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Pythagoras ‘Theorem
In a right-angled triangle, the square of the length to hypotenuse is equalto the sum of the squares of the
lengths of the other two sides.(Pythagoras’ theorem).
if the square of one side of a triangle is equal to the sum of the squares ofthe other two sides then the
triangle is a right angled triangle (converse toPythagoras’ theorem).
1 Motivation
You’re locked out of your house and the only open window is on the second floor, 25 feet above the ground.
You need to borrow a ladder from one of your neighbors. There’s a bush along the edge of the house, so
you’ll have to place the ladder 10 feet from the house. What length of ladder do you need to reach the
window?
Figure 1: Ladder to reach the window
Brief history: Pythagoras lived in the 500’s BC, and was one of the first Greek mathematical thinkers.
Pythagoreans were interested in Philosophy, especially in Music and Mathematics?
The statement of the Theorem was discovered on a Babylonian tablet circa 1900−1600 B.C. Professor
R. Sullying in his book 5000 B.C. and Other Philosophical Fantasies tells of an experiment he ran in one
of his geometry classes. He drew a right triangle on the board with squares on the hypotenuse and legs and
observed the fact the square on the hypotenuse had a larger area than either of the other two squares.
Then he asked, ”Suppose these three squares were made of beaten gold, and you were offered either the
one large square or the two small squares. Which would you choose?” Interestingly enough, about half the
class opted for the one large square and half for the two small squares. Both groups were equally amazed
when told that it would make no difference.
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Pythagorean Theorem
In a right triangle, the square of the length of the hypotenuse is equal to
the sum of the squares of the lengths of the legs.
(c2 = a2 + b2)
It is believed that this theorem was known, in some form,
long before the time of Pythagoras. A thousand years before
Pythagoras, the Babylonians recorded their knowledge of the
theorem on clay tablets. The Egyptians used the concept of
the theorem in the building of the pyramids. In 1100 B.C.
China, Tschou-Gun knew of this theorem.
It was, however, Pythagoras who generalized the theorem to all right triangles and is credited with its
first geometrical demonstration. Consequently, the theorem bears his name.
Geometrical Proof: The Pythagorean Theorem has drawn a good deal of
attention from mathematicians. There are hundreds of geometrical proofs
(or demonstrations) of the theorem, with even a larger number of algebraic
proofs.
Geometrically, the Pythagorean Theorem can be interpreted as discussing
the areas of squares whose sides are the sides of the triangle (as seen in the
picture at the left). The theorem can be rephrased as, "The (area of the)
square described upon the hypotenuse of a right triangle is equal to the sum
of the (areas of )the squares described upon the other two sides."
Converse:
Theorem: If a triangle is a right triangle, the square of the length of the hypotenuse is equal to the
sum of the squares of the lengths of the legs.
Converse: If the square of the length of the longest side of a triangle is equal to the sum of the squares
of the lengths of the other two sides, the triangle is a right triangle.
Pythagorean Triples: There are certain sets of numbers that have a very special property in
connection to the Pythagorean Theorem. Not only do these numbers satisfy the Pythagorean Theorem,
but any multiples of these numbers also satisfy the Pythagorean Theorem.
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For example:
the numbers 3, 4, and 5 satisfy the Pythagorean Theorem. If you multiply all three
numbers by 2 (you will get 6, 8, and 10), these new numbers ALSO satisfy the Pythagorean theorem.
The special sets of numbers that possess this property are called
Pythagorean Triples.
The most common Pythagorean Triples are:
3, 4, 5
5, 12, 13
8, 15, 17
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Answer the following questions dealing with the
Pythagorean Theorem.
1.
Choose:
Princess Fiona is locked in the tower of Castle Kronen.
You have volunteered to rescue the princess. If the tower
window is 36 feet above the ground and you must place
your ladder 10 feet from the base of the castle (because
of the moat), which choice is the shortest length ladder
you will need to reach the tower window?
34feet
35 feet
37 feet
38 feet
Choose:
2.
A baseball diamond is a square
with sides of 90 feet. What is
the shortest distance, to the
nearest tenth of a foot, between
home plate and second base?
90.0
127.3
180.0
180.7
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Choose:
3.
The sides of a triangle measure 2.4, 3.2,
and 4. Is this triangle a right triangle?
YES
NO
Choose:
4.
waddle
swim
Daisy Duck has a nest on the edge of the pond. From her
favorite feeding spot, she can either waddle on land around
the pond to the nest (80 meters by 60 meters), or she can
swim across the pond to the nest. Daisy waddles more
quickly than she swims. She waddles at the rate of 30
m/min and she swims at the rate of 20 m/min. Which route
is quicker to travel from the feeding spot to the nest?
Waddling on land or swimming in the pond?
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Choose:
5.
Using the Pythagorean
Theorem, find the area of an
equilateral triangle whose side
measures 5 units. Find the area
to the nearest tenth of a square
unit.
4.3
6.5
10.8
12.5
Choose:
6.
5 ft
10 ft
25 ft
35 ft
Joe Bean regularly takes a short-cut across Mr. Wilson's
lawn instead of walking on the sidewalk on his way home
from school. How much distance is saved by Joe cutting
across the lawn?
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Choose:
7.
Find x.
Choose:
8.
3
4
Find the value of x.
5
6
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9.
A spider has taken up
Choose:
residence in a small cardboard
box which measures 2 inches
by 4 inches by 4 inches. What
is the length, in inches, of a
straight spider web that will
carry the spider from the lower
right front corner of the box to
the upper left back corner of
the box?
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Theorems Related with Area
Figures on the Same Base and Between the Same Parallels
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1. If BC=QR=10cm AB=4cm find AREA of PQR
2. The area of ∆ABC is given to be 18 cm2. If the altitude DL equals 4.5 cm, find the base of the ∆BCD.
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3. If perimeter of triangle BCD=12cm ,BD=CD=3cm and altitude of triangle ABC =4cm find area of triangle ABC
4. ABCD is parallelogram. The area of ∆ABC is 14cm2, AD=7cm
Find DL?
5. If the area of ∆ABD is 20cm2 and DE= 4cm find the perimeter of ∆PQR? If ∆PQR is equilateral triangle.
6. The area of triangle BCD=44cm,BC =11cm
a. Find altitude of triangle
b. Find area of parallelogram BCED
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Theorem: Parallelograms on the same base and between the same parallelsare equal
in area.
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9.
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Chords of a Circle
 one and only one circle can pass through three non-collinear points
 a straight line drawn from the center of a circle to bisect a chord which isnot a
diameter is perpendicular to the chord
 perpendicular from the centre of a circle on a chord bisects it
 if two chords of a circle are congruent then they will be equidistant fromthe
centre
 two chords of a circle which are equidistant from the centre are congruent
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28.1.2 draw a circle passing through three given non – collinear points;
Circle through 3 Points
Printable instructions worksheet.
After doing this
Your work should look like this
We start with three given points. We will construct a circle that
passes through all three.
1. (Optional*) Draw straight lines to create the line segments
AB and BC. Any two pairs of the points will work.
* We draw the two lines to make it clear when we later draw
their perpendicular bisectors, but it is not strictly necessary for
them to actually be there to do this.
2. Find the perpendicular bisector of one of the lines. See
Constructing the Perpendicular Bisector of a Line Segment.
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After doing this
Your work should look like this
3. Repeat for the other line.
4. The point where these two perpendiculars intersect is the
center of the circle we desire.
5. Place the compass point on the intersection of the
perpendiculars and set the compass width to one of the points
A,B or C. Draw a circle that will pass through all three.
6. Done. The circle drawn is the only circle that will pass
through all three points.
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Activity 2
•A straight line drawn from the centre of a circle to bisect a chord which is
not a diameter is perpendicular to the chord
converse
•perpendicular from the centre of a circle on a chord bisects it
•
Step1. Drown of circle of any radius.
Step 2. Draw a chord of any measurement on this circle .
Step 3. Find the midpoint of chord
Step 4. Join a line segment from centre to mid point.
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Activity 3
If two chords of a circle are congruent then they will be equidistant from
The centre
Converse
•Two chords of a circle which are equidistant from the centre are congruent
Step1. Drown of circle of any radius.
Step 2. Draw two chord of same
Step 3.Join
measurement on this circle .
line segment OA and OB from centre to these two chords .
B
OA=OB
0
A
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Worksheet 7.2: Chord Properties Math 11
1. For each item, find the required lengths, to the nearest tenth.
a. Determine AO and AB.
b. Determine EF and CD.
c. Determine OA and CD.
d. Determine OC and AB.
e. Determine OC and AB.
f. Determine AB and FD.
g. Determine OA and CD.
h. Determine OE.
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2.
A chord that is 10 cm long is 12 cm from the centre of a circle. Find the length of the
radius.
3. The diameter of a circle is 20 cm. A chord is 8 cm from the centre. What is the length of
the chord?
4. A chord is 12 cm long and the diameter of the circle is 16 cm. What is the distance
between the chord and the centre of the circle?
1.
Given: Circle O
CD = 16
AB = 16
OB = 10
marked perpendiculars
Choose:
4.5
6
7.5
Find OF.
2.
10
Choose:
Given: Circle O,
diameter, marked
perpendicular
Find x.
2.5
5
6.5
10
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3.
Choose:
Given: Circle O,
marked
perpendiculars, hash
marks indicating
congruent
Find x.
4
8
10
16
Choose:
4.
Given: Circle O,
AB = CD, marked
perpendiculars
Find x.
6
9
15
16
Choose:
5.
Given: Circle O,
marked perpendicular
26
Find AB.
40
32
64
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Tangent to a Circle
 if a line is drawn perpendicular to a radial segment of a circle at its outer end
point it is tangent to the circle at that point.
 the tangent to a circle and the radial segment joining the point of contact and the
Centre are perpendicular to each other.
 the two tangents drawn to a circle from a point outside it are equal in length.
 if two circles touch externally or internally the distance between their centers is
respectively equal to the sum or difference of their radii
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A tangent to a circle is a line in the plane of the circle that intersects the circle in exactly one
point.
If you spin an object in a circular orbit and
release it, it will travel on a path that is tangent
to the circular orbit.
Theorem:
If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of
tangency.
Tangent segments to a circle from the same external point are congruent.
Theorem:
(You may think of this as the "Hat" Theorem because the diagram looks like a circle wearing
a pointed hat.)
This theorem can be proven using congruent triangles and the previous
theorem. The triangles shown below are congruent by the Hypotenuse Leg
Postulate for Right Triangles. The radii (legs) are congruent and the
hypotenuse is shared by both triangles. By using Corresponding Parts of
Congruent Triangles are Congruent, this theorem is proven true.
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Common Tangents:
Common tangents are lines or segments that are tangent to more than
one circle at the same time.
4 Common Tangents
3 Common Tangents
2 Common Tangents
(2 completely separate circles)
(2 externally tangent circles)
(2 overlapping circles)
2 external tangents (blue)
2 internal tangents (black)
2 external tangents (blue)
1 internal tangent (black)
2 external tangents (blue)
0 internal tangents
1 Common Tangent
(2 internally tangent circles)
0 Common Tangents
(2 concentric circles)
Concentric circles are circles
with the same center.
1 external tangent (blue)
0 internal tangents
0 external tangents
0 internal tangents
(one circle floating inside the
other, without touching)
0 external tangents
0 internal tangents
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Step-by-step Instructions
After doing this
Your work should look like this
We start with a point P somewhere on a given circle, with center
point O.
If the center is not given, you can use: "Finding the center of a circle
with compass and straightedge or ruler",
or
"Finding the center of a circle with any right-angled object".
1. Draw a straight line from the center O, through the given point P
and on beyond P.
.
2. Put the compass point on P and set it to any width less than the
distance OP. Then, on the line just drawn, draw an arc on each side
of P. This creates the points Q and R as shown.
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After doing this
Your work should look like this
3. Set the compass on Q and set it to any width greater than the
distance QP.
4. Without changing the compass width, draw an arc approximately
in the position shown on one side of P.
5. Without changing the compass width, move the compass to R and
make another arc across the first, creating point S.
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After doing this
Your work should look like this
6. Draw a line through P and S.
7. Done. The line PS just drawn is the tangent to the circle O
through point P.
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After doing this
Your work should look like this
We start with a given circle with center O, and a point P outside
the circle.
1. Draw a straight line between the center O of the given circle and the
given point P.
2. Find the midpoint of this line by constructing the line's
perpendicular bisector.
The midpoint may be inside or outside the circle, depending on
the circle size and the location of the given point.
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After doing this
Your work should look like this
3. Place the compass on the midpoint just constructed, and set it's
width to the center O of the circle.
4. Without changing the width, draw an arc across the circle in the two
possible places. These are the contact points J, K for the tangents.
5. Draw the two tangent lines from P through J and K.
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After doing this
Your work should look like this
6. Done. The two lines just drawn are tangential to the given circle and
pass through P.
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`
𝐼𝑛 𝑓𝑖𝑔𝑢𝑟𝑒 1
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Work sheet
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Work sheet
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Answer the following questions dealing with the
tangents and circles.
(Do not assume that diagrams are drawn to scale.)
1.
Choose:
7.5
12
15
30
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2.
Choose:
21
16
8
4
3.
Choose:
15
14
12
9
4.
Choose:
12
8
4
2
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5.
Choose:
5
6
7
9
6.
Choose:
5
10
12
15
Choose:
7.
YES
NO
(OB = 15)
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8.
a = OA; b = OB; c = CB
Find a, b, c.
Choose:
a=
Choose:
b=
Choose:
9
16
12
12
18
14
14
20
16
16
24
20
9.
c=
Choose:
YES
NO
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Chords and Arcs
 If two arcs of a circle (or of congruent circles) are congruent then the
corresponding chords are equal.
 If two chords of circle (or of congruent circles) are equal, then their
corresponding arcs (minor, major or semi – circular) are congruent.
 Equal chords of a circle (or of congruent circles) subtend equal angles atthe
centre (at the corresponding centers).
 If the angles subtended by two chords of a circle (or congruent circles) atthe
centre (corresponding centers) are equal, the chords are equal.
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Arcs in Circles
An arc is part of a circle's circumference.
Definition:
In a circle, the degree measure of an arc is equal to the
measure of the central angle that intercepts the arc.
In a circle, the length of an arc is a portion of the circumference.
Definition:
Remembering that the arc measure is the measure of the central angle, a
definition can be formed as:
Example:
In circle O, the radius is 8, and the measure of minor arc
Find the length of minor arc
is 110 degrees.
to the nearest integer.
Solution:
= 15.35889742 = 15
Understanding how an arc is measured makes the next theorems common sense.
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Theorem:
In the same circle, or congruent circles,
congruent central angles have congruent arcs.
Theorem:
In the same circle, or congruent circles,
congruent arcs have congruent central angles.
(converse)
Remember: In the same circle, or congruent circles, congruent arcs have congruent chords. Knowing
this theorem makes the next theorems seem straight forward.
Theorem:
In the same circle, or congruent circles,
congruent central angles have congruent chords.
Theorem:
In the same circle, or congruent circles,
congruent chords have congruent central angles.
(converse)
Solve the following problems dealing with angles in circles.
(Do not assume diagrams are drawn to scale.)
Choose:
1.
Given the labeled diagram at the
left, with diameter
.
Find x.
59
44
43
34
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Choose:
2.
Given circle with center
indicated.
Find x.
55
70
110
290
Choose:
3.
Given circle with center
indicated.
36
Find x.
90
54
108
Choose:
4.
Given diameter.
Find x.
28
56
62
124
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1.
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Angle in a Segment of a Circle
. Circle Theorems Student Sheet: Angle in centre is twice the angle at circumference
1
2
C
B
A
38
O
Angle BAC is 52º. Find angles BOC and BCO
Angle AOB is 130º
Find angles ACB and OBC
3
4
c
O
71
b
a
OB=BC. Calculate the size of angle CAB.
Find the angles a, b and c.
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5
6
Q
P
O
56
R
Find the angle ADC.
S
Find angle QSR.
Circle Theorems Student Sheet: Angles in the same segment are equal
1
2
A
C
B
44
A
33
O
45
B
52
D
C
D
Find The Angles CBD and ADB.
Find the angles ACD, AOD and ADC.
3
4
N
L
56
C
132
b
O
42
P
M
a
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Find the angles MNO and LMN.
5
Calculate the angles a, b and c.
6
Calculate the angles BCD and CAD.
Calculate the angles PQR and QRS.
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Circle Theorems Student Sheet: Angle in a semi-circle is 90°
1
2
Z
X
Y
46
O
Find the value of y
Find angles XOZ, OZY, OZX and show that
XZY is 90º
3
4
A
M
O
C
Find angles QPT and RTP in terms of
x
26
B
M is the midpoint of AC. Find the angle MOC
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5
6
D
23
A
C
B
O
Find the angles e and d
Calculate the size of the angle ADC.
Circle Theorems Students Sheet: Cyclic Quadrilaterals
1
2
D
X
61
100

Y
E
109
81
38
81
Z
V
W
C
80

Calculate the sizes of angles XWV and
ZXW.
100

B
A
Calculate the sizes of angles CBE and
EBA.
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3
4
Q
A
R
108
44
D
P
44
108
72
T
80
S
56
B
C
Calculate the sizes of angles TPR and
TQR.
5
Calculate the sizes of angles ACB and
CAB.
6
Z
Z
88
63
20
Y
A
64
28
116
20
X
28
234
44
O
126
24
Y
W
117
V
W
Calculate the sizes of angles ZVY, XZY
and YWV.
X
Calculate the sizes of angles WOY, WXY
and OYX.
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