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M2
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
GEOMETRY
Lesson 31: Using Trigonometry to Determine Area
Student Outcomes
๏‚ง
Students prove that the area of a triangle is one-half times the product of two side lengths times the sine of
the included angle and solve problems using this formula.
๏‚ง
Students find the area of an isosceles triangle given the base length and the measure of one angle.
Lesson Notes
Students discover how trigonometric ratios can help with area calculations in cases where the measurement of the
height is not provided. In order to determine the height in these cases, students must draw an altitude to create right
triangles within the larger triangle. With the creation of the right triangles, students can then set up the necessary
trigonometric ratios to express the height of the triangle (G-SRT.C.8). Students carefully connect the meanings of
1
2
formulas to the diagrams they represent (MP.2 and 7). In addition, this lesson introduces the formula Area = ๐‘Ž๐‘ sin ๐ถ
as described by G-SRT.D.9.
Classwork
Opening Exercise (5 minutes)
Opening Exercise
Three triangles are presented below. Determine the areas for each triangle, if possible. If it is not possible to find the
area with the provided information, describe what is needed in order to determine the area.
The area of โ–ณ ๐‘จ๐‘ฉ๐‘ช is
๐Ÿ
๐Ÿ
๐Ÿ
(๐Ÿ“)(๐Ÿ๐Ÿ), or ๐Ÿ‘๐ŸŽ square units, and the area of โ–ณ ๐‘ซ๐‘ฌ๐‘ญ is (๐Ÿ–)(๐Ÿ๐ŸŽ), or ๐Ÿ–๐ŸŽ square units.
๐Ÿ
There is
not enough information to find the height of โ–ณ ๐‘ฎ๐‘ฏ๐‘ฐ and, therefore, the area of the triangle.
Is there a way to find the missing information?
Without further information, there is no way to calculate the area.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
466
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Example 1 (13 minutes)
๏‚ง
What if the third side length of the triangle were provided? Is it possible to determine the area of the triangle
now?
Example 1
Find the area of โ–ณ ๐‘ฎ๐‘ฏ๐‘ฐ.
Allow students the opportunity and the time to determine what they must find (the height) and how to locate it (one
MP.1 option is to drop an altitude from vertex ๐ป to side ฬ…ฬ…ฬ…
๐บ๐ผ ). For students who are struggling, consider showing just the
3 altitude and allowing them to label the newly divided segment lengths and the height.
๏‚ง
How can the height be calculated?
๏ƒบ
By applying the Pythagorean theorem to both of the created right triangles to find ๐‘ฅ,
โ„Ž2 = 49 โˆ’ ๐‘ฅ 2
โ„Ž2 = 144 โˆ’ (15 โˆ’ ๐‘ฅ)2
49 โˆ’ ๐‘ฅ 2 = 144 โˆ’ (15 โˆ’ ๐‘ฅ)2
49 โˆ’ ๐‘ฅ 2 = 144 โˆ’ 225 + 30๐‘ฅ โˆ’ ๐‘ฅ 2
130 = 30๐‘ฅ
13
๐‘ฅ=
3
13
32
๐ป๐ฝ =
, ๐ผ๐ฝ =
3
3
๏ƒบ
The value of ๐‘ฅ can then be substituted into either of the expressions equivalent to โ„Ž2 to find โ„Ž.
13 2
โ„Ž2 = 49 โˆ’ ( )
3
169
โ„Ž2 = 49 โˆ’
9
โ„Ž=
Lesson 31:
4โˆš17
3
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
467
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
๏‚ง
What is the area of the triangle?
1
4โˆš17
Area = ( ) (15) (
)
2
3
Area = 10โˆš17
Discussion (10 minutes)
Scaffolding:
๏‚ง For students who are
ready for a challenge, pose
the unstructured question,
โ€œHow could you write a
formula for area that uses
trigonometry, if we didnโ€™t
know โ„Ž?โ€
Now, consider โ–ณ ๐ด๐ต๐ถ, which is set up similarly to the triangle in Example 1:
๏‚ง For students who are
struggling, use a numerical
example, such as the
6โ€“8โ€“10 triangle in Lesson
24, to demonstrate how
โ„Ž can be found using sin ๐œƒ.
๐‘Ž
๏‚ง
Write an equation that describes the area of this triangle.
๏ƒบ
1
2
Area = ๐‘Žโ„Ž
Write the left-hand column on the board, and elicit the anticipated student response on the right-hand side after writing
the second line; then, elicit the anticipated student response after writing the third line.
๏‚ง
We will rewrite this equation. Describe what you see.
1
๐‘Žโ„Ž
2
1
๐‘
Area = ๐‘Žโ„Ž ( )
2
๐‘
1
โ„Ž
Area = ๐‘Ž๐‘ ( )
2
๐‘
MP.2
&
MP.7
Area =
The statement is multiplied by 1.
The last statement is rearranged, but the value remains the
same.
Create a discussion around the third line of the newly written formula. Modify โ–ณ ๐ด๐ต๐ถ: Add in an arc mark at vertex ๐ถ,
and label it ๐œƒ.
๏‚ง
โ„Ž
What do you notice about the structure of ? Can we think of this newly written area formula in a different
๐‘
way using trigonometry?
๏ƒบ
The value of
โ„Ž
๐‘
is equivalent to sin ๐œƒ; the
newly written formula can be written as
1
2
Area = ๐‘Ž๐‘ sin ๐œƒ.
๏‚ง
1
If the area can be expressed as ๐‘Ž๐‘ sin ๐œƒ, which
2
part of the expression represents the height?
๏ƒบ
โ„Ž = ๐‘ sin ๐œƒ
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
468
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 31
M2
GEOMETRY
๏‚ง
MP.2
&
MP.7
Compare the information provided to find the area of โ–ณ ๐บ๐ป๐ผ in the Opening Exercise, how the information
changed in Example 1, and then changed again in the triangle above, โ–ณ ABC.
๏ƒบ
๏‚ง
Only two side lengths were provided in the Opening Exercise, and the area could not be found because
there was not enough information to determine the height. In Example 1, all three side lengths were
provided, and then the area could be determined because the height could be determined by applying
the Pythagorean theorem in two layers. In the most recent figure, we only needed two sides and the
included angle to find the area.
If you had to determine the area of a triangle and were given the option to have three side lengths of a triangle
or two side lengths and the included measure, which set of measurements would you rather have? Why?
The response to this question is a matter of opinion. Considering the amount of work needed to find the area when
provided three side lengths, students should opt for the briefer, trigonometric solution!
Example 2 (5 minutes)
Example 2
A farmer is planning how to divide his land for planting next yearโ€™s crops. A triangular plot of land is left with two known
side lengths measuring ๐Ÿ“๐ŸŽ๐ŸŽ ๐ฆ and ๐Ÿ, ๐Ÿ•๐ŸŽ๐ŸŽ ๐ฆ.
What could the farmer do next in order to find the area of the plot?
๏‚ง
With just two side lengths known of the plot of land, what are the farmerโ€™s options to determine the area of
his plot of land?
๏ƒบ
๏‚ง
He can either measure the third side length, apply the Pythagorean theorem to find the height of the
triangle, and then calculate the area, or he can find the measure of the included angle between the
known side lengths and use trigonometry to express the height of the triangle and then determine the
area of the triangle.
Suppose the included angle measure between the known side lengths is 30°. What is the area of the plot of
land? Sketch a diagram of the plot of land.
๏ƒบ
1
2
Area = (1700)(500) sin 30
Area = 212โ€†500
The area of the plot of land is
212,500 square meters.
Exercise 1 (5 minutes)
Exercise 1
A real estate developer and her surveyor are searching for their next piece of land to build on. They each examine a plot
of land in the shape of โ–ณ ๐‘จ๐‘ฉ๐‘ช. The real estate developer measures the length of ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ฉ and ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ช and finds them to both be
approximately ๐Ÿ’, ๐ŸŽ๐ŸŽ๐ŸŽ feet, and the included angle has a measure of approximately ๐Ÿ“๐ŸŽ°. The surveyor measures the
length of ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ช and ฬ…ฬ…ฬ…ฬ…
๐‘ฉ๐‘ช and finds the lengths to be approximately ๐Ÿ’, ๐ŸŽ๐ŸŽ๐ŸŽ feet and ๐Ÿ‘, ๐Ÿ’๐ŸŽ๐ŸŽ feet, respectively, and measures
the angle between the two sides to be approximately ๐Ÿ”๐Ÿ“°.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
469
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
a.
Draw a diagram that models the situation, labeling all lengths and angle measures.
b.
The real estate developer and surveyor each calculate the area of the plot of land and both find roughly the
same area. Show how each person calculated the area; round to the nearest hundred. Redraw the diagram
with only the relevant labels for both the real estate agent and surveyor.
๐Ÿ
(๐Ÿ’๐ŸŽ๐ŸŽ๐ŸŽ)(๐Ÿ’๐ŸŽ๐ŸŽ๐ŸŽ) ๐ฌ๐ข๐ง ๐Ÿ“๐ŸŽ
๐Ÿ
๐‘จ โ‰ˆ ๐Ÿ”โ€†๐Ÿ๐Ÿ๐Ÿ–โ€†๐Ÿ‘๐Ÿ“๐Ÿ”
๐Ÿ
(๐Ÿ‘๐Ÿ’๐ŸŽ๐ŸŽ)(๐Ÿ’๐ŸŽ๐ŸŽ๐ŸŽ) ๐ฌ๐ข๐ง ๐Ÿ”๐Ÿ“
๐Ÿ
๐‘จ โ‰ˆ ๐Ÿ”โ€†๐Ÿ๐Ÿ”๐Ÿโ€†๐Ÿ–๐Ÿ—๐Ÿ‘
๐‘จ=
๐‘จ=
The area is approximately ๐Ÿ”, ๐Ÿ๐Ÿ๐Ÿ–, ๐Ÿ‘๐Ÿ“๐Ÿ” square feet.
c.
The area is approximately ๐Ÿ”, ๐Ÿ๐Ÿ”๐Ÿ, ๐Ÿ–๐Ÿ—๐Ÿ‘ square feet.
What could possibly explain the difference between the real estate agentโ€™s and surveyorโ€™s calculated areas?
The difference in the area of measurements can be accounted for by the approximations of the measurements
taken instead of exact measurements.
Closing (2 minutes)
Ask students to summarize the key points of the lesson. Additionally, consider asking students to respond to the
following questions independently in writing, to a partner, or to the whole class.
๏‚ง
For a triangle with side lengths ๐‘Ž and ๐‘ and included angle of measure ๐œƒ, when will we need to use the area
1
2
formula Area = ๐‘Ž๐‘ sin ๐œƒ ?
๏ƒบ
๏‚ง
We will need it when we are determining the area of a triangle and are not provided a height.
1
2
1
2
Recall how to transform the equation Area = ๐‘โ„Ž to Area = ๐‘Ž๐‘ sin ๐œƒ.
Exit Ticket (5 minutes)
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
470
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Name
Date
Lesson 31: Using Trigonometry to Determine Area
Exit Ticket
1.
Given two sides of the triangle shown, having lengths of 3 and 7 and their included angle of 49°, find the area of the
triangle to the nearest tenth.
2.
In isosceles triangle ๐‘ƒ๐‘„๐‘…, the base ๐‘„๐‘… = 11, and the base angles have measures of 71.45°. Find the area of โ–ณ ๐‘ƒ๐‘„๐‘…
to the nearest tenth.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
471
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Exit Ticket Sample Solutions
1.
Given two sides of the triangle shown, having lengths of ๐Ÿ‘ and ๐Ÿ• and their included angle of ๐Ÿ’๐Ÿ—°, find the area of the
triangle to the nearest tenth.
๐€๐ซ๐ž๐š =
๐Ÿ
(๐Ÿ‘)(๐Ÿ•)(๐ฌ๐ข๐ง ๐Ÿ’๐Ÿ—)
๐Ÿ
๐€๐ซ๐ž๐š = ๐Ÿ๐ŸŽ. ๐Ÿ“(๐ฌ๐ข๐ง ๐Ÿ’๐Ÿ—) โ‰ˆ ๐Ÿ•. ๐Ÿ—
The area of the triangle is approximately ๐Ÿ•. ๐Ÿ— square units.
1.
In isosceles triangle ๐‘ท๐‘ธ๐‘น, the base ๐‘ธ๐‘น = ๐Ÿ๐Ÿ, and the base angles have measures of ๐Ÿ•๐Ÿ. ๐Ÿ’๐Ÿ“°. Find the area of
โ–ณ ๐‘ท๐‘ธ๐‘น to the nearest tenth.
ฬ…ฬ…ฬ…ฬ… cuts
Drawing an altitude from ๐‘ท to midpoint ๐‘ด on ๐‘ธ๐‘น
the isosceles triangle into two right triangles with
๐‘ธ๐‘ด = ๐‘ด๐‘น = ๐Ÿ“. ๐Ÿ“. Using tangent, solve the following:
๐ญ๐š๐ง ๐Ÿ•๐Ÿ. ๐Ÿ’๐Ÿ“ =
๐‘ท๐‘ด
๐Ÿ“. ๐Ÿ“
๐‘ท๐‘ด = ๐Ÿ“. ๐Ÿ“(๐ญ๐š๐ง ๐Ÿ•๐Ÿ. ๐Ÿ’๐Ÿ“)
๐Ÿ
๐’ƒ๐’‰
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š = (๐Ÿ๐Ÿ)(๐Ÿ“. ๐Ÿ“(๐ญ๐š๐ง ๐Ÿ•๐Ÿ. ๐Ÿ’๐Ÿ“))
๐Ÿ
๐€๐ซ๐ž๐š =
๐€๐ซ๐ž๐š = ๐Ÿ‘๐ŸŽ. ๐Ÿ๐Ÿ“(๐ญ๐š๐ง ๐Ÿ•๐Ÿ. ๐Ÿ’๐Ÿ“) โ‰ˆ ๐Ÿ—๐ŸŽ. ๐Ÿ
The area of the isosceles triangle is approximately ๐Ÿ—๐ŸŽ. ๐Ÿ square units.
Problem Set Sample Solutions
Find the area of each triangle. Round each answer to the nearest tenth.
1.
๐€๐ซ๐ž๐š =
๐Ÿ
(๐Ÿ๐Ÿ)(๐Ÿ—) (๐ฌ๐ข๐ง ๐Ÿ๐Ÿ)
๐Ÿ
๐€๐ซ๐ž๐š = ๐Ÿ“๐Ÿ’(๐ฌ๐ข๐ง ๐Ÿ๐Ÿ) โ‰ˆ ๐Ÿ๐Ÿ—. ๐Ÿ’
The area of the triangle is approximately ๐Ÿ๐Ÿ—. ๐Ÿ’ square
units.
2.
๐€๐ซ๐ž๐š =
๐Ÿ
(๐Ÿ)(๐Ÿ๐Ÿ)(๐ฌ๐ข๐ง ๐Ÿ‘๐Ÿ’)
๐Ÿ
๐€๐ซ๐ž๐š = ๐Ÿ๐Ÿ(๐ฌ๐ข๐ง ๐Ÿ‘๐Ÿ’) โ‰ˆ ๐Ÿ”. ๐Ÿ
The area of the triangle is approximately ๐Ÿ”. ๐Ÿ square units.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
472
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
3.
๐€๐ซ๐ž๐š =
๐Ÿ
๐Ÿ
(๐Ÿ–) (๐Ÿ” ) (๐ฌ๐ข๐ง ๐Ÿ“๐Ÿ“)
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š = ๐Ÿ๐Ÿ”(๐ฌ๐ข๐ง ๐Ÿ“๐Ÿ“) โ‰ˆ ๐Ÿ๐Ÿ. ๐Ÿ‘
The area of the triangle is approximately ๐Ÿ๐Ÿ. ๐Ÿ‘ square units.
4.
The included angle is ๐Ÿ”๐ŸŽ° by the angle sum of a triangle.
๐€๐ซ๐ž๐š =
๐Ÿ
(๐Ÿ๐Ÿ)(๐Ÿ” + ๐Ÿ”โˆš๐Ÿ‘) ๐ฌ๐ข๐ง ๐Ÿ”๐ŸŽ
๐Ÿ
๐€๐ซ๐ž๐š = ๐Ÿ”(๐Ÿ” + ๐Ÿ”โˆš๐Ÿ‘) (
โˆš๐Ÿ‘
)
๐Ÿ
๐€๐ซ๐ž๐š = (๐Ÿ‘๐Ÿ” + ๐Ÿ‘๐Ÿ”โˆš๐Ÿ‘) (
โˆš๐Ÿ‘
)
๐Ÿ
๐€๐ซ๐ž๐š = ๐Ÿ๐Ÿ–โˆš๐Ÿ‘ + ๐Ÿ๐Ÿ–(๐Ÿ‘)
๐€๐ซ๐ž๐š = ๐Ÿ๐Ÿ–โˆš๐Ÿ‘ + ๐Ÿ“๐Ÿ’ โ‰ˆ ๐Ÿ–๐Ÿ“. ๐Ÿ
The area of the triangle is approximately ๐Ÿ–๐Ÿ“. ๐Ÿ square units.
5.
In โ–ณ ๐‘ซ๐‘ฌ๐‘ญ, ๐‘ฌ๐‘ญ = ๐Ÿ๐Ÿ“, ๐‘ซ๐‘ญ = ๐Ÿ๐ŸŽ, and ๐’Žโˆ ๐‘ญ = ๐Ÿ”๐Ÿ‘°. Determine the area of the triangle. Round to the nearest tenth.
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š = (๐Ÿ๐ŸŽ)(๐Ÿ๐Ÿ“)๐ฌ๐ข๐ง(๐Ÿ”๐Ÿ‘) โ‰ˆ ๐Ÿ๐Ÿ‘๐Ÿ‘. ๐Ÿ•
The area of โ–ณ ๐‘ซ๐‘ฌ๐‘ญ is ๐Ÿ๐Ÿ‘๐Ÿ‘. ๐Ÿ• units2.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
473
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
6.
A landscape designer is designing a flower garden for a triangular area that is bounded on two sides by the clientโ€™s
house and driveway. The length of the edges of the garden along the house and driveway are ๐Ÿ๐Ÿ– ๐Ÿ๐ญ. and ๐Ÿ– ๐Ÿ๐ญ.,
respectively, and the edges come together at an angle of ๐Ÿ–๐ŸŽ°. Draw a diagram, and then find the area of the garden
to the nearest square foot.
The garden is in the shape of a triangle in which the lengths of two sides and the included angle have been provided.
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช) =
๐Ÿ
(๐Ÿ– ๐Ÿ๐ญ. )(๐Ÿ๐Ÿ– ๐Ÿ๐ญ. ) ๐ฌ๐ข๐ง ๐Ÿ–๐ŸŽ
๐Ÿ
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช) = (๐Ÿ•๐Ÿ ๐ฌ๐ข๐ง ๐Ÿ–๐ŸŽ) ๐Ÿ๐ญ๐Ÿ
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช) โ‰ˆ ๐Ÿ•๐Ÿ ๐Ÿ๐ญ ๐Ÿ
7.
A right rectangular pyramid has a square base with sides of length ๐Ÿ“. Each lateral face of the pyramid is an isosceles
triangle. The angle on each lateral face between the base of the triangle and the adjacent edge is ๐Ÿ•๐Ÿ“°. Find the
surface area of the pyramid to the nearest tenth.
Using tangent, the altitude of the triangle to the base of length ๐Ÿ“ is equal to ๐Ÿ. ๐Ÿ“ ๐ญ๐š๐ง ๐Ÿ•๐Ÿ“.
๐Ÿ
๐’ƒ๐’‰
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š = (๐Ÿ“)(๐Ÿ. ๐Ÿ“ ๐ฌ๐ข๐ง ๐Ÿ•๐Ÿ“)
๐Ÿ
๐€๐ซ๐ž๐š =
๐€๐ซ๐ž๐š = ๐Ÿ”. ๐Ÿ๐Ÿ“(๐ฌ๐ข๐ง ๐Ÿ•๐Ÿ“)
The total surface area of the pyramid is the sum of the four lateral faces and the area of the
square base:
๐’๐€ = ๐Ÿ’(๐Ÿ”. ๐Ÿ๐Ÿ“(๐ฌ๐ข๐ง ๐Ÿ•๐Ÿ“)) + ๐Ÿ“๐Ÿ
๐’๐€ = ๐Ÿ๐Ÿ“ ๐ฌ๐ข๐ง ๐Ÿ•๐Ÿ“ + ๐Ÿ๐Ÿ“
๐’๐€ โ‰ˆ ๐Ÿ’๐Ÿ—. ๐Ÿ
The surface area of the right rectangular pyramid is approximately ๐Ÿ’๐Ÿ—. ๐Ÿ square units.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
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Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
8.
The Pentagon building in Washington, DC, is built in the shape of a regular pentagon. Each side of the pentagon
measures ๐Ÿ—๐Ÿ๐Ÿ ๐Ÿ๐ญ. in length. The building has a pentagonal courtyard with the same center. Each wall of the center
courtyard has a length of ๐Ÿ‘๐Ÿ“๐Ÿ” ๐Ÿ๐ญ. What is the approximate area of the roof of the Pentagon building?
Let ๐‘จ๐Ÿ represent the area within the outer perimeter of the
Pentagon building in square feet.
๐‘จ๐Ÿ =
๐’๐’ƒ๐Ÿ
๐Ÿ๐Ÿ–๐ŸŽ
๐Ÿ’ ๐ญ๐š๐ง (
)
๐’
๐‘จ๐Ÿ =
๐Ÿ“ โ‹… (๐Ÿ—๐Ÿ๐Ÿ)๐Ÿ
๐Ÿ๐Ÿ–๐ŸŽ
๐Ÿ’ ๐ญ๐š๐ง (
)
๐Ÿ“
๐‘จ๐Ÿ =
๐Ÿ’โ€†๐Ÿ๐Ÿ’๐Ÿโ€†๐Ÿ๐ŸŽ๐Ÿ“
โ‰ˆ ๐Ÿโ€†๐Ÿ’๐Ÿ“๐Ÿ—โ€†๐Ÿ‘๐Ÿ•๐Ÿ—
๐Ÿ’ ๐ญ๐š๐ง(๐Ÿ‘๐Ÿ”)
The area within the outer perimeter of the Pentagon building is
approximately ๐Ÿ, ๐Ÿ’๐Ÿ“๐Ÿ—, ๐Ÿ‘๐Ÿ•๐Ÿ— ๐Ÿ๐ญ ๐Ÿ.
Let ๐‘จ๐Ÿ represent the area within the perimeter of the
courtyard of the Pentagon building in square feet.
๐‘จ๐Ÿ =
Let ๐‘จ๐‘ป represent the total area of the roof of the
Pentagon building in square feet.
๐’๐’ƒ๐Ÿ
๐Ÿ’ ๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐‘จ๐‘ป = ๐‘จ๐Ÿ โˆ’ ๐‘จ๐Ÿ
๐Ÿ“(๐Ÿ‘๐Ÿ“๐Ÿ”)๐Ÿ
๐‘จ๐Ÿ =
๐Ÿ’ ๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐Ÿ”๐Ÿ‘๐Ÿ‘โ€†๐Ÿ”๐Ÿ–๐ŸŽ
๐‘จ๐Ÿ =
๐Ÿ’ ๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐‘จ๐Ÿ =
๐Ÿ๐Ÿ“๐Ÿ–โ€†๐Ÿ’๐Ÿ๐ŸŽ
โ‰ˆ ๐Ÿ๐Ÿ๐Ÿ–โ€†๐ŸŽ๐Ÿ’๐Ÿ”
๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐‘จ๐‘ป =
๐Ÿ’โ€†๐Ÿ๐Ÿ’๐Ÿโ€†๐Ÿ๐ŸŽ๐Ÿ“ ๐Ÿ๐Ÿ“๐Ÿ–โ€†๐Ÿ’๐Ÿ๐ŸŽ
โˆ’
๐Ÿ’ ๐ญ๐š๐ง(๐Ÿ‘๐Ÿ”)
๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐‘จ๐‘ป =
๐Ÿ’โ€†๐Ÿ๐Ÿ’๐Ÿโ€†๐Ÿ๐ŸŽ๐Ÿ“ ๐Ÿ”๐Ÿ‘๐Ÿ‘โ€†๐Ÿ”๐Ÿ–๐ŸŽ
โˆ’
๐Ÿ’ ๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐Ÿ’ ๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
๐‘จ๐‘ป =
๐Ÿ‘โ€†๐Ÿ”๐ŸŽ๐Ÿ•โ€†๐Ÿ“๐Ÿ๐Ÿ“
โ‰ˆ ๐Ÿโ€†๐Ÿ๐Ÿ’๐Ÿโ€†๐Ÿ‘๐Ÿ‘๐Ÿ‘
๐Ÿ’ ๐ญ๐š๐ง ๐Ÿ‘๐Ÿ”
The area of the roof of the Pentagon building is approximately ๐Ÿ, ๐Ÿ๐Ÿ’๐Ÿ, ๐Ÿ‘๐Ÿ‘๐Ÿ‘ ๐Ÿ๐ญ ๐Ÿ.
9.
A regular hexagon is inscribed in a circle with a radius of ๐Ÿ•. Find the perimeter and area of the hexagon.
The regular hexagon can be divided into six equilateral triangular regions with
each side of the triangles having a length of ๐Ÿ•. To find the perimeter of the
hexagon, solve the following:
๐Ÿ” โ‹… ๐Ÿ• = ๐Ÿ’๐Ÿ , so the perimeter of the hexagon is ๐Ÿ’๐Ÿ units.
To find the area of one equilateral triangle:
๐Ÿ
๐€๐ซ๐ž๐š = (๐Ÿ•)(๐Ÿ•) ๐ฌ๐ข๐ง ๐Ÿ”๐ŸŽ
๐Ÿ
๐€๐ซ๐ž๐š =
๐Ÿ’๐Ÿ— โˆš๐Ÿ‘
( )
๐Ÿ ๐Ÿ
๐€๐ซ๐ž๐š =
๐Ÿ’๐Ÿ—โˆš๐Ÿ‘
๐Ÿ’
The area of the hexagon is six times the area of the equilateral triangle.
๐Ÿ’๐Ÿ—โˆš๐Ÿ‘
๐“๐จ๐ญ๐š๐ฅ ๐€๐ซ๐ž๐š = ๐Ÿ” (
)
๐Ÿ’
๐“๐จ๐ญ๐š๐ฅ ๐€๐ซ๐ž๐š =
๐Ÿ๐Ÿ’๐Ÿ•โˆš๐Ÿ‘
โ‰ˆ ๐Ÿ๐Ÿ๐Ÿ•. ๐Ÿ‘
๐Ÿ
The total area of the regular hexagon is approximately ๐Ÿ๐Ÿ๐Ÿ•. ๐Ÿ‘ square units.
Lesson 31:
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
475
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 31
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
๐Ÿ
๐Ÿ
10. In the figure below, โˆ ๐‘จ๐‘ฌ๐‘ฉ is acute. Show that ๐€๐ซ๐ž๐š (โ–ณ ๐‘จ๐‘ฉ๐‘ช) = ๐‘จ๐‘ช โ‹… ๐‘ฉ๐‘ฌ โ‹… ๐ฌ๐ข๐ง โˆ ๐‘จ๐‘ฌ๐‘ฉ.
Let ๐œฝ represent the degree measure of angle ๐‘จ๐‘ฌ๐‘ฉ, and let ๐’‰ represent the
altitude of โ–ณ ๐‘จ๐‘ฉ๐‘ช (and โ–ณ ๐‘จ๐‘ฉ๐‘ฌ).
๐€๐ซ๐ž๐š(โ–ณ ๐‘จ๐‘ฉ๐‘ช) =
๐ฌ๐ข๐ง ๐œฝ =
๐Ÿ
โ‹… ๐‘จ๐‘ช โ‹… ๐’‰
๐Ÿ
๐’‰
, which implies that ๐’‰ = ๐‘ฉ๐‘ฌ โ‹… ๐ฌ๐ข๐ง ๐œฝ.
๐‘ฉ๐‘ฌ
Therefore, by substitution:
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š(โ–ณ ๐‘จ๐‘ฉ๐‘ช) = ๐‘จ๐‘ช โ‹… ๐‘ฉ๐‘ฌ โ‹… ๐ฌ๐ข๐ง โˆ ๐‘จ๐‘ฌ๐‘ฉ.
11. Let ๐‘จ๐‘ฉ๐‘ช๐‘ซ be a quadrilateral. Let ๐’˜ be the measure of the acute angle formed by diagonals ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ช and ฬ…ฬ…ฬ…ฬ…ฬ…
๐‘ฉ๐‘ซ. Show that
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช๐‘ซ) = ๐‘จ๐‘ช โ‹… ๐‘ฉ๐‘ซ โ‹… ๐ฌ๐ข๐ง ๐’˜.
(Hint: Apply the result from Problem 10 to โ–ณ ๐‘จ๐‘ฉ๐‘ช and โ–ณ ๐‘จ๐‘ช๐‘ซ.)
ฬ…ฬ…ฬ…ฬ… and ๐‘ฉ๐‘ซ
ฬ…ฬ…ฬ…ฬ…ฬ… be called point ๐‘ท.
Let the intersection of ๐‘จ๐‘ช
Using the results from Problem 10, solve the following:
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š(โ–ณ ๐‘จ๐‘ฉ๐‘ช) = ๐‘จ๐‘ช โ‹… ๐‘ฉ๐‘ท โ‹… ๐ฌ๐ข๐ง ๐’˜
and
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š(โ–ณ ๐‘จ๐‘ซ๐‘ช) = ๐‘จ๐‘ช โ‹… ๐‘ท๐‘ซ โ‹… ๐ฌ๐ข๐ง ๐’˜
๐Ÿ
๐Ÿ
๐Ÿ
๐Ÿ
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช๐‘ซ) = ( ๐‘จ๐‘ช โ‹… ๐‘ฉ๐‘ท โ‹… ๐ฌ๐ข๐ง ๐’˜) + ( ๐‘จ๐‘ช โ‹… ๐‘ท๐‘ซ โ‹… ๐ฌ๐ข๐ง ๐’˜)
Area is additive.
๐Ÿ
๐Ÿ
Distributive property
๐Ÿ
๐Ÿ
Distance is additive
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช๐‘ซ) = ( ๐‘จ๐‘ช โ‹… ๐ฌ๐ข๐ง ๐’˜) โ‹… (๐‘ฉ๐‘ท + ๐‘ท๐‘ซ)
๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช๐‘ซ) = ( ๐‘จ๐‘ช โ‹… ๐ฌ๐ข๐ง ๐’˜) โ‹… (๐‘ฉ๐‘ซ)
And commutative addition gives us ๐€๐ซ๐ž๐š(๐‘จ๐‘ฉ๐‘ช๐‘ซ) =
Lesson 31:
๐Ÿ
โ‹… ๐‘จ๐‘ช โ‹… ๐‘ฉ๐‘ซ โ‹… ๐ฌ๐ข๐ง ๐’˜.
๐Ÿ
Using Trigonometry to Determine Area
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M2-TE-1.3.0-08.2015
476
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.