Geometry Semester 2 Review Notes

Unit 4A: Similarity

• If $A ADE \sim A ABC$, solve for x and y.

  • Given ratios: $\frac{3}{3+x} = \frac{2}{2+21} = \frac{y}{24}$
  • Solving for y: $\frac{2}{23} = \frac{y}{24} \Rightarrow 3y = 48 \Rightarrow y = 16$
  • Solving for x: $\frac{2}{23} = \frac{3}{3+x} \Rightarrow 2(3+x) = 3(23) \Rightarrow 6+2x = 69 \Rightarrow 2x = 63 \Rightarrow x = \frac{63}{2} = 31.5$

• If $A PQS \sim A TRS$, solve for x and y.

  • Given ratios: $\frac{3}{12} = \frac{x}{15} = \frac{5}{5+y}$
  • Solving for x: $\frac{3}{12} = \frac{x}{15} \Rightarrow 12x = 45 \Rightarrow x = \frac{45}{12} = \frac{15}{4} = 3.75$
  • Solving for y: $\frac{3}{12} = \frac{5}{5+y} \Rightarrow 3(5+y) = 5(12) \Rightarrow 15+3y = 60 \Rightarrow 3y = 45 \Rightarrow y = 15$

• Complete the proof that $A QRS \sim A QUT$.

  • Statement 1: $RS \parallel TU$ (Given)
  • Statement 2: $\angle Q \cong \angle Q$ (Reflexive Property)
  • Statement 3: $\angle QRS \cong \angle U$ (If lines are parallel, corresponding angles are congruent)
  • Statement 4: $A QRS \sim A QUT$ (AA Similarity)

• Complete the proof that $A RTU \sim A QSU$.

  • Statement 1: $RT \parallel QS$ (Given)
  • Statement 2: $\angle QUS \cong \angle RUT$ (Vertical angles are congruent)
  • Statement 3: $\angle T \cong \angle S$ (If lines are parallel, alternate interior angles are congruent)
  • Statement 4: $A RTU \sim A QSU$ (AA Similarity)

• Prove that $A APQ \sim A ABC$.

  • Statement 1: $PQ \parallel BC$ (Given)
  • Statement 2: $\angle A \cong \angle A$ (Reflexive Property)
  • Statement 3: $\angle AQP \cong \angle ACB$ (If lines are parallel, corresponding angles are congruent)
  • Statement 4: $A APQ \sim A ABC$ (AA Similarity)

Unit 4B: Trigonometry

• What is the value of x? Round your answer to the nearest thousandth.

  • Given: Angle = $55^\circ$, Adjacent = 13 cm, Hypotenuse = x cm
  • Using cosine: $cos(55^\circ) = \frac{13}{x}$
  • Solving for x: $x = \frac{13}{cos(55^\circ)} \approx 22.665$ cm

• What is the measure of $\angle P$?

  • Given: Opposite = 15, Hypotenuse = 22
  • Using sine: $sin(P) = \frac{15}{22}$
  • Solving for P: $P = sin^{-1}(\frac{15}{22}) \approx 42.986^\circ$

• Find the lengths of y and z in the diagram below.

  • Given: Angle = $40^\circ$, Opposite = 4.5
  • Finding z (adjacent): $tan(40^\circ) = \frac{4.5}{z} \Rightarrow z = \frac{4.5}{tan(40^\circ)} \approx 5.363$
  • Finding y (hypotenuse): $sin(40^\circ) = \frac{4.5}{y} \Rightarrow y = \frac{4.5}{sin(40^\circ)} \approx 7.00$

• Solve the following missing pieces of the right triangle.

  • Given: $b=15.6, A=20^\circ$
    • Finding AB: $tan(20) = \frac{15.6}{AB} \Rightarrow AB = \frac{15.6}{tan(20)} \approx 42.870$
  • Finding hypotenuse: $cos(20) = \frac{15.6}{H} \Rightarrow H = \frac{15.6}{cos(20)} \approx 16.601$
  • Finding angle A: <A = 90 - 20 = 70
    • AC = 5.678
    • Solving for H: cos(20) = \frac{15.6}{H} => H = \frac{15.6}{cos(20)} = 16.601
      *Solve the following missing pieces of the right triangle.
  • Given: $a = 6, b = 5.292$
    • $AB = \sqrt{6^2 + 5.292^2} = \sqrt{36 + 28.005} = \sqrt{64.005} = 8$
    • tan(\theta) = \frac{6}{5.292} => tan^{-1}(\frac{6}{5.292}) = 48.400
    • Finding AB: $AB = \sqrt{6^2 + 7^2} = \sqrt{36 + 49} = \sqrt{85} \approx 9.220$
    • Finding angle A: $tan(A) = \frac{6}{7} \Rightarrow A = tan^{-1}(\frac{6}{7}) \approx 40.601^\circ$
    • Finding angle B: $49.399^\circ$

• Find the value of w and x. Round to the nearest thousandth.

  • Given: $\theta = 50^\circ$, H = 10
    • Finding w: sin(\theta) = \frac{w}{10} => w = 10*sin(50) = 7.660
    • Finding x:
    • sin(50) = \frac{10}{X} => X = sin^{-1}(\frac{10}{7.660}) = 41.136
    • sin(x) = \frac{w}{H} -> x = sin^{-1}(\frac{w}{x}) = sin^{-1}(\frac{7.660}{10}) = 49.134

Unit 5: Circles (Central, Inscribed and circumscribed angle)

• Given AE is tangent to circle P. Solve for the length of EC.

  • Given $AP = 6$ and $AD = 8$, then $PD = 6$
  • By Pythagorean theorem, $6^2 + 8^2 = x^2$ where x = DE. Thus, $36 + 64 = x^2 \Rightarrow x^2 = 100 \Rightarrow x = 10$
  • $EC = DE - DC = 10 - 6 = 4$

• $OB$ with $m\angle ABC = 130^\circ$

  • a. $m\stackrel{\frown}{AC} = 130^\circ * 2 = 260^\circ$
  • b. $m\angle CAD = \frac{130}{2} = 65^\circ$
  • Use the diagram to solve for the following missing pieces.
  • a.) $m\stackrel{\frown}{NA} = 46^\circ$
  • b.) $m\stackrel{\frown}{NI} = 82^\circ$
  • c.) $m\stackrel{\frown}{IP} = 46^\circ$
  • d.) $m\stackrel{\frown}{AP} = 134^\circ$
  • e.) $m\angle ITA = 67^\circ$
  • f.) $m\angle ANP = 67^\circ$
  • g.) $m\angle IAT = 23^\circ$
  • h.) $m\angle AIT = 90^\circ$

Unit 6: Measuring Circles, Angles, and Shapes

• Find each value to the nearest tenth:
  • Circumference of circle: $C = 2\pi r = 2 \pi (12) \approx 75.398$ cm
  • Find the area of the smaller sector: $A = \frac{\theta}{360} \pi r^2 = \frac{28}{360} \pi (12^2) \approx 35.186$ cm$</li> <li>The shortest arc length AC:$L = \frac{\theta}{360} 2\pi r = \frac{28}{360} 2 \pi (12) \approx 5.864$cm</li> <li>Find each value in the regular polygon:</li> <li>Measure of one interior angle:$\frac{360}{5} =54 \approx
    108$degrees</li> <li>Perimeter:$6*5 = 30$cm</li> <li>Area:$tan(36) = \frac{3}{A}$</li> <li>$A = \frac{3}{tan(36)} = 4.129$</li> <li>$A = \frac{5}{2}(6 * 4.129) = 61.935$</li></ul></li> </ul> <h3 id="unit7volumesof3dshapes">Unit 7: Volumes of 3D Shapes</h3> <ul> <li>A cylinder with a radius of 5 ft. and a height of 15 ft.<ul> <li>Volume =$\pi r^2 h = \pi (5^2)(15) \approx 1178.097$cubic ft.</li></ul></li> <li>A sphere with a diameter of 22 in.<ul> <li>Radius = 11 in.</li> <li>Volume =$\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (11^3) \approx 5575.279$cubic in.</li></ul></li> <li>A cone with a height of 6 in. and a diameter of 3 in.<ul> <li>Radius = 1.5 in.</li> <li>Volume =$\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.5^2)(6) \approx 14.137 cubic in.

Unit 7: Algebra Review

• Directions: Solve the following equations for x

  • Solve for x. 3x - 6 = 24
    • 3x-6=24 => +6 => 3x = 30 => x = 10
  • Solve for x. 5(x+9) = 95
    • 5(x+9) = 95 => /5 => x+9 = 19 => x = 10
  • Solve for x. \frac{4x}{3} - 5 = 11$<ul> <li>$\frac{4x}{3} = 16 \Rightarrow 4x = 48 \Rightarrow x = 12$</li></ul></li> <li>Solve for x.$\frac{3x+2}{5} = 7$<ul> <li>$3x + 2 = 35 \Rightarrow 3x = 33 \Rightarrow x = 11$</li></ul></li> <li>Solve for x.$\frac{12}{4} = 6$<ul> <li>$12/x = 24 \Rightarrow x = \frac{1}{2}$</li></ul></li> <li>Solve for x. 3x+2y = 7<ul> <li>$3x+2y=7 => 3x = 7-2y => x = \frac{7-2y}{3}$</li></ul></li> <li>Solve for x.$\frac{(x + 3)}{4} = 6$<ul> <li>$x+3 = 24 \Rightarrow x = 21$</li></ul></li> <li>Solve for x. -5y = 11<ul> <li>$-5y=11 => y = -\frac{11}{5}$</li></ul></li> <li>Solve for x. (x + y) = 6<ul> <li>$y/6 = 1 \Rightarrow y = 6$</li></ul></li></ul></li> <li><p>Directions: Simplify the following exponent expressions</p> <ul> <li>$\frac{40x^3y^{-7}z}{6x^{-5}y^2} = \frac{20x^{3+5}z}{3y^{7+2}} = \frac{20x^8z}{3y^9}$</li></ul></li> <li><p>Directions: Factor the following expressions</p> <ul> <li>$x^2 + 10x + 9 = (x+9)(x+1)$</li> <li>$x^2 - 4 = (x+2)(x-2)$</li> <li>$x^2 - 12x + 27 = (x-9)(x-3)$</li> <li>$x^2 - 7x - 30 = (x-10)(x+3)$</li> <li>$2x^2 + 15x + 18 = (2x+3)(x+6)$</li> <li>$3x^2 - 14x + 15 = (3x-5)(x-3)$</li></ul></li> <li><p>Find the value of x. </p> <ul> <li>$Cos(35) = \frac{13}{x} = \frac{Adjacent}{Hypotenuse} => x = \frac{13}{Cos(35)} \approx 15.866$</li></ul></li> <li><p>Find the value of x. </p> <ul> <li>$Cos(40) = \frac{x}{15} = \frac{Adjacent}{Hypotenuse} => x = 15*Cos(40) \approx 11.491$$