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Boundary Boss Workshop

A focused geometry practice session on calculating area and perimeter of rectangles and squares using standard facts up to 12 × 12.

FIFrancis Ibarra

Boundary Boss Practice Worksheet

Area & Perimeter Mastery • Rectangles & Squares (Facts up to 12 × 12)

Name

Date

Formula Vault:

Square: \(P = 4 \times s\) | \(A = s \times s\) Rectangle: \(P = 2 \times (l + w)\) | \(A = l \times w\)

A Part 1: Diagram Practice — Calculate Perimeter & Area

Include units with every answer

1 Square

\(7\text{ cm}\) \(7\text{ cm}\)

Perimeter:

Area:

2 Rectangle

\(9\text{ in}\) \(4\text{ in}\) RECTANGLE

Perimeter:

Area:

3 Rectangle

\(12\text{ m}\) \(5\text{ m}\) RECTANGLE

Perimeter:

Area:

4 Square

\(11\text{ ft}\) \(11\text{ ft}\)

Perimeter:

Area:

5 Rectangle

\(8\text{ cm}\) \(6\text{ cm}\) RECTANGLE

Perimeter:

Area:

6 Rectangle

\(10\text{ yd}\) \(3\text{ yd}\) RECTANGLE

Perimeter:

Area:

Standard Practice • Page 1 of 3 Boundary Boss Worksheet

Boundary Boss • Part 2: Missing Sides & Comparisons

Show work in calculation boxes

Pro Tip: Work backward from the known area or perimeter to find unknown side lengths!

7

A square garden has a perimeter of \(36\text{ feet}\). Find its side length and area.

Side length:

Total area:

8

A rectangular painting has an area of \(48\text{ sq inches}\). Its width is \(6\text{ inches}\). Find the length and perimeter.

Length:

Perimeter:

9

A square rug covers \(64\text{ sq feet}\) of floor space. What is the side length of the rug, and what is its perimeter?

Side length:

Perimeter:

10

A rectangular banner has a length of \(11\text{ meters}\) and a perimeter of \(30\text{ meters}\). Find its width and area.

Width:

Total area:

11

Compare: Shape X is an \(8\text{ cm} \times 3\text{ cm}\) rectangle. Shape Y is a square with side \(5\text{ cm}\).

Greater Area:

Greater Perim:

Missing Dimensions & Comparisons • Page 2 of 3 Boundary Boss Worksheet

Boundary Boss • Part 3: Real-World Word Problems

Read carefully & show your steps

Remember: Perimeter is the outside distance (e.g. \(\text{ft}\)). Area is total surface space (e.g. \(\text{sq ft}\)).

12

The Garden Fence: Maya's garden is \(12\text{ ft}\) long and \(7\text{ ft}\) wide. How many feet of fencing enclose it? What is the total planting area?

Fence:

Area:

13

Classroom Display: Mr. Alvarez decorates a square board with side \(6\text{ ft}\). He puts ribbon on the border and butcher paper on the surface. Find the ribbon length and paper area.

Ribbon:

Paper:

14

Kitchen Floor: A kitchen is \(9\text{ ft}\) long and \(8\text{ ft}\) wide. Each tile covers \(1\text{ sq ft}\). How many tiles are needed to cover it? How many feet of baseboard trim line the perimeter?

Tiles:

Baseboard:

15

Community Dog Park: A rectangular dog park is \(11\text{ yd}\) long and \(6\text{ yd}\) wide. What length of fence surrounds it? What is the total running area?

Fence:

Area:

16

The Sandbox Dilemma: Leo designs Box A (square, side \(6\text{ ft}\)) and Box B (rectangle, \(9\text{ ft} \times 4\text{ ft}\)). Leo says Box B has more play area because it is longer. Is he right?

Box A / B Area:

Is Leo right?

Real-World Word Problems • Page 3 of 3 Boundary Boss Worksheet

Boundary Boss Answer Key

Teacher Reference

Full Solutions & Step-by-Step Scoring Guide • Facts up to 12 × 12

Total Points: 32 pts (2 pts / problem)

Grading Criteria:

1 point for correct numerical calculation 1 point for correct linear or square units

A Part 1: Diagram Practice Solutions

Questions 1 – 6

1 Square (\(s = 7\text{ cm}\))

Perimeter: \(4 \times 7 = \mathbf{28\text{ cm}}\)

Area: \(7 \times 7 = \mathbf{49\text{ cm}^2}\)

\(P = 28\text{ cm}\) \(A = 49\text{ cm}^2\)

2 Rectangle (\(9\text{ in} \times 4\text{ in}\))

Perimeter: \(2(9 + 4) = 2(13) = \mathbf{26\text{ in}}\)

Area: \(9 \times 4 = \mathbf{36\text{ in}^2}\)

\(P = 26\text{ in}\) \(A = 36\text{ in}^2\)

3 Rectangle (\(12\text{ m} \times 5\text{ m}\))

Perimeter: \(2(12 + 5) = 2(17) = \mathbf{34\text{ m}}\)

Area: \(12 \times 5 = \mathbf{60\text{ m}^2}\)

\(P = 34\text{ m}\) \(A = 60\text{ m}^2\)

4 Square (\(s = 11\text{ ft}\))

Perimeter: \(4 \times 11 = \mathbf{44\text{ ft}}\)

Area: \(11 \times 11 = \mathbf{121\text{ ft}^2}\)

\(P = 44\text{ ft}\) \(A = 121\text{ ft}^2\)

5 Rectangle (\(8\text{ cm} \times 6\text{ cm}\))

Perimeter: \(2(8 + 6) = 2(14) = \mathbf{28\text{ cm}}\)

Area: \(8 \times 6 = \mathbf{48\text{ cm}^2}\)

\(P = 28\text{ cm}\) \(A = 48\text{ cm}^2\)

6 Rectangle (\(10\text{ yd} \times 3\text{ yd}\))

Perimeter: \(2(10 + 3) = 2(13) = \mathbf{26\text{ yd}}\)

Area: \(10 \times 3 = \mathbf{30\text{ yd}^2}\)

\(P = 26\text{ yd}\) \(A = 30\text{ yd}^2\)

Worked Solutions • Page 1 of 3 Boundary Boss Answer Key

Boundary Boss • Part 2 Solutions: Missing Sides & Comparisons

Questions 7 – 11

Teaching Note: Verify that students demonstrate inverse operations (division to find unknown factors) rather than guessing.

7

Square garden with perimeter \(P = 36\text{ ft}\).

• Side: \(s = P \div 4 = 36 \div 4 = \mathbf{9\text{ ft}}\)

• Area: \(A = s \times s = 9 \times 9 = \mathbf{81\text{ sq ft}}\)

Side: \(9\text{ ft}\)

Area: \(81\text{ sq ft}\)

8

Rectangular painting with area \(A = 48\text{ sq in}\), width \(w = 6\text{ in}\).

• Length: \(l = A \div w = 48 \div 6 = \mathbf{8\text{ in}}\)

• Perimeter: \(P = 2(8 + 6) = 2(14) = \mathbf{28\text{ in}}\)

Length: \(8\text{ in}\)

Perim: \(28\text{ in}\)

9

Square rug with area \(A = 64\text{ sq ft}\).

• Side: \(s \times s = 64 \implies s = \mathbf{8\text{ ft}}\)

• Perimeter: \(P = 4 \times 8 = \mathbf{32\text{ ft}}\)

Side: \(8\text{ ft}\)

Perim: \(32\text{ ft}\)

10

Rectangular banner with length \(l = 11\text{ m}\), perimeter \(P = 30\text{ m}\).

• Half-perim: \(l + w = 30 \div 2 = 15 \implies w = 15 - 11 = \mathbf{4\text{ m}}\)

• Area: \(A = 11 \times 4 = \mathbf{44\text{ sq m}}\)

Width: \(4\text{ m}\)

Area: \(44\text{ sq m}\)

11

Compare Shape X (\(8\text{ cm} \times 3\text{ cm}\)) & Shape Y (square \(5\text{ cm}\)).

• X: \(P = 2(8+3) = 22\text{ cm}\); \(A = 8 \times 3 = 24\text{ cm}^2\)

• Y: \(P = 4(5) = 20\text{ cm}\); \(A = 5 \times 5 = 25\text{ cm}^2\)

Area: Shape Y (\(25 > 24\))

Perim: Shape X (\(22 > 20\))

Worked Solutions • Page 2 of 3 Boundary Boss Answer Key

Boundary Boss • Part 3 Solutions: Real-World Problems

Questions 12 – 16

Common Error Alert: Watch for students mixing up perimeter (linear units: \(\text{ft}\)) with area (square units: \(\text{sq ft}\)).

12

The Garden Fence (\(12\text{ ft} \times 7\text{ ft}\))

• Fence (Perimeter): \(2 \times (12 + 7) = 2 \times 19 = \mathbf{38\text{ feet}}\)

• Planting Area: \(12 \times 7 = \mathbf{84\text{ sq feet}}\)

Fence: \(38\text{ ft}\)

Area: \(84\text{ sq ft}\)

13

Classroom Display (Square, side \(6\text{ ft}\))

• Ribbon (Perimeter): \(4 \times 6 = \mathbf{24\text{ feet}}\)

• Butcher Paper (Area): \(6 \times 6 = \mathbf{36\text{ sq feet}}\)

Ribbon: \(24\text{ ft}\)

Paper: \(36\text{ sq ft}\)

14

Kitchen Renovation (\(9\text{ ft} \times 8\text{ ft}\))

• Tiles (Area): \(9 \times 8 = 72\text{ sq ft} \implies \mathbf{72\text{ tiles}}\)

• Baseboard (Perimeter): \(2 \times (9 + 8) = 2 \times 17 = \mathbf{34\text{ feet}}\)

Tiles: \(72\text{ tiles}\)

Trim: \(34\text{ ft}\)

15

Community Dog Park (\(11\text{ yd} \times 6\text{ yd}\))

• Chain-link Fence (Perimeter): \(2 \times (11 + 6) = 2 \times 17 = \mathbf{34\text{ yards}}\)

• Running Space (Area): \(11 \times 6 = \mathbf{66\text{ sq yards}}\)

Fence: \(34\text{ yd}\)

Area: \(66\text{ sq yd}\)

16

The Sandbox Dilemma (Box A: \(6\text{ ft}\) square vs. Box B: \(9\text{ ft} \times 4\text{ ft}\))

• Box A: \(A = 6 \times 6 = \mathbf{36\text{ sq ft}}\); \(P = 4 \times 6 = 24\text{ ft}\)

• Box B: \(A = 9 \times 4 = \mathbf{36\text{ sq ft}}\); \(P = 2(9+4) = 26\text{ ft}\)

• Explanation: Leo is incorrect. Both sandboxes provide identical play area (\(36\text{ sq ft}\)).

Both: \(36\text{ sq ft}\)

Leo is Incorrect

Worked Solutions • Page 3 of 3 Boundary Boss Answer Key

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Boundary Boss Workshop

Boundary Boss Practice Worksheet

Boundary Boss Answer Key

Boundary Boss • Part 3 Solutions: Real-World Problems

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