Download Geometry - Semester 2 Unit 4 Circles

Geometry - Semester 2
Unit 4 Circles
Mrs. Day-Blattner
2/8/2017
Agenda 2/8/2017
1)
2)
3)
4)
5)
Bulletin
Lesson 1. Circles Opening Activity
Thales’ theorem and Exercises
Lesson summary review and vocabulary check
Homework and 3B test update
Getting ready to go:
Pick up a colored paper, measure and mark the
angles made by the corners.
Also collect a white paper with points A and B
marked on it.
Opening Activity - step 1
Take the colored paper provided and “push”
(slide) that paper up between points A and B on
the white sheet, so the edges of the paper go
through the middle of each point. (Watch Mrs.
D-B demonstrate this important step.)
Opening Activity - step 2
When you have the colored paper in position
between points A and B, mark on the white paper
the location of the corner of the colored paper,
using a different color than black. Mark that point
C.
See Mrs. D-B mark a sample on the board.
Opening Activity - step 3
Repeat, pushing the corner of the colored paper
up between A and B, but at a different angle this
time. Again mark the location of the corner, but
this time label it D.
Keep going E, F, G, H, etc. (5 minutes)
What shape are the points starting to trace?
Summary
What shape are the points starting to trace?
The points seem to be starting to trace a
semicircle.
Where might the center of the semicircle be?
Summary
What shape are the points starting to trace?
The points seem to trace a semicircle.
Where might the center of the semicircle be?
The midpoint of the line segment connecting
points A and B on the white paper will be the
center point of the semicircle .
What would the radius of this semicircle be?
Summary
The midpoint of the line segment connecting
points A and B on the white paper will be the
center point of the semicircle .
What would the radius of this semicircle be?
The radius would be half the distance between
points A and B - or the distance from A to the
midpoint of the line segment AB.
Draw in the line segment that connects points A and
B and then construct a perpendicular bisector to find
the midpoint of line segment AB.
When we have found the
midpoint of line segment AB,
put your compass on the
midpoint, stretch it out to point
A or B and draw a circle what do you notice about the
points you had created by
pushing the colored paper
“through” A and B?
We have shown that the marked points created
by the corner of the colored paper do indeed lie
on a semicircle.
We have shown that each marked point is the
same distance from the midpoint of the line
segment connecting the original points A and B.
Exploratory Challenge (12 minutes)
Draw the right triangle formed by the line
segment between the 2 original points A and B
and any one of the colored points you created
using the corner of the colored paper.
Then construct a rotated copy of that triangle
underneath. Like this:
Exploratory Challenge cont.
E
A
B
Draw a rotated copy of the triangle underneath
it.
Exploratory Challenge cont.
E
A
x
y
y
x
B
Label the acute angles x and y and label the corresponding
angles in the rotated triangle the same.
Todd says the shape is a rectangle. Maryam say it is
a quadrilateral, but she is not sure it is a rectangle.
E
A
x
y
y
x
B
Todd is right - how should he explain why to Maryam?
E
Exploratory Challenge
A
y
x
y
B
x
E’
a. What composite figure is formed by the two triangles? How
would you prove it? A rectangle is formed. We need to
show that all four angles measure 90 degrees.
E
Exploratory Challenge
A
y
x
y
B
x
E’
i. What is the sum of x and y? The sum of the acute angles in
any right triangle has to be 90 degrees, so the sum of x and y
must be 90 degrees.
E
Exploratory Challenge
A
x
y
y
B
x
ii. How do we know that the figureE’whose vertices are the
colored points (C, D…) and points A and B is a rectangle? All
four angles measure 90 degrees. The colored points are
constructed as right angles, and the angle at points A and B
measure x + y which is also 90 degrees.
E
Exploratory Challenge
A
y
x
y
B
x
E’
b. Draw the two diagonals of the rectangle. Where is the
midpoint of the segment connecting the two original points A
and B? Why?
E
Exploratory Challenge
A
x
y
y
B
x
b. Draw the two diagonals of the rectangle. Where is the
midpoint of the segment connecting
E’ the two original points A
and B? Why? The midpoint of the segment connecting points
A and B is the intersection of the diagonals of the rectangle
because the diagonals of a rectangle are congruent and
bisect each other.
E
Exploratory Challenge -
A
x
y
P
y
B
x
E’
c. Label the intersection of the diagonals as point P. How
does the distance from point P to a colored point (C, D, E…)
compare to the distance from P to points A and B?
E
Exploratory Challenge
A
x
y
P
y
B
x
E’
c. The distances from P to each of the points (A, B or any
colored point) are all equal since they are all the radius of a
circle with center at point P.
Skip to e. How does your drawing demonstrate that
all the colored points you marked do indeed lie on a
circle?
For any of the colored points we can
construct a rectangle with that colored
point and the two original points, A and
B, as vertices.
The diagonals of this rectangle intersect
at the same point P, because diagonals
intersect at their midpoints, and the
midpoint of the diagonal between points
A and B is P.
The distance from P to that colored point
equals the distance from P to points A
and B. By transitivity, the distance from P
to the first colored point C, equals the
distance from P to any other colored
point.
By definition a circle is the set of all
points in the plane that are the same
distance from a given center point.
Therefore, each colored point on the
drawing lies on the circle with center P
and a radius equal to half the length of
the original line segment joining points A
and B.
We have just proved this theorem:
If triangle ABC is a right triangle with angle at C
the right angle, then A, B, and C are three distinct
points on a circle with diameter AB.
This is the Converse of Thales’ theorem, a
famous theorem in geometry.
We started with a right angle (the corner of the
colored paper) and created a circle.
C
B
A
P
If triangle ABC is a
right triangle with
angle at C the
right angle, then
A, B, and C are
three distinct
points on a circle
with diameter AB.
This is the converse of Thales’ theorem
c 624 - 546 BCE
maybe oldest recorded
geometric theorem with
a proof
Example.
Thales’ theorem states: If A, B, and C are three
distinct points on a circle and line segment AB is
a diameter of the circle then angle ACB is right.
Thales’ theorem - start with circle and then
create a right-angle.
a. Draw circle P with distinct points, A, B, and C on
the circle with diameter AB.
Prove that angle ACB is a right angle.
(Draw on white board) Draw a circle, mark center P.
Draw in a diameter, mark ends of diameter A and B.
Pick a point C on the circumference, and label it.
Join A to C and C to B to draw the triangle.
b. Draw a third radius (PC). What types of triangles
are △APC and △BPC
(Draw on white board)
b. Draw a third radius (PC). What types of triangles
are △APC and △BPC
They are isosceles triangles because
both sides of each triangle are radii of
the circle P and therefore are of equal
length.
c. Using the diagram that you just created, develop a
strategy to prove Thales’ theorem.
What do we know about isosceles
triangles?
c. Using the diagram that you just created, develop a
strategy to prove Thales’ theorem.
Isosceles triangles have 2 sides that are
the same length and 2 base angles that
are congruent. How can we use that
information to prove angle ACB is a right
angle?
d. Label the base angles of triangle APC as b
degrees and the bases of triangle BPC as a
degrees. Express the measure of angle ACB in
terms of a° and b°.
d. Label the base angles of triangle APC as b
degrees and the bases of triangle BPC as a
degrees. Express the measure of angle ACB in
terms of a° and b°.
m∠ACB = m∠b + m∠a
m∠ACB = b° + a°
e. How can the previous conclusion be used to prove
that angle ACB is a right angle?
e. How can the previous conclusion be used to prove
that angle ACB is a right angle?
2a° + 2b° = 180°
Since for any triangle the sum of angle
measures has to be 180°
e. How can the previous conclusion be used to prove
that angle ACB is a right angle?
2a° + 2b° = 180°
Using distributive property of
multiplication we can write
2(a° + b°) = 180°
e. How can the previous conclusion be used to prove
that angle ACB is a right angle?
2a° + 2b° = 180°
2(a° + b°) = 180°
And so dividing both sides by 2 gives:
a° + b° = 90°
Hence angle ACB is indeed 90 degrees.
Exercises 1
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
a) Find m ∠C
Exercises 1
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
a) Find m ∠C: AB is a diameter of a
circle and C a point on the circle, so
By Thales’ theorem
m ∠C= 90°
Exercises.
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
b) Find AB
Diameter AB = 2 x radius
Exercises.
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
b) Find AB
Diameter AB = 12.5 cm + 12.5 cm
= 25 cm
Exercises.
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
c) Find BC
Using Pythagorean theorem
AC2 + BC2 = AB2
Exercises.
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
c) Find BC
Using Pythagorean theorem
AC2 + BC2 = AB2
(7cm)2 + BC2 = (25cm)2
BC2 = (25cm)2 - (7cm)2
Exercises.
1. AB is a diameter of the circle shown. The radius is
12.5 cm and AC = 7cm.
c) Find BC
2
2
BC = √(625cm - 49cm )
BC = √576cm2
BC = 24cm
Exercises.
2. In the circle shown, BC is a diameter with center
A.
a) Find m∠DAB =
180° - 18° - 18° = 144°
Exercises.
2. In the circle shown, BC is a diameter with center
A.
b) Find m∠BAE =
180° - 2(26°) = 128°
Exercises.
2. In the circle shown, BC is a diameter with center
A.
c) Find m∠DAE =
360° - 144° - 128° = 88°
Notice m ∠DBE = 18° + 26° = 44°
Read Lesson Summary outloud to neighbor
You read Theorems - they read Relevant
Vocabulary
Mark a chord on circles in Exercises.
Mark a central angle in Circle in Exercise 2.’
Mark Inscribed angle.
Inscribed angle
An inscribed angle is an angle
formed by two chords in a circle
which have a common endpoint.
This common endpoint forms the
vertex of the inscribed angle.
The other two endpoints define
what we call an intercepted arc on
the circle.
Central angle
A central angle (B) is an
angle whose apex (vertex)
is the center of a circle and
whose legs (sides) are radii
intersecting the circle in
two distinct points.
Homework
Thales’ theorem - single sheet Lesson 1. Page 7
Vocabulary Practice page 8
Need to find definitions (use textbook at home)
for tangent line, minor arc and semicircle and
write neatly on the sheet in some of the space.
Problem Set - correct your own work, use color
B
.O
A
C
1. A, B, and C are three points on a circle, and angle ABC is a
right angle. What’s wrong with the picture? Explain your
reasoning. Student’s said angle ABC couldn’t be a right angle
because AC wasn’t a diameter of the circle. That is correct.
Problem Set - correct your own work, use color to
B
show this proof also
.O
A
C
Draw in the 3 radii from O to A, B and C. - isosceles triangles,
label congruent base angles (a°, b°, and c°).
For △ABC
Problem Set - correct your own work, use color
B
.O
A
C
Draw in the 3 radii from O to A, B and C. - isosceles triangles,
label congruent base angles (a°, b°, and c°).
For △ABC 2a° + 2b° + 2c° = 180°
Problem Set - correct your own work, use color
B
.O
C
A
For △ABC 2a° + 2b° + 2c° = 180°
2(a° + b° + c°) = 180°
a° + b° + c° = 90°
BUT, angle B = 90° = b°+ c°
Problem Set - correct your own work, use color
B
.O
A
C
a° + b° + c° = 90° and 90° = b°+ c° can’t both be true at
the same time - we have a contradiction
(a° can’t be 0°)
2. Show that there is something mathematically
wrong with the picture below. Again students
noted that angle ABC couldn’t be a right angle,
B
C
A
.O
since line segment AC is not a diameter of the
circle. Also, can do a proof similar to answer for 2.
C
Use Pythagorean theorem
AC2 + (30mile)2 = (34miles)2
30
miles
AC = sq.rt [(34mile)2 -(30miles)2]
AC = sq.rt [(1156 - 900) miles2]
.
A
17
miles
B
17
miles
AC = sq.rt256 miles
AC = 16 miles
3. In the figure, AB is the diameter of a circle of
radius 17 miles. If BC = 30 miles, what is AC?
4. In the figure below, O is the center of the circle, and
AD is a diameter.
a) Find measure of angle AOB
isosceles triangle BOD, 24° + 24° + angle BOD = 180°
angle BOD = 132°
Angle AOB and BOD are linear pairs,
so Angle AOB = 48°
4. In the figure below, O is the center of the circle, and
AD is a diameter.
b)If measure angle AOB: measure of angle BOC =
3:4, what is measure of angle BOC?
48°
=3
m∠BOC 4
m∠BOC = 4(48°) / 3 = 64°