Ch 9 Note sheet L1 Key.doc
Name ___________________________
Do the Algebra Review for Chapter 9: Radicals
In General
ON ALL PROBLEMS!!
1. State the relationship (the formula)
Order of Operations
Parenthesis
2. Substitute in known values
Exponents and Roots
3. Solve the equation. Undo what is being
done to the variable by using the inverse
operations in the reverse order of operations.
Multiplication and Division (or multiply by reciprocal)
Addition and Subtraction (or add the opposite of)
Lesson 9.1 The Theorem of Pythagoras
Pythagorean Theorem
In a right triangle
The hypotenuse is
hypotenuse
leg 1
a
If you have a right triangle,
c
then the sum of the squares of
• the side opposite the right angle
leg 2
b
• the longest side of the right triangle.
the lengths of the legs equals the
The legs
square of the hypotenuse.
• are the other two sides of the right triangle
c2 = a 2 + b2
• form the sides of the right angle
c MUST BE THE HYPOTENUSE!!!
Example A
Example B
How high up on the wall will a 20-foot ladder
touch if the foot of the ladder is placed 5 feet
from the wall?
Find the area of the rectangular rug if the width
is 12 feet and the diagonal measures 20 feet.
c = a +b
2
2
20 = 5 + h
2
2
2
400 = 25 + h
375 = h
c2 = a2 + b2
2
202 = 12 2 + h 2
h
400 = 144 + h 2
20 f t
2
h
256 = h 2
2
h = 375 ≈ 19.4
wall
5 ft
12
h = 256 = 16
A = bh
A = (12 )(16 ) = 192 ft 2
S. Stirling
20
Page 1 of 12
Ch 9 Note sheet L1 Key.doc
Name ___________________________
Additional Examples: Approximate your final answers only!!
8
Example 1:
Example 2:
Find p.
Find the perimeter of a rhombus
with diagonals measuring 16 m
and 12 m.
For the left-hand right triangle:
c2 = a2 + b2
142 = 82 + b 2
196 = 64 + b 2
Diagonals are perpendicular bisectors of each other so
the parts of the diagonals measure 8 and 6 and you have
4 congruent right triangles.
c2 = a2 + b2
132 = b 2
b = 132 ≈ 11.49
c 2 = 82 + 6 2
c 2 = 64 + 36
For the right-hand right triangle:
c 2 = 100
c2 = a2 + b2
c = 100 = 10
p 2 = 212 + 132
2
p 2 = 441 + 132 = 573
p = 573 ≈ 23.94
Example 3:
Since the side of the rhombus is 10,
The perimeter is 4 * 10 or 40 m.
Example 4:
Base area = 16π cm2 and slant height = 3 cm.
What’s wrong with this information?
Find the area.
c2 = a2 + b2
7 2 = 32 + h 2
(label diagram)
49 = 9 + h 2
π r 2 = 16π
40 = h
2
h = 40 = 2 10 ≈ 6.32
(
)
1
A = ( 6 ) 2 10
2
= 6 10
A = 6 10 ≈ 18.97 ft2
S. Stirling
6
3
r = 16 = 4
4
In the right triangle, the hypotenuse
is 3, but the leg is 4.
The hypotenuse must be the longest side!
Page 2 of 12
Ch 9 Note sheet L1 Key.doc
Name ___________________________
9.2 The Converse of the Pythagorean Theorem
Pythagorean Triple: Three positive integers that work in the Pythagorean Theorem.
Pythagorean Triples
Converse of the Pythagorean Theorem
st
“Primitives” are the 1 row, then below are multiples of the primitives.
3:4:5
6 : 8 : 10
9 : 12 : 15
12 : 16 : 20
5 : 12 : 13
10 : 24 : 26
15 : 36 : 39
8 : 15 : 17
16 : 30 : 34
7 : 24 : 25
14 : 48 : 50
Note: The multiples listed are the ones which are seen most often.
If the lengths of the 3 sides of a triangle
satisfies the Pythagorean equation, T
then the triangle is a right triangle.
2
R
c
If c = a + b ,
2
a
2
b
then ∆TRI is a right triangle.
I
Hint: See if the given sides of the triangle are multiples of one of
the primitives. If so, use the multiples of the primitives.
If not, just use Pythagorean Theorem.
Example 1:
Example 2:
Determine whether a triangle with side lengths 80, 82,
18 is a right triangle.
Find x. Explain your answers. Then find
AB to two decimal places.
First check to see if it makes a triangle. Sum of the two
shortest sides must be longer than the third side.
18 + 80 > 82
They can all be divided by 2, so... reorder small to large and
reduce.
18 : 80 : 82 → 9 : 40 : 41
Not a triple we know so, may not be a right triangle.
Do not assume that m∠ADC = 90 nor that ∆CDB
is a right triangle!
∆ADC is right, 7:24:25 ratio.
Since m∠ADC = 90 then m∠CDB = 90
because linear pair supp.
Check with the converse of the Pythagorean Theorem.
c2 = a2 + b2
Pythagorean Theorem
By Pythagorean Theorem
822 = 802 + 182
Substitute (longest side = c!)
45 2 = 7 2 + x 2
6724 = 6400 + 324 Simplify
6724 = 6724
Simplify
2025 = 49 + x 2
1976 = x 2
x = 1976 ≈ 44.45
AB ≈ 44.45 + 24 ≈ 68.45
So it is a right triangle and it is another primitive!
S. Stirling
Page 3 of 12
Ch 9 Note sheet L1 Key.doc
A
Name ___________________________
B
17.06
12
8
8
D 8.94
Example 5:
10
32
17.06
6
8
C
Determine whether Quadrilateral ABCD is a
rectangle and justify your answer. If not enough
information is given, write “cannot be determined.”
Firstly, recall your properties
Example 4:
Find the area of trapezoid ABCD .
Need to find the other base of the trapezoid, so draw
heights from A and B, forming right triangles.
of rectangles.
Need a parallelogram with one right angle…or
a parallelogram with equal diagonals.
Let x = base AB.
For the left-hand right triangle, c = 12, a = 8:
c = a +b
2
2
1.
2
122 = 82 + b 2
AB = 3, BC = 4,
and AC = 6.
3
4
∆ABC is not right.
144 = 64 + b 2
3:4:5 makes a right ∆. So NOT a rectangle.
80 = b 2
b = 80 ≈ 8.94
2.
For the right-hand right triangle, c = 10, a = 8:
3:4:5
Pythagorean Triple
6 : 8 : 10
Multiply by 2.
So b = 6.
Now the bottom base measures 32 and subtract off the two
sides of the right triangles (to find the top base):
AB = 3, BC = 4,
DA = 4, and AC = 5. 3
4
4
∆ABC is right, 3:4:5.
Know BC = AD, but need both pairs of opposite
sides equal. ABCD is a may not be a rectangle.
“cannot be determined.”
x = 32 – 6 – 8.94 = 17.06
So using formula for area of a trapezoid:
A=
A=
1
h(b1 + b2 )
2
1
2
(8)(32 + 17.06) = 196.24 cm
2
3. AB = 3, BC = 4, CD = 3,
3
DA = 4, and AC = BD.
4
3
4
Since opposite sides are equal, ABCD is a
parallelogram.
And AC = BD so diagonals are equal and ABCD is
a rectangle.
S. Stirling
Page 4 of 12
Ch 9 Note sheet L1 Key.doc
9.3 Two Special Right Triangles
Name ___________________________
Isosceles Right Triangle Conjecture
(or 45°-45°-90° Triangle Conjecture)
30°-60°-90° Triangle Conjecture.
Also found in sides and diagonals of rectangles and
in equilateral triangles.
Also found in sides and diagonals of a square.
Hypotenuse = short leg • 2
Hypotenuse = leg • 2
45
h
l
Or h = l 2
Long leg = short leg • 3
30
h
45
l
Or h = 2 s and l = s 3
l
x
60
x
or as a ratio x : x : x 2
s
2x
x
x 2
or as a ratio x : x 3 : 2 x
Example 1:
x 3
Example 2: Can you use Pythagorean Theorem? NO
45
hyp = leg 2
a = 14 2
long = sh 3
hyp
45 leg
12 3 = a 3
long 30
hyp
a = 12
Compare to using Pythagorean Theorem.
a 2 = 142 + 142
hyp = 2i sh
a 2 = 196 + 196
b = 2i12 = 24
sh
a = 392
2
a = 392 = 4i2i49
= 2i7 2 = 14 2
Example 4:
Or as a ratio multiply 1:1: 2 by 14
hyp = 2i sh
14 :14 :14 2
16 = 2i h
h=8
Example 3:
hyp
long = sh 3
b=6 3
60
30
sh
hyp
long = sh 3
sh
b=8 3
long
hyp = 2i sh
A = bh
a = 2i6 = 12
A = 8 3 • 8 = 64 3
S. Stirling
long
Page 5 of 12
Ch 9 Note sheet L1 Key.doc
Name ___________________________
Example 6:
Example 5:
Find the area of an equilateral
Find the perimeter and area of KLMN. Round
your final answer to the nearest hundredth.
E
triangle EQU with side length 8 m.
30 30
Since the angles of the triangle
h
l
are 60° , draw altitude EA to
form two congruent 30-60-90
U
triangles:
8.49
60
60
s
A
Q
6
h = 2s
30-60-90 Triangle Conjecture
6
6
7
10.39
Draw altitudes from N and M to form two right triangles:
8 = 2s
For the right-hand right triangle, hyp = 12:
4=s
h = 2s
30-60-90 Triangle Conjecture
12 = 2s
l=s 3
30-60-90 Triangle Conjecture
l=4 3
6=s
So, height of trapezoid = 6.
l=s 3
30-60-90 Triangle Conjecture
l=6 3
So, part of base = 6 3 .
Use the formula for area of a triangle.
A=
1
1
bh and A = (8)(4 3) = 16 3 m2
2
2
For the left-hand right triangle, l = 6:
Since the legs of the right isosceles triangle are equal,
another part of the base = 6.
Example 7:
Find the area of a right isosceles triangle with a
hypotenuse of 5 6 .
So now the second base of the trapezoid is:
KL = 6 + 7 + 6 3 = 13 + 6 3 ≈ 23.39
h=l 2
45-45-90 Triangle Conjecture
5 6 =l 2
5 6 l 2
so l = 5 3
=
2
2
Use the formula for area of a triangle.
A=
1
bh
2
A=
1
5 3 5 3
2
=
(
Perimeter = 7 + 12 + 6 + 7 + 6 3 + 6 2 ≈ 50.88 or
32 + 6 3 + 6 2
Using formula for area of a trapezoid:
A=
)(
1
(6)(7 + 23.39) ≈ 91.17 square units.
2
)
1
75
• 25 9 =
2
2
S. Stirling
Page 6 of 12
Ch 9 Note sheet L1 Key.doc
Chapter 9 Applications: 9.4 Story Problems
Name ___________________________
Example 1:
Example 2:
What is the longest stick that will fit inside a 15-by16-by-12 foot box?
Find the area of the triangle.
Draw a diagram.
15
25
20
30
Draw the altitude.
Since the triangle is
equilateral it is also
equilangular and each
3√3
60
3
angle measures 60°.
Also the altitude from the vertex angle of an
isosceles triangle bisects the base and the
vertex angle. So 30-60-90 triangles are
formed.
The diagonal of the base of the box.
3: 4: 5 → multiply by 4 → 12: 16: 20
h = 2s
So the diagonal is 20.
30-60-90 Triangle Conjecture
6 = 2s
The right triangle that is formed by the diagonal of the
stick, the base and the height of the box gives…
3=s
l=s 3
30-60-90 Triangle Conjecture
l =3 3
s 2 = 20 2 + 152
s 2 = 625
s = 625 = 25
Use the formula for area of a triangle.
A=
OR
3: 4: 5 → multiply by 5 → 15: 20: 25
(
)
1
1
bh and A = ( 6 ) 3 3 = 9 3 in2.
2
2
Example 3:
Find the area of an isosceles triangle with a base of length 14 cm and congruent sides of length 25 cm.
Draw a diagram.
Then draw an altitude.
25 cm
13
25
13 cm
cm
24
The altitude from the
vertex angle of an
isosceles triangle bisects
the base and the vertex angle.
7
14
10 cm
cm
Forms a right triangle with hypotenuse 25
7: 24: 25 is a primitive so h = 24.
Use the formula for area of a triangle.
A=
1
1
bh and A = (14 )( 24 ) = 168 cm 2
2
2
and leg 7, now find h, the height.
S. Stirling
Page 7 of 12
Ch 9 Note sheet L1 Key.doc
Name ___________________________
Chapter 9 Applications: 9.6 Circles and the Pythagorean Theorem
Example 2:
Example 1:
Find the area of the shaded
region. Write your answers in
8
6
5
5
terms of π and rounded to
Find the area of the shaded
12√3
region. Write your answers in
12
30
terms of π and rounded to
12
12
12
2
2
the nearest 0.1 cm .
the nearest 0.1 cm .
Given AO = 5 cm and AC = 8 cm .
Given tangent PT , QM = 12 and m∠P = 30° .
The triangle is a right triangle since ∠ACB is
inscribed in a semicircle.
Hypotenuse = 10 (diameter) and leg = 8.
Draw radius TM. The triangle PTM is a right
triangle since a tangent is perpendicular to the
radius. m∠P = 30° so m∠PMT = 60° .
QM = 12 so the other radius TM = 12 , the short
leg of 30-60-90 triangle.
3: 4: 5 → multiply by 2 → 6: 8: 10
So the other leg is 6.
l = s 3 so PT = 12 3
Use the formula for area of a triangle.
Use the formula for area of a triangle.
A=
1
1
bh and A = ( 6 )( 8 ) = 24 cm2.
2
2
(
)
1
1
bh and A = (12 ) 12 3 = 72 3 cm2.
2
2
A=
Use the formula for area of a circle.
Use the formula for area of a sector.
A = π r 2 and A = π ( 5 ) = 25π cm2.
A=
2
degree
60
2
• π r 2 and A =
• π (12 ) = 24π cm2.
360
360
Area of shaded = area of circle minus area of triangle.
A = 25π − 24 ≈ 54.54 cm2.
Area of shaded = area of triangle minus area of sector.
A = 72 3 − 24π ≈ 49.31 cm2.
Example 3:
Find the area of the shaded region.
Use the formula for area of a triangle.
A=
The triangle is a right triangle since
∠LMN is inscribed in a semicircle.
45-45-90 triangle
LN = 4 2 and radius is 2 2 .
1
1
bh and A = ( 4 )( 4 ) = 8
2
2
Use the formula for area of a sector.
A=
(
1
•π 2 2
2
)
2
=
1
• π • 8 = 4π
2
Area of shaded = area of sector minus area of triangle.
A = 4π − 8 cm2.
Note:
(2 2 )
2
= 2 2 • 2 2 = 4 • 4 = 8 !!!
Additional Examples are on pages 508 and 509 of your book.
S. Stirling
Page 8 of 12
Ch 9 Note sheet L1 Key.doc
EXERCISES Page 498 #1, 4, 5, 7, 6.
Name ___________________________
See your book for a full explanation!
#1: Does the 34 in. baseball bat fit in a box that
#5: A-frame front isosceles triangles base = 10 m
and legs 13m. Glass 1 cm thick costs $120/m2 .
Cost?
measures 24-by-20-by-12 inch box?
Draw a diagram.
12
y
Draw a diagram.
12 in.
13 m
cm
13
13 m
cm
12
x
20 in.
5:12:13 right triangles
24 in.
x 2 = 20 2 + 24 2
y 2 = x 2 + 12 2
x 2 = 400 + 576
y 2 = 976 + 144
x 2 = 976
y 2 = 1120 ≈ 33.47
5
1
A = bh
2
1
A = • 10 • 12 = 60 m 2
2
Cost =
So the bat is too long!
#4: 25-ft. ladder against a building w/ bottom 7 ft.
away. If top slides down 4 ft., how far will it slide
out?
10 m
cm
60 m 2 $120
•
= $7200
1
1 m2
#7: Find the perimeter of an equilateral triangle
whose median measures 6 cm.
Draw a diagram.
Draw a diagram.
30
wall
20 fftt
25
Original triangle:
7:24:25 right triangle so h = 24.
Slides down 4 ft. so h now 20
252 = 20 2 + y 2
625 = 400 + y 2
y 2 = 225
y = 225 = 15
x
25
3:4:5 → 15:20:25
So y = 15
So the ladder slid out 15 – 7 = 8 feet out.
S. Stirling
x
•
2
2 6 x
• = •
3 1 2
12
=x
3
6=
20
x
6
long = sh 3
5
7 fftt
y
or
x
h
60
x/2
3
1
3 2
•
1
3
3 12 36
Perim = •
=
≈ 20.785
1
3
3
Page 9 of 12
x/2
Ch 9 Note sheet L1 Key.doc
Name ___________________________
#7: Regular hexagonal prism in side cylindrical box with diameter 6 cm and
6
height 10 cm. Find surface areas.
long = sh 3
6
p = 2π ( 3 ) = 6π
3
3 3 3
a= •
=
2 1
2
B = π ( 3 ) = 9π
2
10
Cylinder:
SA = 2 B + ph
SA = 2 • 9π + 6π • 10
SA = 18π + 60π
10
3
p = 6 • 3 = 18
1
B = ap
2
1 3 3 18
= •
•
2
2
1
27 3
=
2
3
60
30
3
a
60
3/2
3/2
Prism:
SA = 2 B + ph
= 78π ≈ 245 cm 2
2 27 3
•
+ 18 • 10
1
2
SA = 27 3 + 180
SA =
≈ 226.8 cm 2
EXERCISES Page 509 #1, 2, 3, 5, 7, 4
2) Square SQRE, with SQ = 4 m
1)
6
Diagonals form 45:45:90 ∆s
To find the radius:
6
3: 4: 5 → multiply by 4 → 12: 16: 20
So d = 12 and r = 6
x
hyp = leg 2
4
4= x 2
4
=x
2
A = A○ − Asquare
2
1
2
Asemi ○ = π ( 6 )
2
= 18π cm 2
≈ 56.55 cm 2
S. Stirling
 4 
2
=π 
 −4
 2
 16 
= π   − 16
 2
= 8π − 16 ≈ 9.132 m 2
Page 10 of 12
Ch 9 Note sheet L1 Key.doc
Name ___________________________
EXERCISES Page 509 Cont.
3) OD = 24 cm
4) TA = 12 3
Kite OBYD:
360 – 285 = 75
30:60:90 triangle:
12
60
long = sh 3
75
Circle:
360 – 75 = 285
12 3
12 3 = sh 3
24
AN = 12
Reflex angle
Circle:
360 – 60 = 300 Reflex angle
285
2
• π ( 24 )
360
19 576
π
A=
•
24 1
A = 456π ≈ 1432.57 cm 2
A=
300
2
• π (12 )
360
5 144
π
A= •
6 1
A = 120π ≈ 376.991
A=
5) HT = 8 3
7) HO = 8 3
90
30:60:90
triangle:
P
8 3
8
30
8 3
=4 3
2
long = sh 3
Split into 30:60:90 triangles:
8 3 = sh 3
long = sh 3
RT = 8
4
4 3 = sh 3
4 3
sh = 4
hyp = sh • 2
hyp = 4 • 2 = 8
Ashaded = Asemi ○ − A∆
(
1
1
2
• π (8) − (8 ) 8 3
2
2
= 32π − 32 3
=
≈ 45.105
S. Stirling
)
Asegment = Asec tor − A∆
( )
( )
120
1
2
• π (8 ) − 8 3 ( 4 )
360
2
1
= • 64π − 2 8 3
3
64
= π − 16 3 ≈ 39.31
3
=
Page 11 of 12
60
8
30
4 3
Ch 9 Note sheet L1 Key.doc
EXTRA EXERCISES Chapter 9
Name ___________________________
1. If the leg of a right isosceles triangle measures 6
5. Find the area of an equilateral triangle with side
feet, how long is its hypotenuse?
length 10 m.
7
h2 = l 2 + l 2
h 2 = 62 + 62
10 = 5 + h
2
So 8.485 ft
h = 72
2
2
2
100 = 25 + h 2
h = 72 = 6 2 ≈ 8.485
75 = h 2
h = 75 = 5 3
1
bh
2
1
= (10 ) 5 3
2
A = 25 3 m 2
A=
(
)
2. Find the area of an isosceles triangle with a base
of 16 and legs measuring 17 inches.
17 = 8 + x
2
2
2
289 = 64 + x 2
225 = x 2
x = 15
1
bh
2
1
A = • 16 • 15
2
A = 120 in 2
A=
6. If the hypotenuse of a right isosceles triangle
measures 24 feet, what is its area?
24 = x 2
3. The sides of a rhombus measure 10 in. and the
shorter diagonal measures 12 in, what is the area
of the rhombus? Hint: use right triangles.
10 = 6 + x
2
2
2
100 = 36 + x
2
64 = x 2
x=8
1
2 • A = 2 • bh
2
1
= 2 • (12 )( 8 )
2
A = 96 in 2
4. Find the area of an equilateral triangle with side
24
=x
2
7. If the diagonal of a square measures 10 m, what is
its area?
10 = x 2
10
=x
2
length 16 m.
16 = 8 + h
2
2
2
256 = 64 + h 2
192 = h 2
h = 192 = 8 3
S. Stirling
1
bh
2
1
= (16 ) • 8 3
2
= 64 3
A=
A = 110.848 m
1
bh
2
1  24   24 
= 
2  2   2 
A = 144 m 2
A=
2
A = bh
 10   10 
=


 2  2 
A = 50 ft 2
8. If the diagonal of a rectangle measures 25 m and
the height measures 7 m, what is its perimeter?
252 = 7 2 + x 2
625 = 49 + x 2
576 = x 2
p = 2 ( 24 ) + 2 ( 7 )
= 62 m
x = 576 = 24
Page 12 of 12
Ch 9 Note sheet L1 Key.doc
S. Stirling
Name ___________________________
Page 13 of 12