Boundary Masters Worksheet
Level 1: Grid Boxes Mathematics • Perimeter & Area
Perimeter & Area Workshop
Count boundary units and square units for each rectangular box.
Name:
Date:
Class:
Score: / 12
P
Perimeter (Distance Around): Sum of all 4 outer edges (units)
A
Area (Covered Space): Total unit squares inside (sq units)
A Rectangular Grid Boxes
Each grid square = 1 unit × 1 unit
1. Rectangle Box A 5 units × 3 units
Perimeter Work & Solution:
P = _____ + _____ + _____ + _____ P = _____ u
Area Work & Solution:
A = _____ rows × _____ cols A = _____ sq u
2. Square Box B 4 units × 4 units
Perimeter Work & Solution:
P = _____ + _____ + _____ + _____ P = _____ u
Area Work & Solution:
A = _____ rows × _____ cols A = _____ sq u
3. Rectangle Box C 6 units × 2 units
Perimeter Work & Solution:
P = _____ + _____ + _____ + _____ P = _____ u
Area Work & Solution:
A = _____ rows × _____ cols A = _____ sq u
4. Rectangle Box D 7 units × 3 units
Perimeter Work & Solution:
P = _____ + _____ + _____ + _____ P = _____ u
Area Work & Solution:
A = _____ rows × _____ cols A = _____ sq u
Perimeter & Area Workshop • Page 1 of 3 Remember: Perimeter = distance around (units) | Area = total surface inside (square units)
Level 2
Rectangles & Formulas
Name:
Perimeter Box P = 2l + 2w or P = 2(l + w)
Area Box A = length × width (A = l × w)
B Dimensioned Rectangles & Missing Sides
Show complete formula steps with proper units
5. Rectangle Box EFGH 12 cm × 5 cm
12 cm 12 cm 5 cm 5 cm
Perimeter Formula & Work:
Final Perimeter: P = ________ cm
Area Formula & Work:
Final Area: A = ________ cm²
6. Square Box JKLM Side s = 8 m
8 m 8 m 8 m 8 m
Perimeter Formula & Work:
Final Perimeter: P = ________ m
Area Formula & Work:
Final Area: A = ________ m²
7. Detective Box: Perimeter P = 32 in
A rectangular display has a length of 10 in and a perimeter of 32 in.
Show Your Work (Find Missing Width w):
Missing Width (w): w = _____ in
Calculated Area (A): A = _____ in²
8. Detective Box: Area A = 54 ft²
A rectangular rug covers an area of 54 ft² and has a width of 6 ft.
Show Your Work (Find Missing Length l):
Missing Length (l): l = _____ ft
Calculated Perimeter (P): P = _____ ft
Perimeter & Area Workshop • Page 2 of 3 Linear Units: cm, m, in, ft | Square Units: cm², m², in², ft²
Level 3
Real-World Box Applications
Name:
C Rectangular Real-World Scenarios
Solve both boundary (perimeter) and coverage (area)
9. The Community Playground Yard 16 m by 9 m
A neighborhood park is building a rectangular playground that measures 16 meters long and 9 meters wide. A wooden fence will enclose the outer perimeter, and protective rubber turf will cover the entire play area inside.
Part A: Outer Fence (Perimeter) Show formula setup and addition:
Total Fencing: P = ________ m
Part B: Rubber Turf Surface (Area) Show multiplication calculation:
Total Turf: A = ________ m²
10. Art Gallery Poster Frame 24 inches by 18 inches
An artist needs custom framing for a rectangular painting that measures 24 inches long and 18 inches wide. The gold wood molding goes along the outer edge, and a sheet of clear glass covers the front.
Part A: Wood Frame Molding (Perimeter) Show calculation steps:
Frame Length: P = ________ in
Part B: Glass Cover (Area) Show multiplication calculation:
Glass Area: A = ________ in²
11. Classroom Reading Rug & Border Tape 11 feet by 7 feet
A rectangular reading rug is 11 feet long and 7 feet wide. Colored boundary tape costs $2 per foot around the perimeter, and deep carpet cleaning costs $1 per sq foot.
Part A: Tape Around Boundary (P & Cost)
Perimeter: _____ ft Cost: $________
Part B: Rug Cleaning Area (A & Cost)
Area: _____ ft² Cost: $________
Perimeter & Area Workshop • Page 3 of 3 Challenge Complete! All problems are standard rectangular boxes with full working space.
Boundary Masters Answer Key
Teacher Answer Key Page 1 • Level 1 Solutions
Perimeter & Area Workshop
Rectangular grid solutions with full scoring guidelines.
Grading Guide 1 pt perimeter + 1 pt area Total: 8 pts on Page 1
A Rectangular Grid Solutions (Problems 1–4)
Verify units: linear u vs sq u
1. Rectangle Box A (5 × 3) 2 pts
Perimeter:
P = 5 + 3 + 5 + 3 = 2(5) + 2(3)
P = 16 units
Area:
A = 5 columns × 3 rows
A = 15 sq units
2. Square Box B (4 × 4) 2 pts
Perimeter:
P = 4 + 4 + 4 + 4 = 4 × 4
P = 16 units
Area:
A = 4 columns × 4 rows
A = 16 sq units
3. Rectangle Box C (6 × 2) 2 pts
Perimeter:
P = 6 + 2 + 6 + 2 = 2(6) + 2(2)
P = 16 units
Area:
A = 6 columns × 2 rows
A = 12 sq units
4. Rectangle Box D (7 × 3) 2 pts
Perimeter:
P = 7 + 3 + 7 + 3 = 2(7) + 2(3)
P = 20 units
Area:
A = 7 columns × 3 rows
A = 21 sq units
Teacher Discussion Note Notice Problems 1, 2, and 3:
Boxes A, B, and C all have the exact same perimeter of 16 units, but their areas are completely different: Box B (square) = 16 sq u, Box A = 15 sq u, and Box C (narrow) = 12 sq u. This is an excellent moment to show students that square boxes enclose the maximum possible area for a given perimeter.
Perimeter & Area Workshop • Teacher Key • Page 1 of 3 Page 1 Points: 8
Teacher Answer Key
Level 2 Solutions: Rectangles & Formulas
Total: 8 pts on Page 2
B Dimensioned Rectangles Solutions (Problems 5–8)
Deduct 0.5 pt if units are omitted
5. Rectangle Box EFGH 2 pts
Perimeter:
P = 2(12) + 2(5) = 24 + 10
P = 34 cm
Area:
A = 12 × 5
A = 60 cm²
6. Square Box JKLM 2 pts
Perimeter:
P = 4 × 8 = 32
P = 32 m
Area:
A = 8 × 8
A = 64 m²
7. Detective: Given Perimeter 2 pts
2(10) + 2w = 32 → 20 + 2w = 32 → 2w = 12
Width (w): 6 in Area (A): 60 in²
Area calculation: A = 10 × 6 = 60 in²
8. Detective: Given Area 2 pts
l × 6 = 54 → l = 54 ÷ 6 = 9 ft
Length (l): 9 ft Perimeter (P): 30 ft
Perimeter calculation: P = 2(9) + 2(6) = 18 + 12 = 30 ft
Instructional Tip for Handwriting & Space: Encourage students to always label their diagrams with all 4 side lengths before calculating. For #7, students who write 10 on both top and bottom can easily see that \(32 - 20 = 12\), and dividing by the 2 remaining sides gives \(6\text{ in}\).
Perimeter & Area Workshop • Teacher Key • Page 2 of 3 Page 2 Points: 8
Teacher Answer Key
Level 3 Solutions: Real-World Scenarios
Total: 6 pts on Page 3
C Real-World Applications Solutions (Problems 9–11)
2 pts per scenario (1 pt perimeter + 1 pt area)
9. Playground Yard (16 m × 9 m) 2 pts
Part A: Fencing (Perimeter)
P = 2(16) + 2(9) = 32 + 18
P = 50 meters
Part B: Rubber Turf (Area)
A = 16 × 9
A = 144 m²
10. Art Poster Frame (24 in × 18 in) 2 pts
Part A: Molding (Perimeter)
P = 2(24) + 2(18) = 48 + 36
P = 84 inches
Part B: Glass Sheet (Area)
A = 24 × 18
A = 432 in²
11. Reading Rug & Tape (11 ft × 7 ft) 2 pts
Part A: Tape Perimeter & Cost
P = 2(11) + 2(7) = 36 ft
Cost = 36 ft × $2/ft = $72
Part B: Cleaning Area & Cost
A = 11 × 7 = 77 ft²
Cost = 77 ft² × $1/sq ft = $77
Perimeter & Area Workshop • Teacher Key • Page 3 of 3 Total Score: 22 points across all 3 pages
