Ch 8 Notesheet L1 Key
In General
Name ___________________________
Simplify
ON ALL PROBLEMS!!
Order of Operations
1. State the relationship (or the formula).
Parenthesis
2. Substitute in known values.
Exponents and Roots
3. Simplify or Solve the equation. Use the
order of operations in the correct order.
Multiplication and Division (or
multiply by reciprocal)
Addition and Subtraction (or add
the opposite of)
Solve
Lesson 8.1 Areas of Rectangles and Parallelograms
Read pages 422 – 424 in the book, up to the investigation. Pay close attention to
Example A and how to identify the height. How can you count squares quickly?
Area
The measure of the size of the interior of a figure, expressed in square units.
Write the units as in2, or sq. in, NOT
2
in!
base (of a polygon) Any side of the polygon used for reference to determine an altitude or other features.
altitude Any perpendicular segment from a base to the opposite “part” of the figure.
An altitude for a parallelogram would go from the line containing one parallel base to the line containing the
other parallel base.
In a triangle, an altitude goes from a vertex perpendicular to the line containing the opposite side.
height The length of an altitude. Does NOT need to be a part of the figure!!
Rectangle Area Conjecture
The area of a rectangle is given by
Parallelogram Area Conjecture
The area of a parallelogram is given by
A = bh
where
A is the area
b is the length of the base
h is the height (or width)
A = bh
b
h
b
where
A is the area
b is the length of the base
h is the height
h
Page 424 Investigation: Deriving the Area Formula for Parallelograms
Follow Steps 1 and 2 of the investigation in your book. In
Step 2, each new shape you form has the same area as
the original parallelogram because you have simply
rearranged pieces, without adding or removing any
cardboard. Form a rectangle with the two pieces.
Notice that the base and height of the rectangle are the
same as the base and height of the original
parallelogram. Because the area of the rectangle and the
parallelogram are the same, the area of the parallelogram
is bh. The side length, s, has nothing to do with finding
the area.
S. Stirling
Page 1 of 12
Ch 8 Notesheet L1 Key
Name ___________________________
Examples (Note: drawing are not necessarily to scale!)
A represents area and P represents perimeter.
Ex. B
Find the height of a parallelogram that has an
area 7.13 m2 and a base length 2.3 m.
A = bh
Ex. 1
A=?
24 m
6m
11 m
7.13 = 2.3h
7.13
h=
= 3.1
2.3
10 m
Abig = 24 • 6 = 144
Asm = 10 • 5 = 50
Abig = 24 •11 = 264
144 + 50 = 194 m2
Ashape = 264 − 76 = 194
Asm = 14 • 5 = 76
Or try subtraction!
Ex. 2 Given the rectangle with P = 42 ft, A = ?
x
Ex. 3 Given the parallelogram, find the shaded
area = ?
9 ft
5 ft
3 ft
6 ft
P = 42 = x + 9 + x + 9
42 = 2 x + 18
24 = 2x So x = 12
A = 12 • 9 = 108 ft
2
Atri = ½ base ● height
Atri =
1
• 6 • 3 = 9 ft 2
2
Half the area of the parallelogram.
Note: Choose the height 1st! Look for perpendiculars!
Use algebra!!!
S. Stirling
Page 2 of 12
Ch 8 Notesheet L1 Key
Name ___________________________
Lesson 8.2 Areas of Triangles, Trapezoids and Kites
Triangle Area Conjecture
The area of a triangle is given by
1
A = bh
2
Trapezoid Area Conjecture
The area of a trapezoid is given by
A=
b
1
( b1 + b2 ) h ,
2
b1
A is the area
b is base length
h is the height
A is the area
b1 and b2 are the
base lengths
h is the height
h
Deriving the Triangle Area Conjecture
Apara = bh
Kite Area Conjecture
The area of a kite is given by
1
A = d1d2 ,
2
b2
h
A is the area
d1 and d2 are the
lengths of
the diagonals
Deriving the Trapezoid Area Conjecture
Put two trapezoids together, forms a parallelogram…
1
Atri = bh
2
b
h
Apara = ( b1 + b2 ) h
Trapezoid is half…
Atrap =
1
( b1 + b2 ) h
2
Note: base must be a side of the triangle!
The height does not.
Deriving the Kite Area Conjecture
Kite makes two congruent triangles, so
1⎛1
1
⎞
Atri = ⎜ d1d2 ⎟ = d1d2
2⎝ 2
4
⎠
⎛1
⎞ 1
Akite = 2 • ⎜ d2 d1 ⎟ = d1d2
⎝4
⎠ 2
Examples (Note: drawing are not necessarily to scale!):
Ex. 1 Given A = 40 cm, find h.
h
16 cm
A = 1 bh
2
40 = 1 (16)h
2
40 = 8h
40
h=
= 5 cm
8
S. Stirling
Ex. 2
Given A = 245.25 m2, b = ?
1
A = h ( b1 + b2 )
2
1
245.25 = (9) ( b + 36 )
2
245.25 = 4.5 ( b + 36 )
54.5 = b + 36
9m
36 m
18.5 meters = b
Page 3 of 12
b
Ch 8 Notesheet L1 Key
Name ___________________________
Ex. 3 Find the area of a kite with diagonals
7 feet and 16 feet.
Ex. 4 Sketch and label two different triangles each with an
area of 54 cm2.
A = 1 d1d2
2
A = 1 ( 7 )(16 )
2
A = 8 ( 7 ) = 56 feet
A = 1 bh
2
54 = 1 bh
2
108 = bh
Sketch any triangles that fit the equation. For example,
base = 13.5 and height = 8 or
base = 4 and height = 27 or
base = 9 and height = 12, etc…
8.2 Page 433 Exercise #30
56
68
34
56
56
inscribed
angle ∠EDC
56
68
90
m = 1 i112
2
AB = AC
Tangents =
56
OB = OC Radii =
Kite ABOC so
56
∠BOA ≅ ∠AOC
so d = e = 56
O
90
Tangent
mDE = 68
S. Stirling
so mBC = 112
34
Semicircle
180 – 112
m∠BEC = 56°
f = 90
34
Page 4 of 12
⊥ radius
Ch 8 Notesheet L1 Key
Lesson 8.3 Area Problems
Name ___________________________
On all word problems:
You must show all work, as you have been doing, and also label the sections of your work!
For example, you need to write: “Area of walls and ceiling:”, then show your equations and solutions.
Note: Be careful with unit conversions!!! If 1 yard = 3 feet, then 1 square yard = 9 square feet. Draw a square to
prove this to yourself!
Geometrically:
Algebraically:
3 ft
1 yard = 3 feet
2
2
2
If you square both sides… 1 yard = 3 feet
3 ft
1 yd
1 yd
2
1 yard 2 = 9 feet 2
So
3
3
Likewise: 1 yard = 27 feet
Example:
You are painting your 15 ft by 13 ft room and your 20 ft by 17 ft family room. The ceilings in your house are 9.5
ft high. You want to do a good job, so you will do one coat on the ceiling and two coats on the walls. (You may
ignore the area taken up by windows and doors.) The product information you got from the website. How much is
it going to cost to paint your room?
Area ceilings:
Abed = 15i13 = 195 ft 2
Afamily = 20i17 = 340 ft
BEHR Premium Plus 1-Gal.
Ceiling Paint Premium Plus
Interior $23.98
1 container covers up to
300 sq. ft.
2
Atotal = 195 + 340 = 535 ft 2
gallons = 535 ÷ 300 = 1.783
So need 2 gallons ceiling paint.
BEHR Premium Plus 1-Gal.
Swiss Coffee Semi-Gloss
Enamel Zero VOC Interior
Paint $28.96
1 container covers up to
400 sq. ft.
Area walls:
Abed = (15i2 + 13i2 )i9.5 = 532 ft 2
Afamily = ( 20i2 + 17i2)i9.5 = 703 ft 2
Atotal = 532 + 703 = 1235 ft 2
gallons = 1235 ÷ 400 = 3.0875
So need 4 gallons ceiling paint, probably.
Cost:
Cost = 2i23.98 + 4i28.96 = 163.80
$163.80 to paint both rooms.
Caution: Say you are painting something objects measured in inches: 13 in by 45 in, 57 in by 29 in and 66 in by
98 in. To figure out how much paint, you will need square feet! The easiest way is to find the square inches and
convert to square feet.
A = 13i45 + 57i29 + 66i98 = 8706 in
S. Stirling
2
now
8706 in 2 1 ft 2
i
= 60.49 ft 2
2
1
144 in
Page 5 of 12
Ch 8 Notesheet L1 Key
Name ___________________________
Lesson 8.4 Areas of Regular Polygons
center (of a regular polygon) The point that is the center of the circle that is circumscribed about the polygon.
radius (of a regular polygon) A segment from the center to a vertex of the polygon. Also, the length of that
segment.
apothem (of a regular polygon) A perpendicular segment from the center of the polygon’s circumscribed circle to a
side of the polygon. Also, the length of that segment.
Deriving the Regular Polygon Area Conjecture
Try to find a formula for a regular polygon with n-sides. Hint: make congruent triangles from the center!
⎛1
⎝2
⎞
⎠
Apoly = 3 ⎜ as ⎟
⎛1
⎝2
⎞
⎠
⎛1
⎝2
Apoly = 4 ⎜ as ⎟
r
s
⎞
⎠
Apoly = 5 ⎜ as ⎟
a
a
r
n = # sides, #
angles, or #
congruent triangles
⎛1 ⎞
A = n ⎜ as ⎟
⎝2 ⎠
a
r
s
s
Regular Polygon Area Conjecture
The area of a regular polygon is given by
1
A = ans
2
1
A
=
ap
OR
2
A is the area, p is the perimeter,
a is the apothem, s is the length of each side,
and n is the number of sides.
P 443 #8
Find the approximate length of each side of a regular ngon if a = 80 feet, n = 20, and A ≈ 20,000 square feet.
1
A = asn
2
1
20000 = (80)(20)s
2
20000 = 800i s
200
s=
≈ 25 ft
8
Ex.
Find area of a regular n-gon with an apothem of 21.9 cm
and a perimeter of 140 cm.
1
A = ap
2
1
A = (21.9)(140)
2
= 1533 cm2
S. Stirling
Page 6 of 12
Ch 8 Notesheet L1 Key
Lesson 8.5 Areas of Circles
Name ___________________________
Read page 449 in the book, including the investigation. Then…
Deriving the Circle Area Conjecture
Dimensions of the
“parallelogram”?
Cut up the circle into
16 congruent wedges.
Using parallelogram area:
Apara = base * height
base = ½ Circ.
r=
radius
= ½ Circ * height
1
i2π r ir
2
2
= πr
height = radius of
circle
=
½ Circumference
WARNING! It is very easy to confuse the formulas for area and circumference. Just remember that area is
measured in square units so the formula for area contains squaring.
Circle Area Conjecture
The area of a circle is given by
A = π r2 ,
A is the area
r is the radius
Circle Circumference Conjecture
The circumference of a circle is
given by
C = 2π r = dπ
,
C is the circumference
r is the radius & d is the diameter
As before, when we studied circumference, if the problem asks for an exact answer, DO NOT substitute in for π.
Only use the π key or 3.14 or 22/7 if they want an approximate answer. Look for ≈, which means approximate!
P 450 EXAMPLE A
The small apple pie has a diameter of 8 inches, and
the large cherry pie has a radius of 5 inches. How
much (what percent) larger is the large pie?
P 450 EXAMPLE B
The area of a circle is 256π m2, what is the
circumference of the circle?
Start with area:
Asm = π (4) = 16π ≈ 50.2
2
Alg = π (5) = 25π ≈ 78.5
2
25π
= 1.56 or 100% + 56% so large is 56% larger
16π
Now:
A = π r2
256π = π r 2
r 2 = 256
2
so r = 256 = 16
C = 2π r
C = 2π (16)
C = 32π meters
than the small.
On calculator: Use the
⎡⎣ x ⎤⎦ key. Type
2
S. Stirling
key. Above the
256 .
Page 7 of 12
Ch 8 Notesheet L1 Key
Name ___________________________
Ex. 1
Find the area of the shaded region. All measures in
inches.
(6, 8)
8
(0, 0)
6
(10, 0)
Use location (x, y), to
find distances.
radius = 10
height = 16
width = 12
Ex. 2
Find the area of the
shaded region.
All measures in cm.
Area Shaded =
Area Square – Area 4 ○
A = (12 ) − 4 • π (3)
2
3
2
= 144 − 36π cm.
= 30.90 in2
Area Shaded = Area Circle – Area of rectangle
2
= π (10) − (12)(16)
100π − 192 or approx. 122.159 in2
Lesson 8.6 Any Way You Slice It
sector of a circle The region between two radii and an arc of the circle. (aka a slice of pizza)
segment of a circle The region between a chord and an arc of the circle. (aka just keep a slice of the crust)
annulus The region between two concentric circles of unequal radius. (aka just the crust, cut out the center)
Sector of a Circle
A fraction of the area of a
circle = a piece of pizza.
Just like arc length,
but its part of the
area.
S. Stirling
Segment of a Circle
Area of sector minus area of triangle =
area of “the crust”.
Annulus or “Washer”
Area of the whole “big” circle
minus the area if the inner
“small” circle.
CAUTION: The choice for the base and the
height of the triangle will depend on the type
of triangle. If it is a right triangle, the radii
will be the base and the height.
Page 8 of 12
Ch 8 Notesheet L1 Key
Name ___________________________
Ex. 1
Ex. 2 (P 454 Example B)
Find the area of the shaded sector.
Find the area of the shaded segment.
Asec tor = fract ○i Acircle
270
A=
π (8)2
360
3
A = π (8i8) = 192π cm2
4
Ashaded = Asec tor − Atriangle
90
1
π (6)2 − (6)(6)
360
2
A = 9π − 18 ≈ 10.27 ft2
A=
Think: How
much do you
want to keep?
Ex 3.
Find x. The area of the shaded region is 3π cm2. The large circle has a radius 10 cm and the inner
circle has a radius 8 cm.
Ashaded = fract ○• ( Awasher )
Ashaded = fract ○• ( Aoutside ○ − Ainside ○ )
x
⎡⎣π (10)2 − π (8)2 ⎤⎦
360
x
πx
3π =
• 36π and 3π =
360
10
3π =
x = 30 degrees
Also examine Example A and Example C on page 454 of the book.
8.6 Page 456 Exercise #15 In a - c, the shaded is half the square. (Move A to B)
10/16 or 5/8 shaded
1B
2B
1B
2B
1B
2B
2B
3B
1B
5B
1A
2A
S. Stirling
1A
2A
1A
4B
2A
1A
4A
2A
Page 9 of 12
3A
5A
Ch 8 Notesheet L1 Key
Lesson 8.7 Surface Area
Name ___________________________
surface area The sum of the areas of all the surfaces, faces, of a solid.
base (of a solid) A polygon or circle used for reference to determine an altitude or other feature of the solid, or used
to classify the solid.
lateral face A face of a solid other than a base. Does not have to be vertical!
slant height The height of each triangular lateral face of a pyramid. They usually use l for slant height.
Prism
height of prism
Pyramid
pentagonal
base
b
b
rectangular
lateral face
height
of
prism
b
l
b
Surface Area = (base area) + (lateral surface area)
= (base area) + (area one face * n)
Surface Area
= 2 (base area) + (lateral surface area)
= 2 (base area) + (perimeter base * height)
Use appropriate
area formula.
Is the large rectangle
that wraps around the
base.
= (base area) + (½ bl * n)
= (base area) + (½ pl )
n = number of sides of regular polygon base
b = length of base edge
l = slant height
p = perimeter of the base
USE THESE FORMULAS TO GET THE FORMULAS FOR A CYLINDER AND A CONE!!
Cylinder
Same as a prism with circular bases.
Cone
Same as a pyramid with circular bases.
circular base
l
l
h=
height
rectangular lateral face
“the can’s label”
Circumference
triangular-ish
lateral face
circular base
Surface Area = (base area) + (lateral surface area)
Surface Area
= 2 (base area) + (perimeter base * height)
= 2 (π r2) + (2 π r * h)
S. Stirling
= (base area) + (½ pl )
= (π r2) + (½ * 2 π r * l )
= π r2 + π r l
Page 10 of 12
Ch 8 Notesheet L1 Key
Name ___________________________
EXAMPLE A
EXAMPLE B
Find the surface area of a rectangular
prism with dimensions
3 m, 6 m and 8 m.
Find the surface area of a cylinder with base diameter 10
in. and height 12 in.
h=3
SA = 2B + lateral face area
10
8
6
SA = 2(6i8) + 3(6 + 8 + 6 + 8)
= 2(48) + 3(28)
= 96 + 84 = 180 m2
h = 12
SA = 2B + lateral face area
SA = 2 (π 52 ) + 12(2iπ i5)
SA = 50π + 120π = 170π
≈ 534.07 in2
OR
pbase = 6 + 8 + 6 + 8 = 28
OR
pbase = 2iπ i5 = 10π
Abase = 6i8 = 48
SA = 2B + ph
SA = 2 ( 48) + ( 28)( 3) = 96 + 84 = 180 m2
Abase = π 52 = 25π
SA = 2B + ph
SA = 2 ( 25π ) + (10π )12 = 50π + 120π = 170π
EXAMPLE D
EXAMPLE C
Find the surface area of a cone with base radius 5 cm. and
slant height 10 cm.
The surface area of the right square pyramid is 95 in2 with
base edge of 5 in. What is the measure of the slant height?
SA = B + lateral face area
SA = B + lateral face area
SA = π 52 + π i5i10
SA = 25π + 50π = 75π
2
≈ 235.62 cm
OR
Cbase = 2iπ i5 = 10π
Abase = π 52 = 25π
1
SA = B + pl
2
1
SA = 25π + (10π )10 = 25π + 50π = 75π
2
S. Stirling
1
95 = 52 + 4i i5il
2
10
5
l
(4 cong. triangles)
95 = 25 + 10il
70 = 10il
l = 7 in.
5
OR
pbase = 4i5 = 20
Abase = 5i5 = 25
1
SA = B + pl
2
1
95 = 25 + ( 20 ) l
2
70 = 10il
l = 7 in
Page 11 of 12
5
Ch 8 Notesheet L1 Key
8.7 Page 467 Exercise #12
Name ___________________________
6.5
height of Δ:
38.5 – 36 = 2.5
15
height
trapezoid
40
30
height
rectangle
24
30
End View Paint Area:
1
Atri = i2.5i12 = 15
2
1
Atrap = (12)(12 + 30) = 252
2
Arect = 30i24 = 720
One face “end view” =
15 + 252 + 720 = 987 ft2
Total Area:
= 2 ends + 2 rectangular sides
TA = 2(987) + 2(24 • 40)
TA = 3894 ft2
Paint:
one gallon = 250 ft2
3894 ft 2 1 gallon 3894
i
=
≈ 15.576 gallon
1
250 ft 2
250
Cost: $25 per gallon
(Need to round up to the nearest gallon = 16 gallons.)
$25 16 gallon
i
= $400
1 gallon
1
Roof Area:
Anext = 40(15) = 600
Atop rect = 40(6.5) = 260
( 600 + 260) • 2 = 1720 ft2
1720 ft 2 1 bundle 1720
i
=
≈ 17.2 bundles
1
100 ft 2
100
(Need to round up to the nearest bundle = 18 bundles.)
Cost: $65 per bundle
$65 18 bundles
i
= $1170
1 bundle
1
Total Cost = $400 + $1170 = $1570.00
S. Stirling
Page 12 of 12