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Slide 1 / 152
Geometry
Area of Figures
2015-10-27
www.njctl.org
Slide 2 / 152
Table of Contents
Click on the topic to go to that sectionArea of Rectangles
Area of Triangles
Area of Parallelograms
Area of Regular Polygons
Area of Circles & Sectors
Law of Sines
PARCC Sample Questions
Area of Other Quadrilaterals
Area & Perimeter of Figures in the Coordinate Plane
Slide 3 / 152
Throughout this unit, the Standards for Mathematical Practice are used.
MP1: Making sense of problems & persevere in solving them.MP2: Reason abstractly & quantitatively.MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics.MP5: Use appropriate tools strategically.MP6: Attend to precision.MP7: Look for & make use of structure.MP8: Look for & express regularity in repeated reasoning.
Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used.
If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.
Slide 4 / 152
Area of Rectangles
Return to Table of Contents
Slide 5 / 152
The area of a rectangle is defined to be the number of squares of area "1" that can fit within it.
In the below drawing, 6 unit squares fit within the below rectangle of height 2 and based 3.
Area of a Rectangle
3
2
Slide 6 / 152
In general, the area of a rectangle is equal to its base times its height. This can also be referred to as its length times its width.
Arectangle = length x width (lw) = base x height (bh)
Area of a Rectangle
b
h
Slide 7 / 152
Area of a Rectangle
Sometimes, the dimensions will not be given, so you will need to calculate them before calculating the area.
Since a rectangle is a quadrilateral with 4 right angles, 2 right triangles can be formed when drawing one of its diagonals. Therefore, Pythagorean Theorem can be a helpful formula.
Another helpful formula is the perimeter formula for a rectangle: P = 2l + 2w
There might also be questions asking you about the population density of a town, state, or country. It is the ratio that represents the number of people living per square mile and can be found by dividing the total population by the total area.
Slide 8 / 152
ExampleThe diagonal of a rectangle is 34 feet and its length is 14 feet more than its width. Find the length, width, and area of the rectangle.
34 ftx
x + 14We know that the width is unknown, so let's call it "x".
Therefore, the length will be "x + 14".
Using the Pythagorean Theorem & our Algebra skills, we can solve for x.
x2 + (x + 14)2 = 342
x2 + x2 + 28x + 196 = 1,1562x2 + 28x - 980 = 02(x2 + 14x - 480) = 02(x + 30)(x - 16) = 0x + 30 = 0 or x - 16 = 0x = -30 or x = 16
Since x is the length of a side, it has to be positive. Therefore, our final answer is x = 16 ft = widthlength = 30 ftArea = 480 ft2
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1 What is the area of a rectangle that has a length of 8.4 cm and a width of 3.7 cm?
Slide 10 / 152
2 Televisions, are advertised using the length of the diagonal. For example, a 26" TV could have a length of 24" and a width of 10", as shown below.
What is the area of an 80" TV if the length is 69.3"?
26"
24"
10"
Slide 11 / 152
3 The population density is the amount of people living per square mile. If the town of Geometryville is a rectangular town that has a length of 24 miles and a width of 13 miles, and its population is 2,500 people, what is the population density of the town?
Slide 12 / 152
4 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the length of the rectangle?
A 4 feet
B 6 feet
C 8 feet
D 9 feet
Slide 13 / 152
5 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the width of the rectangle?
A 4 feet
B 6 feet
C 8 feet
D 9 feet
Slide 14 / 152
6 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the area of the rectangle?
A 80 ft2
B 60 ft2
C 48 ft2
D 24 ft2
Slide 15 / 152
Area of Triangles
Return to Table of Contents
Slide 16 / 152
Clearly, the area of each of the below right triangles is equal to half the area of the rectangle they comprise.
Since the area of the rectangle is bh, the area of each right triangle is 1/2 bh.
Area of a Right Triangle
b
hAΔ = 1/2 bh
Slide 17 / 152
Now, let's find the area formula for an arbitrary scalene triangle ABC with base "b" and height "h." Keep in mind that the height is always measured perpendicular to the base that is opposite the vertex.
Area of Any Triangle
b
h
A
B C
Slide 18 / 152
Let's draw right triangle ACD such that the new triangle plus the original triangle form a larger right triangle ABD. ΔADC is a right Δ with base "b' " and height "h". ΔADB is a right Δ with base "b' + b" and height "h".
Area of Any Triangle
b
h
b'
A
B C D
Slide 19 / 152
AreaΔADC = 1/2 b'h
AreaΔADB = 1/2 (b' + b) h
AreaΔABC = AreaΔADB - AreaΔADC
AreaΔABC = 1/2 (b' + b) h - 1/2 b'h
AreaΔABC = 1/2 b'h + 1/2bh - 1/2b'h
AreaΔABC = 1/2 bh
Area of Any Triangle
b
h
A
B C D b'
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Slide 20 / 152
So, the area of any triangle, not just right triangles, is given by :
AreaΔ = 1/2 bh
Area of Any Triangle
b
h
A
B C
Slide 21 / 152
While the formula is the same for any triangle: AreaΔ = 1/2 bh
For triangles that are not right triangles, the height is usually not directly given.
Instead, you may be given the lengths of one or more sides and the measures of one or more angles.
Area of Any Triangle
4
A
B C
5 8For instance, in this case how would you find the height...and then the area? 40°
Slide 22 / 152
Draw an altitude from the vertex to the base, creating a right triangle with the same height as our original triangle.
ΔABD has the same height as our original triangle ΔABC.
And both triangles share a side, AB, and an angle, ∠B.
Which trig function would allow us to find the value of "h"?
Area of Any Triangle
h
4
A
B C
5 8
D 40°
Slide 23 / 152
h
4
A
B C
5 8
D 40°
sin θ = opposite / hypotenuse = opp / hyp
opp = (hyp)(sin θ)
h = (8)(sin 40°)
h = (8)(0.64)
h = 5.1 units
Area of Any Triangle
Now that you have the base and the height of ΔABC, how would you find the area?
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Slide 24 / 152
A = 1/2 bh
A = 1/2 (4)(8)(sin 40°)
A = 1/2 (4)(8)(0.64)
A = 10.2 units2
Area of Any Triangle
h
4
A
B C
5 8
40°D
Slide 25 / 152
A = 1/2 bh
A = 1/2 (4)(8)(sin 40°)
A = 1/2 (4)(8)(0.64)
A = 10.2 units2
Area of Any Triangle
h
4
A
B C
5 8
40°D
Examining our solution, we can see that the area of the triangle is given by half the product of two sides multiplied by the sine of the angle between them.
In this case, the sides were of length 4 and 8 and the angle is 40°.
Slide 26 / 152
AΔ = 1/2 ac sinB
Area of Any Triangle
a
A
B C
b c
Replacing the numbers by labeling the angles with upper case letters and the sides opposite them with matching lower case letters yields this formula.
But, there's nothing special about those sides and that angle, so it will also be true that:
AΔ = 1/2 ab sinC
Or
AΔ = 1/2 bc sinA
The area of a triangle is equal to the product of any two sides and the sineof the included angle.
Slide 27 / 152
Find the area of ΔABC.
In this case, the altitude you draw will be within the triangle.
Example
4
A
B C 9
8
50°
Slide 28 / 152
7 Find the area of ΔDEF. Round your answer to the nearest hundredth.
34°
8 in.
12 in.D
E
F
Slide 29 / 152
8 Find the area of ΔGHI. Round your answer to the nearest hundredth.
28°
16 in.
9 in.G
H
I
Slide 30 / 152
9 Find the area of ΔJKL. Round your answer to the nearest hundredth.
32°6 in.
15 in.
J
K
L
Slide 31 / 152
10 Find the area of ΔPQR. Round your answer to the nearest hundredth.
52°
9 in.
14 in.P
Q
R
Slide 32 / 152
Law of Sines
Return to Table of Contents
Slide 33 / 152
We just learned that we can find the area of any triangle with any of these three formulas:
AΔ = 1/2 ac sinBAΔ = 1/2 ab sinCAΔ = 1/2 bc sinA
Since the area of a triangle will be the same regardless of which formula we use, these three formulas must be equal.
Setting them equal to one another will give usa general relationship between the sides and the angles of a triangle.
Law of Sines
a
A
B C
b c
Slide 34 / 152
AΔ = 1/2 ac sinB = 1/2 ab sinC = 1/2 bc sinA
Let's look at one pair at a time and simplify.
1/2 ac sinB = 1/2 ab sinC AND 1/2 ab sinC = 1/2 bc sinA
ac sinB = ab sinC ab sinC = bc sinA
c sinB = b sinC a sinC = c sinA
c / sinC = b / sinB a / sinA = c / sinC
Law of Sines
= = a b c sinA sinB sinC
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Law of SinesThis relationship between the sides and angles of a triangle will be true for all triangles.
a
A
B C
b c
= = a b c sinA sinB sinC
= = a b c sinA sinB sinC
OR
Slide 36 / 152
Find the missing segment lengths and angle measures in ΔABC if m∠B = 28°, m∠C = 103°, and b = 26 cm.
Example
A
B
C
28°
103°26 cm
ca
Since we have a triangle, we know that the sum of the interior angles is 180°. Therefore, we can find the m∠A using the Triangle Sum Theorem.m∠A = 180 - 103 - 28 = 49°
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Slide 37 / 152
Find the missing segment lengths and angle measures in ΔABC if m∠B = 28°, m∠C = 103°, and b = 26 cm.
Example
A
B
C
28°
103°26 cm
ca
Now that we have the measurements of all of the angles and one side length given, we can find the remaining sides using the Law of Sines.
26 sin28°
a sin49°
26sin49° = asin28°
a =
a = 41.80 cm
=
26sin49° sin28°
26 sin28°
c sin103°
26sin103° = csin28°
c =
c = 53.96 cm
=
26sin103° sin28°
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Slide 38 / 152
( )
Find the missing segment lengths and angle measures in ΔABC if m∠C = 122°, a = 12 cm and c = 18 cm.
Example
A
B
C122°
b
18 cm12 cm
In this problem, we need to determine one of the missing angles first. However, we can't use the Triangle Sum Theorem yet, since we only know the measurement of 1 angle. Therefore, we need to use the Law of Sines.
18 sin122°
12 sinA
18sinA = 12sin122°
sinA =
m∠A = sin-1
m∠A = 34.43°
=
12sin122° 18
12sin122° 18
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Slide 39 / 152
Find the missing segment lengths and angle measures in ΔABC if m∠C = 122°, a = 12 cm and c = 18 cm.
Example
A
B
C122°
b
18 cm12 cm
Now, that we know 2 angle measures, we can calculate the measurement of our final missing angle.m∠B = 180 - 122 - 34.43 = 23.57°
Finally, we can calculate the length of our final side.
18 sin122°
b sin23.57°
18sin23.57° = bsin122°
b =
b = 8.49 cm
=
18sin23.57° sin122°
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11 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd.What is m∠C?
A 36°
B 46°
C 56°
D 136° A
B
C70°
b
5 yd a64°
Slide 41 / 152
12 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd.What is the value of a? Round your answer to the nearest hundredth.
A
B
C70°
b
5 yd a64°
Slide 42 / 152
13 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd.What is the value of b? Round your answer to the nearest hundredth.
A
B
C70°
b
5 yd a64°
Slide 43 / 152
14 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in.What is m∠B? Round your answer to the nearest hundredth.
A
B
C111°
a5 in.
3 in.
Slide 44 / 152
15 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in.What is m∠A? Round your answer to the nearest hundredth.
A 124.93°
B 44.93°
C 34.93°
D 24.93°A
B
C111°
a5 in.
3 in.
Slide 45 / 152
16 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in.What is the value of a? Round your answer to the nearest hundredth.
A
B
C111°
a5 in.
3 in.
Slide 46 / 152
Area of Parallelograms
Return to Table of Contents
Slide 47 / 152
From these results, we can find the area formula for any parallelogram by dividing the parallelogram into two triangles.
Area of a Parallelogram
b
h
Slide 48 / 152
We can see below that a parallelogram is comprised of two triangles.
Since the area of each triangle is 1/2 bh.
The area of a parallelogram is bh.
Area of a Parallelogram
b
hAparallelogram = bh
Slide 49 / 152
As with triangles, the height of a parallelogram is often not given.
The height must be found by drawing an altitude and using trigonometry to find the height of the parallelogram.
Area of a Parallelogram
b
hAparallelogram = bh
Slide 50 / 152
5
8110°
20°
5
Find the area of ABCD.
This time, use the cosine rather than the sine function, just for practice. Either can be used depending on the angle you choose.
Example
To use cosine, our 1st step will be to drop down the altitude, or height.
Since we know that the entire obtuse angle is 110° and the right angle measures 90°, the remaining acute angle is 20°
Slide 51 / 152
Find the area of ABCD.
This time, use the cosine rather than the sine function, just for practice. Either can be used depending on the angle you choose.
Example
5
8110°
20°
5
Now, let's find the height of our parallelogram.
cos20° = h 5
h = 5cos20°h = 4.70 units
And its area
A = 8(4.70)A = 37.60 units2
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17 A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75°. Find the height of the parking space.
A 2.4 ft
B 9 ft
C 9.7 ft
D 34.8 ft
20.5 ft
9 ft 4 in. 9 ft 4 in.75° 75°
Slide 53 / 152
18 A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75°. Find the area of the parking space.
A 184.5 ft2
B 191.3 ft2
C 198.85 ft2
D 324.8 ft2
20.5 ft
9 ft 4 in. 9 ft 4 in.75° 75°
Slide 54 / 152
19 A window frame is in the shape of a parallelogram. Its base is 3 feet long, its other side length is 2.5 feet long, and the obtuse angle created between two of the sides is 114°. How much glass would be required to fill the window? Round your answer to the nearest hundredth.
Slide 55 / 152
20 Mrs. Polygon is creating a quilt for her grandson. The quilt is created by stitching together parallelograms that are different colors. Each parallelogram is 4 inches long its other side is 3 inches long, and the acute angle created between the two sides is 32°. What is the area of each parallelogram used to make the quilt? Round your answer to the nearest hundredth.
Slide 56 / 152
21 If the quilt requires 5 parallelograms horizontally and 10 parallelograms vertically, how much material will Mrs. Polygon need to make the quilt?
Slide 57 / 152
Area of Regular Polygons
Return to Table of Contents
Slide 58 / 152
A regular polygon is a polygon that has all its sides and angles congruent: it is both equilateral & equiangular.
Area of Regular Polygons
Slide 59 / 152
rC
Circumscribing a Regular Polygon
A polygon is circumscribed by drawing around it the smallest circle on which lie all the vertices of the polygon.
That circle is called the circumcircle of the polygon.
Slide 60 / 152
rC
The radius of a regularpolygon is the radius of thecircumscribed circle, shown here as "r."
This is called the circumradius.
Circumcenter and Circumradius
The center of a regular polygon is the center of its circumscribed circle, shown here as "C."
This is called the circumcenter of the polygon.
Slide 61 / 152
Central Angle of a Regular Polygon
rC
r
The degrees of the central anglecan be found using the formula
Central angle =
where n is the number of sidesin the regular polygon.
360 n
A central angle of a regular polygon is an angle with one vertex at the circumcenter and two vertices on the circumcircle.
The sides of the central angle are radii of the circle.
Slide 62 / 152
rC
ra
The Apothem a Regular Polygon The apothem of a Regular Polygon is the shortest distance from the center of the polygon to one of its sides.
An apothem is perpendicular to a side of the polygon.
An apothem is also the altitude (height) of the isosceles triangle formed by the sides of a central angle and the side of the polygon that is opposite the central angle.
Slide 63 / 152
A polygon of n sides is comprised of n triangles.
Each triangle has a height equal to the apothem, "a."
The base of each triangle is equal to the perimeter, P, divided by the number of sides, n: b = P/n.
Area of Regular Polygons
What is the area of a triangle whose base is P/n and whose height is a?
ra
Slide 64 / 152
AΔ = 1/2 bh
AΔ = 1/2(P/n)a
There are n triangles in a regular polygon, so the area of the polygon is given by:
A = nAΔ
A = (n)(1/2)(P/n)a
A = 1/2Pa
Area of Regular Polygons
rr
r r
a
aaa
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Slide 65 / 152
The formula for the area of a polygon is simple: A = 1/2Pa
But, often we are not given both P and a.
We have to find one or both of them using trigonometry.
Area of Regular Polygons
rr
r r
a
aaa
Slide 66 / 152
Let's find the area of a regular pentagon whose sides have a length of 7.
Example
Slide 67 / 152
Example
A = 1/2 Pa
For a pentagon, n = 5, so it's perimeter will be 5 times the length of a side.
In this case, P = (5)(7) = 35.
But, how do we find the apothem?
Let's look at one of the five triangles that comprise the pentagon.
7
Let's find the area of a regular pentagon whose sides have a length of 7.
Slide 68 / 152
Example
A = 1/2 Pa
For a pentagon, n = 5, so it's perimeter will be 5 times the length of a side.
In this case, P = (5)(7) = 35.
But, how do we find the apothem?
Let's look at one of the five triangles that comprise the pentagon.
7
Let's find the area of a regular pentagon whose sides have a length of 7.
Slide 69 / 152
Example
The central angle is given by:
m∠C = 360/n = 360/5 = 72°
Since the ∠s must add to 180° and the measures of the base ∠s of an isosceles triangle are equal, those base ∠s must = 54°.
The apothem, a, is the altitude of this triangle.
7
72°
54° 54°
r r
7
C
Let's find the area of a regular pentagon whose sides have a length of 7.
Slide 70 / 152
7
36°
54° 54°
r r
3.5
a
Example
The two legs of the right triangle shown are a and 3.5.
Tangent θ = opposite / adjacent
tan θ = opp / adj
opp = adj (tan θ)
a = 3.5 (tan 54°)
a = 3.5 (1.38)
a = 4.8 units
Let's find the area of a regular pentagon whose sides have a length of 7.
Slide 71 / 152
Example
We now know that P = 35 units and a = 4.8 units.
A = 1/2 Pa
A = 1/2 (35)(4.8)
A = 84 units2
7
72°
54° 54°
r r
7
Let's find the area of a regular pentagon whose sides have a length of 7.
Slide 72 / 152
Find the area of a regular octagon whose sides have a length of 8.
Example
8What is the question asking? To find the area of the octagon
How can you represent the problem with symbols and numbers? A = 1/2 Pa
How could you start this problem? Is there anything that you can find right away? P = 8(8) = 64 units
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Find the area of a regular octagon whose sides have a length of 8.
Example
8How do we find the apothem? How could you use a drawing to show your way of thinking?
Create an equation to find the apothem.
8
45°
44
22.5°
67.5° 67.5°67.5°
a
tan 67.5° = a 4
a = 4 tan 67.5°a = 9.66 units
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Slide 74 / 152
Find the area of a regular octagon whose sides have a length of 8.
Example
8Now that we know all of the needed measurements, find the area of the regular octagon.
a = 9.66 unitsP = 8(8) = 64 unitsA = 1/2 (9.66)(64)A = 309.12 units2
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Slide 75 / 152
22 Calculate the apothem of the regular polygon shown in the figure below.
16
Slide 76 / 152
23 Calculate the side length of the regular polygon shown in the figure below.
16
Slide 77 / 152
24 Calculate the perimeter & area of the regular polygon shown in the figure below.
16
Slide 78 / 152
25 Calculate the side length of the regular polygon shown in the figure below.
15
Slide 79 / 152
26 Calculate the perimeter & area of the regular polygon shown in the figure below.
15
Slide 80 / 152
27 Calculate the apothem of the regular polygon shown in the figure below.
7
Slide 81 / 152
28 Calculate the perimeter of the regular polygon shown in the figure below.
7
Slide 82 / 152
29 Calculate the area of the regular polygon shown below.
7
Slide 83 / 152
Area of Circles and Sectors
Return to Table of Contents
Slide 84 / 152
Interestingly, the formula, A = 1/2Pa, also leads to the formula for the area of a circle.
If you let n go to infinity, the regular polygon approaches the shape of a circle.
The apothem, a, approaches the radius of the circle, r.
And, the perimeter of the polygon approaches the circumference of the circle: 2πr.
Then, A = 1/2Pa approaches
A = 1/2(2πr)(r)
A = πr2
Area of a Circle
r a
Slide 85 / 152
30 Find the area of a circle that has a radius of 8 in.
A 4π in2
B 8π in2
C 16π in2
D 64π in2
Slide 86 / 152
31 Find the area of a circle that has a diameter of 17 in.
A 8.5π in2
B 17π in2
C 72.25π in2
D 289π in2
Slide 87 / 152
32 Find the area of a circle that has a circumference of 18π in.
A 324π in2
B 81π in2
C 18π in2
D 9π in2
Slide 88 / 152
A sector of a circle is the portion of the circle enclosed by two radii and the arc that connects them.
Minor Sector
Major SectorA
B
C
Sectors of Circles
Slide 89 / 152
33 Which arc borders the minor sector?
A Arc AB
B Arc AC
C Arc ADB
A
BC
D
Slide 90 / 152
34 Which arc borders the major sector?
A Arc AB
B Arc AC
C Arc ADB
A
BC
D
Slide 91 / 152
The Area of a Sector
A sector is part of a circle.
The area of a complete circle is given by Acircle = πr2
A
BC
D
If the central angle of the sector is given in degrees, that's just the measure of that angle divided by 360 degrees, yielding:Asector = (θ/360°)(πr2)when θ is the central angle of the sector measured in degrees
Similar to the arc length, we have to find the fraction of the circle in the sector and multiply this fraction by the area of the entire circle to find the area of a sector
θr
Slide 92 / 152
The Area of a Sector
Using our formula, we know that θ = 110°, and r = 20 cm.
A
BC
D
110°20 cm
If we substitute these numbers into our formula, it will give us the area of our sector.Asector = (110°/360°)π(20)2
Asector = 383.97 cm2
In this case, we know that the measure of our central angle, or m∠C = 110°.
But, if we are also told that the radius of the circle is 20 cm, we can determine the area of the sector enclosed by AB, AC & CB.
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The Area of a Sector
Alternatively, we could just set this up as a proportion and solve it in one step.
Area of SectorArea of Circle
Central angle360
=
Asector
πr2110360=
Asector = (πr2)110360
= π(20)2110360Asector
Asector = π cm2 = 383.97 cm2 11009
A
BC
D
110°20 cm
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Slide 94 / 152
35 Find the area of the sector. Leave your answer in terms of π.
45°
AB
C
3
Asector = (θ/360°)(πr2)
Slide 95 / 152
36 Find the area of the major sector. Leave your answer in terms of π.
Asector = (θ/360°)(πr2)
C
A
T
8 cm
60°
Slide 96 / 152
37 Find the area of the minor sector of the circle. Round your answer to the nearest hundredth.
C
A
T5.5 cm 30°
Slide 97 / 152
38 Find the Area of the major sector for the circle. Round your answer to the nearest thousandth.
C
A
T
12 cm
85°
Slide 98 / 152
39 What is the central angle for the major sector of the circle?
C
A
G
15 cm
120°
Slide 99 / 152
40 Find the area of the major sector. Round to the nearest thousandth.
C
A
G
15 cm
120°
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41 If a circle is divided into 2 sectors, one major and one minor, then the sum of the areas of the 2 sectors is equal to the total area of the circle.
True
False
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42 A group of 10 students order pizza. They order 5 12" pizzas, that contain 8 slices each. If they split the pizzas equally, how many square inches of pizza does each student get (to the nearest hundredth)?
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43 You have a circular sprinkler in your yard. The sprinkler has a radius of 25 ft. How many square feet does the sprinkler water if it only rotates 120°? Round your answer to the nearest hundredth.
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Area of Other Quadrilaterals
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Lab: Area of Other Quadrilaterals
Slide 104 / 152
So far, we have discussed calculating the area of rectangles, triangles, parallelograms, circles, sectors and regular polygons, and the relationships between each formula.
In this section, we are going to determine how 3 additional area formulas are related to those that we already know and to each other.
Area of Other Quadrilaterals
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Area of a TrapezoidLet's start off by deriving the area formula of a trapezoid.
If you will recall from our unit on Quadrilaterals, a trapezoid is a quadrilateral with 1 pair of opposite sides that are parallel, called bases and 1 pair of opposite sides that are not parallel, called legs.
If we label our bases as b1 and b2 and draw an altitude, or the height (h), we have the figure given to the right.
b1
b2
h
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Area of a Trapezoid
b1
b2
h
We can split the trapezoid into 2 triangles by drawing the diagonal from the upper left corner to the lower right corner. Click on the bases to reveal each new triangle (given in red).
Now, we can separate these two triangles.
Using the triangles, how could you calculate the area of a trapezoid?
Add up the triangle areasclick
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Area of a Trapezoidb1
b2
hh
b2
b1
h+ =
We know that the area of a triangle is 1/2 bh and that the sum of these two triangles will be the area of the trapezoid.
Create an equation using this fact and the variables provided.
ATrapezoid = 1/2 b1h + 1/2 b2h
ATrapezoid = 1/2 h(b1 + b2)
What algebra facts can be used to simplify the equation? How does it simplify?
click
click
Distributive Property or Factoringclick
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Area of a RhombusNext, we are going to derive the area formula of a rhombus.
If you will recall from our unit on Quadrilaterals, a rhombus is a parallelogram with congruent sides and diagonals that are perpendicular and bisect each other.
Since it is a parallelogram, the formula A = bh will work for this shape. But what if we are given the diagonal lengths instead? What connections do you see? Can we figure out a formula for this case?
If we label our diagonals as d1 and d2, we have the figure given to the right.
d1
d2
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Area of a RhombusThe diagonals split the rhombus into 4 congruent triangles. If we move two the bottom triangles to the top, but on opposite sides (i.e. move the bottom right triangle to the upper left corner), we will create another shape and determine the area formula.
d1
d2
Use the following steps to show the animation in the diagram:· Click on the hash marks in the 2 bottom triangles to reveal each new
triangle (given in red). · Move the new red triangles to the opposite corners of the shape. · Click the bottom of the original rhombus (shown in black) to hide it.
Slide 110 / 152
d1
d2
What shape has been made?
Rectangle
What is the area formula for this shape?
A = bh or A = lw
Using these connections and the variables given above, create an equation to represent the area of a rhombus.
ARhombus = 1/2 d1d2
Area of a Rhombus
click
click
click
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Area of a KiteSince the shape of a kite is very similar to a rhombus, you are going to explain how the Area of a Kite formula is the same as the Area of a Rhombus formula for homework.
d1
d2AKite = 1/2 d1d2
Slide 112 / 152
You are constructing a desk to fit into your bedroom using wood for the flat top surface and metal bars for the legs. In order to save space, you determine that an isosceles trapezoid would be the best shape. The bases of the trapezoid will be 3 feet and 7 feet and the angle formed by the short base and each leg of the trapezoid is 135°. How much wood is required to make the flat top surface of your desk?
Example
7 ft
3 ft135° 135°
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44 In order to make a kite, you need to cut enough wrapping paper, based on the lengths of the spars, or the sticks used as the frame of the kite (see diagram to the right).
If the longest spar is 36 inches & the shortest spar is 24 inches, how much wrapping paper do you need to make your kite?
Spars
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45 One diagonal of a rhombus is 1/3 times as long as the other. The area of the rhombus is 0.24m2. What are the lengths of the diagonals?
A 1.2 m and 1.2 m
B 1.2 m and 0.4 m
C 1.8 m and 0.6 m
D 0.6 m and 0.4 m
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46 A glass window in the shape of an isosceles trapezoid has bases that measure 8 inches and 12 inches. If the angle between the longest base and the legs is 60°, what is the area of the window? Round your answer to the nearest hundredth.
12 in.
8 in.
60°
Slide 116 / 152
Area of Complex FiguresIn most real-world problems, you will need to calculate the area of complex figures, or shapes that are a combination of 2 or more shapes.
In order to calculate the area of these types of shapes, we need to either add or subtract the areas of the primary shapes involved.
Add the area of a rectangle& 1/2 the area of a circle.
Subtract the area of the rectangles from the area of the trapezoid.
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Slide 118 / 152
Example
In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color.
· Find the amount of wall space that will be painted.
· If one quart-size can of paint covers 87.5 ft2, then how many cans are required to paint 2 coats of the accent color?
· If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall?
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Example
In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color.
· Find the amount of wall space that will be painted.
Wall Area: A = 13.5(8.5) A = 114.75 ft2
Doorway Area: A = 7(3) = 21 ft2- 2 doorways, so total area notincluded is 42 ft2
Painted Area: A = 114.75 - 42 A = 72.75 ft2
click
click
click
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Example
In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color.
· If one quart-size can of paint covers 87.5 ft2, then how many cans are required to paint 2 coats of the accent color?
Total painted Area = 72.75(2) = 145.5 ft2
Quart-sizedCans of Paint = 145.5/87.5 = 1.66 ≈ 2 cans
click
click
Slide 121 / 152
Example
In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color.
· If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall?
$15(2) = $30click
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47 What is the area of the shaded region?
A 20 cm2
B 30 cm2
C 100 cm2
D 200 cm2
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48 What is the area of the entire figure? Round your answer to the nearest tenth.
A 25.1 yd2
B 28.9 yd2
C 54 yd2
D 79.1 yd2
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49 A plant vase is formed by combining a square base with 4 isosceles trapezoids (see figure below). The square base has an area of 25 in2, the long bases of the trapezoids are 7 in., and the heights of the trapezoids are 4 in. How much glass was needed to make the entire vase?
7 in.
4 in.
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50 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles.
What is the area of the entire roof?
A 700 ft2
B 600 ft2
C 250 ft2
D 100 ft2
20 ft30 ft
20 ft
10 ft
10 ft
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51 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles.
Each bundle of shingles can cover approximately 40 ft2, and shingles must be purchased in full bundles. How many bundles of shingles are required to cover the roof?
A 15
B 17C 18
D 2020 ft
30 ft
20 ft
10 ft
10 ft
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52 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles.
Each bundle of shingles costs $28.50. How much should the Bisect Building Company budget for the shingles?
20 ft30 ft
20 ft
10 ft
10 ft
Slide 128 / 152
Area & Perimeter of Figures in the Coordinate Plane
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Slide 129 / 152
Some problems will show your shapes in a coordinate plane. In these cases, you will need to calculate the side lengths, or diagonal lengths, of your shapes using the distance formula.
Remember that the distance formula is
Area & Perimeter of Figures in the Coordinate Plane
d =
After calculating the desired distances, you will either need to add them to calculate the perimeter or multiply them to calculate the area, using the appropriate formula.
Slide 130 / 152
ExampleCalculate the perimeter and area of rhombus JKLM.
4
2
0 x
y
6
8
42 6 8
J
K
L
M
10
-4
-6
-2
Since JKLM is a rhombus, all of the sides are congruent.
How would we calculate the perimeter?Calculate one side length using the distance formula & then multiply that value by 4 to calculate our perimeter.
JK = √(7 - 3)2 + (3 - 1)2
= √42 + 22
= √16 + 4 = √20 = 2√5 ≈ 4.47Perimeter = 4(2√5) = 8√5 units ≈ 17.89 units
click
click
click
click
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ExampleCalculate the perimeter and area of rhombus JKLM.
4
2
0 x
y
6
8
42 6 8
J
K
L
M
10
-4
-6
-2
Since JKLM is a rhombus, A = 1/2 d1d2
Therefore, we need to calculate the lengths of the diagonals using the distance formula.
JL = √(5 - 3)2 + (-1 - 1)2
= √22 + (-2)2
= √4 + 4 = √8 = 2√2KM = √(7 - 1)2 + (3 - (-3))2
= √62 + 62
= √36 + 36 = √72 = 6√2A = 1/2 (2√2)(6√2)A = 12 units2
click
click
click
click
click
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2
1
0 x
y
3
4
21 3 4
E F
GHM
iles
(hun
dred
s)
Miles (hundreds)
The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.
Example
· Based on the information given, how many miles is the perimeter of North Dakota?
· At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010?
Slide 133 / 152
2
1
0 x
y
3
4
21 3 4
E F
GHM
iles
(hun
dred
s)
Miles (hundreds)
The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.
Example
· Based on the information given, how many miles is the perimeter of North Dakota?
How do we calculate the perimeter?
add up all of the side lengths
What strategies are you going to use?
counting & distance formulaclick
click
Slide 134 / 152
2
1
0 x
y
3
4
21 3 4
E F
GHM
iles
(hun
dred
s)
Miles (hundreds)
The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.
Example
· Based on the information given, how many miles is the perimeter of North Dakota?
EF = 3.3 = 330 milesGH = 3.4 = 340 milesEH = 2.11 = 211 milesFG = √(3.4 - 3.3)2 + (0 - 2.11)2
= √(0.1)2 + (-2.11)2
= √0.01 + 4.4521 = √4.4621 ≈ 2.11 = 211 miles
Perimeter:330 + 340 + 211 + 211 = 1092 miles
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click
click
clickclick
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2
1
0 x
y
3
4
21 3 4
E F
GHM
iles
(hun
dred
s)Miles (hundreds)
The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.
Example
· At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010?
What does population density mean?ratio that represents the number of people living per square mile.
How do we find it? dividing the total population by the total area
click
click
Slide 136 / 152
2
1
0 x
y
3
4
21 3 4
E F
GHM
iles
(hun
dred
s)
Miles (hundreds)
The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.
Example
· At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010?
Area = 1/2(211)(340 + 330) = 1/2(211)(670) = 70,685 square miles
Population Density:672,591 70,685
= 9.51 people per square mile
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53 Calculate the perimeter of square PQRS.
A 10 units
B 20 units
C 25 units
D 28.28 units4
2
0 x
y
6
8
42 6 8
P
QS
R
10
-4
-6
-2
Slide 138 / 152
54 Calculate the area of square PQRS.
A 28.28 units2
B 20 units2
C 25 units2
D 50 units24
2
0 x
y
6
8
42 6 8
P
QS
R
10
-4
-6
-2
Slide 139 / 152
Series of SMART Response QuestionsThe picture below shows the yard of the Fractal family.
House
Driveway
garden
(0, 150)
(200, 0)
(200, 150)
(0, 20)(50, 25)
(50, 75) (180, 75)(180, 25)
(180, 5)
(140, 25)
(50, 150)
(0, 1
00)
(140, 10)
· What is the perimeter of their yard?
· Every spring, Mr. Fractal fertilizes the lawn. What is the area of it?
· This year, the Fractals are going to have a fence installed, represented by the red dotted line. Determine the amount of fencing required.
· In the future, the Fractals want to install a circular pool centered at (115, 105). If the pool must be at least 10 feet away from the house, which of the listed measurements could be the area of the pool? Select all that apply.
A 2,827.43 ft2 B 1,963.50 ft2
C 1,256.64 ft2
D 706.86 ft2E 314.16 ft2
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55 The picture below shows the yard of the Fractal family. · What is the perimeter of their yard?
House
Driveway
garden
(0, 150)
(200, 0)
(200, 150)
(0, 20)(50, 25)
(50, 75) (180, 75)
(180, 25)
(180, 5)
(140, 25)
(50, 150)
(0, 1
00)
(140, 10)
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56 The picture below shows the yard of the Fractal family.· Every spring, Mr. Fractal fertilizes the lawn. What
is the area of it?
House
Driveway
garden
(0, 150)
(200, 0)
(200, 150)
(0, 20)(50, 25)
(50, 75) (180, 75)
(180, 25)
(180, 5)
(140, 25)
(50, 150)
(0, 1
00)
(140, 10)
Slide 142 / 152
57 The picture below shows the yard of the Fractal family.· This year, the Fractals are going to have a fence
installed, represented by the red dotted line. Determine the amount of fencing required.
House
Driveway
garden
(0, 150)
(200, 0)
(200, 150)
(0, 20)(50, 25)
(50, 75) (180, 75)
(180, 25)(180, 5)
(140, 25)
(50, 150)
(0, 1
00)
(140, 10)
Slide 143 / 152
58 The picture below shows the yard of the Fractal family.· In the future, the Fractals want to install a circular pool
centered at (115, 105). If the pool must be at least 10 feet away from the house, which of the listed measurements could be the area of the pool? Select all that apply.
A 2,827.43 ft2
B 1,963.50 ft2
C 1,256.64 ft2
D 706.86 ft2
E 314.16 ft2
House
Driveway
garden
(0, 150)
(200, 0)
(200, 150)
(0, 20)(50, 25)
(50, 75) (180, 75)
(180, 25)
(180, 5)
(140, 25)
(50, 150)
(0, 1
00)
(140, 10)
Slide 144 / 152
PARCC Sample Questions
The remaining slides in this presentation contain questions from the PARCC Sample Test. After finishing this unit, you should be able to answer these questions.
Good Luck!
Return to Table of Contents
Slide 145 / 152
59 A steel pipe in the shape of a right circular cylinder is used for drainage under a road. The length of the pipe is 12 feet and its diameter is 36 inches. The pipe is open at both ends.
A wire screen in the shape of a square is attached at one end of the pipe to allow water to flow through but to keep people from wandering into the pipe. The length of the diagonals of the screen are equal to the diameter of the pipe. The figure represents the placement of the screen at the end of the pipe.
A 72B 201C 125
D 324E 648F 1,018
and the area of the screen is _________ square inches.
Select from each set of choices to correctly complete the sentence. The perimeter of the screen is approximately ___________ inches,
PARCC Sample Question - EOY
Topic: Area of a RectangleQuestion 13/25: Part B
Slide 146 / 152
60 The figure shows rectangle ABCD in the coordinate plane with point A at (0, 2.76), B at (3.87, 2.76), C at (3.87, 0) and D at the origin. Rectangle ABCD can be used to approximate the size of the state of Colorado with x and y scales representing hundreds of miles.
PARCC Sample Question - EOY
Topic: Area & Perimeter of a Figures in the Coordinate Plane
Question 15/25
Part ABased on the information given, how many miles is the perimeter of Colorado?
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61 The figure shows rectangle ABCD in the coordinate plane with point A at (0, 2.76), B at (3.87, 2.76), C at (3.87, 0) and D at the origin. Rectangle ABCD can be used to approximate the size of the state of Colorado with x and y scales representing hundreds of miles.
A 25 people per square mile
B 47 people per square mile
C 2,269 people per square mile
D 7,586 people per square mile
PARCC Sample Question - EOY
Topic: Area & Perimeter of a Figures in the Coordinate Plane
Question 15/25
Part BAt the end of 2010, the population of Colorado was 5,029,196 people. Based on the information given, what was the population density at the end of 2010?
Slide 148 / 152
62 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet.
Part AWhat is the perimeter of the plot of land. Express your answer to the nearest tenth of a foot.
Topic: Area & Perimeter of a Figures in the Coordinate PlaneQuestion 24/25
PARCC Sample Question - EOY
Slide 149 / 152
63 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet.
Part BWhat is the area of the plot of land that does not include the warehouse and the parking area?
Topic: Area & Perimeter of a Figures in the Coordinate Plane
Question 24/25
PARCC Sample Question - EOY
Slide 150 / 152
64 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet.
Part CLuke is planning to put a fence along two interior sides of the parking area. The sides are represented in the plan by the legs of the trapezoid. What is the total length of fence needed? Express your answer to the nearest tenth of a foot.
Topic: Area & Perimeter of a Figures in the Coordinate Plane
Question 24/25
PARCC Sample Question - EOY
Slide 151 / 152
65 Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet.
A 10 feetB 15 feet
C 18 feet
D 22 feet
E 25 feet
Part DIn the future, Luke has plans to construct a circular storage bin centered at coordinates (50, 40) on the plan. Which of the listed measurements could be the diameter of the bin that will fit on the plot and be at least 2 feet away from the warehouse?Select all that apply.
Topic: Area & Perimeter of a Figures in the Coordinate PlaneQuestion 24/25
PARCC Sample Question - EOY
Slide 152 / 152