Name Class Date 10.5Perimeter and Area on the
Coordinate Plane
Essential Question: H
ow do you find the perimeter and area of polygons in the coordinate
plane?
Resource
Resource
Locker
Locker
Explore Finding Perimeters of Figures on the Coordinate
Plane
Recall that the perimeter of a polygon is the sum of the lengths of the polygon’s sides. You can
use the Distance Formula to find perimeters of polygons in a coordinate plane.
Follow these steps to find the perimeter of a pentagon with vertices A​(−1, 4)​, B​(4, 4)​, C​(3, −2)​,
D​(−1, −4)​, and E​(−4, 1)​. Round to the nearest tenth.
A
Plot the points. Then use a straightedge to draw the pentagon that is
determined by the points.
4
y
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2
B
Are there any sides for which you do not need to use the Distance
Formula? Explain, and give their length(s). C
Use the Distance Formula to find the remaining side lengths. Round
your answers to the nearest tenth.
D
Find the sum of the side lengths.
x
-4
-2
0
2
4
-2
-4
Reflect
1.
Explain how you can find the perimeter of a rectangle to check that your answer is
reasonable.
Module 10
549
Lesson 5
Explain 1
Finding Areas of Figures on the Coordinate Plane
You can use area formulas together with the Distance Formula to determine areas of figures such
as triangles, rectangles and parallelograms.
triangle
1 bh
A=_
2
h
rectangle
A = bh
parallelogram
A = bh
h
h
b
rhombus
1d d
A=_
2 1 2
d2
d1
b
b
kite
1d d
A=_
2 1 2
trapezoid
1 b +b h
A=_
(
2)
2 1
d2
d1
b1
h
b2
Example 1
A
Find the area of each figure.
Step 1 Find the coordinates of the vertices of △ABC.
A(-4, -2), B(-2, 2), C(5, 1)
-4
2
-2
0
A
y
C
x
2
4
-2
y
-4
B
2
-2
0
A
D
C
x
2
4
Find the length of the base and the height.
AC =
――――――――――
―――――――――― 10
+ (1 - (-2)) = √―
90 = 3 √―
10 ; BD = √(-1 - (-2)) + (-1 - 2) = √―
√(5 - (-4))
2
Step 3 Determine the area of △ABC.
2
2
2
― ―
1 (AC)(BD) = _
1 ∙ (3 √10 )( √10 ) = _
1 bh = _
1 ∙ 30 = 15 square units
Area = _
2
2
2
2
Module 10
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Lesson 5
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Step 2 Choose a base for which you can easily find the height of the
triangle.
_
Use AC as the base. A segment from the opposite vertex, B,
to
_
point D(-1, -1) appears to be perpendicular to the base AC.
Use slopes to check.
_ 1 - (-2)
-1 - 2 = -3
1 ; slope of _
slope of AC = _ = _
BD = _
3
5 - (-4)
-1 - (-2)
_
1
_
The product of the slopes is ∙ (-3) = -1. BD is perpendicular
3
_
_
_
to AC, so BD is the height for the base AC.
B
B
D
Step 1 Find the coordinates of the vertices of DEFG.
D​(-2, 6)​, E​(4, 3)​, F​(2, -1)​, G​(-4, 2)​
6
y
4
E
2
G
x
Step 2 DEFG appears to be a rectangle. Use slopes to check that
adjacent sides are perpendicular.
_
 -   ​ = ​ _
 ​ 
 =
​   
  
slope of DE​
​  : _
6
4 - (​ -2)​
-2
0
F
4
_ _
- 3 ​ 
; slope of EF​
​  : ​ 
 
=_
​ 
 ​ 
 =
2-
_ _
slope of FG​
​  : ​  2  
 ​ 
=_
​ 
 ​ 
 
=
-4
_ __
; slope of DG​
​  : ​  2  
 ​ 
=_
​ 
 ​ 
 =
-
-4 so DEFG is a .
――――――――― _
――
√
√
b = FG = ​  √​​(   
 2 - )​ ​ ​ + (​​  - 2)​ ​ ​ ​ = ​  ​ = ​     ​ 
――――――――――― _
――

√
(
)
√
  - (​ -4)​)​​ ​ + ​​ 6 ​​ ​ ​ = ​
 ​ 
=
​   
 ​ 
h = GD = ​ ​​(   
――
――
(
)
(
√
√
Area of DEFG: A = bh = ​
​   
 ​ 
​​
​   
 ​) 
​=
square units
Step 3 Find the area of DEFG.
2
2
2
2
Reflect
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2.
In Part A, is it possible to use another side of △ABC as the base? If so, what length
represents the height of the triangle?
3.
Discussion In Part B, why was it necessary to find the slopes of the sides?
Module 10
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Lesson 5
Your Turn
Find the area of quadrilateral JKLM with vertices J(−4, −2), K(2, 1), L(3, 4),
M(−3, 1).
4.
2
-4
Explain 2
y
4
-2
0
2
-2
Finding Areas of Composite Figures
A composite figure is made up of simple shapes, such as triangles, rectangles, and parallelograms.
To find the area of a composite figure, find the areas of the simple shapes and then use the Area
Addition Postulate. You can use the Area Addition Postulate to find the area of a composite figure.
Area Addition Postulate
The area of a region is equal to the sum of the areas of its nonoverlapping parts.
A
Find the area of each figure.
Possible solution: ABCDE can be divided up into a rectangle and
two triangles, each with horizontal bases.
area of rectangle AGDE: A = bh = (DE)(AE) = (6)(4) = 24
1 bh = _
1 (AC)(BF) = _
1 (8)(2) = 8
area of △ABC: A = _
2
2
2
1 bh = _
1 (CG)(DG) = _
1 (2)(4) = 4
area of △CDG: A = _
2
2
2
area of ABCDE: A = 24 + 8 + 4 = 36 square units
Module 10
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4
A
-4
E
2
-2
0
-2
yB
F
G
2
C
x
4
D
Lesson 5
© Houghton Mifflin Harcourt Publishing Company
Example 2
y
BPQRST can be divided into a parallelogram and a triangle.
_
△PQT appears to be a right triangle. Check that PT​
​   
and
perpendicular:
P
are
Q
2
-4
-3 B
R
T
_ 1slope of PT​
​  
: ​  _
 
 ​ 
=_
​ 
 ​ 
 =
4
0
S
2
x
4
- 3 
 ​ 
 =
: ​  _
 ​ 
=_
​ 
slope of
-3 -
_
△PQT is a right triangle with base PT​
​   
and height
____
PT
=
√​ (​​ -3 - )​ ​ ​ + (​​ 1 - )​ ​ ​ ​ = √​
2
_
2
√(
) (
) = ​√
2
_
2
(√
_
)(
√​
 ​ 
 ​ ​=
_
, so QT​
​   
||
-1 
= ​ _
 ​ 
=_
​ 
 ​ 
=
parallelogram.
)
√​
_
area of △PQT: A = _
​  1 ​  bh = _
​  1 ​​  ​
 ​   ​​
2
2
_
_
​ QR​ 
|| TS​
​   since both sides are vertical.
slope of
_
=
 ​ 
​ ​​   
-3 ​ ​ ​ + ​​
-3 ​ ​ ​ ​
 ​ 
   
____
=
.
. Therefore, QRST is a
-3 - © Houghton Mifflin Harcourt Publishing Company
_
_ _
​ RT​ 
is an of △QRST and is horizontal. Because RT​
​   
⊥ RQ​
​  , △QRT is a right
triangle with base
RT =
, QR =
and height
area of PQRST: A =
( )​ (​ )​ = 6.
, so the area of △QRT = _
​  1 ​  ​
2
△QRST = 2 ∙ (​ area of QRT)​= 2 ∙
. Therefore, the area of △QRST = 2 ∙ (​ area of △QRT)​.
+
= 12
=
square units
Reflect
5.
Discussion How could you use subtraction to find the area of a figure on the coordinate plane?
Module 10
553
Lesson 5
Your Turn
6.
Find the area of the polygon by addition.
Find the area of polygon by subtraction.
y
4 E
D
W y
2
J
-4
7.
F
G
0
x
x
2
-4
4
-2
-4
I
-2
0
2
4
Z -2
X
-4
H
Y
Using Perimeter and Area in Problem Solving
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Brozova/Shutterstock
Explain 3
You can use perimeter and area techniques to solve problems.
Example 3
Miguel is planning and costing an ornamental garden in the a shape of an
irregular octagon. Each unit on the coordinate grid represents one yard.
He wants to lay the whole garden with turf, which costs $3.25 per square
yard, and surround it with a border of decorative stones, which cost
$7.95 per yard. What is the total cost of the turf and stones?
y
A
4
H
G
2
B
x
-4
-2
F
0
2
4
C
-2
E
-4
D
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Lesson 5
Analyze Information
Identify the important information.
(
) ( ), C(6, ), D(4, ), E(−1, −3),
), and H( , 2).
• The vertices are A
F
(
) (
, 5 , B 1,
, −3 , G −5,
per square yard.
• The cost of turf is $
• The cost of the ornamental stones is $
per yard.
Formulate a Plan
• Divide the garden up into
• Add up the
.
of the smaller figures.
• Find the cost of turf by
the total area by the cost per square yard.
• Find the perimeter of the garden by adding the
• Find the cost of the border by
of the sides.
the perimeter by the cost per yard.
• Find total cost by adding the
and
.
Solve
Divide the garden into smaller figures.
The garden can be divided into square BCDE, kite ABEH, and
parallelogram EFGH.
Find the area of each smaller figure.
4
G
H
0
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-2
_
slope of BC: _ =
6 F
4
E
-4
4-
Also, BC =
√(
) (
2
-1 + 0-
√(4 - ) + (
) =√
_
2
_____
2
-
) =√
(√
_
So BCDE is a square, with area A = s =
and
_
2
2
Module 10
2
D
-0
:_=
____
CD =
B
x
area of BCDE:
slope of
y
A
555
.
)
2
yd =
yd 2.
Lesson 5
C
area of kite ABEH:
√
√ ( ( )) ( ) √
――――――――――
――
(
)
√
√
)
√(   
 
   
――――――――――― ―――― ――
√(   
  ( )) ( ) √   √   
_____
2
__
2
- 2 ​ ​ ​ ​= ​ 4 +
HA = ​ ​​   
-1 - ​
​ ​ ​ +
​ ​​
- (​ -1)​ ​ ​ ​+ ​ 2 -
AB =​  ​​
__
2
2
​  ​ = ​
2
=​
 ​ 
≈
 ​ 
+ 9 ​ 
= ​ 
 ​ ≈
- 2 ​​ ​ ​ = ​ 
+ 25 ​ = ​ 
―――――――――― ―――― ――
=​  √ 
 ​ 
≈
BE = ​  √(​​   
  -1)​​ ​ + (​​ -3 - )​​ ​ ​ = ​  √4  +  ​ 
2
So,
≅
;
;
2
​ ​​ ​ + ​​
HE = ​  ​​ -1 - ​
_
 ​ ≈
;
2
≅
and
. Therefore ABEH is a kite.
b = ​d 1​ ​ = 8, h = ​d 2​ ​ = 4
( )​(​ )​ =
​  1 ​​ 
A=_
​  1 ​​ d ​1​​d ​2​ = _
2
2
​yd ​2​
area of parallelogram EFGH:
and GH​
​ ¯ 
are both horizontal, so are parallel;
_ 2slope of EH​
​  : ​  _
 
 ​ 
=_
​ 
 ​ 
 
=
-3 -
So EFGH is a parallelogram, with base
(
)(
area of EFGH: A = bh = ​
2: ​  __
 
 ​ 
=_
​ 
  
​ 
=
-
; slope of
)
=
and height. FH =
 yd ​ =
 yd ​ ​
.
​yd 2​​
Find the total area of the garden and the cost of turf.
area of garden: A =
) ( 
​yd 2​​ =
)
Find the perimeter of the garden.
EF = 2 yd, GH = 2 yd
_
√
From area calculations, BC = CD = DE = ​
) (
)
_____
√(
​yd ​2​
/​yd ​2​  ​ = $
​yd ​2​ ​ ​ $ 
cost of turf: ​
​yd 2​​ +
2
2
 ​ ≈
_
√
FG = ​ ​​   
- (​-3)​ ​ ​ ​  ​= ​
​ ​ +
​ ​​
yd, and AB = AH =
yd
 ​, 
perimeter of garden = GH + HA + AB + BC + CD + DE + EF + FG
+
=
+
Find the cost of the stones for the border.
(
cost of stones: ​
Find the total cost.
total cost: $
Module 10
)(
 yd  ​ ​ $
+$
+
+
+
+
+
=
yd
)
per yd ​ = $
=$
556
Lesson 5
© Houghton Mifflin Harcourt Publishing Company
(
​yd ​2​ +
Justify and Evaluate
a
The area can be checked by subtraction:
( )-( )
( ) ( )( )
- (5)(
) - _21 ( )(2) - _21 ( )(5)
1
-_
)( ) - ( )( ) - _21 ( )( )
2(
area = (11)
=
_
-
-
H
G
area of large rectangle = (11)(10) = 110 square units
1 (2)
(3) - _
2
1
-_
2
-
-
b
-
-
F
,
0
e
2
x
4
-2
-
-
f
=
y
A
H
B
2
x
-2
F
0
2
4
-2
E-4
D
Your Turn
A designer is making a medallion in the shape of the letter “L.” Each
unit on the coordinate grid represents an eighth of an inch, and the
medallion is to be cut from a 1-in. square of metal. How much metal
is wasted to make each medallion? Write your answer as a decimal.
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8.
4
557
y
P
Q
2
x
-4
-2
U
Module 10
C
D
G
so the answer is reasonable.
d
B
E
h -4 g
The perimeter is approximately the perimeter of the polygon shown:
The perimeter of the polygon shown is
4
y
c
2
-2
i
A
0
-2
R
2
4
S
T
Lesson 5
C
Elaborate 9.
y
B
Create a flowchart for the process of finding the area of the
polygon ABCDEFG. Your flowchart should show when, and
why, the Slope and Distance Formulas are used.
D
C
A
2
-4
F
-2 0
-2
E
2
x
4
-4
G
10. Discussion If two polygons have approximately the same area, do they have approximately the same
perimeter? Draw a picture to justify your answer.
11. Essential Question Check-In What formulas might you need to solve problems involving the perimeter
and area of triangles and quadrilaterals in the coordinate plane?
Module 10
558
Lesson 5
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Evaluate: Homework and Practice
• Online Homework
• Hints and Help
• Extra Practice
Find the perimeter of the figure with the given vertices.
Round to the nearest tenth.
1.
D(0, 1), E(5, 4), and F(2, 6)
2.
P(2, 5), Q(-3, 0), R(2, -5), and S(6, 0)
3.
M(-3, 4), N(1, 4), P(4, 2), Q(4, -1), and R(2, 2)
4.
A(-5, 1), B(0, 3), C(5, 1), D(4, -2), E(0, -4),
and F(-2, -4)
Find the area of each figure.
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5.
y
K
-4
6.
y
4
4
2
2
0
-2
J
S
M x
2
-4
4
-2
P
T
x
0
2
4
Q
L
-4
-4
R
Module 10
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Lesson 5
Find the area of each figure by addition.
y
7.
A
y
8.
B
4
4
N
2
F
C
-2
E
0
-2
2
D
M
2
L
x
-4
4
0
2
-4
K
-2
x
4
J
-4
Find the area of each figure by subtraction.
9.
Q
P
-4
0
W -2 V
2
4 R
Q
R
U
2
S
x
4
S
-4
Y
-4
x
U
P
T
T
0
© Houghton Mifflin Harcourt Publishing Company
2
y
10.
y
2
4
-2
V
X
W
Module 10
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Lesson 5
11. Fencing costs $1.45 per yard, and each unit on the grid represents 50 yd. How much
will it cost to fence the plot of land represented by the polygon ABCDEF?
y
A
B
2
F
-4
-2
C x
0
E
-2
2
D
12. A machine component has a geometric shaped plate, represented on the
coordinate grid. Each unit on the grid represents 1 cm. Each plate is punched from
an 8-cm square of alloy. The cost of the alloy is $0.43/c​m ​2​, but $0.28/c​m ​2​can be
recovered on wasted scraps of alloy. What is the net cost of alloy for each component?
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
©micheldenijs/iStockPhoto.com
4
y
2
x
-4
2
4
-2
13. △ABC with vertices A(1, 1) and B(3, 5) has an area of 10 u
​ nits ​​. What is the location
of the third vertex? Select all that apply.
2
A. C(−5, 5)
B. C(3, −5)
C. C(−2, 5)
D. C(6, 1)
E. C(3, −3)
Module 10
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Lesson 5
14. Pentagon ABCDE shows the path of an obstacle course, where each
unit of the coordinate plane represents 10 meters. Find the length of
the course to the nearest meter.
y
4
B
C
2
A
x
-4
-2
0
E
2
4
D
Algebra Graph each set of lines to form a triangle. Find the area and perimeter.
15. y = 2, x = 5, and y = x
5
16. y = −5, x = 2, and y = −2x + 7
y
4
2
3
x
1
x
1
-2
3
0
2
4
6
-2
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Steve
Skjold/Alamy
0
y
-4
-6
17. Prove that quadrilateral JKLM with vertices J(1, 5), K(4, 2), L(1, −4), and M(−2, 2) is a
kite, and find its area.
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Lesson 5
H.O.T. Focus on Higher Order Thinking
18. Explain the Error Wendell is trying to prove that ABCD is a rhombus and to find
its area. Identify and correct his error. (Hint: A rhombus is a quadrilateral with four
congruent sides.)
2
y B
2
  = 5,
AB = ​  ​​ 2 - (​ -2)​ ​​ ​ + (​​ 5 - 2)​​ ​ ​ = ​  √25 ​
―――――――― ― 
―――――――
   = 5
25 ​
BC = ​  √​​(  
 6 - 2)​​ ​ + (​​ 2 - 5)​​ ​ ​ = ​  √―
――――――――
   = 5,
25 ​
CD = ​  √​​(  
 2 - 6)​​ ​ + (​​ -1 - 2)​​ ​ ​ = ​  √―
――――――――――
   = 5
 2 - (​ -2)​ ​​ ​ + ​​ -1 - (​ 2)​ ​​ ​ ​ = ​  √―
25 ​
AD = ​  ​​   
 
)
√(   
2
2
√(
4
2
2
)
2
(
)
2
A
2
C
x
-2
_ _ _ _
≅ BC​
​   
≅ CD​
​   
≅ AD​
​  , and therefore ABCD is a rhombus.
So AB​
​   
D
4
area of ABCD: b = AB = 5 and h = BC = 5, so A = bh = (​ 5)​​(5)​ = 25
19. Communicate Mathematical Ideas Using the figure, prove that the area of a kite
is half the product of its diagonals. (Do not make numerical calculations.)
A
B
D
© Houghton Mifflin Harcourt Publishing Company
C
20. Justify Reasoning Use the Trapezoid Midsegment Theorem to show that the area
of a trapezoid is the product of its midsegment and its height.
b2
h
x
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563
b1
y
Lesson 5
Lesson Performance Task
The coordinate plane shows the floor plan of two
rooms in Fritz’s house. Because he enjoys paradoxes,
Fritz has decided to entertain his friends with one by
drawing lines on the floor of his tiled kitchen, on the
left, and his tiled recreation room, on the right. The
four sections in the kitchen are congruent to the four
sections in the recreation room. Each square on the
floor plan measures 1 yard on a side.
y
Kitchen
2
1
Rec Room
6
4
4
3
1.Find the area of each of the four sections
of the kitchen. Add the four areas to find
the total area of the kitchen.
-8
2.Find the area of the kitchen by finding the
product of the length and the width.
-6
1
4
2
x
-4
-2
0
-2
-4
2
4
6
8
3
2
-6
3.Find the area of the recreation room by
finding the product of the length and the
width.
4. Describe the paradox.
5. Explain the paradox.
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Module 10
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Lesson 5