1. Lesson 4: The Area ofObtuseTriangles Using Height and Base
Date: 5/12/15 53
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
Lesson 4: The Area of Obtuse Triangles Using Height and
Base
Student Outcomes
 Students constructthe altitudefor three different cases: an altitudethat is a sideof a rightangle, an altitude
that lies over the base,and an altitudethat is outsidethe triangle.
 Students deconstruct triangles to justify thatthe area of a triangleis exactly one half the area of a
parallelogram.
Lesson Notes
Students will need the attached templates, scissors,a ruler,and glue to complete the Exploratory Challenge.
Classwork
OpeningExercise (5 minutes)
Opening Exercise
Draw and label the height ofeach trianglebelow.
1.
2.
height
height

2. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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3.
Discussion(3 minutes)
 The lastfew lessons showed that the area formula for triangles is 𝐴 =
1
2
× 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡. Today we are going
to showthat the formula works for three different types of triangles.
 Examine the triangles in the Opening Exercise. Whatis different about them?
 The height or altitude is in a different location for each triangle. The first triangle has an altitude inside
the triangle. The second triangle has a side length that is the altitude, and the third triangle has an
altitude outside of the triangle.
 If we wanted to calculatethe area of these triangles,whatformula do you think we would use? Explain.
 We will use 𝐴 =
1
2
× 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 because that is the area formula we have used for both right
triangles and acute triangles.
Exploratory Challenge (22 minutes)
Students work in small groups to show that the area formula is the same for all three types of triangles shown in the
Opening Exercise. Each group will need the attached templates, scissors,a ruler,and glue. Each exercisecomes with
steps that might be useful to providefor students who work better with such scaffolds.
Exploratory Challenge
1. Use rectangle “x” and thetrianglewiththe altitude inside(triangle“x”)to show theareaformulafor the triangleis
𝑨 =
𝟏
𝟐
× 𝒃𝒂𝒔𝒆× 𝒉𝒆𝒊𝒈𝒉𝒕.
a. Step One: Find the areaofrectangle x.
𝑨 = 𝟑 in. × 𝟐.𝟓 in. = 𝟕.𝟓 in2
b. Step Two: What ishalfthe areaofrectangle x?
Halfofthearea oftherectangleis 𝟕.𝟓 in2 ÷ 𝟐 = 𝟑.𝟕𝟓in2.
height
MP.1

3. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
c. Step Three: Prove, by decomposing triangle x, that it isthesame ashalfofrectangle x. Please glue your
decomposed triangleonto aseparate sheetofpaper. Glueit nextto rectangle x.
Students will cut theirtriangleand glueit intohalfoftherectangle. This may takemorethanonetry, so
extra copies ofthetriangles may benecessary. Becausethetrianglefits insidehalfoftherectangle, weknow
thetriangle’s area is halfoftherectangle’s area.
2. Use rectangle “y” and thetrianglewithaside that isthe altitude(triangle“y”)to show theareaformulafor the
triangle is 𝑨 =
𝟏
𝟐
× 𝒃𝒂𝒔𝒆× 𝒉𝒆𝒊𝒈𝒉𝒕.
a. Step One: Find the areaofrectangle y.
𝑨 = 𝟑in. × 𝟑in. = 𝟗in2
b. Step Two: What ishalfthe areaofrectangle y?
Halfthearea oftherectangleis 𝟗 in2 ÷ 𝟐 = 𝟒.𝟓 in2.
c. Step Three: Prove, by decomposing triangle y, that it isthe same ashalfofrectangle y. Please glueyour
decomposed triangleonto aseparate sheetofpaper. Glueit nextto rectangle y.
Students will againcut triangle“y” and glueit intotherectangle. This may takemorethanonetry, so extra
copies ofthetriangles may benecessary. Students will seethat theright trianglealsofits in exactly halfof
therectangle, so thetriangle’s area is onceagain halfthesizeoftherectangle’s area.
3. Use rectangle “z” and the trianglewith thealtitudeoutside (triangle“z”)to show the area
formulafor the triangle is 𝑨 =
𝟏
𝟐
× 𝒃𝒂𝒔𝒆× 𝒉𝒆𝒊𝒈𝒉𝒕.
a. Step One: Find the areaofrectangle z.
𝑨 = 𝟑in. × 𝟐.𝟓 in. = 𝟕.𝟓 in2
b. Step Two: What ishalfthe areaofrectangle z?
Halfofthearea oftherectangleis 𝟕.𝟓 in2 ÷ 𝟐 = 𝟑.𝟕𝟓in2.
c. Step Three: Prove, by decomposing triangle z, that it isthe same ashalfofrectangle
z. Please glue your decomposed triangleonto aseparatesheetofpaper. Glueit next
to rectangle z.
Students will cut theirtriangleand glueit ontotherectangleto show that obtusetriangles also havean area
that is halfthesizeofa rectanglethat has thesamedimensions. This may takemorethanonetry, so extra
copies ofthetriangles may benecessary.
NOTE: In order for students to fit an obtusetriangleinto halfofa rectangle, they willneed to cutthetriangle
into threeseparate triangles.
4. When finding the areaofatriangle, doesit matter wherethe altitude islocated?
It does not matter wherethealtitudeis located. To find thearea ofa triangletheformulais always
𝑨 =
𝟏
𝟐
× 𝒃𝒂𝒔𝒆 × 𝒉𝒆𝒊𝒈𝒉𝒕.
5. How can you determinewhich part ofthe triangleisthe base and theheight?
The baseand theheight ofany triangleform a rightanglebecausethealtitudeis perpendicular to thebase.
MP.1
Scaffolding:
 Students may struggle
with this step sincethey
have yet to see an obtuse
angle. The teacher may
want to model this step if
he or she feels students
may become confused.
 After modeling, the
students can then try this
step on their own.

4. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
Take time to show how other groups may have calculated the area of the triangleusinga different sidefor the base and
how this still results in thesame area.
After discussinghowany sideof a trianglecan be labeled the base, have students write a summary to explain the
outcomes of the Exercise.
Exercises(5 minutes)
Calculate the areaofeach triangle. Figuresare not drawn to scale.
6.
𝑨 =
𝟏
𝟐
( 𝟐𝟒 𝒊𝒏. )( 𝟖 𝒊𝒏. ) = 𝟗𝟔 𝒊𝒏 𝟐
7.
𝑨 =
𝟏
𝟐
( 𝟏𝟐
𝟑
𝟒
𝒇𝒕. )( 𝟗
𝟏
𝟐
𝒇𝒕. ) =
𝟏
𝟐
(
𝟓𝟏
𝟒
𝒇𝒕. )(
𝟏𝟗
𝟐
𝒇𝒕. ) =
𝟗𝟔𝟗
𝟏𝟔
𝒇𝒕 𝟐
= 𝟔𝟎
𝟗
𝟏𝟔
𝒇𝒕 𝟐
8. Draw three triangles(acute, right, and obtuse)that have thesame area. Explain how you know they have thesame
area.
Answers will vary.
Closing(5 minutes)
 Have different groups sharetheir Exploratory Challengeand discuss theoutcomes.
 Why does the area formula for a trianglework for every triangle?
 Every type of triangle fits inside exactly half of a rectangle that has the same base and height lengths.
Exit Ticket (5 minutes)
𝟒𝟐in.
𝟖in.
𝟐𝟒in. 𝟔in.
𝟏𝟎in.
𝟗
𝟏
𝟐
ft. 𝟏𝟒
𝟏
𝟖
ft.
𝟑𝟒
𝟓
𝟔
ft.
𝟏𝟐
𝟑
𝟒
ft.

5. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
Name Date
Lesson 4: The Area of Obtuse Triangles Using Height and Base
Exit Ticket
Find the area of each triangle. Figures are not drawn to scale.
1.
2.
3.
12.6 cm
21 cm
16.8 cm
17 in.
8 in.
25 in.
15 in.
20 in.
8 ft.
29 ft.
21 ft.
12 ft.

6. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
𝟖ft.
𝟐𝟗ft.
𝟐𝟏ft.
𝟏𝟐ft.
𝟏𝟕in.
𝟖in.
𝟐𝟓in.
𝟏𝟓in.
𝟐𝟎in.
𝟏𝟐.𝟔cm
𝟐𝟏cm
𝟏𝟔.𝟖
Exit Ticket Sample Solutions
Find the areaofeach triangle. Figuresare notdrawn to scale.
1.
𝑨 =
𝟏
𝟐
( 𝟏𝟐.𝟔 𝒄𝒎)( 𝟏𝟔.𝟖 𝒄𝒎) = 𝟏𝟎𝟓.𝟖𝟒 𝒄𝒎 𝟐
2.
𝑨 =
𝟏
𝟐
( 𝟐𝟖 𝒊𝒏. )( 𝟏𝟓 𝒊𝒏. )= 𝟐𝟏𝟎 𝒊𝒏 𝟐
3.
𝑨 =
𝟏
𝟐
( 𝟏𝟐 𝒇𝒕. )( 𝟐𝟏 𝒇𝒕. )= 𝟏𝟐𝟔 𝒇𝒕𝟐
Problem Set Sample Solutions
Calculate the areaofeach trianglebelow. Figuresare not drawn to scale.
1.
𝑨 =
𝟏
𝟐
( 𝟐𝟏 𝒊𝒏. )( 𝟖 𝒊𝒏. ) = 𝟖𝟒 𝒊𝒏 𝟐
𝟏𝟎in.𝟖in.
𝟏𝟕in.
𝟏𝟓in. 𝟔in.

7. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
2.
𝑨 =
𝟏
𝟐
( 𝟕𝟐 𝒎)( 𝟐𝟏 𝒎)= 𝟕𝟓𝟔 𝒎 𝟐
3.
𝑨 =
𝟏
𝟐
( 𝟕𝟓.𝟖 𝒌𝒎)( 𝟐𝟗.𝟐 𝒌𝒎)= 𝟏𝟏𝟎𝟔.𝟔𝟖 𝒌𝒎 𝟐
4.
𝑨 =
𝟏
𝟐
( 𝟓 𝒎)( 𝟏𝟐 𝒎) = 𝟑𝟎 𝒎 𝟐
𝑨 =
𝟏
𝟐
( 𝟕 𝒎)( 𝟐𝟗 𝒎) = 𝟏𝟎𝟏.𝟓 𝒎 𝟐
𝑨 = 𝟏𝟐 𝒎( 𝟏𝟗 𝒎)= 𝟐𝟐𝟖 𝒎 𝟐
𝑨 = 𝟑𝟎 𝒎 𝟐
+ 𝟑𝟎 𝒎 𝟐
+ 𝟏𝟎𝟏.𝟓 𝒎 𝟐
+ 𝟐𝟐𝟖 𝒎 𝟐
𝑨 = 𝟑𝟖𝟗.𝟓 𝒎 𝟐
𝟐𝟏m
𝟕𝟐m
𝟕𝟓m
𝟐𝟏.𝟗km
𝟏𝟎𝟎.𝟓km
𝟕𝟓.𝟖km
𝟐𝟗.𝟐km

8. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
5. The Andersonswere going on along sailing tripduring the summer. However, oneofthe sailson their sailboat
ripped, and they have to replace it. The sail ispictured below.
If the sailboat sailsare on sale for $𝟐asquare foot, how much will thenew sail cost?
𝑨 =
𝟏
𝟐
𝒃𝒉
=
𝟏
𝟐
( 𝟖𝒇𝒕)( 𝟏𝟐𝒇𝒕)
= 𝟒𝟖 𝒇𝒕 𝟐
$𝟐 × 𝟒𝟖 𝒇𝒕 𝟐
= $𝟗𝟔
6. Darnell and Donovan are both trying to calculatetheareaofan obtuse triangle. Examinetheir calculationsbelow.
Darnell’sWork Donovan’sWork
𝑨 =
𝟏
𝟐
× 𝟑 in. × 𝟒in.
𝑨 = 𝟔 in2
𝑨 =
𝟏
𝟐
× 𝟏𝟐in. × 𝟒in.
𝑨 = 𝟐𝟒in2
Which student calculatedthe areacorrectly? Explain why theother student isnot correct.
Donovan calculated theareacorrectly. Although Darnell did usethealtitudeofthetriangle, heused thelength
between thealtitudeand thebase, rather than theactual base.
7. Russell calculated the areaofthe trianglebelow. Hiswork isshown.
𝑨 =
𝟏
𝟐
× 𝟒𝟑cm × 𝟕 cm
𝑨 = 𝟏𝟓𝟎.𝟓 cm2
Although Russell wastold hiswork iscorrect, hehad ahard time explaining why it iscorrect. Help Russellexplain
why hiscalculationsare correct.
The formula for thearea ofthea triangleis 𝑨 =
𝟏
𝟐
𝒃𝒉. Russell followed this formulabecause 𝟕cm is theheight of
thetriangle, and 𝟒𝟑cm is thebase ofthetriangle.
𝟐𝟓cm
𝟕cm
𝟐𝟒cm
𝟐𝟓cm
𝟒𝟑cm

9. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4
8. The larger triangle below hasabase of 𝟏𝟎.𝟏𝟒m; the gray triangle hasan areaof 𝟒𝟎.𝟑𝟐𝟓m2.
a. Determinethe areaofthe larger triangleifit hasaheight of 𝟏𝟐.𝟐 m.
𝑨 =
𝟏
𝟐
( 𝟏𝟎.𝟏𝟒 𝒎)( 𝟏𝟐.𝟐 𝒎)
= 𝟔𝟏.𝟖𝟓𝟒 𝒎 𝟐
b. Let 𝑨 be the areaofthe unshaded (white)triangle insquare meters. Writeand solve an equation to
determinethe valueof 𝑨, using the areasofthe larger triangle and thegray triangle.
𝟒𝟎.𝟑𝟐𝟓 𝒎 𝟐
+ 𝑨 = 𝟔𝟏.𝟖𝟓𝟒 𝒎 𝟐
𝟒𝟎.𝟑𝟐𝟓 𝒎 𝟐
+ 𝑨 − 𝟒𝟎.𝟑𝟐𝟓 𝒎 𝟐
= 𝟔𝟏.𝟖𝟔𝟓 𝒎 𝟐
− 𝟒𝟎.𝟑𝟐𝟓 𝒎 𝟐
𝑨 = 𝟐𝟏.𝟓𝟐𝟗 𝒎 𝟐

10. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4

11. Lesson 4: The Area ofObtuseTriangles Using Height and Base
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NYS COMMON CORE MATHEMATICS CURRICULUM 6•5Lesson 4