1. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is an isosceles right...
19
14 12 10 8 6 4 2 -5 5 10 15 A B A C 2.83 cm 4.00 cm 5.71 cm 8.08 cm 4.00 cm 5.66 cm 9.25 cm 13.08 cm 7.46 cm 10.54 cm 3.64 cm 5.14 cm AC = 5.14 cm AB = 3.64 cm A B C 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? d. Using your prior knowledge of right triangles, explain how you could have found the slope without graphing any points. 2
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Transcript of 1. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is an isosceles right...
14
12
10
8
6
4
2
-5 5 10 15
AB AC
2.83 cm 4.00 cm
5.71 cm 8.08 cm
4.00 cm 5.66 cm
9.25 cm 13.08 cm
7.46 cm 10.54 cm
3.64 cm 5.14 cm
AC = 5.14 cm
AB = 3.64 cm
A
B
C
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?
d. Using your prior knowledge of right triangles, explain how you could have found the slope without graphing any points.
2
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
b. Conjecture a relationship between the length of altitude and the product (AP)(PB).
c. Prove your conjecture using similar triangles and proportions.
2. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is a right triangle and is an altitude on hypotenuse . CP AB
PC2 = (AP)(PB)
PB
PC
PC
AP
means
extremes
Cross multiplication – The product of the means equals the product of the extremes.
When the altitude is drawn to the hypotenuse of a right triangle, the altitude is the geometric mean between the segments of the hypotenuse.
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
b. Conjecture a relationship between the length of altitude and the product (AP)(PB).
c. Prove your conjecture using similar triangles and proportions.
2. Shown below is a Geometer’s Sketchpad screen. In the diagram, ABC is a right triangle and is an altitude on hypotenuse . CP AB
PC2 = (AP)(PB)
PB
PC
PC
AP
Cross multiplication – The product of the means equals the product of the extremes.
When the altitude is drawn to the hypotenuse of a right triangle, the altitude is the geometric mean between the segments of the hypotenuse.
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
PB
PC
PC
AP
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~ CPB (AA~)
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)
APC BPC (right angles)
PB
PC
PC
AP Definition of similar triangles
PAC BCP because they are complementary to the same angle.
PC2 = (AP)(PB)
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.
b
h
h
a abh 2
or
a b
h
E
C
AD
B
4. In the diagram, ABCD is a rectangle and is perpendicular to .
Prove:
CE BD
CD
BD
DE
AB
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
PB
PC
PC
AP
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
AP
AC
AC
AB
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
From APC ~ ACB
BP
BC
BC
AB
From BPC ~ BCA
We can also prove that each smaller triangle is similar to ABC.
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
PB
PC
PC
AP
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
AP
AC
AC
AB
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
From APC ~ ACB
BP
BC
BC
AB
From BPC ~ BCA
We can also prove that each smaller triangle is similar to ABC.
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
PB
PC
PC
AP
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
AP
AC
AC
AB
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
From APC ~ ACB
BP
BC
BC
AB
From BPC ~ BCA
We can also prove that each smaller triangle is similar to ABC.
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
PB
PC
PC
AP
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
AP
AC
AC
AB
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
From APC ~ ACB
BP
BC
BC
AB
From BPC ~ BCA
We can also prove that each smaller triangle is similar to ABC.
AP PB APPB 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
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
APPB = 35.99 cm2
CP = 6.00 cm
PB = 11.05 cm
AP = 3.26 cm
PA
C
B
PB
PC
PC
AP
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
AP
AC
AC
AB
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
From APC ~ ACB
BP
BC
BC
AB
From BPC ~ BCA
We can also prove that each smaller triangle is similar to ABC.
The 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.
222 cba
B
C
A
a
c
b
on theon the
B
C
A
a
c
b
P
Proof of the Pythagorean Theorem
Prove: 222 cba
B
C
A
a
c
b
c
P
Proof of the Pythagorean Theorem
Prove: 222 cba
AP
AC
AC
AB
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
BP
BC
BC
AB
PB
BC
BC
ABandAPAC
ACAB ==
AC2 = (AB)(AP) and BC2 = (AB)(PB)
AC2 + BC2 = (AB)(AP) + (AB)(PB)
AC2 + BC2 = (AB)(AP+ PB)
AC2 + BC2 = AB2
222 cab
AC2 + BC2 = (AB)(AB)
Proof of the Pythagorean Theorem
Prove: 222 cba
B
C
A
a
c
b
c
P
AP
AC
AC
AB
BP
BC
BC
AB
B
C
A
a
c
b
The Converse of the Pythagorean Theorem is also true:
If the side lengths of a triangle are a, b, and c, and , …. , then the triangle is a right triangle.
222 cba
Pythagorean Theorem applications
D
B A
C
8 cm
15 cm
3. What is the length of the diagonal of the rectangle shown?
E
B
A
C
13 in
5 in
16 in
BE is an altitude of ABC. Find the perimeter and area of ABC.
4.
In questions 1 and 2, find the length of the side marked x to the nearest tenth.
A
B C
1.
x7
24A
B
C
5
8
2.x
25 6.2
17 cm
P = 54 inches
A = 126 sq inches
30
60
a2a
3a
2a
45
45
a
a
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.
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?
282.8 sq cm
C
D
B
A
60
6. What is the area of quadrilateral ABCD? Answer to the nearest tenth of a square inch.
10"
10"
136.6 sq "
40 inches
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