Chapter 9 Polygons Notes and Homework

Chapter Nine: Polygons

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Index

• A: Interior & Exterior Angles
• B: Perimeter & Area
• C: Volume
• D: Volume (Day 2)
• E: Volume (Day 3) and Density
• F: Volume (Day 3) and Density Day 2
• G: Cross Sections and Multiple Choice Practice

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Chapter 9 Polygons

Perimeter

The continuous line forming the boundary of a closed geometric figure.

• Any polygon: P = sum of all sides
• Circle: C = πd = 2πr
• Rectangle: P = 2l + 2w
• Square: P = 4 × side

Area

The amount of space inside the boundary of a flat (2-dimensional) object such as a triangle or circle.

• Triangle: A = ½ × b × h
• Rectangle: A = l × w
• Square: A = (side)²
• Circle: A = πr²
• Trapezoid: A = ½ × h × (b₁ + b₂)
• Parallelogram: A = b × h

Volume

Volume is described as the amount of space inside a figure. It is used in three-dimensional objects and can be represented as the amount of space an object takes up. Volume always uses cubic units (or units³).

• General Prisms: V = B × h (where B is the area of the shape)
• Cone: V = ⅓ × πr² × h
• Sphere: V = ⁴⁄₃ × πr³
• Cylinder: V = πr² × h
• Pyramid: V = ⅓ × B × h = ⅓ × l × w × h

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Geometry Interior/Exterior Angles 9A

Introduction to Polygons

A Polygon is a closed figure that is formed by 3 or more coplanar line segments. The line segments are the sides of the polygon and the endpoints are the vertices. Polygons are named by its vertices starting with any vertex and going in order in either direction. Polygons are classified by their number of sides.

Regular Polygons are both Equilateral and Equiangular.

Note: 7 sides can also be a Septagon. 9 sides can also be a Novagon.

A Diagonal of a polygon is a line segment connecting one vertex to any other non-consecutive vertex. (When connecting the diagonals, notice that triangles are formed!)

Types of Angles

• Interior Angles: the angles inside the polygon, each angle is less than 180°.
• Exterior Angles: the angles formed by extending one of the two adjacent sides.

Special Relationship: Each interior and adjacent exterior angle is supplementary.

Formulas

# of Sides	Interior Sum	ONE interior	Exterior Sum	ONE exterior
n	180° × (n – 2)	[180° × (n – 2)] / n	360°	360° / n

Practice

1.) If the measure of one exterior angle of a regular polygon is 72°, find the number of sides that the polygon has.

Solution: The measure of each exterior angle of a regular polygon is 360° / n. Given 360° / n = 72°, so n = 360° / 72° = 5. The polygon has 5 sides.

2.) What is the measure of an interior and an exterior angle of a regular 46-gon? Round to the nearest tenth if necessary.

Solution: Interior angle = 180° × (46 – 2) / 46 = (180 × 44) / 46 ≈ 172.2°. Exterior angle = 180° – 172.2° = 7.8°.

3.) If the measure of one interior angle of a regular polygon is 144°, find the number of sides that the polygon has. Classify this polygon.

Solution: Interior angle = 180° × (n – 2) / n = 144°. Solving: 180(n – 2) = 144n → 180n – 360 = 144n → 36n = 360 → n = 10. A decagon.

4.) If the sum of the interior angles of a regular polygon is 2160°, how many sides does the polygon have?

Solution: Sum = 180° × (n – 2) = 2160 → n – 2 = 12 → n = 14 sides.

5.) Using the provided diagram, find the value of x.

a) Quadrilateral: 60° + x + x + x = 360 → 60 + 3x = 360 → 3x = 300 → x = 100°.

b) Exterior angles: x + 2x + 3x + 4x = 360 → 10x = 360 → x = 36°.

c) Pentagon: Interior angles: 138°, 100°, 110°, 2x°, x°. Sum = 138+100+110+2x+x = 348+3x = 540 → 3x = 192 → x = 64°, 2x = 128°.

Geometry Interior/Exterior Angles 9A HW

1.) Using hexagon HIJKLM:
a) Angles consecutive to ∠H: ∠I and ∠M
b) Diagonals: HJ, IK, MJ, HK
c) Consecutive sides: HI, IJ, LK, KJ

2.) Interior angle = 150° → 180(n-2)/n = 150 → 180n – 360 = 150n → 30n = 360 → n = 12 sides.

3.) 18 sides: Interior = 180×16/18 = 160°, Exterior = 20°.

4.) Exterior angle = 24° → n = 360/24 = 15 sides.

5.) Sum = 3060° → 180(n-2) = 3060 → n-2 = 17 → n = 19 sides.

6.) Pentagon: 98+104+120+90+x = 540 → 412+x = 540 → x = 128°.

7.) Exterior sum: x+47+102+50+95 = 360 → x+294 = 360 → x = 66°.

8.) Exterior sum: 3m+2m+m+2+100 = 360 → 6m+102=360 → 6m=258 → m=43°.

9.) Hexagon: (2x-50)+(x+40)+80+(x+20)+x+150 = 720 → 5x+240 = 720 → 5x=480 → x=96°; angles: 142°, 136°, 80°, 116°, 96°, 150°.

10.) Regular hexagon: Interior y = 180×4/6 = 120°, Exterior x = 60°.

11.) a) perpendicular, b) parallel, c) parallel, d) perpendicular.

12.) Rotation of rectangle about side w → [4] cylinder.

13.) Sequence: reflection then rotation → [1].

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Geometry Perimeter/Area 9B

Formulas Review

• Any polygon: P = sum of all sides
• Circle: C = 2πr, A = πr²
• Rectangle: P = 2l+2w, A = lw
• Square: P = 4s, A = s²
• Triangle: A = ½ bh
• Trapezoid: A = ½ h(b₁+b₂)
• Parallelogram: A = bh

Practice

1.) Rectangle: width 5mm, length 13mm. Perimeter = 2(13+5) = 36 mm.

2.) Pentagon perimeter = 30: (y+3)+(2y+7)+(3y-10)+15+3 = 30 → 6y+18 = 30 → 6y=12 → y=2.

3.) Circle area = 36π → πr² = 36π → r²=36 → r=6 → diameter = 12.

4.) Area = 49π → r=7 → Circumference = 2π×7 = 14π.

5.) Square diagonal = 58 → side = 58/√2 ≈ 41.01 cm. Frame width 4 cm each side → total side ≈ 49.01 cm → area ≈ 2401 cm².

6.) Walking path area: two rectangles (2×90×10=1800 ft²) + two semicircles (area of annulus: π(30²-20²)=500π≈1571.42 ft²) → total ≈ 3371 ft².

Geometry Perimeter/Area 9B HW

1.) C=10π → r=5 → Area = 25π in².

2.) Pentagon side 16 → perimeter 80 → square side = 20 m.

3.) Triangle area = 135 cm², height 18 → ½×b×18 = 135 → b = 15 cm.

4.) Countertop area: (calculate from diagram) → 19 ft².

5.) Congruent triangles: BC ≅ ST → [3].

6.) Pentagon interior sum = 540°, exterior sum = 360°, difference = 180° → [4].

7.) Similar triangles proportion: AE/BE = AC/BD → [2].

8.) RS/XY = 6/9 = 2/3, ST/YZ = 14/21 = 2/3, ∠S ≅ ∠Y → SAS similarity → Yes.

9.) Construction of equilateral triangle inscribed in circle T. (Leave marks.)

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Geometry Volume 9C

Volume Formulas

• General Prism: V = B×h
• Cylinder: V = πr²h
• Cone: V = ⅓πr²h
• Sphere: V = ⁴⁄₃πr³
• Pyramid: V = ⅓Bh = ⅓(l×w×h)

Examples

1.) Cylinder: diameter 12 in → r=6 in, h=15 in → V = π×36×15 = 540π in³.

2.) Cone: r=3 in, h=8 in → V = ⅓×π×9×8 = 24π in³.

3.) Pyramid: square base, V=256 cm³, h=12 → ⅓×B×12=256 → B=64 → side=8 cm.

4.) Fish tank: V=3360 in³, l=14 in, w=12 in → h = 3360/(14×12) = 20 in.

5.) Triangular prism: base right triangle legs 8 cm and 2.5 cm, length 5.3 cm → B=½×8×2.5=10 cm² → V=10×5.3=53 cm³.

6.) Sphere: diameter 12 in → r=6 in → V=⁴⁄₃×π×216 = 288π ≈ 905 in³.

7.) Soda can: diameter 4 in → r=2 in, h=8 in → V=π×4×8=32π in³.

Geometry Volume 9C HW

1.) Pyramid: square base side 6 in, h=12 in → B=36 → V=⅓×36×12=144 in³ → [2].

2.) Fish tank: 14×16×10=2240 in³, water 1680 in³ → empty = 560/2240=0.25=25% → [2].

3.) Great Pyramid: V=2,592,276 m³, h=146.5 m → B = 3V/h ≈ 53,088 → side≈230 m → [4].

4.) Rectangular prism: V=144 in³, h=8 in → B=18 in² → options: 2.5×7.2=18 → [2].

5.) Cone candles: diameter 3 in → r=1.5 in, h=8 in → V=⅓×π×2.25×8=6π≈18.85 in³ each ×100 = 1885 in³.

6.) Cylinder: diameter 10 cm → r=5 cm, h=7 cm → V=π×25×7=175π cm³.

7.) Sphere: diameter 15 in → r=7.5 in → V=⁴⁄₃×π×421.875≈1767.1 in³.

8.) Triangle angles: x+(2x+2)+(3x+4)=180 → 6x+6=180 → x=29 → angles: 29°, 60°, 91° → obtuse scalene.

9.) Construction of square inscribed in circle T. (Leave marks.)

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Geometry Volume (Day 2) 9D

1.) Rectangular pyramid: l=13 cm, w=6 cm, h=14 cm → B=78 cm² → V=⅓×78×14=364 cm³.

2.) Cone: slant height √137, h=11 → r² = (√137)² - 11² = 137-121=16 → r=4 → V=⅓×π×16×11≈184.7 units³.

3.) Square pyramid: slant height 13, base side 10 → half diagonal = 5√2 → h² = 13² - (5√2)² = 169-50=119 → h≈10.91 → B=100 → V≈363.7 units³.

4.) Cone: diameter 16 → r=8, slant height √185 → h² = 185-64=121 → h=11 → V=⅓×π×64×11 = (704/3)π.

5.) Two prisms equal volume equal height → B₁ = B₂ → 5² = 10×w → 25=10w → w=2.5 in.

6.) Sphere inscribed in cube edge 6 cm → r=3 cm → V=⁴⁄₃×π×27=36π cm³.

7.) Cone r=3, slant=5 → h=4 → V=⅓×π×9×4=12π. Cylinder r=4, h=6 → V=π×16×6=96π. Number of cones = 96π/12π = 8 cones.

Geometry Volume (Day 2) 9D HW

1.) Basketball diameter 9.5 in → r=4.75, tennis ball diameter 2.5 in → r=1.25. Volume ratio = (4.75³)/(1.25³) = (107.17)/(1.95)≈55 → [3].

2.) Sphere diameter 15 in → r=7.5 → V=⁴⁄₃×π×421.875≈1767.1 → [2].

3.) Sphere in cube edge 6 cm → r=3 → V=36π → [2].

4.) Tennis ball diameter 6.7 cm → r=3.35, stack of 4 → h=26.8 → V=π×3.35²×26.8≈945 cm³ → [4].

5.) Square pyramid: slant √193, base side 14 → half diagonal = 7√2 → h² = 193 - 98 = 95 → h≈9.75 → B=196 → V≈637 units³.

6.) Hemisphere diameter 18 ft → r=9 → sphere V=⁴⁄₃×π×729=972π → hemisphere V=486π ft³.

7.) Cone: diameter 30 → r=15, slant height √(r²+h²) needed to find h (problem incomplete).

8.) Similar triangles: (m∠B)/(m∠E) = (m∠C)/(m∠F) → [4].

9.) Pond distance: using similar triangles → CB≈164 yards.

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Geometry Volume (Day 3) & Density 9E

Cavalieri’s Principle

If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Density

Density = mass/volume

Examples

1.) Hemispherical tank diameter 10 ft → r=5 ft → V = ½×(⁴⁄₃×π×125)=⅔×π×125≈261.8 ft³. Weight = 261.8×62.4≈16336 lbs.

2.) Wooden cube edge 6 cm → V=216 cm³, mass=137.8 g → density=137.8/216≈0.638 g/cm³ → Ash.

More Difficult Volume Questions

1.) Water tower: hemisphere (r=8.5 ft), cylinder (r=8.5, h=25), cone (r=8.5, h=9.1). V≈7649 ft³. 85% volume weight ≈ 85%×7649×62.4 ≈ 405,000 lbs > 400,000 limit → cannot fill to 85%.

Geometry Volume (Day 3) & Density 9E HW

1.) Two cylinders: same radius and height → Cavalieri’s Principle → Yes.

2.) Sphere r=4 in → V=268.08 in³, weight=268.08×0.075≈20 lbs.

3.) Brick 8×3.5×2.25=63 in³, weight=63×1.055≈66 oz.

4.) Population density: Niagara 420.4 → [3].

5.) Parallel lines → [2].

6.) Rotation: ∠L=47°, ∠N=57° → ∠M=76°.

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Geometry Volume (Day 3) & Density 9F

1.) Silo: hemisphere r=8 ft, cylinder r=8, h=40. V=⅔×π×512 + π×64×40 = (1024π/3)+2560π = (8704π/3) ft³. Truck 8×5×4=160 ft³. Truckloads = (8704π/3)/160≈57 loads.

2.) Cone r=3 cm, h=12 → V=⅓×π×9×12=36π cm³. Sphere r=3 → V=⁴⁄₃×π×27=36π cm³. Same volume → no overflow.

3.) Water glass (frustum): top diameter 3, bottom 2, height 5. Larger cone height = 15 in. Volume ≈ 24.9 in³.

Geometry Volume (Day 3) & Density 9F HW

1.) Trophy: prism 2×6×10=120 in³, pyramid ⅓×6×10×9=180 in³, total 300 in³. Weight=300×2.5=750 g.

2.) Storage tank: cylinder r=4 m, h=9 m, hemisphere r=4 m. V=π×16×9 + ⅔×π×64 = 144π + (128π/3) = (560π/3)≈586 m³.

3.) Cone container: r=12, h=18, water depth 15. Similar triangles → water radius=10. Container V=864π≈2714 in³, water V=500π≈1571 in³, unfilled=364π≈1143 in³.

4.) Scale factor calculation: OQ/ON is not correct → [2].

5.) Sequence of transformations to map ∆ABC onto ∆DEF: translation, reflection, rotation (list and explain).

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Geometry Cross Sections & Multiple Choice 9G

Cross Section

A cross section is the face you get when you make one slice through an object. The polygon formed by the slice is the cross section. The cross section cannot contain any piece of the original face; it all comes from "inside" the solid.

Examples

1.) Which figure can have the same cross section of a sphere? → Cone.

2.) Cone cross sections: parallel to base = circle; slanted = ellipse → [3].

3.) Cross section of regular pyramid containing altitude → triangle → [3].

4.) Postcard folded and rotated → cylinder.

5.) Rectangle rotated about side w → cylinder → [4].

Geometry Cross Sections & Multiple Choice 9G HW

1.) Cross section of sphere is a circle → [1].

2.) Which drawing cannot be a cross section of a cone? → [1].

3.) NOT a cross section of square pyramid → rectangle (circle also not possible) → [1].

4.) Right triangle rotated about leg → cone → [4].

5.) Shaded figure rotated about line m → cylinder → [1].

6.) Right triangle rotated about segment BC → cone → [1].

7.) Perpendicular bisector: DE and CE not necessarily congruent → [4].

8.) Right triangle altitude theorem: AC = 2√10 → [2].

9.) Translation (5 right, 2 up) then reflection over y=0: A''(6,-3), B''(9,-3), C''(9,-7). (Graph with labels.)

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End of Chapter Nine: Polygons