Honors Geometry Comprehensive Spring Semester Review Study Notes

Similarity and Triangle Review

• Internal Proportion Problem (Problem 1):

  • Given segments: $EA=14$, $HA=34$, $HE=22$.

  • Solving for segments using Triangle Angle Bisector/Similarity logic: $EW = 5.5$ and $WA = 8.5$.

• Special Segments and Ratios (Problem 3):

  • Equation set up: $\frac{x}{4} = \frac{5+x}{6}$.

  • Calculation: $6x = 20 + 4x \rightarrow 2x = 20 \rightarrow x = 10$.

• Coordinate Geometry in Triangles (Problem 4):

  • Given vertices: $A(-4, 9)$, $B(2, 1)$, $C(10, 7)$.

  • Mid-segment $DE$ connects sides of right $\triangle ABC$.

  • Coordinate of mid-point $D$: $D(-1, 5)$.

  • Perimeter of quadrilateral $DECB$: 20 + 5\text{\sqrt{2}}.

• Triangle Proportionality Theorem (Problem 5):

  • Given values: $WK = 4x$, $YZ = 16$, $KZ = 13.5$, $KX = 31.5$.

  • Solving for $x$: $x = 7$.

• Ratios and Proportions in Similar Figures (Problems 6-9):

  • Problem 6: $x = 2$.

  • Problem 7: $LM = 12.5$, $MO = 20$, $LP = x + 2$, $NO = 2x$. Solving gives $x = 8$.

  • Problem 8: $x = 8.4$, $y = 5.6$.

  • Problem 9: $x = 6$, $y = 8$.

• True Proportions (Problem 10):

  • For the figure shown with transversals, the true proportion is: $\frac{XY}{XZ} = \frac{WK}{ZK}$.

Special Right Triangles and Trigonometric Ratios

• 45-45-90 Triangle Properties ($x, x, x\sqrt{2}$):

  • Problem 11: Hypotenuse is $16$, so legs are $8\sqrt{2}$. $x = 8\sqrt{2}$.

  • Problem 16: Leg is $3$, so hypotenuse is $3\sqrt{2}$. $z = 3\sqrt{2}$.

• 30-60-90 Triangle Properties ($x, x\sqrt{3}, 2x$):

  • Problem 12: Long leg is $12$, short leg is $4\sqrt{3}$. $x = 6\sqrt{3}$ (Note: Transcript specifies $x=6\sqrt{3}$ for Problem 13 based on a side of $12$ in an equilateral triangle).

  • Problem 14: Hypotenuse is $12$, legs are $6$ and $6\sqrt{3}$. $x = 6$, leg = $6\sqrt{3}$.

  • Problem 15: Side is $7\sqrt{3}$. Solving yields $n = 3$, $y = 3\sqrt{3}$.

• Basic Trigonometric Values (Problem 17):

  • $\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$

  • $\sin(30^{\circ}) = \frac{1}{2}$

  • $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

• Advanced Triangle Solving (Problems 18-24):

  • Problem 18: $x = 24\sqrt{2}$.

  • Problem 19: Classifying triangle with sides $9, 12, 12$. Because 9^2 + 12^2 > 12^2 (81 + 144 > 144), the triangle is ACUTE.

  • Problem 20: Using Pythagorean triples ($8, 15, 17$), $x = 17$.

  • Problem 21: In a right triangle with legs $10$ and some other side, $m\angle R = 50^{\circ}$.

  • Problem 22: $x = 16.565$, $y = 11.190$.

  • Problem 23: $x = 8.203$, $y = 11.190$.

  • Problem 24: Solving for altitude and base segments: altitude = $13.247$, $x = 30.477$, $y = 17.730$.

Trigonometry Application Problems

• Solar Elevation (Problem 27):

  • Angle of elevation = $62^{\circ}$, shadow length = $18\,m$.

  • Height of building: $\tan(62^{\circ}) = \frac{height}{18} \rightarrow height = 33.853\,m$.

• Guy Wire (Problem 28):

  • Distance from base = $35\,m$, angle with ground = $65^{\circ}$.

  • Wire length: $\cos(65^{\circ}) = \frac{35}{wire} \rightarrow wire = 82.817\,m$.

• Ladder Problems (Problem 29):

  • Scenario A: $20\,ft$ ladder at $75^{\circ}$. Distance from base involves $\cos(75^{\circ})$: $x = 5.176\,ft$.

  • Scenario B: $10\,ft$ ladder at same angle. Height reached involves $\sin(75^{\circ})$: $x = 9.659\,ft$.

• Angle of Depression (Problem 30):

  • Building height = $26\,m$, distance to car = $50\,m$.

  • $\theta = \tan^{-1}(\frac{26}{50}) \rightarrow \theta = 27.474^{\circ}$.

Polygonal Properties and Classifications

• Internal and External Angles:

  • Problem 32: Regular polygon with exterior angle = $72^{\circ}$. Calculation: $\frac{360}{72} = 5$. Name: Pentagon.

  • Problem 34: Interior angle is 11 times the exterior angle ($11y + y = 180$). $y = 15^{\circ}$. Sides $n = \frac{360}{15} = 24$.

  • Problem 35: In pentagon $ASHLE$, $m\angle A = 60^{\circ}$, $m\angle S = 130^{\circ}$. Total sum = $540^{\circ}$. If $H = L = 3E$, then $m\angle H = 150^{\circ}$.

• Always, Sometimes, Never (Problem 36):

  • The sum of exterior angles of a polygon is Always $360^{\circ}$.

  • If diagonals of a parallelogram are perpendicular, it is Always a rhombus.

  • Opposite angles of a trapezoid are Sometimes congruent (only if isosceles).

  • Consecutive angles of a trapezoid are Sometimes supplementary (only 2 pairs between parallel bases).

  • If diagonals of a rhombus are congruent, it is Always a square.

  • If opposite sides of a quadrilateral are congruent, it is Sometimes a square (could be a generic parallelogram or rectangle).

  • A trapezoid Never has exactly one right angle (it must have at least two if it has one).

  • If consecutive angles of a parallelogram are congruent ($90^{\circ}$ each), it is Always a rectangle.

  • A square is Always a rhombus.

  • A rhombus is Sometimes a square.

Areas of Polygons

• Rhombus (Problem 33):

  • Diagonals: $18\,mm$ and $80\,mm$.

  • Area: $A = \frac{1}{2} \times 18 \times 80 = 720\,mm^2$.

  • Perimeter: Side = $\sqrt{9^2 + 40^2} = 41$. $P = 4 \times 41 = 164\,mm$.

• Trapezoids (Problems 37, 38, 43):

  • Problem 37: Area = $52.402\,m^2$.

  • Problem 38: Legs = $15$, height derived from $45^{\circ}$ angle. Area = $375\,in^2$.

  • Problem 43 (Isosceles): Base 1 = $12$, Leg = $12$, Angle = $60^{\circ}$. Area in radical form: 63\text{\sqrt{3}}.

• Kites and Parallelograms (Problems 39-41):

  • Problem 39 (Kite): Area = $420$.

  • Problem 40 (Kite): Area = $120$.

  • Problem 41 (Parallelogram): Base = $15$, height = $8$. Area = $120$.

• Regular Polygons (Problems 44-46):

  • Problem 44: 9-sided polygon inscribed in circle ($r=10$). Area \approx $289.3$.

  • Problem 45: Regular hexagon, perimeter = $24\,cm$ ($side = 4$). Area: $24\sqrt{3}\,cm^2$.

  • Problem 46: Regular octagon, apothem = $14.5\,ft$. Shaded portion area = $435.442\,ft^2$.

Circle Geometry: Arcs, Sectors, and Segments

• Arc Length and Sector Area:

  • Problem 47: Circle $P$, radius = $5$, angle = $240^{\circ}$. Arc length $ABC = \frac{20\pi}{3}$.

  • Problem 48: Shaded area = $5\pi$, arc = $72^{\circ}$. Radius $r = 5$. Length of arc $AB = 2\pi$.

  • Problem 50: Center $C$. Area of segment with $120^{\circ}$ angle: $\frac{256\pi-64\sqrt{3}}{1}\text{ cm}^2$.

• Angle and Segment Relationships (Problems 51-61):

  • Problem 51 (Circle C): Angles: $a=144^{\circ}, b=18^{\circ}, c=36^{\circ}, d=144^{\circ}, e=36^{\circ}, f=90^{\circ}, g=18^{\circ}, h=144^{\circ}$.

  • Problem 52: Tangent rays. Arc is $245^{\circ}$. Solving for angle $x = 65^{\circ}$.

  • Problem 53: Chord segments. $x = 5$.

  • Problem 54: Chord $AB = 30$.

  • Problem 55: Trigonometry used to find $m\angle COB = 53.778^{\circ}$.

  • Problem 56: Segment lengths. $x = 3$, $AE = 8$.

  • Problem 57: Circle $P$. $m\text{arc BD} = 36^{\circ}$, $m\angle BED = 18^{\circ}$, $m\angle EBA = 114^{\circ}$.

  • Problem 59: Inscribed angle. $x = 61^{\circ}$.

  • Problem 60: Hexagon perimeter with inscribed circle: $P = 65$.

  • Problem 61: $AB = 10.57$.

Surface Area and Volume of Solids

• Pyramids and Prisms:

  • Problem 62 (Square Pyramid): Base = $10$, slant height derived. $SA = 10\sqrt{2} + 420$, $V = 350$.

  • Problem 63: Base triangle Area = $16\sqrt{3}$. Total $V = 32\sqrt{3}$, $SA = 162 + 16$.

  • Problem 64: Regular base. $SA = 48\sqrt{3} + 120$, $V = 120\sqrt{3}$.

  • Problem 70 (Triangular Prism): $V = 120$, $SA = 64 + 40\sqrt{10}$.

• Curved Surfaces (Cylinders, Cones, Spheres):

  • Problem 65: Cylinder with $r=7, h=10$. $SA = 238\pi$, $V = 490\pi$.

  • Problem 66 (Hollow Pipe): Total Surface Area = $242\pi$.

  • Problem 67 (Ice Cream Cone/Hemisphere): Cone height = $9\,in$, radius = $4\,in$. Total $V = 232\pi\,in^3$.

  • Problem 68: Composite figure surface area = $740$.

  • Problem 69 (Recasting): Lead cylinder with diameter $26\,in$ and height $11\,in$ recast as a cone with same base. Calculation: $h_{\text{cone}} = 3 \times 11 = 33\,in$.

• Egyptian Pyramid Logic (Problem 71):

  • Base = $80\,m$, slant height = $50\,m$.

  • Height calculation (Pythagorean theorem on the cross-section): $h^2 + 40^2 = 50^2 \rightarrow h = 30\,m$.

Coordinate Proofs

• Proving a Rhombus (Problem 72):

  • Given Parallelogram $ABCD$ with vertices like $B(6, 8)$ and $D(10, 0)$.

  • Coordinate of $C = (16, 8)$.

  • Diagonal Slopes: Used to determine if diagonals are perpendicular.

  • Conclusion: The figure is a rhombus because the diagonals are perpendicular and bisect each other.

• Logic Proofs:

  • P1: Given $\angle 1 \cong \angle 6$ and $AD \parallel CB$. Prove $ABCD$ is a parallelogram using congruent triangles (ASA or SAS post-congruence).

  • P3: Prove quadrilateral $NICE$ is a parallelogram. Coordinates: $N(-4, 2)$, $I(1, 6)$, $C(7, -1)$, $E(2, -5)$. Proof based on both pairs of opposite sides being parallel (slope equality).

  • P4: Prove quadrilateral $MATH$ is a rectangle. Coordinates: $M(-4, 5)$, $A(-1, 9)$, $T(7, 3)$, $H(4, -1)$. Proof shown through both pairs of opposite sides being congruent and diagonals being congruent.