Matching tick marks indicate geometric congruence and equal numerical angle measures.
Problem 3B Solution:
Congruence: \(\angle 3 \cong \angle 5\)
Measure: \(m\angle 5 = 62^\circ\)
Equality refers to numbers (\(m\angle 3 = m\angle 5 = 62^\circ\)); congruence refers to shapes (\(\angle 3 \cong \angle 5\)).
Topic 4 Setting Up Equations & Solving for \(x\) (Standard G.GF.1) Algebra Steps
Problem 4A Step-by-Step:
1. \(DE + EF = DF\)
2. \(4x + (3x + 7) = 9x - 1\)
3. \(7x + 7 = 9x - 1\)
4. Subtract \(7x\): \(7 = 2x - 1\)
5. Add 1: \(8 = 2x \implies \mathbf{x = 4}\)
Final Answer: \(x = 4\)
Problem 4B Step-by-Step:
1. \(JK + KL = JL\)
2. \((5x - 3) + (2x + 8) = 8x - 1\)
3. \(7x + 5 = 8x - 1\)
4. Subtract \(7x\): \(5 = x - 1\)
5. Add 1: \(\mathbf{x = 6}\)
Final Answer: \(x = 6\)
Common Algebra Error: When combining like terms on one side (\(4x + 3x = 7x\)), ensure students do not accidentally subtract or cancel terms across the equals sign until like terms are combined.
Teacher Guide & Key • Day 1 Page 2 of 5 Topics 3 & 4 Solutions
Teacher Key • Day 1 of 2
Standard G.GF.1 Multi-Step Practice
Instructional Focus: Checking algebraic answers by substituting back into geometric expressions and correctly ordering collinear points before forming equations.
Topic 5 Substituting Back to Find Total Length (Standard G.GF.1) Substitution
Problem 5A Solution:
Substitute \(x = 4\) into \(DF = 9x - 1\):
\(DF = 9(4) - 1 = 36 - 1 = \mathbf{35}\)
Check: \(DE = 4(4) = 16\), \(EF = 3(4) + 7 = 19\)
\(16 + 19 = 35\) ✓ Verification confirmed.
Final Answer: \(DF = 35\)
Problem 5B Solution:
Substitute \(x = 6\):
\(JK = 5(6) - 3 = 30 - 3 = \mathbf{27}\)
\(JL = 8(6) - 1 = 48 - 1 = \mathbf{47}\)
Check: \(KL = 2(6) + 8 = 20\)
\(27 + 20 = 47\) ✓
Final Answer: \(JK = 27, \; JL = 47\)
Topic 6 Re-Ordered Collinear Points (Standard G.GF.1) Point Order
Problem 6A Solution:
Point V is between W and U \(\implies WV + VU = WU\)
\((11x + 2) + (4x + 6) = 17x - 4\)
\(15x + 8 = 17x - 4 \implies 2x = 12 \implies \mathbf{x = 6}\)
\(WV = 11(6) + 2 = 66 + 2 = \mathbf{68}\)
Final Answer: \(x = 6, \; WV = 68\)
Problem 6B Solution:
\(GH + HI = GI \implies (5x - 1) + (3x + 7) = 46\)
\(8x + 6 = 46 \implies 8x = 40 \implies \mathbf{x = 5}\)
\(GH = 5(5) - 1 = \mathbf{24}\)
\(HI = 3(5) + 7 = \mathbf{22}\)
Since \(24 \neq 22\), H does not bisect GI.
Final Answer: [X] No, \(GH \neq HI\)
Read Carefully: Students who do not read that "V is between W and U" often set up \(WU + UV = WV\). Always have students underline the word "between" and draw a labeled line segment before solving.
Teacher Guide & Key • Day 1 Page 3 of 5 Topics 5 & 6 Solutions
Teacher Key • Day 1 of 2
Standard G.GF.2 Inductive Reasoning
Core Skill: Recognizing inductive patterns: recursive summation (each term equals the sum of the prior two) and second-difference arithmetic progressions.
Topic 7 Recursive & Fibonacci-Style Sequences (Standard G.GF.2) Recursive Sums
Problem 7A Solution:
Sequence: 2, 3, 5, 8, 13, 21, ...
Each term is the sum of the preceding two terms:
\(13 + 21 = \mathbf{34}\) (Next term)
\(21 + 34 = \mathbf{55}\) (Following term)
Final Answer: 34 and 55
Problem 7B Solution:
Sequence: 3, 4, 7, 11, 18, 29, ...
Rule: Add the preceding two terms.
7th term: \(18 + 29 = \mathbf{47}\)
Final Answer: 7th term = 47
Topic 8 Sequences with Growing Differences (Standard G.GF.2) Difference Rows
Problem 8A Solution:
Sequence: 3, 10, 18, 27, 37, ...
Differences: \(+7, +8, +9, +10\)
Next difference must be \(+11\):
\(37 + 11 = \mathbf{48}\)
Final Answer: 48
Problem 8B Solution:
Sequence: 4, 9, 15, 22, 30, ...
Differences: \(+5, +6, +7, +8\)
Next difference is \(+9\):
\(30 + 9 = \mathbf{39}\)
Final Answer: 39
Strategy Tip: Instruct students to write out the differences directly beneath the numbers in the sequence with a \(+\) sign. This makes growing patterns instantly visual.
Teacher Guide & Key • Day 1 Page 4 of 5 Topics 7 & 8 Solutions
Teacher Key • Day 1 of 2
Standard G.GF.2 Conditional Logic
Core Skill: Isolating hypothesis and conclusion clauses cleanly without "If" or "Then", and switching the order of clauses to form the converse statement (\(Q \to P\)).
Topic 9 Isolating Hypothesis & Conclusion (Standard G.GF.2) Clause Separation
Problem 9A Solution:
Hypothesis (\(p\)): "an angle measures greater than \(90^\circ\) and less than \(180^\circ\)"
Conclusion (\(q\)): "it is an obtuse angle"
Notice that "If" and "Then" are completely omitted from the answer.
Problem 9B Solution:
Hypothesis (\(p\)): "two coplanar lines do not intersect"
Conclusion (\(q\)): "the lines are parallel"
Pure clauses only.
Topic 10 Writing the Converse (\(Q \to P\)) (Standard G.GF.2) Converse Logic
Problem 10A Solution:
"If a person lives in Texas, then they live in Dallas."
Truth Value: [X] False / Cannot be determined (a person could live in Houston, Austin, or San Antonio).
Problem 10B Solution:
"If a geometric figure has four right angles, then it is a square."
Truth Value: [X] No, false (a non-square rectangle has 4 right angles but unequal adjacent sides).
Converse Pitfall: Students often assume that if a conditional statement is true, its converse must also be true. Emphasize counterexamples to disprove converses.
Teacher Guide & Key • Day 1 Page 5 of 5 Topics 9 & 10 Solutions
Problem 14A (Using Figure A):
Given: \(\overline{CE} \perp \overline{AB}\) at point \(E\).
Conclusion Reason
\(\angle CEA\) is a right angle
What is \(m\angle CEA\)?
Problem 14B (Perpendicular Deduction Check):
Given: Line \(u \perp\) Line \(v\) intersecting at point \(Z\). Check the valid deduction:
They intersect to form \(90^\circ\) right angles (Def. \(\perp\))
Point \(Z\) is the midpoint of line \(u\) (Def. Midpoint)
Why can't an entire line have a midpoint? Lines extend infinitely.
Geometry Benchmark Review • Day 2 Page 2 of 5 Topics 13 & 14: Midpoints & Perpendicular Lines
Geometry Benchmark Review • Day 2 of 2
Standard G.GF.3 Proof Justifications
Figure B: Subtraction
\(\triangle MPN\) with interior segments \(PK\) and \(PL\).
P M N K L
Figure C: Halves
Two congruent angles with interior bisecting rays.
E J
Topic 15 Segment Subtraction Postulate (Standard G.GF.3) Standard G.GF.3
Problem 15A (Collinear Subtraction):
Given collinear points \(A, B, C, D\) where \(AC = BD\). Segment \(BC\) is subtracted from both sides:
\(AC - BC = AB\) and \(BD - BC = CD\).
Conclusion: \(\overline{AB} \cong \overline{CD}\) Reason:
Problem 15B (Using Figure B):
In Figure B, Given: \(\overline{PM} \cong \overline{PN}\) and \(\overline{PK} \cong \overline{PL}\).
Select the valid deduction table:
Conclusion: \(\overline{KM} \cong \overline{LN}\) | Reason: Segment Subtraction Postulate
Conclusion: \(\overline{KM} \cong \overline{LN}\) | Reason: Segment Addition Postulate
Wholes are equal; congruent parts are subtracted.
Topic 16 Halves Postulate & Angle Bisectors (Standard G.GF.3) Standard G.GF.3
Problem 16A (Deduction Table):
Given: \(\angle E \cong \angle J\), ray \(\overrightarrow{EG}\) bisects \(\angle E\), and ray \(\overrightarrow{JK}\) bisects \(\angle J\).
Complete the deduction:
Conclusion Reason
The halves of \(\angle E\) and \(\angle J\) are \(\cong\)
Options: [Halves Postulate] or [Segment Addition Postulate]
Problem 16B (Define & Apply Halves Postulate):
State the Halves Postulate in your own words:
Application: If two \(110^\circ\) angles are bisected, each half measures .
Geometry Benchmark Review • Day 2 Page 3 of 5 Topics 15 & 16: Subtraction & Halves Postulates
Geometry Benchmark Review • Day 2 of 2
Standard G.GF.7 Coordinate Formulas
Midpoint Formula: Average of \(x\)-coordinates, Average of \(y\)-coordinates
\(M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right)\)
Topic 17 Midpoint Formula Calculations (Standard G.GF.7) 2 Practice Problems
Problem 17A (Original Coordinates):
Find the midpoint of segment \(\overline{AB}\) given endpoints \(A(6, 9)\) and \(B(2, -1)\):
Midpoint \(M = \)
Problem 17B (Decimal Coordinates):
Find the midpoint of segment \(\overline{CD}\) given endpoints \(C(3, 8)\) and \(D(8, 2)\):
Midpoint \(M = \)
Topic 18 Finding a Missing Endpoint from Midpoint (Standard G.GF.7) Algebraic Setup
Problem 18A (Working Backwards):
The midpoint of \(\overline{EF}\) is \(M(5, 3)\). Endpoint \(E\) is \((1, -2)\). Find the coordinates of endpoint \(F(x, y)\):
Endpoint \(F = \)
Problem 18B (Negative Coordinates):
The midpoint of \(\overline{GH}\) is \(M(0, 4)\). Endpoint \(G\) is \((-6, 1)\). Find the coordinates of endpoint \(H(x, y)\):
Endpoint \(H = \)
Geometry Benchmark Review • Day 2 Page 4 of 5 Topics 17 & 18: Midpoint Formula & Missing Endpoints
Geometry Benchmark Review • Day 2 of 2
Standard G.GF.7 Distance Formula
Distance Formula: Subtract coordinates, square each, add together, take square root.
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Topic 19 Distance Formula Calculations (Standard G.GF.7) 2 Practice Problems
Problem 19A (Original Coordinates):
Find the exact distance between points \(P(1, 2)\) and \(Q(7, 10)\). Show all substitution steps:
Distance \(d = \)
Problem 19B (Original Coordinates):
Find the exact distance between points \(R(2, 3)\) and \(S(6, 6)\). Show all substitution steps:
Distance \(d = \)
Topic 20 Distance with Negative Coordinates (Standard G.GF.7) Double Negatives
Problem 20A (Double Negatives):
Find the distance between \(A(-3, 4)\) and \(B(5, -2)\). Careful with minus signs:
Distance \(d = \)
Problem 20B (Negative Coordinates):
Find the distance between \(C(-4, -1)\) and \(D(2, 7)\):
Distance \(d = \)
Geometry Benchmark Review • Day 2 Page 5 of 5 Topics 19 & 20: Distance Formula Mastery
Problem 13B Solution:
Reason: Definition of Midpoint
\(SK = KT \implies 4x - 3 = 2x + 9\)
\(2x = 12 \implies \mathbf{x = 6}\)
Check: \(SK = 21, KT = 21\) ✓
Final Answer: \(x = 6\)
Topic 14 Definition of Perpendicular Lines (Standard G.GF.3) Perpendicular Proofs
Problem 14A Solution:
Conclusion: \(\angle CEA\) is a right angle
Reason: Definition of Perpendicular Lines
Measure: \(m\angle CEA = 90^\circ\)
Perpendicularity directly defines right angles.
Problem 14B Solution:
Correct Deduction: [X] They intersect to form \(90^\circ\) right angles (Def. \(\perp\))
Incorrect option: Points cannot bisect infinite lines; only finite segments can have midpoints.
Terminology Precision: Ensure students understand that perpendicularity (\(\perp\)) produces right angles, while the definition of right angles states that their measure is \(90^\circ\). Many textbooks combine these into the Definition of Perpendicular Lines.
Teacher Guide & Key • Day 2 Page 2 of 5 Topics 13 & 14 Solutions
Teacher Key • Day 2 of 2
Standard G.GF.3 Postulate Justifications
Key Deductive Postulates: Segment Subtraction: If equal parts are subtracted from equal wholes, the remainders are equal.
Halves Postulate: If two figures are congruent, their respective halves are congruent.
Topic 15 Segment Subtraction Postulate (Standard G.GF.3) Subtraction Proofs
Problem 15A Solution:
Conclusion: \(\overline{AB} \cong \overline{CD}\)
Reason: Segment Subtraction Postulate
Since \(BC\) is common to both segments, subtracting it from equal wholes leaves equal remainders.
Problem 15B Solution:
Correct Deduction: [X] Conclusion: \(\overline{KM} \cong \overline{LN}\) | Reason: Segment Subtraction Postulate
Whole segments \(\overline{PM} \cong \overline{PN}\); subtracting \(\overline{PK} \cong \overline{PL}\) leaves \(\overline{KM} \cong \overline{LN}\).
Topic 16 Halves Postulate & Angle Bisectors (Standard G.GF.3) Halves Postulate
Problem 16A Solution:
Conclusion: The halves of \(\angle E\) and \(\angle J\) are \(\cong\)
Reason: Halves Postulate (or Division Property of Congruence)
Whenever two congruent entities are divided into equal halves, the halves are congruent.
Problem 16B Solution:
Definition: "If two angles (or segments) are congruent, their halves are congruent."
Application: If two \(110^\circ\) angles are bisected, each half measures \(55^\circ\).
Addition vs. Subtraction Postulate: Remind students: when two smaller pieces combine to make a larger piece, it's the Addition Postulate. When a piece is removed from a larger whole to leave a remaining segment, it's the Subtraction Postulate.
Teacher Guide & Key • Day 2 Page 3 of 5 Topics 15 & 16 Solutions
Teacher Key • Day 2 of 2
Standard G.GF.7 Coordinate Midpoint
Midpoint Formula: \(M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right)\) — remember to add the coordinates, not subtract!
Topic 17 Midpoint Calculations (Standard G.GF.7) Midpoint Formula
Problem 17A Solution:
\(A(6, 9)\) and \(B(2, -1)\)
\(x_m = \frac{6 + 2}{2} = \frac{8}{2} = \mathbf{4}\)
\(y_m = \frac{9 + (-1)}{2} = \frac{8}{2} = \mathbf{4}\)
Final Answer: Midpoint \(M = (4, 4)\)
Problem 17B Solution:
\(C(3, 8)\) and \(D(8, 2)\)
\(x_m = \frac{3 + 8}{2} = \frac{11}{2} = \mathbf{5.5}\)
\(y_m = \frac{8 + 2}{2} = \frac{10}{2} = \mathbf{5}\)
Final Answer: Midpoint \(M = (5.5, 5)\)
Topic 18 Finding a Missing Endpoint from Midpoint (Standard G.GF.7) Working Backwards
Problem 18A Solution:
\(M(5, 3)\), Endpoint \(E(1, -2)\), Target \(F(x, y)\)
\(\frac{1 + x}{2} = 5 \implies 1 + x = 10 \implies \mathbf{x = 9}\)
\(\frac{-2 + y}{2} = 3 \implies -2 + y = 6 \implies \mathbf{y = 8}\)
Final Answer: Endpoint \(F = (9, 8)\)
Problem 18B Solution:
\(M(0, 4)\), Endpoint \(G(-6, 1)\), Target \(H(x, y)\)
\(\frac{-6 + x}{2} = 0 \implies -6 + x = 0 \implies \mathbf{x = 6}\)
\(\frac{1 + y}{2} = 4 \implies 1 + y = 8 \implies \mathbf{y = 7}\)
Final Answer: Endpoint \(H = (6, 7)\)
Frequent Missing Endpoint Error: Students frequently calculate the midpoint between the given endpoint and the midpoint! Remind them: the midpoint is the center, so they must set up the formula \(\frac{x_1 + x_2}{2} = x_m\) and multiply both sides by 2 first.
Teacher Guide & Key • Day 2 Page 4 of 5 Topics 17 & 18 Solutions
Teacher Key • Day 2 of 2
Standard G.GF.7 Distance Formula
Distance Formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) — remember: squaring any real number always yields a positive value!
Topic 19 Distance Calculations (Standard G.GF.7) Distance Formula
Problem 19A Solution:
\(P(1, 2)\) and \(Q(7, 10)\)
\(d = \sqrt{(7 - 1)^2 + (10 - 2)^2}\)
\(d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64}\)
\(d = \sqrt{100} = \mathbf{10}\)
Final Answer: \(d = 10\)
Problem 19B Solution:
\(R(2, 3)\) and \(S(6, 6)\)
\(d = \sqrt{(6 - 2)^2 + (6 - 3)^2}\)
\(d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9}\)
\(d = \sqrt{25} = \mathbf{5}\)
Final Answer: \(d = 5\)
Topic 20 Distance with Negative Coordinates (Standard G.GF.7) Double Negatives
Problem 20A Solution:
\(A(-3, 4)\) and \(B(5, -2)\)
\(d = \sqrt{(5 - (-3))^2 + (-2 - 4)^2}\)
\(d = \sqrt{(8)^2 + (-6)^2} = \sqrt{64 + 36}\)
\(d = \sqrt{100} = \mathbf{10}\)
Final Answer: \(d = 10\)
Problem 20B Solution:
\(C(-4, -1)\) and \(D(2, 7)\)
\(d = \sqrt{(2 - (-4))^2 + (7 - (-1))^2}\)
\(d = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64}\)
\(d = \sqrt{100} = \mathbf{10}\)
Final Answer: \(d = 10\)
Double Negative & Squaring Alert: Two of the most common errors on benchmark exams: 1) forgetting that \(x_2 - (-x_1) = x_2 + x_1\), and 2) entering \(-6^2 = -36\) on a calculator instead of \((-6)^2 = +36\). Remind students that distance inside the radical is always a sum of positive squares!
Teacher Guide & Key • Day 2 Page 5 of 5 Topics 19 & 20 Solutions
Conditional
\(P \to Q\) If P, then Q
Swap Order
Converse
\(Q \to P\) If Q, then P
Negate Both
Inverse
\(\sim P \to \sim Q\) If not P, then not Q
Swap & Negate
Contrapositive
\(\sim Q \to \sim P\) If not Q, then not P
Day 1/2 Review • Slide 5 of 10 Conditional and Contrapositive have identical truth values!
Day 2 • Topics 13 & 14 Standard G.GF.3
A midpoint divides a line segment into two congruent segments:
\(M \text{ is midpoint of } \overline{GH} \implies \mathbf{\overline{GM} \cong \overline{MH}}\)
Algebraic setup: Set the two equal pieces equal! (\(4x - 3 = 2x + 9\)).
Two lines or segments are perpendicular if and only if they intersect to form right angles (\(90^\circ\)):
\(\overline{CE} \perp \overline{AB} \implies \mathbf{\angle CEA \text{ is a right angle } (90^\circ)}\)
Visual cue: Look for the square right-angle mark!
Day 2 Review • Slide 6 of 10 Deductive Reasoning & Justifications
Day 2 • Topics 15 & 16 Postulate Proofs
If equal segments are subtracted from equal wholes, the remainders are congruent!
Whole: \(AC = BD\)
Subtract \(BC\): \(AC - BC = BD - BC\)
Result: \(\mathbf{\overline{AB} \cong \overline{CD}}\)
If two angles or segments are congruent, their respective halves are congruent!
Given: \(\angle E \cong \angle J\)
Both are bisected into 2 equal halves
Result: \(\mathbf{\text{Halves are congruent!}}\)
Day 2 Review • Slide 7 of 10 Proof Justification Checklist
Day 2 • Topics 17 & 18 Coordinate Midpoint
Midpoint Formula (Average the coordinates!)
\(M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right)\)
Find midpoint of \(A(6, 9)\) and \(B(2, -1)\):
\(x_m = \frac{6+2}{2} = \frac{8}{2} = \mathbf{4}\)
\(y_m = \frac{9+(-1)}{2} = \frac{8}{2} = \mathbf{4}\)
Midpoint = \((4, 4)\)
Given Midpoint \(M(5, 3)\) & Endpoint \(E(1, -2)\):
\(\frac{1+x}{2} = 5 \implies 1+x = 10 \implies \mathbf{x = 9}\)
\(\frac{-2+y}{2} = 3 \implies -2+y = 6 \implies \mathbf{y = 8}\)
Target Endpoint = \((9, 8)\)
Day 2 Review • Slide 8 of 10 Remember to add (not subtract) when finding the midpoint!
Day 2 • Topics 19 & 20 Distance Formula
Distance Formula (Pythagorean Theorem on a Grid!)
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\(d = \sqrt{(7 - 1)^2 + (10 - 2)^2}\)
\(d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64}\)
\(d = \sqrt{100} = \mathbf{10}\)
Points \(A(-3, 4)\) & \(B(5, -2)\):
\(d = \sqrt{(5 - (-3))^2 + (-2 - 4)^2}\)
\(d = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36}\)
\(d = \sqrt{100} = \mathbf{10}\)
Day 2 Review • Slide 9 of 10 Crucial: Squaring a negative number always produces a positive result!
Benchmark Ready Exam Strategies
1
Never assume alphabetical order! Draw the segment and label which point is between the other two.
2
When asked for the hypothesis or conclusion, exclude trigger words completely.
3
Midpoint adds coordinates and divides by 2. Distance subtracts, squares, and takes the square root.
4
Did the problem ask for \(x\) or the segment length? Always verify what the prompt actually asked for!
Good Luck on Your Benchmark! You are prepared!
Find midpoint of \(A(4, 7)\) and \(B(10, -1)\):
\(M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) \(M = \)
2. Distance Formula (G.GF.7):
Find distance between \(C(1, 2)\) and \(D(4, 6)\):
\(d = \sqrt{\Delta x^2 + \Delta y^2}\) \(d = \)
3. Deduction & Contrapositive (G.GF.2 & G.GF.3):
Given: \(M\) is midpoint of \(\overline{PQ}\).
Conclusion: \(\overline{PM} \cong \overline{MQ}\) Reason:
"If you live in Boston, then you live in MA."
Contrapositive:
How confident do you feel on Day 2 skills? [ ] 1-Need Help [ ] 2-Getting There [ ] 3-Got It [ ] 4-Ready to Ace!
