Geometry Hunt Wall Stations
Previous Answer Check Did your previous station equal:
26
STATION 01 Proportional Segments
Loop Item 1/20
Triangle Proportionality Theorem
In \(\triangle ABC\) below, segment \(DE\) is parallel to base \(AC\). Using the given segment lengths, set up a proportion to solve for \(x\).
B A C D E 8 12 10 x
Find the value of x.
Side-Splitter: \(\frac{BD}{DA} = \frac{BE}{EC}\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 1 Box
Previous Answer Check Did your previous station equal:
15
STATION 02 Special Right Triangles
Loop Item 2/20
45°-45°-90° Triangle Relationships
In the isosceles right triangle shown below, the hypotenuse has a length of \(14\sqrt{2}\). Find the exact length of leg \(x\).
45° x leg 14√2
Find the length of leg x.
Rule: \(\text{hypotenuse} = \text{leg} \cdot \sqrt{2}\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 2 Box
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14
STATION 03 Angle Bisector Theorem
Loop Item 3/20
Triangle Angle Bisector Theorem
In \(\triangle ABC\), segment \(AD\) bisects vertex angle \(\angle BAC\). An angle bisector in a triangle divides the opposite side into segments proportional to the other two sides.
A B C D 9 15 6 x
Find the length of segment x (\(DC\)).
\(\frac{AB}{AC} = \frac{BD}{DC}\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 3 Box
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10
STATION 04 Parallel Lines & Transversals
Loop Item 4/20
Alternate Interior Angles
Parallel lines \(u\) and \(v\) are intersected by transversal \(t\). Identify the angle relationship for the two indicated angles and solve for \(x\).
u v t (7x − 4)° (4x + 32)°
Find the value of x.
Alternate Interior Angles are Congruent
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 4 Box
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12
STATION 05 Three Parallel Lines
Loop Item 5/20
Transversal Proportionality
Three parallel lines intersect two distinct transversals. By the Three Parallel Lines Theorem, the transversals are divided proportionally.
6 15 10 x
Find the length of unknown segment x.
\(\frac{6}{15} = \frac{10}{x}\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 5 Box
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25
STATION 06 Special Right Triangles
Loop Item 6/20
30°-60°-90° Triangle Properties
In a \(30^\circ-60^\circ-90^\circ\) right triangle, the shorter leg (opposite the \(30^\circ\) angle) has a length of 9. What is the length of hypotenuse \(c\)?
30° 60° 9 9√3 c
Find the length of hypotenuse c.
\(\text{hypotenuse} = 2 \cdot (\text{shorter leg})\)
Solve for c → Search the room for the station with your answer in the top banner!
Record in Station 6 Box
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18
STATION 07 Polygons
Loop Item 7/20
Exterior Angles & Number of Sides
The sum of the exterior angles of any convex polygon is always \(360^\circ\). A certain regular polygon has an exterior angle measuring \(45^\circ\). How many sides (\(n\)) does this polygon have?
45° Regular n-gon
Find the number of sides n.
\(\text{Exterior Angle} = \frac{360^\circ}{n}\)
Solve for n → Search the room for the station with your answer in the top banner!
Record in Station 7 Box
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8
STATION 08 Angle Relationships
Loop Item 8/20
Linear Pairs & Supplementary Angles
Two adjacent angles form a linear pair on a straight line. By the Linear Pair Postulate, their measures add up to \(180^\circ\). Solve for \(x\).
(5x + 10)° (3x + 10)°
Find the value of x.
\((5x + 10) + (3x + 10) = 180\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 8 Box
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20
STATION 09 Triangle Properties
Loop Item 9/20
Triangle Exterior Angle Theorem
The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles. Set up an equation and solve for \(x\).
(8x − 2)° 40° (5x + 9)°
Find the value of x.
\((8x - 2) = 40 + (5x + 9)\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 9 Box
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17
STATION 10 Coordinate Geometry
Loop Item 10/20
Distance Formula
Find the straight-line distance between the points \(A(-3, 7)\) and \(B(9, 2)\) in the Cartesian coordinate plane.
A(−3, 7) B(9, 2) Δx = 12 Δy = 5 d = ?
Find the distance d between points A and B.
\(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Solve for d → Search the room for the station with your answer in the top banner!
Record in Station 10 Box
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13
STATION 11 Side-Splitter Algebra
Loop Item 11/20
Algebraic Proportions in Triangles
In \(\triangle PQR\), segment \(ST\) is parallel to base \(QR\). Using the Side-Splitter Theorem, write a proportion to solve for \(x\).
P Q R S T x + 3 6 21 9
Find the value of x.
\(\frac{x + 3}{6} = \frac{21}{9}\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 11 Box
Previous Answer Check Did your previous station equal:
11
STATION 12 Isosceles Triangles
Loop Item 12/20
Isosceles Triangle Base Angles
In \(\triangle KLM\), side \(KL \cong KM\). The vertex angle \(\angle K\) measures \(48^\circ\). Base angle \(\angle L\) is represented by \((2x + 6)^\circ\). Find the value of \(x\).
K 48° L M (2x + 6)°
Find the value of x.
\(\text{Base Angle} = \frac{180^\circ - 48^\circ}{2} = 66^\circ\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 12 Box
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30
STATION 13 Linear Equations
Loop Item 13/20
Perpendicular Lines & Y-Intercept
Line \(L\) passes through the point \((-2, 1)\) and is perpendicular to the line:
\(y = -\frac{1}{3}x + 5\)
Write the equation of line \(L\) in slope-intercept form (\(y = mx + b\)). What is the value of the y-intercept, \(b\)?
y = −1/3 x + 5 Line L (⊥) (−2, 1) (0, b)
Find the value of the y-intercept, b.
\(m_\perp = 3 \implies y - 1 = 3(x + 2)\)
Solve for b → Search the room for the station with your answer in the top banner!
Record in Station 13 Box
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7
STATION 14 Angle Addition
Loop Item 14/20
Angle Addition Postulate
Point \(W\) lies in the interior of \(\angle HIJ\). The measures are \(m\angle HIW = (3x + 4)^\circ\), \(m\angle WIJ = (5x - 2)^\circ\), and the total measure is \(m\angle HIJ = 130^\circ\). Solve for \(x\).
I J H W (3x + 4)° (5x − 2)° Total ∠HIJ = 130°
Find the value of x.
\((3x + 4) + (5x - 2) = 130\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 14 Box
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16
STATION 15 Triangle Inequality
Loop Item 15/20
Possible Measures for the Third Side
Two known sides of a triangle measure 14 and 23. By the Triangle Inequality Theorem, the third side \(x\) must satisfy the compound inequality \(a < x < b\). What is the strict lower bound, \(a\)?
14 23 x (Third Side) a < x < b
Find the strict lower bound a.
\(\text{Difference} < x < \text{Sum}\)
Solve for a → Search the room for the station with your answer in the top banner!
Record in Station 15 Box
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9
STATION 16 Angle Bisector Algebra
Loop Item 16/20
Angle Bisector Theorem with Expressions
Ray \(AD\) bisects vertex angle \(\angle A\) of \(\triangle ABC\). Set up the ratio of adjacent sides to divided base segments and solve for \(x\).
A B C D x − 3 35 9 15
Find the value of x.
\(\frac{x - 3}{35} = \frac{9}{15}\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 16 Box
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24
STATION 17 Triangle Congruence
Loop Item 17/20
Corresponding Parts of Congruent Triangles (CPCTC)
Given that \(\triangle PQR \cong \triangle STU\), corresponding sides are equal in length. Side \(PQ = 3x - 4\) and corresponding side \(ST = 59\). Solve for \(x\).
R P Q PQ = 3x − 4 ≅ U S T ST = 59
Find the value of x.
\(3x - 4 = 59\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 17 Box
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21
STATION 18 Regular Polygons
Loop Item 18/20
Interior Angle of a Regular Polygon
What is the measure (in degrees) of ONE interior angle of a regular octagon (8 sides)?
Regular Octagon (n = 8) Interior ∠ = ?
Find the measure of each interior angle in degrees.
\(\frac{(n - 2) \cdot 180^\circ}{n}\)
Solve for the angle → Search the room for the station with your answer in the top banner!
Record in Station 18 Box
Previous Answer Check Did your previous station equal:
135
STATION 19 Consecutive Interior Angles
Loop Item 19/20
Consecutive Interior Angles Theorem
Parallel lines \(u\) and \(v\) are intersected by transversal \(t\). The two consecutive interior (same-side interior) angles measure \((5x - 8)^\circ\) and \((3x + 12)^\circ\). What is the value of \(x\)?
u v (5x − 8)° (3x + 12)°
Find the value of x.
\((5x - 8) + (3x + 12) = 180\)
Solve for x → Search the room for the station with your answer in the top banner!
Record in Station 19 Box
Previous Answer Check Did your previous station equal:
22
STATION 20 Pythagorean Theorem
Loop Item 20/20
Finding the Hypotenuse
In the right triangle shown below, the two legs measure 10 and 24. Use the Pythagorean Theorem (\(a^2 + b^2 = c^2\)) to solve for the length of the hypotenuse, \(c\).
10 24 c
Find the length of hypotenuse c.
\(10^2 + 24^2 = c^2 \implies c = \sqrt{676}\)
Solve for c → Search the room for the station with your answer in the top banner!
Record in Station 20 Box
Geometry Hunt Facilitation Guide
Teacher Facilitation Guide
Geometry Loop Showdown
20-Station Circular Scavenger Hunt • High School Geometry Review
Recommended Time: 50–60 Min
Group Size: Pairs or Trios
Total Stations: 20 Circular Loop
Instructional Intent & Activity Mechanics
The Geometry Loop Showdown synthesizes core geometric reasoning from two key curricular units: Proportional Parts (Side-Splitter, Angle Bisector, 3-Parallel Lines) and Mid-Semester Review (Parallel Line Angle Theorems, Polygon Angles, Special Right Triangles, Congruence, and Coordinate Geometry). Because the 20 stations form an unbroken circular loop, students can begin at any station. Finding their computed answer in the top banner of their next card provides immediate, self-correcting feedback without revealing the answers in advance.
Standards Alignment & Source Worksheet Mapping
Unit 7 Proportional Parts CW 4 Stations
- • Station 01: Triangle Proportionality / Side-Splitter
- • Station 03: Angle Bisector Theorem (Numeric)
- • Station 05: Three Parallel Lines Transversal Theorem
- • Station 11 & 16: Algebraic Side-Splitter & Bisector
Mid-Semester Review #4 Modified 15 Stations
- • Stations 04 & 19: Parallel Lines (Alt. Int., Cons. Int.)
- • Stations 02 & 06: Special Right Triangles (\(45^\circ\) & \(30^\circ\))
- • Stations 07 & 18: Polygon Angle Theorems (\(n\)-gons)
- • Stations 08, 09, 10, 12, 13, 14, 15, 17, 20: Angles, Lines & \(\Delta\)s
Suggested Lesson Timeline & Room Setup
1 Launch (5–7 min)
Distribute workspaces. Assign pairs to starting stations (1–2 pairs per station). Explain the loop rule and the 4-tier reflection rubric.
2 The Hunt (35–40 min)
Students move around the perimeter. Circulate with this guide. If a student cannot find their answer, guide them to re-check their algebra.
3 Debrief (8–10 min)
Students tally their 1s, 2s, 3s, and 4s. Use hand raises to identify the top 2 misconception stations for whole-class mini-remediation.
Teacher Questioning Prompts for Stuck Students
Station 13 (Perpendicular Lines):
"What is the negative reciprocal of \(-1/3\)? Did you write \(y - y_1 = m(x - x_1)\) or solve for \(b\) using \((-2, 1)\)?"
Station 19 (Consecutive Interior):
"Look at the angle types: are they alternate (equal) or on the same side inside (supplementary, add to \(180^\circ\))?"
Part 1: Facilitation Blueprint • Turn over for Master Loop & Worked Solutions Page 1 of 3
Master Key
Loop Verification & Worked Solutions (01 – 10)
Loop: Circular • Length: 20
Quick Loop Order: 01 (15) → 02 (14) → 03 (10) → 04 (12) → 05 (25) → 06 (18) → 07 (8) → 08 (20) → 09 (17) → 10 (13)...
| Station |
Topic |
Prev. Ans |
Step-by-Step Worked Solution |
Final Ans |
Next Card |
| 01 |
Side-Splitter |
26 |
\(\frac{8}{12} = \frac{10}{x} \implies 8x = 120 \implies x = \frac{120}{8} = 15\) |
15 |
St. 02 |
| 02 |
45°-45°-90° \(\Delta\) |
15 |
\(\text{hyp} = \text{leg}\sqrt{2} \implies 14\sqrt{2} = x\sqrt{2} \implies x = 14\) |
14 |
St. 03 |
| 03 |
Angle Bisector |
14 |
\(\frac{9}{15} = \frac{6}{x} \implies 9x = 90 \implies x = 10\) |
10 |
St. 04 |
| 04 |
Alternate Interior |
10 |
\(7x - 4 = 4x + 32 \implies 3x = 36 \implies x = 12\) |
12 |
St. 05 |
| 05 |
3 Parallel Lines |
12 |
\(\frac{6}{15} = \frac{10}{x} \implies 6x = 150 \implies x = 25\) |
25 |
St. 06 |
| 06 |
30°-60°-90° \(\Delta\) |
25 |
\(\text{hyp} = 2 \cdot (\text{shorter leg}) \implies c = 2(9) = 18\) |
18 |
St. 07 |
| 07 |
Exterior Angle \(n\) |
18 |
\(\text{ext} = \frac{360^\circ}{n} \implies 45^\circ = \frac{360^\circ}{n} \implies n = \frac{360}{45} = 8\) |
8 |
St. 08 |
| 08 |
Linear Pair |
8 |
\((5x + 10) + (3x + 10) = 180 \implies 8x + 20 = 180 \implies 8x = 160 \implies x = 20\) |
20 |
St. 09 |
| 09 |
Exterior Angle \(\Delta\) |
20 |
\(8x - 2 = 40 + (5x + 9) \implies 8x - 2 = 5x + 49 \implies 3x = 51 \implies x = 17\) |
17 |
St. 10 |
| 10 |
Distance Formula |
17 |
\(d = \sqrt{(9 - (-3))^2 + (2 - 7)^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\) |
13 |
St. 11 |
Check for Understanding Shortcut: If a student has solved Station 05 and says they arrived at Station 06, verify their Station 05 work area shows \(x = 25\). Because Station 06's previous answer banner is 25, the loop mechanics confirm accuracy immediately.
Part 2: Stations 01–10 • Turn over for Stations 11–20 & Debrief Rubric Page 2 of 3
Master Key
Loop Verification & Worked Solutions (11 – 20)
Loop Closes: 20 → 01
Quick Loop Order: 11 (11) → 12 (30) → 13 (7) → 14 (16) → 15 (9) → 16 (24) → 17 (21) → 18 (135) → 19 (22) → 20 (26 → St. 01)
| Station |
Topic |
Prev. Ans |
Step-by-Step Worked Solution |
Final Ans |
Next Card |
| 11 |
Side-Splitter Alg. |
13 |
\(\frac{x+3}{6} = \frac{21}{9} \implies 9(x+3) = 126 \implies 9x = 99 \implies x = 11\) |
11 |
St. 12 |
| 12 |
Isosceles \(\Delta\) Angles |
11 |
\(\text{Base } \angle = \frac{180 - 48}{2} = 66^\circ \implies 2x + 6 = 66 \implies 2x = 60 \implies x = 30\) |
30 |
St. 13 |
| 13 |
Perpendicular \(y\)-int |
30 |
\(m_\perp = 3;\; y - 1 = 3(x + 2) \implies y = 3x + 6 + 1 = 3x + 7 \implies b = 7\) |
7 |
St. 14 |
| 14 |
Angle Addition |
7 |
\((3x + 4) + (5x - 2) = 130 \implies 8x + 2 = 130 \implies 8x = 128 \implies x = 16\) |
16 |
St. 15 |
| 15 |
\(\Delta\) Inequality |
16 |
\( |
23 - 14 |
< x < 23 + 14 \implies 9 < x < 37 \implies \text{lower bound } a = 9\) |
| 16 |
Angle Bisector Alg. |
9 |
\(\frac{x - 3}{35} = \frac{9}{15} \implies \frac{x-3}{35} = \frac{3}{5} \implies x - 3 = 21 \implies x = 24\) |
24 |
St. 17 |
| 17 |
\(\Delta\) Congruence CPCTC |
24 |
\(\triangle PQR \cong \triangle STU \implies PQ = ST \implies 3x - 4 = 59 \implies 3x = 63 \implies x = 21\) |
21 |
St. 18 |
| 18 |
Regular Octagon \(\angle\) |
21 |
\(\text{int } \angle = \frac{(8-2)\cdot 180^\circ}{8} = \frac{1080^\circ}{8} = 135^\circ\) |
135 |
St. 19 |
| 19 |
Consecutive Interior |
135 |
\((5x - 8) + (3x + 12) = 180 \implies 8x + 4 = 180 \implies 8x = 176 \implies x = 22\) |
22 |
St. 20 |
| 20 |
Pythagorean Theorem |
22 |
\(10^2 + 24^2 = c^2 \implies 100 + 576 = 676 \implies c = \sqrt{676} = 26\) |
26 |
St. 01 (Loop!) |
Actionable Post-Hunt Differentiation Plan
Collect or review the Confidence Tally at the bottom of Page 2 of the student workspace to form data-driven groups for the next instruction block:
Target Group A (High '1s' or '2s' in Stations 01, 03, 05, 11, 16):
Focus on setting up part-to-part ratios vs. part-to-whole ratios in triangles and parallel transversals using color highlighters.
Target Group B (High '1s' or '2s' in Stations 02, 06, 07, 18):
Anchor Special Right Triangles (\(1:1:\sqrt{2}\) and \(1:\sqrt{3}:2\)) and polygon angle sum templates \((n-2)180^\circ\) with reference index cards.
Part 3: Master Loop & Differentiation • End of Facilitation Guide Page 3 of 3
