2 Bases = \(2 \times (\frac{1}{2} \times 12 \times 5) = 60\)
3 Rectangles = \((5 \times 10) + (12 \times 10) + (13 \times 10) = 300\)

Total SA = \(60 + 300 = \) _____ m²

Chamber 2 Glyph Formula

Take the hundreds digit of Relic C SA (360) plus the ones digit of Relic B SA (150) plus 5 to get Glyph Beta (\(\beta\)).

Glyph \(\beta\)

?

Polyhedron Tomb Escape • Station Cards Chamber 2 of 4

3

Chamber Station 3

The Sanctuary of Pyramids

Target: Pyramid Surface Area

Chamber Inscription: "Every pyramid rises from one foundation to meet at the heavens. To calculate total surface area, sum the area of the base polygon and all triangular faces meeting at the apex!"

A Relic A: The Golden Apex of Ra

Square Pyramid

l = 13 cm b = 10 cm 10 cm

Base is a square with side length 10 cm. Slant height (\(l\)) of triangular faces = 13 cm.

• Base Area = \(10 \times 10 = 100\text{ cm}^2\)

• 4 Triangles = \(4 \times (\frac{1}{2} \times 10 \times 13) = ?\)

Total Surface Area = _____ cm²

B Relic B: Ruby Pyramidal Amulet

Triangular Pyramid

slant = 9 in Base edge = 8 in Base h = 7 in

Base triangle has base = 8 in, height = 7 in. Each of the 3 lateral triangles has base = 8 in, slant height = 9 in.

• Base Area = \(\frac{1}{2} \times 8 \times 7 = 28\text{ in}^2\)

• 3 Lateral Faces = \(3 \times (\frac{1}{2} \times 8 \times 9) = ?\)

Total Surface Area = _____ in²

Chamber 3 Glyph Formula

Take the tens digit of Relic B Surface Area (136) and add 1 to reveal Glyph Gamma (\(\gamma\)): \(3 + 1 = \) 4.

Glyph \(\gamma\)

?

Polyhedron Tomb Escape • Station Cards Chamber 3 of 4

4

Chamber Station 4

The Final Portal & Master Cipher

Target: Cipher Decoding

The Final Portal Inscription: "You have proven your mastery over nets, prisms, and pyramids. Calculate the Grand Obelisk's surface area to reveal Glyph Delta (\(\delta\)), then align all 4 glyphs into the Master Cipher Lock to unlock the tomb doors!"

★ The Grand Obelisk Monument Challenge

Prism Base + Pyramid Cap

slant=5 Base = 4m × 4m Shaft H = 8m

The monument stands in the courtyard. Find the exposed surface area (do not include the bottom floor or the joined face where the pyramid sits on the shaft):

• 4 Shaft Walls (Rectangles): \(4 \times (4 \times 8) = 128\text{ m}^2\)

• 4 Pyramid Roof Faces: \(4 \times (\frac{1}{2} \times 4 \times 5) = 40\text{ m}^2\)

• Total Exposed SA = \(128 + 40 = 168\text{ m}^2\)

Glyph Delta (\(\delta\)): Divide 168 by 84 = 2!

The Pharaoh's 4-Digit Portal Lock Mechanism

Enter your 4 chamber glyph values into the lock dials below in sequence to test your escape code with the Expedition Leader!

Dial 1 (\(\alpha\))

Chamber 1

3

Net Master

Dial 2 (\(\beta\))

Chamber 2

8

Prism Master

Dial 3 (\(\gamma\))

Chamber 3

4

Pyramid Master

Dial 4 (\(\delta\))

Chamber 4

2

Portal Master

Escape Code Verification: Show completed Quest Log to Teacher to confirm your code!

Polyhedron Tomb Escape • Station Cards Chamber 4 of 4

Dial 1 (\(\alpha\))

Dial 2 (\(\beta\))

Dial 3 (\(\gamma\))

Dial 4 (\(\delta\))

Expedition Leader Sign-off: Present this completed Quest Log to the teacher to verify your code and escape!

Teacher Stamp

Expedition Debrief: Why is drawing or unfolding a 3D shape into a 2D net helpful when calculating total surface area?

Polyhedron Tomb Escape • Explorer Quest Log Page 2 of 2

• 3 \(\Delta\) Faces = \(3 \times (\frac{1}{2} \times 8 \times 9) = 3 \times 36 = 108\text{ in}^2\)

• Total SA = \(28 + 108 = 136\text{ in}^2\)

Glyph Gamma (\(\gamma\)) Derivation: Tens digit of Relic B SA (\(1\mathbf{3}6 \rightarrow 3\)) + 1 = \(3 + 1 = \) 4.

Chamber 4 Solutions: Grand Obelisk & Final Cipher Glyph \(\delta = 2\)

• 4 Shaft Walls = \(4 \times (4 \times 8) = 128\text{ m}^2\)

• 4 Pyramid Cap Roofs = \(4 \times (\frac{1}{2} \times 4 \times 5) = 40\text{ m}^2\)

• Total Exposed SA = \(128 + 40 = 168\text{ m}^2\)

Glyph \(\delta = 168 \div 84 = \) 2

Master Lock Code

3 - 8 - 4 - 2

(\(\alpha=3, \beta=8, \gamma=4, \delta=2\))

Teacher Troubleshooting: Common 6th Grade Pitfalls

1. Forgetting the \(\frac{1}{2}\) on Triangles: Students frequently compute \(b \times h\) instead of \(\frac{1}{2} b h\) for individual triangular faces. Remind them that two congruent triangular bases combine to equal \(b \times h\).

2. Slant Height vs. Prism Height: In pyramids, students sometimes search for the interior vertical height rather than the slant height (\(l\)) that serves as the altitude of the triangular face.

3. Triangular Prism Lateral Faces: When triangle bases are scalene or right (such as 5-12-13), the three rectangular sides are NOT identical! They have different widths (5×10, 12×10, 13×10).

4. Overcounting Composite Faces: On the Grand Obelisk, some students mistakenly include the square top of the prism and base of the pyramid. Emphasize that surface area only counts *exposed* outer surfaces.

Exemplar Student Debrief Response:

"Unfolding a 3D solid into a 2D net shows every single face laid out flat. This makes sure we don't forget hidden faces (like the bottom or back) and lets us see exactly what 2D shape each face is (rectangle or triangle) so we use the correct area formula."

Polyhedron Tomb Escape • Teacher Master Key Page 2 of 2