Transcript Slide 1

1. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is an isosceles
right triangle with right angle B. The points plotted are data from the table (AB, AC).
c.
The slope of the line you graphed in question b is not precise (it has been
rounded off by Sketchpad). What is the exact value of the slope?
2
d.
Using your prior knowledge of right triangles, explain how you could have found
the slope without graphing any points.
AB = 3.64 cm
AC = 5.14 cm
AB
2.83 cm
5.71 cm
4.00 cm
9.25 cm
7.46 cm
3.64 cm
AC
4.00 cm
8.08 cm
5.66 cm
13.08 cm
10.54 cm
5.14 cm
14
12
10
8
A
6
4
B
2
C
-5
5
10
15
2. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is a right
triangle and CP is an altitude on hypotenuse A B.
When
the
altitude
is drawn
to the hypotenuse
aequals
rightand the product (AP)(PB).
Cross
multiplication
– The between
product
of the
means
b. Conjecture
a relationship
the
length
ofofaltitude
triangle,
the altitude
is the geometric mean between the
the product
of the extremes.
c. Prove your conjecture using similar triangles and proportions. PC2 = (AP)(PB)
segments of the hypotenuse.
extremes
AP PC
means
AP = 3.26 cm
PB = 11.05 cm
CP = 6.00 cm
PC
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm 2
C
2.92 cm 8.03 cm 23.42 cm2 4.84 cm
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B

PB
2. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is a right
triangle and CP is an altitude on hypotenuse A B.
When
the
altitude
is drawn
to the hypotenuse
aequals
rightand the product (AP)(PB).
Cross
multiplication
– The between
product
of the
means
b. Conjecture
a relationship
the
length
ofofaltitude
triangle,
the altitude
is the geometric mean between the
the product
of the extremes.
c. Prove your conjecture using similar triangles and proportions. PC2 = (AP)(PB)
segments of the hypotenuse.
AP PC

PC PB
AP = 3.26 cm
PB = 11.05 cm
CP = 6.00 cm
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm 2
C
2.92 cm 8.03 cm 23.42 cm2 4.84 cm
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B
AP PC

PC PB
When the altitude is drawn to the hypotenuse of a right triangle, the altitude
is the geometric mean between the segments of the hypotenuse
APC  BPC (right angles)
Because ACB is a right angle, ACP is complementary to BCP
PAC is complementary to ACP (they are the acute angles of a right triangle)
PAC  BCP because they are complementary to the same angle.
AP = 3.26 cm
APC~ CPB
PB = 11.05 cm
CP = 6.00 cm
APPB
AP= 35.99
PCcm 2
PC

PB
A
(AA~)
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
8.03 cm 23.422 cm2 4.84 cm
 PC = (AP)(PB)
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.92 cm
Definition of similar triangles
C
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
P
B
3. Infinitely many rectangles with different dimensions have an area of 36 square units
(e.g. 3x12, 4x9, 6x6, 8x4½, 10x3.6, 10x1.6, 15x2.4 to name a few). Use Geometer’s
Sketchpad to construct a rectangle whose area is 36, and which retains that area when
the dimensions are changed by dragging its vertices. (Hint – question 2 above can help
you in this construction.)
When the altitude is drawn to the hypotenuse of a right triangle, the
altitude is the geometric mean between the segments of the hypotenuse.
a h

h b
h
a
b
or
h2  ab
4. In the diagram, ABCD is a rectangle and CE is perpendicular to BD .
Prove:
AB BD

DE CD
C
B
E
A
D
AP PC

PC PB
When the altitude is drawn to the hypotenuse of a right triangle, the altitude
is the geometric mean between the segments of the hypotenuse
From APC ~ CPB
AB AC From APC ~ ACB

AC AP When the altitude is drawn to the hypotenuse of a right triangle, then each
leg is the geometric mean between the hypotenuse and the segment of the
hypotenuse adjacent to that leg
AB BC

AP = 3.26 cm
BC BP
From BPC ~ BCA
PB = 11.05 cm
CP = 6.00 cm
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm
We can
also prove that each smaller2.92
triangle
is similar
to cm
ABC.
cm 8.03 cm
23.42 cm2 4.84
2
C
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B
AP PC

PC PB
When the altitude is drawn to the hypotenuse of a right triangle, the altitude
is the geometric mean between the segments of the hypotenuse
From APC ~ CPB
AB AC From APC ~ ACB

AC AP When the altitude is drawn to the hypotenuse of a right triangle, then each
leg is the geometric mean between the hypotenuse and the segment of the
hypotenuse adjacent to that leg
AB BC

AP = 3.26 cm
BC BP
From BPC ~ BCA
PB = 11.05 cm
CP = 6.00 cm
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm
We can
also prove that each smaller2.92
triangle
is similar
to cm
ABC.
cm 8.03 cm
23.42 cm2 4.84
2
C
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B
AP PC

PC PB
When the altitude is drawn to the hypotenuse of a right triangle, the altitude
is the geometric mean between the segments of the hypotenuse
From APC ~ CPB
AB AC From APC ~ ACB

AC AP When the altitude is drawn to the hypotenuse of a right triangle, then each
leg is the geometric mean between the hypotenuse and the segment of the
hypotenuse adjacent to that leg
AB BC

AP = 3.26 cm
BC BP
From BPC ~ BCA
PB = 11.05 cm
CP = 6.00 cm
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm
We can
also prove that each smaller2.92
triangle
is similar
to cm
ABC.
cm 8.03 cm
23.42 cm2 4.84
2
C
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B
AP PC

PC PB
When the altitude is drawn to the hypotenuse of a right triangle, the altitude
is the geometric mean between the segments of the hypotenuse
From APC ~ CPB
AB AC From APC ~ ACB

AC AP When the altitude is drawn to the hypotenuse of a right triangle, then each
leg is the geometric mean between the hypotenuse and the segment of the
hypotenuse adjacent to that leg
AB BC

AP = 3.26 cm
BC BP
From BPC ~ BCA
PB = 11.05 cm
CP = 6.00 cm
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm
We can
also prove that each smaller2.92
triangle
is similar
to cm
ABC.
cm 8.03 cm
23.42 cm2 4.84
2
C
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B
AP PC

PC PB
When the altitude is drawn to the hypotenuse of a right triangle, the altitude
is the geometric mean between the segments of the hypotenuse
From APC ~ CPB
AB AC From APC ~ ACB

AC AP When the altitude is drawn to the hypotenuse of a right triangle, then each
leg is the geometric mean between the hypotenuse and the segment of the
hypotenuse adjacent to that leg
AB BC

AP = 3.26 cm
BC BP
From BPC ~ BCA
PB = 11.05 cm
CP = 6.00 cm
AP
PB
APPB
CP
2.85 cm 7.85 cm 22.38 cm2 4.73 cm
3.55 cm 9.78 cm 34.72 cm2 5.89 cm
APPB = 35.99 cm
We can
also prove that each smaller2.92
triangle
is similar
to cm
ABC.
cm 8.03 cm
23.42 cm2 4.84
2
C
4.25 cm 11.70 cm 49.72 cm2 7.05 cm
2.42 cm 6.62 cm 16.00 cm2 4.00 cm
2.87 cm 8.72 cm 25.00 cm2 5.00 cm
3.26 cm 11.05 cm 35.99 cm2 6.00 cm
A
P
B
The Pythagorean Theorem:
In a right triangle, the square of theonlength
the of the hypotenuse
is equal to the sum of the squares of theonlengths
the of the legs.
C
b
A
a
c
a b  c
2
B
2
2
Proof of the Pythagorean Theorem
C
b
a
A
P
Prove: a  b  c
2
2
2
c
B
Proof of the Pythagorean Theorem
AB AC

AC AP
When the altitude is drawn to the hypotenuse of a right triangle, then each
leg is the geometric mean between the hypotenuse and the segment of the
hypotenuse adjacent to that leg
AB BC

BC BP
C
b
a
A
P
c
c
Prove: a  b  c
2
2
2
B
Proof of the Pythagorean Theorem
AB AC

AC AP
AC
AB
=
AP
AC
AB BC

BC BP
and
AB = BC
BC PB
AC2 = (AB)(AP) and BC2 = (AB)(PB)
AC2 + BC2 = (AB)(AP) + (AB)(PB)
C
AC2 + BC2 = (AB)(AP+ PB)
b
a
AC2 + BC2 = (AB)(AB)
AC2 + BC2 = AB2
b a c
2
A
P
c
c
Prove: a  b  c
2
2
2
B
2
2
The Converse of the Pythagorean Theorem is also true:
If the side lengths of a triangle are a, b, and c, and
a 2  b 2  c 2, then the triangle is a right triangle.
….
C
b
A
a
c
B
,
Pythagorean Theorem applications
In questions 1 and 2, find the length of the side marked x to the nearest tenth.
C
A
1.
7
B
25
2.
3. What is the length of the diagonal
of the rectangle shown?
B
E 5 in
16 in
15 cm
C
B
D
17 cm
8 cm
B
13 in
4.
8
A
C
24
C
A
x
x
6.2
5
A
BE is an altitude of ABC. Find the
perimeter and area of ABC.
P = 54 inches
A = 126 sq inches
60
2a
a
30
a 3
45
a 2
a
a
45
5. One of the angles of a rhombus measures 45, and its sides are 20 cm long.
What is the area of the rhombus? Answer to the nearest tenth of a square cm.
282.8 sq cm
C
60
B
6. What is the area of quadrilateral ABCD?
Answer to the nearest tenth of a square inch.
10"
136.6 sq "
A
10"
D
7. The hypotenuse of a right triangle is 1 inch longer than its longer leg.
If the shorter leg is 9 inches long, what is the length of the longer leg?
40 inches