Coordinate Geometry, Quadrilateral Properties, and Conservation of Momentum

Coordinate Geometry: Polygon Area and Perimeter

• Bounding Box Subtraction Method for Area:

  • To calculate the area of an irregular polygon plotted on a Cartesian coordinate plane, construct an enclosing bounding rectangle whose edges align with the maximum horizontal and vertical coordinates of the figure, then subtract the surplus exterior geometric shapes (triangles and rectangles).

  • Enclosing Bounding Rectangle Dimensions:

  • Length ($L$) = $11$

  • Width ($W$) = $10$

  • Total Area of Bounding Box ($A_{\text{total}}$) = $L \times W = 11 \times 10 = 110$

  • Surrounding Boundary Regions Subtracted:

  • Region I (Right Triangle): $A_I = \frac{B \times H}{2} = \frac{4 \times 7}{2} = 14$

  • Region II (Right Triangle): $A_{II} = \frac{B \times H}{2} = \frac{6 \times 2}{2} = 6$

  • Region III (Right Triangle): $A_{III} = \frac{B \times H}{2} = \frac{5 \times 1}{2} = 2.5$

  • Region IV / VI (Right Triangle): $A_{VI} = \frac{B \times H}{2} = \frac{7 \times 1}{2} = 3.5$

  • Region V / VII (Right Triangle / Small Rectangle): $A_V = \frac{B \times H}{2} = \frac{2 \times 2}{2} = 2$

  • Total Polygon Area Calculation:

  • $A = A_{\text{total}} - A_I - A_{II} - A_{III} - A_{VI} - A_V$

  • $A = 110 - 14 - 6 - 2.5 - 3.5 - 2 = 82$

• Perimeter Calculation via Euclidean Distance Formula:

  • The perimeter of a polygon is the sum of the lengths of all bounding line segments. For each segment between coordinates $(x_1, y_1)$ and $(x_2, y_2)$, length is given by $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$.

  • Individual Side Lengths:

  • Side 1: $d_1 = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.708$

  • Side 2: $d_2 = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} \approx 8.062$

  • Side 3: $d_3 = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.325$

  • Side 4: $d_4 = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} \approx 7.071$

  • Side 5: $d_5 = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.099$

  • Total Perimeter Calculation:

  • $\text{Perimeter} = \sqrt{45} + \sqrt{65} + \sqrt{40} + \sqrt{50} + \sqrt{26}$

  • $\text{Perimeter} \approx 6.708 + 8.062 + 6.325 + 7.071 + 5.099 = 33.265$

  • Rounded to the nearest hundredth: $33.27$

Foundations of Euclidean Geometry: Lines, Angles, and Transformations

• Transversal and Parallel Line Angle Relationships:

  • When two parallel lines are intersected by a transversal line:

  • Alternate interior angles are congruent: $\angle_{\text{alt int 1}} \cong \angle_{\text{alt int 2}}$

  • Corresponding angles are congruent: $\angle_{\text{corr 1}} \cong \angle_{\text{corr 2}}$

  • Consecutive interior angles are supplementary: $m\angle_1 + m\angle_2 = 180^\circ$

  • Vertical angles are congruent across intersecting lines: $x = 116^\circ$, $y = 116^\circ$

• Triangle Angle Sum and Exterior Angle Theorems:

  • The interior angles of any triangle in Euclidean space sum to $180^\circ$:

  • Equation: $(4x + 4) + 79 + 65 = 180$

  • Combine constants: $4x + 148 = 180$

  • Subtract $148$: $4x = 32$

  • Solve for $x$: $x = 8$

  • Exterior Angle Theorem:

  • The measure of an exterior angle of a triangle equals the sum of the measures of its two remote interior angles:

  • Equation: $144 = 4x$

  • Solve for $x$: $x = 36$

  • Algebraic Angle Systems:

  • Interior angle system: $(3x - 7) + (7x - 8) + (2x - 9) = 180$

  • Combine like terms: $12x - 24 = 180$

  • Add $24$: $12x = 204$

  • Solve for $x$: $x = 17$

• The Triangle Inequality Theorem:

  • For any triangle with side lengths $a$, $b$, and $c$, the sum of the lengths of any two sides must be strictly greater than the length of the remaining side ($a + b > c$):

  • Candidate set $\{18, 22, 34\}$: $18 + 22 = 40 > 34$ (Valid triangle)

  • Candidate set $\{8, 22, 30\}$: $8 + 22 = 30 \ngtr 30$ (Invalid triangle)

• Candidate set $\{7, 12, 20\}$: $7 + 12 = 19 \ngtr 20$ (Invalid triangle)

• Candidate set $\{6, 9, 15\}$: $6 + 9 = 15 \ngtr 15$ (Invalid triangle)

• Regular Polygons:

  • The sum of the exterior angles of any convex polygon is $360^\circ$.

  • Each exterior angle of an $n$-sided regular polygon is given by $E = \frac{360^\circ}{n}$.

  • For a regular pentagon ($n = 5$):

  • Exterior angle: $E = \frac{360^\circ}{5} = 72^\circ$

  • Interior angle: $I = 180^\circ - 72^\circ = 108^\circ$

• Geometric Transformations:

  • Translations:

  • A translation shifts every point of a geometric figure by the same displacement distance in a specified direction: $(x, y) \rightarrow (x + h, y + k)$.

  • Rotations:

  • A clockwise rotation of $90^\circ$ about the origin follows the coordinate transformation mapping rule:

  • $(x, y) \rightarrow (y, -x)$

  • Composition of Transformations:

  • Applying sequential transformation mappings to coordinate pairs:

  • $(1, 7) \rightarrow (7, 0) \rightarrow (6, 10)$

• Coordinate Formulas: Midpoints, Distance, and Slopes:

  • Midpoint Formula:

  • The midpoint $M$ of a directed line segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ is:

  • $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

  • Worked evaluation: for points $(-1, 5)$ and $(5, -11)$, $M = \left(\frac{-1 + 5}{2}, \frac{5 + (-11)}{2}\right) = (2, -3)$.

  • Determining an Endpoint from a Given Endpoint and Midpoint:

  • Let known endpoint be $A(-8, 5)$ and midpoint be $M(-2, -3)$, solve for unknown endpoint $B(x_2, y_2)$.

  • Horizontal: $\frac{-8 + x_2}{2} = -2 \rightarrow -8 + x_2 = -4 \rightarrow x_2 = 4$

  • Vertical: $\frac{5 + y_2}{2} = -3 \rightarrow 5 + y_2 = -6 \rightarrow y_2 = -11$

  • Unknown endpoint coordinates: $B(4, -11)$

  • Slope Formula:

  • $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$

  • For points $(9, 5)$ and $(21, -5)$: $m = \frac{-5 - 5}{21 - 9} = \frac{-10}{12} = -\frac{5}{6}$

• Equations of Lines and Perpendicular Bisectors:

  • Slope-Intercept Form: $y = mx + b$

  • Point-Slope Form: $y - y_1 = m(x - x_1)$

  • For a line with slope $m = -\frac{1}{2}$ passing through $(11, 1)$: $y - 1 = -\frac{1}{2}(x - 11)$

  • For a line passing through $(-6, -1)$ with slope $m = \frac{3}{2}$: $y + 1 = \frac{3}{2}(x + 6)$

  • Parallel and Perpendicular Lines:

  • Parallel lines have identical slopes: $m_{\parallel} = m_1$

  • Perpendicular lines have negative reciprocal slopes: $m_{\perp} = -\frac{1}{m_1}$

  • Example: Line $3x - y = 0 \rightarrow y = 3x$. Parallel slope is $3$; perpendicular slope is $-\frac{1}{3}$.

  • Equation of a Perpendicular Bisector:

  • 1. Calculate midpoint of segment: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (-4, -1)$

  • 2. Calculate slope of original segment: $m = \frac{\Delta y}{\Delta x} = \frac{6 - (-4)}{-9 - 1} = \frac{10}{-10} = -1$

  • 3. Determine perpendicular slope: $m_{\perp} = -\left(\frac{1}{-1}\right) = 1$

  • 4. Construct point-slope equation: $y + 1 = 1(x + 4)$

Quadrilateral Properties: Parallelograms

• Fundamental Properties of Parallelograms:

  • A parallelogram is a quadrilateral with both pairs of opposite sides parallel ($AB \parallel CD$ and $BC \parallel DA$).

  • Opposite sides are congruent ($AB \cong CD$ and $BC \cong DA$).

  • Opposite angles are congruent ($\angle A \cong \angle C$ and $\angle B \cong \angle D$).

  • Consecutive angles are supplementary ($m\angle A + m\angle B = 180^\circ$, $m\angle B + m\angle C = 180^\circ$, etc.).

  • Diagonals bisect each other.

• Algebraic Applications to Parallelogram Sides and Slopes:

  • Parallel Slopes Property:

  • In parallelogram $LMNO$, if the slope of side $LO$ is $-\frac{1}{3}$, the slope of opposite side $MN$ is $-\frac{1}{3}$, because opposite sides are parallel.

  • Congruent Opposite Sides Property:

  • In quadrilateral $LMNO$, where $ON = 7x - 5$, $LM = 6x + 4$, and $NM = x + 6$:

  • Set opposite sides equal: $7x - 5 = 6x + 4$

  • Subtract $6x$: $x - 5 = 4$

  • Add $5$: $x = 9$

• Algebraic Applications to Parallelogram Angles:

  • Consecutive Supplementary Angles:

  • In parallelogram $ABCD$, where $m\angle A = 2x + 10$ and $m\angle B = 3x$:

  • Set sum equal to $180^\circ$: $(2x + 10) + 3x = 180$

  • Combine terms: $5x + 10 = 180$

  • Subtract $10$: $5x = 170$

  • Solve for $x$: $x = 34$

  • Evaluate angle measure: $m\angle B = 3(34) = 102^circ$

  • Congruent Opposite Angles:

  • In parallelogram $ABCD$, where $m\angle A = 3x + 10$ and $m\angle C = 100^\circ$:

  • Set opposite angles equal: $3x + 10 = 100$

  • Subtract $10$: $3x = 90$

  • Solve for $x$: $x = 30$

• Parallelogram Diagonals Bisecting Each Other:

  • In parallelogram $ABCD$, diagonals $AC$ and $DB$ intersect at point $E$. Segments along diagonal $AC$ are given as $AE = 3x - 4$ and $EC = x + 12$:

  • Set segments equal: $3x - 4 = x + 12$

  • Subtract $x$: $2x - 4 = 12$

  • Add $4$: $2x = 16$

  • Solve for $x$: $x = 8$

• Comprehensive Parallelogram Geometry Problems:

  • Problem 1 (Parallelogram FRED):

  • Side $ED$ is extended to point $A$, segment $AF$ is drawn such that $AF \cong DF$, and $m\angle R = 124^\circ$.

  • Opposite angles are congruent: $m\angle D = m\angle R = 124^\circ$ (within the parallelogram interior).

  • Line segment $ADE$ forms a linear pair: $m\angle ADF = 180^\circ - 124^\circ = 56^\circ$.

  • Since triangle $\triangle AFD$ is isosceles with $AF \cong DF$, base angles are congruent: $m\angle FAD = m\angle ADF = 56^\circ$.

  • Calculate vertex angle: $m\angle AFD = 180^\circ - (56^\circ + 56^\circ) = 180^\circ - 112^\circ = 68^\circ$.

  • Problem 2 (Parallelogram QRST):

  • Diagonal $TR$ is drawn, points $U$ and $V$ lie on segments $TS$ and $QR$ respectively, and line segment $UV$ intersects $TR$ at point $W$.

  • Given: $m\angle S = 60^\circ$, $m\angle SRT = 83^\circ$, and $m\angle TWU = 35^\circ$.

  • In triangle $\triangle RST$, sum of interior angles is $180^\circ$:

    • $m\angle STR = 180^\circ - (60^\circ + 83^\circ) = 180^\circ - 143^\circ = 37^\circ$

  • Since $TS \parallel QR$, alternate interior angles are congruent: $m\angle QRT = m\angle STR = 37^\circ$.

  • In triangle $\triangle TWU$, interior angle sum yields:

    • $m\angle TUW = 180^\circ - (35^\circ + 37^\circ) = 180^\circ - 72^\circ = 108^\circ$

  • By angle relationships across transversals and intersecting lines: $m\angle WVQ = 72^\circ$.

  • Problem 3 (Parallelogram ABCD with Exterior Extension):

  • Side $AB$ is extended through $B$ to point $E$, and segment $CE$ is drawn such that $CE \cong BE$. Given $m\angle D = 112^\circ$.

  • In parallelogram $ABCD$, opposite angles are congruent ($m\angle B = 112^\circ$) and consecutive angles are supplementary ($m\angle A = 180^\circ - 112^\circ = 68^\circ$).

  • At vertex $B$, the exterior angle to the line extension forms a linear pair: $m\angle CBE = 180^\circ - 112^\circ = 68^\circ$.

  • In triangle $\triangle CBE$, since $CE \cong BE$, base angles are equal: $m\angle BCE = m\angle CBE = 68^\circ$.

  • Interior angle sum of $\triangle CBE$: $m\angle E = 180^\circ - (68^\circ + 68^\circ) = 180^\circ - 136^\circ = 44^\circ$.

  • Problem 4 (Parallelogram ROCK):

  • Given $m\angle C = 70^\circ$ and $m\angle ROS = 65^\circ$.

  • Consecutive angles are supplementary: $m\angle R = 180^\circ - 70^\circ = 110^\circ$.

  • In triangle $\triangle ROS$, angle sum yields: $m\angle RSO = 180^\circ - (110^\circ + 65^\circ) = 180^\circ - 175^\circ = 5^\circ$.

  • Triangle and line calculations at transversal intersection give: $m\angle KSO = 135^\circ$.

  • Problem 5 (Parallelogram LMNO with Diagonal LN):

  • Diagonal $LN$ is drawn, $m\angle M = 118^\circ$, and $m\angle LNO = 22^\circ$.

  • Explanation for why $m\angle NLO = 40^\circ$:

    • Opposite angles of a parallelogram are congruent, so $m\angle O = m\angle M = 118^\circ$.

    • In triangle $\triangle LNO$, the sum of interior angles is $180^\circ$: $m\angle NLO = 180^\circ - (m\angle O + m\angle LNO) = 180^\circ - (118^\circ + 22^\circ) = 180^\circ - 140^\circ = 40^\circ$.

    • Alternatively, consecutive angles are supplementary: $m\angle OLN + m\angle MLN = 180^\circ - 118^\circ = 62^\circ$. By alternate interior angles, $m\angle MLN = m\angle LNO = 22^\circ$, hence $m\angle NLO = 62^\circ - 22^\circ = 40^\circ$.

  • Problem 6 (Parallelogram ABCD Diagonals):

  • Diagonals $AC$ and $BD$ intersect. Given $m\angle 1 = 45^\circ$ and $m\angle DCB = 120^\circ$.

  • By angle relationships and triangle decompositions: $m\angle 2 = 180^\circ - 120^\circ - 45^\circ = 15^\circ$.

Quadrilateral Properties: Trapezoids and Isosceles Trapezoids

• Trapezoid Definition and Baseline Properties:

  • A trapezoid is a quadrilateral possessing exactly one pair of parallel sides (called bases).

  • The non-parallel sides are called legs.

  • Upper and lower base angles along each transversal leg are supplementary ($m\angle_{\text{upper}} + m\angle_{\text{lower}} = 180^\circ$).

  • An altitude of a trapezoid is a perpendicular line segment dropped from one base to the opposite base.

• Median (Midsegment) of a Trapezoid Theorem:

  • The median connects the midpoints of the two legs, is parallel to both bases, and its length equals the arithmetic average of the lengths of the two bases:

  • $\text{Median} = \frac{\text{Base}_1 + \text{Base}_2}{2}$

  • Problem 1 (Base Angle Evaluation):

  • In trapezoid $ABCD$ with bases $AB$ and $CD$, if $m\angle B = 123^\circ$:

  • $m\angle C = 180^\circ - 123^\circ = 57^\circ$

  • Problem 2 (Base Length Evaluation):

  • In trapezoid $RSTU$, $RS \parallel TU$, $X$ is the midpoint of leg $RU$, and $V$ is the midpoint of leg $ST$. Given base $RS = 30$ and median $XV = 44$:

  • $\frac{30 + TU}{2} = 44$

  • Multiply by $2$: $30 + TU = 88$

  • Subtract $30$: $TU = 58$

  • Problem 3 (Algebraic Midsegment Formulation):

  • In trapezoid $ABCD$, median $EF = 2x + 2$, base $AB = 5x - 9$, and base $DC = x + 3$:

  • $\frac{(x + 3) + (5x - 9)}{2} = 2x + 2$

  • Simplify numerator: $\frac{6x - 6}{2} = 2x + 2$

  • Divide: $3x - 3 = 2x + 2$

  • Subtract $2x$ and add $3$: $x = 5$

• Isosceles Trapezoids:

  • Condition of Classification: Never assume a trapezoid is isosceles unless explicitly given or proven.

  • An isosceles trapezoid satisfies all baseline properties of a general trapezoid, plus:

  • Non-parallel legs are congruent ($\text{Leg}_1 \cong \text{Leg}_2$).

  • Diagonals are congruent ($d_1 \cong d_2$).

  • Both pairs of base angles are congruent (lower base angles congruent, upper base angles congruent).

  • Opposite angles are supplementary ($m\angle_{\text{opp 1}} + m\angle_{\text{opp 2}} = 180^\circ$).

  • Problem 4 (Congruent Legs):

  • In isosceles trapezoid $DEFG$, $DE \parallel GF$, leg $DG = 10x - 7$, and leg $EF = 3 + 5x$:

  • Set legs equal: $10x - 7 = 3 + 5x$

  • Subtract $5x$: $5x - 7 = 3$

  • Add $7$: $5x = 10$

  • Solve for $x$: $x = 2$

  • Problem 5 (Congruent Diagonals):

  • In isosceles trapezoid $ABCD$, diagonals are $AC = 5x + 13$ and $BD = 11x - 5$:

  • Set diagonals equal: $5x + 13 = 11x - 5$

  • Subtract $5x$: $13 = 6x - 5$

  • Add $5$: $18 = 6x$

  • Solve for $x$: $x = 3$

  • Problem 6 (Congruent Base Angles):

  • In isosceles trapezoid $QRST$, bases are $QR$ and $TS$. Given lower base angles $m\angle Q = 5x + 3$ and $m\angle R = 7x - 15$:

  • Set angles equal: $5x + 3 = 7x - 15$

  • Subtract $5x$: $3 = 2x - 15$

  • Add $15$: $18 = 2x$

  • Solve for $x$: $x = 9$

  • Evaluate $m\angle Q$: $m\angle Q = 5(9) + 3 = 45 + 3 = 48^\circ$

  • Problem 7 (Supplementary Opposite Angles):

  • In trapezoid $ABCD$, $AB \parallel DC$, $AD \cong BC$, $m\angle A = 4x + 20$, and $m\angle C = 3x - 15$:

  • Set sum equal to $180^\circ$: $(4x + 20) + (3x - 15) = 180$

  • Combine terms: $7x + 5 = 180$

  • Subtract $5$: $7x = 175$

  • Solve for $x$: $x = 25$

  • Evaluate $m\angle D$: Since base angles are congruent, $m\angle D = m\angle C = 3(25) - 15 = 75 - 15 = 60^\circ$

• Right Trapezoids:

  • A right trapezoid satisfies all general trapezoid properties plus the addition of exactly two adjacent right angles ($90^\circ$).

• Trapezoid Altitude and Area Practice Problems:

  • Quadrilateral Diagonal Non-Bisection:

  • If the diagonals of a quadrilateral do not bisect each other, the figure cannot be a parallelogram, rectangle, rhombus, or square; it could be a trapezoid.

  • Equal Area Triangles in a Trapezoid:

  • In trapezoid $RSTV$ with bases $RS$ and $VT$, diagonals $RT$ and $SV$ intersect at point $Q$.

  • Triangles $\triangle RSV$ and $\triangle RST$ share the identical base $RS$ and share identical perpendicular heights between the parallel bases; thus, $\text{Area}(\triangle RSV) = \text{Area}(\triangle RST)$.

  • Subtracting shared area $\triangle RSQ$ yields equal areas for non-overlapping lateral triangles: $\text{Area}(\triangle RQV) = \text{Area}(\triangle SQT)$.

  • Altitude Calculation 1 (Trapezoid LATE):

  • Isosceles trapezoid $LATE$ has bases $LA = 24$ and $ET = 40$, with leg $AT = 10$. Altitudes $LF$ and $AG$ are dropped to base $ET$.

  • Horizontal projection on base: $d = \frac{ET - LA}{2} = \frac{40 - 24}{2} = \frac{16}{2} = 8$

  • Apply Pythagorean theorem to right triangle $\triangle AGT$: $x^2 + 8^2 = 10^2$

  • $x^2 + 64 = 100 \rightarrow x^2 = 36 \rightarrow x = 6$

  • Length of altitude $LF = 6$

  • Altitude Calculation 2 (Trapezoid ABCD):

  • Isosceles trapezoid $ABCD$ has legs $AB = CD = 25$, base $BC = 12$, and base $AD = 26$.

  • Horizontal projection on base: $d = \frac{AD - BC}{2} = \frac{26 - 12}{2} = \frac{14}{2} = 7$

  • Apply Pythagorean theorem: $x^2 + 7^2 = 25^2$

  • $x^2 + 49 = 625 \rightarrow x^2 = 576 \rightarrow x = 24$

  • Length of altitude = $24$

  • Altitude Calculation 3 (Trapezoid ABCD):

  • Isosceles trapezoid $ABCD$ has legs $AB = CD = 17$, base $BC = 20$, and base $AD = 36$.

  • Horizontal projection on base: $d = \frac{AD - BC}{2} = \frac{36 - 20}{2} = \frac{16}{2} = 8$

  • Apply Pythagorean theorem: $x^2 + 8^2 = 17^2$

  • $x^2 + 64 = 289 \rightarrow x^2 = 225 \rightarrow x = 15$

  • Length of altitude = $15$

Quadrilateral Properties: Rhombuses and Squares

• Properties of a Rhombus:

  • A rhombus is a parallelogram with four congruent sides.

  • Diagonals are perpendicular ($AC \perp BD$).

  • Diagonals bisect the interior angles.

  • All general parallelogram properties apply (opposite angles congruent, consecutive angles supplementary, diagonals bisect each other).

• Rhombus Problem Applications:

  • Problem 1 (Equilateral Sides and Perimeter):

  • In rhombus $ABCD$, side $AB = 3x + 2$ and side $BC = x + 12$:

  • Set sides equal: $3x + 2 = x + 12$

  • Subtract $x$: $2x + 2 = 12$

  • Subtract $2$: $2x = 10$

  • Solve for $x$: $x = 5$

  • Side length: $s = 3(5) + 2 = 17$

  • Total perimeter: $P = 4 \times s = 4 \times 17 = 68$

  • Problem 2 (Diagonals Bisecting Angles):

  • In rhombus $ABCD$, $m\angle C = 100^\circ$. Find $m\angle BDA$:

  • Consecutive angles are supplementary: $m\angle ADC = 180^\circ - 100^\circ = 80^\circ$

  • Diagonal $BD$ bisects $\angle ADC$: $m\angle BDA = \frac{80^\circ}{2} = 40^\circ$

  • Problem 3 (Perpendicular Diagonals and Pythagorean Theorem):

  • In rhombus $ABCD$, diagonals $AC$ and $BD$ intersect perpendicularly at $E$. Given half-diagonals $AE = 5$ and $BE = 12$. Find side length $AB$:

  • Apply Pythagorean theorem: $AE^2 + BE^2 = AB^2$

  • $5^2 + 12^2 = c^2 \rightarrow 25 + 144 = c^2 \rightarrow 169 = c^2 \rightarrow c = 13$

  • Side length $AB = 13$

  • Problem 4 (Perimeter from Full Diagonals):

  • In rhombus $PQRS$, diagonals intersect at $T$ with full lengths $PR = 16$ and $QS = 30$.

  • Determine half-diagonal lengths: $a = \frac{16}{2} = 8$, $b = \frac{30}{2} = 15$

  • Calculate side length ($s$): $s = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$

  • Total perimeter: $P = 4 \times s = 4 \times 17 = 68$

• Properties of a Square:

  • A square is a regular quadrilateral that simultaneously satisfies all properties of a rectangle, a rhombus, and a parallelogram.

  • Four congruent sides.

  • Four congruent right angles ($90^\circ$).

  • Diagonals are congruent, perpendicular bisectors of each other, and bisect interior angles at $45^\circ$.

• Square Problem Applications:

  • Problem 1 (Diagonal Bisector Equivalence):

  • In square $ABCD$, diagonals intersect at $E$. Given segments $DE = 7y - 10$ and $CE = 5y + 10$:

  • Since diagonals of a square are congruent and bisect each other, half-diagonals are equal: $7y - 10 = 5y + 10$

  • Subtract $5y$: $2y - 10 = 10$

  • Add $10$: $2y = 20$

  • Solve for $y$: $y = 10$

  • Problem 2 (Diagonal Length from Known Side):

  • If side length $s = 4$, find diagonal $d$ in simplest radical form:

  • Using Pythagorean theorem ($s^2 + s^2 = d^2$):

  • $4^2 + 4^2 = d^2 \rightarrow 16 + 16 = d^2 \rightarrow d^2 = 32$

  • $d = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

  • Problem 3 (Diagonal Length from Perimeter):

  • A square has a perimeter of $28$. Determine its diagonal in simplest radical form:

  • Calculate side length: $s = \frac{28}{4} = 7$

  • Apply Pythagorean theorem: $7^2 + 7^2 = d^2$

  • $49 + 49 = d^2 \rightarrow d^2 = 98$

  • $d = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$

  • Problem 4 (Side Length Expression from Polynomial Perimeter):

  • A square has a perimeter of $16x^4 + 4$. Write an expression representing each side:

  • Since a square has four congruent sides, divide the total perimeter by $4$:

  • $\text{Side} = \frac{16x^4 + 4}{4} = 4x^4 + 1$

Classical Mechanics: Principles of Momentum and Impulse

• Core Definitions and Vector Formulations:

  • System:

  • A system is defined as the specific object or collection of interacting objects chosen for physical analysis.

  • Linear Momentum ($p$):

  • Momentum is a vector quantity defined as the product of an object's mass ($m$) and its velocity ($v$):

  • $p = mv$

  • Standard SI unit: $kg\,m/s$

  • Sign Conventions:

  • Motion directed rightward or eastward is defined as positive ($+$).

  • Motion directed leftward or westward is defined as negative ($-$).

  • Scalar Invariance of Mass:

  • Mass is an intrinsic scalar physical quantity and must strictly be positive ($m > 0$).

  • A negative sign in momentum indicates vector direction (negative velocity), never negative mass.

• The Law of Conservation of Momentum:

  • The total linear momentum of an isolated system remains constant over time unless acted upon by a net external force:

  • $p_{\text{before}} = p_{\text{after}}$

  • $\sum p_i = \sum p_f$

  • A net external force is the only mechanism that can change a system's total velocity and momentum (e.g., friction acting on a vehicle).

  • In any internal two-body interaction or collision, the momentum gained by one object is identical in magnitude and opposite in sign to the momentum lost by the second object:

  • $\Delta p_1 = -\Delta p_2$

• Impulse ($J$) and Force-Time Integrals:

  • Impulse is defined as the product of the average net force acting on an object and the time duration ($\Delta t$) over which it acts, which equals the object's change in linear momentum:

  • $J = \Delta p = F_{\text{net}} \Delta t$

  • Standard SI units: $N\,s$ or $kg\,m/s$

  • Graphical Impulse Determination:

  • The magnitude of impulse equals the geometric area beneath a force-time graph ($F \text{ vs. } t$).

  • For a triangular force pulse rising from zero to peak force $F_{\text{peak}}$ and returning to zero over duration $\Delta t$:

  • $J = \frac{1}{2} \times \Delta t \times F_{\text{peak}}$

• Classification of Collision Typologies:

  • Elastic Collision:

  • Total linear momentum is conserved: $p_i = p_f$

  • Total kinetic energy ($KE = \frac{1}{2}mv^2$) is strictly conserved: $KE_i = KE_f$

  • Inelastic Collision:

  • Total linear momentum is conserved: $p_i = p_f$

  • Total kinetic energy is not conserved ($KE_f < KE_i$); mechanical energy dissipates into internal thermal energy, acoustic energy, or structural deformation.

  • Perfectly Inelastic Collision:

  • Total linear momentum is conserved: $p_i = p_f$

  • Colliding objects lock together upon impact and move with a unified, common final velocity ($v_f$).

  • This collision class exhibits the maximum possible kinetic energy loss consistent with conservation of momentum.

  • Explosions / Push-Aparts:

  • Internal stored energy (e.g., spring potential or chemical potential) converts into kinetic energy, propelling initially united or adjacent resting bodies into opposing directions while keeping total system momentum constant.

Collision Dynamics and Energy Classification

• The Kinetic Sculpture Experiment (Two $3.0\,kg$ Steel Spheres):

  • Setup: Two identical $3.0\,kg$ spheres on a straight, horizontal track inside an enclosure with negligible external impulse.

  • Initial Conditions across all tests:

  • Sphere A: $m_A = 3.0\,kg$, $v_{Ai} = +2.0\,m/s$

  • Sphere B: $m_B = 3.0\,kg$, $v_{Bi} = 0\,m/s$

  • Total initial system momentum:

    • $p_i = m_A v_{Ai} + m_B v_{Bi} = (3.0\,kg)(+2.0\,m/s) + (3.0\,kg)(0\,m/s) = +6.0\,kg\,m/s$

  • Total initial system kinetic energy:

    • $KE_i = \frac{1}{2} m_A v_{Ai}^2 + \frac{1}{2} m_B v_{Bi}^2 = \frac{1}{2}(3.0)(2.0)^2 + 0 = 6.0\,J$

  • Test A: Polished Steel Contact Surface:

  • Observed motion: Sphere A comes to a complete rest ($v_{Af} = 0\,m/s$), and Sphere B moves right at $v_{Bf} = +2.0\,m/s$.

  • Final kinetic energy:

    • $KE_f = \frac{1}{2}(3.0\,kg)(0\,m/s)^2 + \frac{1}{2}(3.0\,kg)(+2.0\,m/s)^2 = 0 + 6.0 = 6.0\,J$

  • Collision Classification: Elastic (Kinetic energy is fully conserved: $KE_i = KE_f = 6.0\,J$).

  • Test B: Soft Polymer Contact Surface:

  • The polymer coating briefly compresses during impact. Sphere A rebounds with a recorded final velocity of $v_{Af} = +0.50\,m/s$.

  • Determine missing final velocity of Sphere B ($v_{Bf}$) using conservation of momentum:

    • $p_{\text{before}} = p_{\text{after}}$

    • $m_A v_{Ai} + m_B v_{Bi} = m_A v_{Af} + m_B v_{Bf}$

    • $+6.0\,kg\,m/s = (3.0\,kg)(+0.50\,m/s) + (3.0\,kg)(v_{Bf})$

    • $6.0 = 1.5 + 3.0 v_{Bf}$

    • $4.5 = 3.0 v_{Bf} \rightarrow v_{Bf} = +1.5\,m/s$

  • Final kinetic energy calculation:

    • $KE_f = \frac{1}{2}(3.0)(0.50)^2 + \frac{1}{2}(3.0)(1.5)^2 = \frac{1}{2}(3.0)(0.25) + \frac{1}{2}(3.0)(2.25)$

    • $KE_f = 0.375\,J + 3.375\,J = 3.75\,J$

  • Energy loss: $\Delta KE = 6.0\,J - 3.75\,J = 2.25\,J$

  • Collision Classification: Inelastic (Kinetic energy decreased by $2.25\,J$, and the spheres did not lock together).

  • Test C: Magnetic Connectors:

  • Spheres connect and latch together upon impact, moving as a single combined mass.

  • Determine shared final velocity ($v_f$):

    • $p_i = (m_A + m_B) v_f$

    • $+6.0\,kg\,m/s = (3.0\,kg + 3.0\,kg) v_f$

    • $6.0 = 6.0 v_f \rightarrow v_f = +1.0\,m/s$

  • Collision Classification: Perfectly Inelastic.

  • Fundamental Conservation Rationale:

  • Total linear momentum is conserved across all three simulations because the spheres exert internal forces on one another, while the net external impulse on the two-sphere system is negligible.

• Floating Solar Panels Protective Bumper Experiment:

  • System model: Module A ($m_A = 150\,kg$) drifts east at $v_{Ai} = +1.2\,m/s$. Module B (unknown mass $m_B$) drifts west at $v_{Bi} = -0.40\,m/s$.

  • Following a flexible bumper collision: Module A moves west at $v_{Af} = -0.40\,m/s$, and Module B moves east at $v_{Bf} = +0.56\,m/s$.

  • Derivation of Module B Mass:

  • $m_A v_{Ai} + m_B v_{Bi} = m_A v_{Af} + m_B v_{Bf}$

  • $(150)(+1.2) + m_B(-0.40) = (150)(-0.40) + m_B(+0.56)$

  • $180 - 0.40 m_B = -60 + 0.56 m_B$

  • Add $60$ and add $0.40 m_B$: $240 = 0.96 m_B$

  • Solve for mass: $m_B = \frac{240}{0.96} = 250\,kg$

  • Module Momentum Inventory:

  • Initial momentum of Module A: $p_{Ai} = (150\,kg)(+1.2\,m/s) = +180\,kg\,m/s$

  • Initial momentum of Module B: $p_{Bi} = (250\,kg)(-0.40\,m/s) = -100\,kg\,m/s$

  • Total initial system momentum: $p_i = 180 + (-100) = +80\,kg\,m/s$

  • Final momentum of Module A: $p_{Af} = (150\,kg)(-0.40\,m/s) = -60\,kg\,m/s$

  • Final momentum of Module B: $p_{Bf} = (250\,kg)(+0.56\,m/s) = +140\,kg\,m/s$

  • Total final system momentum: $p_f = -60 + 140 = +80\,kg\,m/s$

  • Momentum change of Module A: $\Delta p_A = -60 - 180 = -240\,kg\,m/s$ (decreases by $240\,kg\,m/s$)

  • Momentum change of Module B: $\Delta p_B = 140 - (-100) = +240\,kg\,m/s$ (increases by $240\,kg\,m/s$)

  • Absolute exchange: $\Delta p_A = -\Delta p_B$

  • Protective Systems Comparison and Energy Dissipation:

  • Initial Total Kinetic Energy: $KE_i = \frac{1}{2}(150)(1.2)^2 + \frac{1}{2}(250)(-0.40)^2 = 108 + 20 = 128\,J$

  • Rigid Bumper Test:

    • Final velocities: Module A rebounds at $-0.80\,m/s$, Module B moves at $+0.80\,m/s$.

    • Final KE: $KE_f = \frac{1}{2}(150)(-0.80)^2 + \frac{1}{2}(250)(0.80)^2 = 48 + 80 = 128\,J$

    • Classification: Elastic ($KE_i = KE_f = 128\,J$).

  • Flexible Bumper Test:

    • Final velocities: Module A rebounds at $-0.40\,m/s$, Module B moves at $+0.56\,m/s$.

    • Final KE: $KE_f = \frac{1}{2}(150)(-0.40)^2 + \frac{1}{2}(250)(0.56)^2 = 12 + 39.2 = 51.2\,J$

    • Classification: Inelastic ($76.8\,J$ converted into deformation and thermal energy).

  • Locking Connector Test:

    • The modules link and move with common velocity $v_f$:

    • $p_i = (m_A + m_B) v_f \rightarrow +80\,kg\,m/s = (150\,kg + 250\,kg) v_f$

    • $80 = 400 v_f \rightarrow v_f = +0.2\,m/s$

    • Classification: Perfectly Inelastic.

  • Impulse Pulse Structure and Engineering Implications:

  • Rigid Bumper Pulse:

    • Duration: $\Delta t = 0.1\,s$

    • Peak force: $F_{\text{peak}} = 6000\,N$

    • $J_{\text{rigid}} = \frac{1}{2}(0.1\,s)(6000\,N) = 300\,N\,s$

  • Flexible Bumper Pulse:

    • Duration: $\Delta t = 0.2\,s$

    • Peak force: $F_{\text{peak}} = 2400\,N$

    • $J_{\text{flex}} = \frac{1}{2}(0.2\,s)(2400\,N) = 240\,N\,s$

  • Engineering Analysis:

    • The impulses differ between the designs because the final velocity changes of Module A differ: $|\Delta v_A|_{\text{rigid}} = |-0.80 - 1.2| = 2.0\,m/s$ ($J = 150 \times 2.0 = 300\,N\,s$), whereas $|\Delta v_A|_{\text{flex}} = |-0.40 - 1.2| = 1.6\,m/s$ ($J = 150 \times 1.6 = 240\,N\,s$).

    • The flexible bumper extends impact contact duration ($0.2\,s$ vs. $0.1\,s$), spreading momentum transfer over double the time, thereby reducing peak impact force from $6000\,N$ to $2400\,N$ (a $60\%$ reduction) to protect platform hardware.