Geometry Unit Review Notes

Geometry: Unit 1 Volume Review

Volume Calculations

  • Cylinder (Problem 1):

    • Given: Diameter = 10 ft, Height = 21 ft

    • Formula: V=πr2hV = \pi r^2 h

    • Radius: r=102=5r = \frac{10}{2} = 5 ft

    • V=π(52)(21)=π(25)(21)=525π≈1649.34 ft3V = \pi (5^2)(21) = \pi (25)(21) = 525\pi \approx 1649.34 \text{ ft}^3

  • Rectangular Prism (Problem 2):

    • Given: Sides = 10 in, 6 in, 4 in

    • V=lwh=(10)(6)(4)=240 in3V = lwh = (10)(6)(4) = 240 \text{ in}^3

  • Cone (Problem 3):

    • Given: Radius = 18.3 inches, Height = 48.6 inches

    • Formula: V=13πr2hV = \frac{1}{3} \pi r^2 h

    • V=13π(18.3)2(48.6)=13π(334.89)(48.6)≈17071.83 in3V = \frac{1}{3} \pi (18.3)^2 (48.6) = \frac{1}{3} \pi (334.89)(48.6) \approx 17071.83 \text{ in}^3

  • Triangular Prism (Problem 4):

    • Missing information: Assuming base is a right triangle

    • V=12bhlV = \frac{1}{2}bhl

    • V=12(3)(4)(21)=6(21)=126unit3V=\frac{1}{2}(3)(4)(21) = 6(21) = 126 unit^3

  • Rectangular Prism (Problem 5):

    • Given: Sides = 5.5 ft, 5.5 ft, 6.2 ft

    • V=lwhV=lwh

    • V=(5.5)(5.5)(6.2)=187.55 ft3V = (5.5)(5.5)(6.2) = 187.55 \text{ ft}^3

  • Sphere (Problem 6):

    • Given: Diameter = 20 yd, so Radius = 10 yd

    • Formula: V=43πr3V = \frac{4}{3} \pi r^3

    • V=43π(103)=43π(1000)=4000π3≈4188.79 yd3V = \frac{4}{3} \pi (10^3) = \frac{4}{3} \pi (1000) = \frac{4000\pi}{3} \approx 4188.79 \text{ yd}^3

  • Using the Volume to Find Missing Dimensions

Volume and Radius

  • Sphere Volume to Radius (Problem 8):

    • Given: Volume = 64 cubic meters

    • Formula: V=43πr3V = \frac{4}{3} \pi r^3

    • 64=43πr364 = \frac{4}{3} \pi r^3

    • 64⋅34=πr364 \cdot \frac{3}{4} = \pi r^3

    • 48=πr348 = \pi r^3

    • r3=48πr^3 = \frac{48}{\pi}

    • r=48π3≈2.47 mr = \sqrt[3]{\frac{48}{\pi}} \approx 2.47 \text{ m}

    • 700=(30)w(7)700 = (30)w(7)

    • 700=210w700 = 210w

    • w=700210w = \frac{700}{210}

    • w=103≈3.33 ftw = \frac{10}{3} \approx 3.33 \text{ ft}

Oblique Figures

  • Oblique Cylinder (Problem 10):

    • Given: Radius = 2 in., Height = 5 in.

    • Formula: V=πr2hV = \pi r^2 h

    • V=π(22)(5)=20π≈62.83 in3V = \pi (2^2) (5) = 20\pi \approx 62.83 \text{ in}^3

  • Oblique Rectangular Prism (Problem 11):

    • Given: Length = 6 mm, Width = 4 mm, Height = 3 mm

    • Formula: V=lwhV = lwh

    • V=(6)(4)(3)=72 mm3V = (6)(4)(3) = 72 \text{ mm}^3

Solving for Missing Variable

Volume

  • Cone Volume to Height (Problem 12):

    • Given: Volume = 64 in³, Radius = 4 in.

    • Formula: V=13πr2hV = \frac{1}{3} \pi r^2 h

    • 64=13π(42)h64 = \frac{1}{3} \pi (4^2) h

    • 64=163πh64 = \frac{16}{3} \pi h

    • h=64⋅316πh = \frac{64 \cdot 3}{16\pi}

    • h=19216π≈3.82 inh = \frac{192}{16\pi} \approx 3.82 \text{ in}

  • Cone Volume to Height Without PI (Problem 13):

    • Assume V=192 cm3V=192 \text{ cm}^3

    • Formula: V=13BhV=\frac{1}{3}Bh

    • V=13πr2hV = \frac{1}{3} \pi r^2 h

    • 192=13(π62)h192 = \frac{1}{3} (\pi 6^2) h

    • 192=13(36π)h192 = \frac{1}{3} (36\pi) h

    • 192=12πh192 = 12\pi h

    • h=19212πh = \frac{192}{12\pi}

    • h=16π≈5.09h = \frac{16}{\pi} \approx 5.09

    • If Volume is 3.64 instead of 192, then
      3.64=13<em>B</em>h3.64 = \frac{1}{3}<em>B</em>h
      3.64=13<em>(8)</em>(6)∗h3.64 = \frac{1}{3}<em>(8)</em>(6)*h
      3.64=16h3.64 = 16h
      h=3.6416=.2275h = \frac{3.64}{16} = .2275

Rotations about Sides

  • Rotating a Square (Problem 14):

    • Rotating square MATH around side AT creates a right cylinder.

    • The radius of the cylinder is the side length of the square (7 in).

    • Therefore, the answer is (d) a right cylinder with a radius of 7 in.

  • Rotating a Right Triangle (Problem 15):

    • Rotating a right triangle around side AB (length 6) creates a cone.

    • Volume of a cone: V=13πr2hV = \frac{1}{3} \pi r^2 h

    • Here, r=4r = 4 and h=6h = 6

    • V=13π(42)(6)=13π(16)(6)=32πV = \frac{1}{3} \pi (4^2) (6) = \frac{1}{3} \pi (16)(6) = 32\pi

    • The answer is a) 32π

  • Right Hexagonal Prism (Problem 16):

    • A two-dimensional cross-section perpendicular to the base is a rectangle.

  • Density (Problem 17):

    • Density formula: d=MVd = \frac{M}{V}

    • Given: Volume V=30 cm3V = 30 \text{ cm}^3, Mass M=60 gramsM = 60 \text{ grams}

    • d=6030=2gcm3d = \frac{60}{30} = 2 \frac{\text{g}}{\text{cm}^3}

Simplifying Expressions

  • Addition (Problem 1):

    • (4k2−6k+3)+(−4k2+8k+4)=2k+7(4k^2 - 6k + 3) + (-4k^2 + 8k + 4) = 2k + 7

  • Subtraction (Problem 2):

    • (7x3+8x−1)−(2x2−x+5)=7x3−2x2+9x−6(7x^3 + 8x - 1) - (2x^2 - x + 5) = 7x^3 - 2x^2 + 9x - 6

  • Addition and Subtraction (Problem 3):

    • (3x2+11x+6)−(1−2x)=3x2+13x+5(3x^2 + 11x + 6) - (1 - 2x) = 3x^2 + 13x + 5

  • Addition (Problem 4):

    • (5b4−3b2+3b)+(b4+3b2)=6b4+3b(5b^4 - 3b^2 + 3b) + (b^4 + 3b^2) = 6b^4 + 3b

  • Multiplication (Problem 5):

    • (9x+5)(8x+9)=72x2+81x+40x+45=72x2+121x+45(9x + 5)(8x + 9) = 72x^2 + 81x + 40x + 45 = 72x^2 + 121x + 45

  • Multiplication (Problem 6):

    • (6m−11)(m+2)=6m2+12m−11m−22=6m2+m−22(6m - 11)(m + 2) = 6m^2 + 12m - 11m - 22 = 6m^2 + m - 22

  • Closure Property (Problem 7):

    • Division is not closed under polynomial operations. For example, x+2x2\frac{x+2}{x^2} is not a polynomial.

  • Area of Shaded Region (Problem 8):

    • (2x+3)(2x+3)−(2x2+9x)=4x2+6x+6x+9−2x2−9x=2x2+3x+9(2x + 3)(2x + 3) - (2x^2 + 9x) = 4x^2 + 6x + 6x + 9 - 2x^2 - 9x = 2x^2 + 3x + 9

  • Perimeter of Rectangle (Problem 9):

    • Area =x2+8x+15=(x+3)(x+5)= x^2 + 8x + 15 = (x + 3)(x + 5)

    • Perimeter =2(x+3)+2(x+5)=2x+6+2x+10=4x+16= 2(x + 3) + 2(x + 5) = 2x + 6 + 2x + 10 = 4x + 16

Triangle

  • Area of Triangle (Problem 10):

    • Area =12(8x−6)(x+2)=12(8x2+16x−6x−12)=4x2+5x−6= \frac{1}{2} (8x - 6)(x + 2) = \frac{1}{2} (8x^2 + 16x - 6x - 12) = 4x^2 + 5x - 6

  • Perimeter of given trapezoid

  • Length of Missing Side (Problem 11):

    • Let Missing Side =S= S

    • P=8x4−7x3+9x2−5x−1P = 8x^4 - 7x^3 + 9x^2 - 5x - 1

    • S=(8x4−7x3+9x2−5x−1)−((2x4−6x3−1)+(5x4−x+3)+(x2+6x2−2))S = (8x^4 - 7x^3 + 9x^2 - 5x - 1) - ((2x^4-6x^3-1) + (5x^4 -x+3) + (x^2+6x^2-2))

    • S=(8x4−7x3+9x2−5x−1)−(7x4−5x3+6x2−x)S = (8x^4 - 7x^3 + 9x^2 - 5x - 1) - (7x^4 - 5x^3 + 6x^2 -x)

    • S=x4−2x3+3x2−4x−1S = x^4 - 2x^3 + 3x^2 - 4x - 1

Garden

Area

  • Area of Square Garden (Problem 12):
    Total width: 2x+32x+3
    *Area of garden: (2x+3)(2x+3)=4x2+12x+9(2x+3)(2x+3) = 4x^2 + 12x + 9

  • Area of Walkway and Garden(Problem 12): Total width: 4x+34x+3

    • Area =(4x+3)(4x+3)=16x2+24x+9= (4x+3)(4x+3) = 16x^2 + 24x + 9

Length of Rectangle(Problem 13):
Formula: Area/Width is Length
2x2+22x+562x+8=(x+7)\frac{2x^2 + 22x + 56}{2x+8} = (x+7)

Line and Point Relationships

  • Lines and Points (Problem 1):

    • a) Name a line that contains point A: Line l or AD

    • b) Another name for line m: BD

    • c) Name a point not on AC: E or B

    • d) Name a point collinear with point D: B

  • Planes and Lines (Problem 2):

    • a) Three line segments that intersect at point A: AB, AG, AD

    • b) Intersection of planes GAB and FEH: GH

    • c) Do planes GFE and HBC intersect? Yes, they meet at FE.

    • d) Planes for top of the box: plane GFH or plane ABD

Angle Types

  • Angles and Lines (Problem 3):

    • a) (\angle 1) and (\angle 9): Corresponding angles

    • b) (\angle 10) and (\angle 14): Alternate interior angles

    • c) (\angle 8) and (\angle 9): Consecutive interior angles
      Missing Angles

  • Angle measures (Problem 3):

    \begin{aligned}
    m\angle 1 &= 130^\circ \
    m\angle 5 &= 130^\circ \
    m\angle 13 &= 130^\circ \
    m\angle 4 &= 50^\circ \
    m\angle 6 &= 50^\circ \
    m\angle 8 &= 50^\circ \
    m\angle 10 &= 50^\circ \
    m\angle 12 &= 50^\circ \
    m\angle 14 &= 50^\circ
    \end{aligned}

    Solve for Lines
    Problem 4:
    6x+4=8x−86x+4=8x-8
    12=2x12=2x
    x=6x=6
    angles: alternate angles
    Missing Sides
    Problem 5:
    Missing Angle is 60∘60^\circ
    Other Angles
    115∘and105∘115^\circ and 105^\circ
    Lines are not parallel
    angle relationship: consecutive interior

Lines are parallel and the angle relationship is corresponding
Find Sides
Problem 8:
5y−4+3y=1805y-4+3y=180
8y=1848y=184
y=23y=23
3(23)=2x+133(23)=2x+13
69=2x+1369=2x+13
x=28x=28
Exterior Angle Sum
Problem 9:
4x+2+3x−7=1004x+2+3x-7=100
7x−5=1007x-5=100
7x=1057x=105
x=15x= 15
M angle ABC=(4(15)+2)=62ABC= (4(15)+2) = 62
Radius Calculation
Missing Sides
Problem 10:
3x+1=2623x+1= \frac{26}{2}
3x+1=133x+1=13
3x=123x=12
x=4x= 4
Base angles of Triangle
Formula that the Angle = \frac{180-146}{2}$$
Problem 12:
Missing Angle = 17
Problem 12 Statements and reasons proof (Provided)
missing angles for the diagram
Problem 13:
Angle measures are shown in the sides!
Sides of Parallelogram
Problem 14:
Using slope to identify the parallelograms!
Quadrilaterals
Problem 15:
Using Slope and finding length of sides
Transformation
Problem 1
Translation Add and Subtracts to the coordinates
Problems 2&4
Finding the coordinates after translation
Problems 3&4
Identifications of Axis
Rotation of figures:
Problem 5&6
Rotation of the figure 180CCW or 90CCW:
Composites
Problems:
Find the order and importance of doing transformations after one another
Triangle Congruence
Find proofs
Find Reason
Reasons for the Triangle congruence
Problem:1 Find Reasons and statements
Problem 2&3 Find If N and find reason
Triangle Congruence
Using SSS
Missing value Problem 7:
Using Side lengths to find missing values
DF value can be calculated with 12,9
Using Cross Multiplication
Used AA formula:
17 & 18
Using AA formula using cross multiplication formula
Triangle are Similar:
Missing Sides
Using Cross multiplication
Missing Angle Measure
1-Using Trig Function
tan 32:
1 Find Angle measures
Finding trigonometric Ratio:
Problem 2
Cos
Problem 3:
tan Function
Exact Form Sides:
Problem4&5
Finding Angle Degrees
Solving Angle Degrees
Find Angle A Using Trig Functions
Height Of the TREE:
Problems 14
Height of the three with the given sides tan = x/60
Airport: Problem 16
Sine function and is x high
Wheelchair Problem: 17
tan(9) is = 1/x
Problem: 18 Height of the height
tan37 degrees or = x
expressions for the sides
Geometry Probability:
cards : 1 a
die is rolled
The table is rolled determine what is required
card deck what is the probability for items
is the event A and B if it is independent
What is the independent value
Complete the two way table for probabilities
the results are shown the survey complete it: problem 7