

41. About Triangles
* Similar Triangles have same angles and
proportional sides

42. About Triangles
* Similar Triangles have same angles and
proportional sides
Two shapes are called “similar” if they have the same
shape but different sizes.

43. About Triangles
* Similar Triangles have same angles and
proportional sides
Two shapes are called “similar” if they have the same
shape but different sizes. For similar triangles,
this means the corresponding sides are proportional.

44. About Triangles
* Similar Triangles have same angles and
proportional sides
Following are two cases of
similar triangles that are
useful to keep in mind.
Two shapes are called “similar” if they have the same
shape but different sizes. For similar triangles,
this means the corresponding sides are proportional.

45. About Triangles
* Similar Triangles have same angles and
proportional sides
Following are two cases of
similar triangles that are
useful to keep in mind.
Two shapes are called “similar” if they have the same
shape but different sizes. For similar triangles,
this means the corresponding sides are proportional.
I. The “tip” of a triangle is
similar to the original triangle.

46. About Triangles
* Similar Triangles have same angles and
proportional sides
Following are two cases of
similar triangles that are
useful to keep in mind.
parallel to b
b
Two shapes are called “similar” if they have the same
shape but different sizes. For similar triangles,
this means the corresponding sides are proportional.
I. The “tip” of a triangle is
similar to the original triangle.

47. About Triangles
* Similar Triangles have same angles and
proportional sides
Following are two cases of
similar triangles that are
useful to keep in mind.
parallel to b
b
a
cb
x
zy
Two shapes are called “similar” if they have the same
shape but different sizes. For similar triangles,
this means the corresponding sides are proportional.
I. The “tip” of a triangle is
similar to the original triangle.

48. About Triangles
* Similar Triangles have same angles and
proportional sides
Two shapes are called “similar” if they have the same
shape but different sizes. For similar triangles,
this means the corresponding sides are proportional.
Following are two cases of
similar triangles that are
useful to keep in mind.
I. The “tip” of a triangle is
similar to the original triangle.
parallel to b
b
a
cb
x
zy
Hence as shown here,
we have that:
a cb
x zy= =

49. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15

50. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15
From the similar triangles we have the proportion
10 15
x x + (x – 2)=

51. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15
From the similar triangles we have the proportion
10 15
x x + (x – 2)= or
10 15
x 2x – 2=

52. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15
From the similar triangles we have the proportion
10 15
x x + (x – 2)= or
10 15
x 2x – 2= cross-multiply
10(2x – 2) = 15x

53. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15
From the similar triangles we have the proportion
10 15
x x + (x – 2)= or
10 15
x 2x – 2= cross-multiply
10(2x – 2) = 15x
20x – 20 = 15x
5x = 20 → so x = 4

54. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15
From the similar triangles we have the proportion
10 15
x x + (x – 2)= or
10 15
x 2x – 2= cross-multiply
10(2x – 2) = 15x
20x – 20 = 15x
5x = 20 → so x = 4
Il. Two parallel line-segments and their diagonals
form two similar triangles.
a
x
b
y
c
z

55. Example C.
Given the following figure, find x.
About Triangles
x
x–2
10
15
From the similar triangles we have the proportion
10 15
x x + (x – 2)= or
10 15
x 2x – 2= cross-multiply
10(2x – 2) = 15x
20x – 20 = 15x
5x = 20 → so x = 4
Il. Two parallel line-segments and their diagonals
form two similar triangles.
a cb
x zy= =
a
x
b
y
c
z
So in the figure here we’ve
that

56. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
1
3
P
2
Base
Base

57. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
1
3
P
2
h
H
Base
Base

58. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
We’ve h 1
H
= so that H = 3h.3
1
3
P
2
h
H
Base
Base

59. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
We’ve
Since h + H = 2 so h + 3h = 2
so that H = 3h.
1
3
P
2
h
H
Base
Base
h 1
H
=
3

60. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
We’ve
Since h + H = 2 so h + 3h = 2
or 4h = 2 → h = ½ and that H = 3/2
so that H = 3h.
1
3
P
2
h
H
Base
Base
h 1
H
=
3

61. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
We’ve
Since h + H = 2 so h + 3h = 2
or 4h = 2 → h = ½ and that H = 3/2
so that H = 3h.
1
3
P
2
h
H
Note the distances from P
to two parallel lines are proportional
to the lengths of the two lines. P
Base
Base
h 1
H
=
3

62. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
We’ve
Since h + H = 2 so h + 3h = 2
or 4h = 2 → h = ½ and that H = 3/2
so that H = 3h.
1
3
P
2
h
H
Note the distances from P
to two parallel lines are proportional
to the lengths of the two lines.
So if line A is k times the length of line B,
1
k
P
Base
Base
h 1
H
=
3

63. Example D.
Given the following figure, find the
distances from P to the base-lines.
About Triangles
The question is asking for the
heights of the triangles as shown.
Let’s label them as h and H.
We’ve
Since h + H = 2 so h + 3h = 2
or 4h = 2 → h = ½ and that H = 3/2
so that H = 3h.
1
3
P
2
h
H
Note the distances from P
to two parallel lines are proportional
to the lengths of the two lines.
So if line A is k times the length of line B,
then P is k times as far to line A
then to line B.
1
k
P
h
kh
Base
Base
h 1
H
=
3

64. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles

65. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles
Area of a Parallelogram
Let B = base and H = height of a parallelogram.
H=height
B=base

66. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles
Area of a Parallelogram
Let B = base and H = height of a parallelogram.
By cutting and pasting as shown,
we get a rectangle. H=height
B=base

67. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles
Area of a Parallelogram
Let B = base and H = height of a parallelogram.
By cutting and pasting as shown,
we get a rectangle. So the area
of a parallelogram is A = BH.
H=height
B=base

68. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles
Area of a Parallelogram
Let B = base and H = height of a parallelogram.
By cutting and pasting as shown,
we get a rectangle. So the area
of a parallelogram is A = BH.
H=height
B=baseArea of a Triangle
Given a triangle with base B and height H,
H=height
B=base

69. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles
Area of a Parallelogram
Let B = base and H = height of a parallelogram.
By cutting and pasting as shown,
we get a rectangle. So the area
of a parallelogram is A = BH.
H=height
B=baseArea of a Triangle
Given a triangle with base B and height H, placing two
copies of it as shown,
we get a parallelogram.
H=height
B=base

70. L
W A = LW
Area of a Rectangle
The area of a rectangle A = LW
where L = length and W = width.
About Triangles
Area of a Parallelogram
Let B = base and H = height of a parallelogram.
By cutting and pasting as shown,
we get a rectangle. So the area
of a parallelogram is A = BH.
H=height
B=baseArea of a Triangle
Given a triangle with base B and height H, placing two
copies of it as shown,
we get a parallelogram.
Let A be the area of the triangle,
then 2A = HB or that A = ½ HB.
H=height
B=base

71. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem

72. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.

73. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.
Given a right triangle as shown.
a
b
c

74. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.
Given a right triangle as shown.
a
b
Taking four copies and arranging
them into a (a + b) x (a + b) square,b
a
c
c

75. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.
Given a right triangle as shown.
a
b
Taking four copies and arranging
them into a (a + b) x (a + b) square,
we have the following relation:
b
a
b
4 a
+
c
c
x =
a + b
a + bc
c

76. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.
Given a right triangle as shown.
a
c
b
Taking four copies and arranging
them into a (a + b) x (a + b) square,
we have the following relation:
b
a
b
4 a
+
c
c
x =
a + b
a + b
In terms of areas 4(ab/2) + c2 = (a + b)2
c

77. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.
Given a right triangle as shown.
a
b
Taking four copies and arranging
them into a (a + b) x (a + b) square,
we have the following relation:
b
a
b
4 a
+
c
c
x =
a + b
a + b
In terms of areas 4(ab/2) + c2 = (a + b)2
Simplify and expand 2ab + c2 = a2 + 2ab + b2
c
c

78. For any right triangle a2 + b2 = c2.
Pythagorean Theorem
The Pythagorean Theorem
Here is a proof of the theorem.
Given a right triangle as shown.
a
b
Taking four copies and arranging
them into a (a + b) x (a + b) square,
we have the following relation:
b
a
b
4 a
+
c
c
x =
a + b
a + b
In terms of areas 4(ab/2) + c2 = (a + b)2
Simplify and expand 2ab + c2 = a2 + 2ab + b2
so c2 = a2 + b2
c
c

79. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem hypotenuse
leg
leg

80. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem
Example E. a. Given the diameter
of a circle is 10. How far is a cord
of length 6 from the center?
10
6
distance = ?5
hypotenuse
leg
leg

81. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem
Example E. a. Given the diameter
of a circle is 10. How far is a cord
of length 6 from the center?
10
6
distance = ?
Let d be the distance.
The radius of the circle is 5 and
half of the cord is 3,
5
hypotenuse
leg
leg

82. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem
Example E. a. Given the diameter
of a circle is 10. How far is a cord
of length 6 from the center?
10
6
distance = ?
Let d be the distance.
The radius of the circle is 5 and
half of the cord is 3, so 32 + d2 = 52.
Hence d2 = 16 or d = 4.
5
hypotenuse
leg
leg

83. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem
Example E. a. Given the diameter
of a circle is 10. How far is a cord
of length 6 from the center?
10
6
distance = ?
Let d be the distance.
The radius of the circle is 5 and
half of the cord is 3, so 32 + d2 = 52.
Hence d2 = 16 or d = 4.
5
b. The cord that’s ½ way up the semi-
circle as shown has length 12.
What’s the radius of the circle? rr/2
hypotenuse
leg
leg

84. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem
Example E. a. Given the diameter
of a circle is 10. How far is a cord
of length 6 from the center?
10
6
distance = ?
Let d be the distance.
The radius of the circle is 5 and
half of the cord is 3, so 32 + d2 = 52.
Hence d2 = 16 or d = 4.
5
6
b. The cord that’s ½ way up the semi-
circle as shown has length 12.
What’s the radius of the circle? rr/2
From the right triangle, we’ve that
r2= (r/2)2 + 62 .
6
hypotenuse
leg
leg

85. We use the Pythagorean Theorem
to solve for the hypotenuse or the legs.
The Pythagorean Theorem
Example E. a. Given the diameter
of a circle is 10. How far is a cord
of length 6 from the center?
10
6
distance = ?
Let d be the distance.
The radius of the circle is 5 and
half of the cord is 3, so 32 + d2 = 52.
Hence d2 = 16 or d = 4.
5
6
b. The cord that’s ½ way up the semi-
circle as shown has length 12.
What’s the radius of the circle? rr/2
From the right triangle, we’ve that
r2= (r/2)2 + 62 . Hence r2 = 48 or r = 4√3 ≈ 6.93.
6
hypotenuse
leg
leg

86. Right Triangles
Right triangles are central to trigonometry because
by slicing from the tips to the bases vertically,
every triangle may be split into two right triangles.
Many proofs in geometry
take advantages of this fact.
Cutting perpendiculars
right triangles actually produces
two smaller triangles
which are “similar” to the original.

87. 1. a. Two sides of a triangle are a = 18 and b = 15,
what are the possible measurements for the third side c?
About Triangles
b. Suppose 18 is the longest side, what can c be?
c. Suppose 15 is the shortest side, what can c be?
f. Suppose 18 is the longest side and 15 is the shortest
what can c be?
2. Answer the same questions in 1 if a = 14.7 and b = 25.6.
3. Following are the measurements of the sides of triangles,
{2, 5, 4}, {7, 3, 3}, {12, 25, 34}, {28, 75, 44}, {13.9, 7.2, 5.8}
which are possible which are not?
d. Suppose 18 is a medium side what can c be?
e. Suppose 15 is a medium side what can c be?
4. Following are the measurements of the quadrilaterals
or 4–sided polygons, {2, 5, 4,11}, {7, 9, 3,1}, {12, 25, 34, 75},
{2, 8, 53, 44}, which are possible which are not?
5. An isosceles triangle has a = 18 and b = 8, what is c?
Exercise 5. A.

88. 6. a. Two sides of a triangle are a = 9.8 and b = 6.4,
what are the possible measurements for the third side c?
About Triangles
b. Suppose c is the longest side, what can c be?
d. Suppose c is the shortest side, what can c be?
f. Suppose c is a medium side what can c be?
g. Suppose c is not the a medium side what can c be?
Exercise 5. A.
Given the information of two sides a and b of a triangle the
length of the 3rd side c is restricted by the triangle inequality.
If in addition, the order of the lengths is known, then we may
refine further the restriction on c.
Following are logical questions given the length–order of side c.
c. Suppose c is not the longest side, what can c be?
e. Suppose c is not the shortest side, what can c be?

89. About Triangles
Ex. B. Find the angle x.
1. 2. 3. 4.
5. 6. 7. 8.
5
x
9
6
5
x+8
9
x
5
x
x+3
6
x
6 4
14
x
6 x–3
14
x
6 4
x + 6
1. 2. 3.
4. 5. 6.
Ex C. Solve for x.

90. About Triangles
Ex. D. Find x and the area of each right triangle below
assuming the unit is ft.
x
5
x
13
1. 3.
3
12x
10
2.
6
x
5
4.
x – 1
x + 2
10
5.
6
x + 4
8
6.
x

91. x
E. Find x for 7–12
8
10
x
8.
6
8
9.
x
10.
8
10
x
11.
6
8
x
12.
4
3
7.
About Triangles
x
4
2

92. About Triangles
Exercise 5.
Answers:
A. 1. a. 3 < c < 33 b. 3 < c < 18 c. 15 < c < 33 d. 18 ≤ c < 33
e. 3 < c ≤ 15 f. 15 < c < 18
3. {2, 5, 4}, {12, 25, 34} are possible,
{7, 3, 3}, {28, 75, 44} and {13.9, 7.2, 5.8} are not
5. c = 18
B. 1. x = 83 3. x = 65 5. x = 17 7. x = 33
C. 1. x = 28/3 3. x = 12 5. x = 10
D. 1. x = 4 3. x = 5 5. x = 6 7. x = 3 9. x = √3 11. x = √23

93. F2F1
P Q
R
Let P, Q, and R be points on an ellipse as shown,
The Pythagorean Theorem

94. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2
The Pythagorean Theorem

95. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2
q1
q2
The Pythagorean Theorem

96. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2 = r1 + r2
= a constant. q1
q2
r2
r1
The Pythagorean Theorem

97. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2 = r1 + r2
= a constant. q1
q2
r2
r1
So an ellipse is the set of
all the points whose sum
of the distances to two foci
F1 and F2 , is a constant.
The Pythagorean Theorem

98. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2 = r1 + r2
= a constant. q1
q2
r2
r1
So an ellipse is the set of
all the points whose sum
of the distances to two foci
F1 and F2 , is a constant.
The Pythagorean Theorem
(x – h)2 (y – k)2
a2 b2+ = 1
The standard form of an ellipses is

99. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2 = r1 + r2
= a constant. q1
q2
r2
r1
So an ellipse is the set of
all the points whose sum
of the distances to two foci
F1 and F2 , is a constant.
The Pythagorean Theorem
(x – h)2 (y – k)2
a2 b2 with (h, k) as the center+ = 1
The standard form of an ellipses is
F2
F1
(h, k)

100. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2 = r1 + r2
= a constant. q1
q2
r2
r1
So an ellipse is the set of
all the points whose sum
of the distances to two foci
F1 and F2 , is a constant.
The Pythagorean Theorem
(x – h)2 (y – k)2
a2 b2
x-radius=a y-radius = b
with (h, k) as the center and+ = 1
The standard form of an ellipses is
F2
F1
a
b
(h, k)

101. F2F1
P Q
R
p1
p2
Let P, Q, and R be points on an ellipse as shown,
then p1 + p2 = q1 + q2 = r1 + r2
= a constant. q1
q2
r2
r1
So an ellipse is the set of
all the points whose sum
of the distances to two foci
F1 and F2 , is a constant.
The Pythagorean Theorem
(x – h)2 (y – k)2
a2 b2
x-radius=a y-radius = b
with (h, k) as the center and+ = 1
The standard form of an ellipses is
F2
F1
a
b
From these, let’s calculate
the distance d from the center (h, k) to F1.
(h, k)
d

102. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
F2F1

103. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d. F2F1 d

104. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d.
Points P and Q on the ellipse
shown here must have the same
distance-sums to F1 and F2.
F2F1 d
X X
P
Q

105. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d.
Points P and Q on the ellipse
shown here must have the same
distance-sums to F1 and F2.
F2F1 d
X X bX X
P
Q Q
P
By moving the short segment as shown,

106. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d.
Points P and Q on the ellipse
shown here must have the same
distance-sums to F1 and F2.
F2F1 d
X X bX X
P
Q Q
P
By moving the short segment as shown, we see that
x + x =

107. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d.
Points P and Q on the ellipse
shown here must have the same
distance-sums to F1 and F2.
F2F1 d
X X bX X
P
Q Q
P
By moving the short segment as shown, we see that
x + x =
aa

108. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d.
Points P and Q on the ellipse
shown here must have the same
distance-sums to F1 and F2.
F2F1 d
X X bX X
P
Q Q
P
By moving the short segment as shown, we see that
x + x = a + a or that x = a.
aa

109. The Pythagorean Theorem
a
b
Assuming the ellipse has the a longer
x-radius as shown,
we want to find d.
Points P and Q on the ellipse
shown here must have the same
distance-sums to F1 and F2.
F2F1 d
X X bX X
P
Q Q
P
By moving the short segment as shown, we see that
x + x = a + a or that x = a. Hence d = √a2 – b2.
d a

110. The Pythagorean Theorem
In particular for ellipses centered at (0, 0),
we have the coordinates of the following points:
x2 y2
a2
b2+ = 1
F2F1
(√a2–b2, 0)(–√a2–b2, 0) (a, 0)(–a, 0)
(0, b)
(0, –b)