geometry 11 - andrews universityrwright/geometry/powerpoints/11 measur… · 𝑖 =6 1 2 j p n h j...
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![Page 1: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/1.jpg)
Geometry 11
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![Page 2: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/2.jpg)
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![Page 3: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/3.jpg)
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![Page 4: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/4.jpg)
Rectangle can be divided into b by h unit squares.
Parallelogram can be cut apart and built into a rectangle.
Triangle is ½ a parallelogram.
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![Page 5: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/5.jpg)
P = 17 + 10 + 21 = 48
𝐴 =1
221 8 = 84
𝑃 = 20 + 30 + 20 + 30 = 100𝐴 = 30 20 = 600
𝑎2 + 𝑏2 = 𝑐2
52 + 𝑏2 = 132
25 + 𝑏2 = 169𝑏2 = 144𝑏 = 12
𝑃 = 5 + 12 + 13 = 30
𝐴 =1
25 12 = 30
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![Page 6: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/6.jpg)
𝐴 = 𝑏ℎ153 = 𝑏17
𝑏 = 9
Length of right rectangle is 6𝐴 = 7 6 + 3 6 = 60
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![Page 7: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/7.jpg)
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![Page 8: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/8.jpg)
Trapezoid is a triangle + parallelogram
𝐴 =1
2(𝑏2 − 𝑏1)ℎ + 𝑏1ℎ
𝐴 =1
2𝑏2ℎ −
1
2𝑏1ℎ + 𝑏1ℎ
𝐴 =1
2𝑏2ℎ +
1
2𝑏1ℎ
𝐴 =1
2ℎ 𝑏1 + 𝑏2
Rhombus is four small triangles
𝐴 = 41
2
1
2𝑑1
1
2𝑑2
𝐴 =4
8𝑑1𝑑2
𝐴 =1
2𝑑1𝑑2
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![Page 9: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/9.jpg)
A Kite is two triangles with base d2 and height ½ d1
𝐴 = 21
2𝑑2
1
2𝑑1
𝐴 =2
4𝑑1𝑑2
𝐴 =1
2𝑑1𝑑2
Trapezoid
𝐴 =1
24 6 + 8 = 28
Kite
𝐴 =1
26 14 = 42
Rhombus
𝐴 =1
280 60 = 2400
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![Page 10: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/10.jpg)
Kite
𝐴 =1
2𝑑1𝑑2
𝑑1 = 4𝑑2
80 =1
24𝑑2 𝑑2
80 = 2𝑑22
40 = 𝑑22
𝑑2 = 2 10
𝑑1 = 8 10
RhombusDiagonals are 𝑑1 = 8, 𝑑2 = 4
𝐴 =1
28 4 = 16
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![Page 11: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/11.jpg)
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![Page 12: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/12.jpg)
P = 4; A = 1
P = 12; A = 9
12/4 = 3
9/1 = 9 = 32
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![Page 13: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/13.jpg)
Lengths16
12
Areas 162
122=
256
144=
16
9
Area of ΔDEF16
9=64
𝐴16𝐴 = 576𝐴 = 36
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![Page 14: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/14.jpg)
Lengths 𝑎
𝑏
Areas𝑎2
𝑏2
𝑎𝑟𝑒𝑎𝑠 =20
36
𝑙𝑒𝑛𝑔𝑡ℎ𝑠 =20
36=2 5
6
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![Page 15: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/15.jpg)
Convert 66 inches to feet
66𝑖𝑛1𝑓𝑡
12𝑖𝑛= 5.5𝑓𝑡
Find perimeter of Rectangle II𝑃 = 35 + 35 + 20 + 20 = 110𝑓𝑡
Find ratio of perimeters5.5
110=
1
20Find ratio of areas
12
202=
1
400Find the area of Rectangle II
35 20 = 700𝑓𝑡2
Use the ratio to find the area of Rectangle I1
400=
𝐴
700400𝐴 = 700
𝐴 =7
4= 1.75𝑓𝑡2
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![Page 16: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/16.jpg)
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![Page 17: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/17.jpg)
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![Page 18: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/18.jpg)
𝐶 = 𝜋𝑑𝐶 = 𝜋5 = 15.7 𝑖𝑛
17 = 𝜋𝑑17
𝜋= 𝑑
𝑑 = 5.41 𝑓𝑡
28 𝑖𝑛 = 21
3𝑓𝑡
𝐶 = 𝜋 21
3= 7.3304 𝑓𝑡
𝑅𝑒𝑣𝑜𝑢𝑙𝑡𝑖𝑜𝑛𝑠 =500
7.3304= 68.2 𝑟𝑒𝑣
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𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ 𝐴𝐵 =𝑚 𝐴𝐵
360°⋅ 2𝜋𝑟
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![Page 20: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/20.jpg)
𝑟 = 4.5 𝑦𝑑
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ 𝑃𝑄 =75
360⋅ 2𝜋4.5 𝑦𝑑 = 5.89 𝑦𝑑
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = 61.26 𝑚
61.26 𝑚 =270
360⋅ 2𝜋𝑟
61.26 𝑚 = .75 ⋅ 2𝜋𝑟81.7 𝑚 = 2𝜋𝑟 = 𝐶
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The two ends make a circle𝐶 = 2𝜋44.02 𝑚 = 276.59 𝑚
Add the two straight stretches276.59 𝑚 + 2 84.39 𝑚 = 445.4 𝑚
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![Page 23: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/23.jpg)
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![Page 24: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/24.jpg)
𝐴 = 𝜋142 = 615.8 𝑓𝑡2
𝐴 =120
360𝜋142 = 205.3 𝑓𝑡2
𝐴 =240
360𝜋142 = 410.5 𝑓𝑡2
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![Page 25: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/25.jpg)
Semicircle
𝐴 =1
2𝜋3.52 = 19.2423 𝑚2
Triangle
𝐴 =1
27 7 = 24.5 𝑚2
Total19.2423 𝑚2 + 24.5 𝑚2 = 43.7 𝑚2
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![Page 26: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/26.jpg)
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![Page 27: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/27.jpg)
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It has 8 sides, I’ll call them s. If we draw lines connecting opposite vertices, we have 8 identical triangles. Draw the altitudes from the center of the sign and call it a. The area of each triangle is ½ sa. The area of the sign then is 8(½ sa). But the perimeter, P, is 8s, so the Area = ½ Pa.
![Page 28: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/28.jpg)
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![Page 30: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/30.jpg)
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Pentagon• Pythagorean theorem to find side
6.52 + 𝑥2 = 82
42.25 + 𝑥2 = 64𝑥2 = 21.75
𝑥 =87
2= 4.6637
𝑠 = 2𝑥 = 9.3274• Area
𝐴 =1
29.3274 ⋅ 5 6.5 = 151.6
Decagon• Find ½ central angle
1
2
360
10= 18°
• Find apothem
tan 18° =3.5
𝑎𝑎 ⋅ tan 18° = 3.5𝑎 = 10.7719
• Find area
![Page 31: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/31.jpg)
𝐴 =1
27 ⋅ 10 10.7719 = 377.0
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![Page 32: Geometry 11 - Andrews Universityrwright/geometry/powerpoints/11 Measur… · 𝑖 =6 1 2 J P N H J = 1 2 360 6 =30 Apothem tan30=6 ⋅tan30=6 =10.3923 Area 1 2 12⋅6 10.3923=374.1230](https://reader034.vdocuments.net/reader034/viewer/2022042212/5eb46d0ed24a262bb85b7e61/html5/thumbnails/32.jpg)
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Find the area of the hexagon and subtract the area of the triangle.
Hexagon: 1
2𝑠𝑖𝑑𝑒 = 6
1
2𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑎𝑛𝑔𝑙𝑒 =
1
2
360
6= 30
Apothem tan 30 =6
𝑎
𝑎 ⋅ tan 30 = 6𝑎 = 10.3923
Area1
212 ⋅ 6 10.3923 = 374.1230
Triangle: Find the length of the segment from the center to the vertex.
sin 30 =6
𝑟𝑟 sin 30 = 6𝑟 = 12
Apothem sin 30 =𝑎Δ
12
𝑎Δ = 6
Side cos 30 =𝑥
12
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𝑥 = 10.3923𝑠 = 20.7846
Area 𝐴 =1
220.7846 ⋅ 3 6 = 187.0615
Subtract the areas 374.12 – 187.06 = 187.06
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Area of A = 22 = 12.566Area of B = 52 – 12.566 = 65.974Area of C = 102 – 52 = 235.619Area of Board = 102 = 314.159P(A) = 12.566/314.159 = .040 = 4%P(B) = 65.974/314.159 = .21 = 21%P(C) = 235.619/314.159 = .75 = 75%
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