Ever tried to finish a Unit 13 homework packet and felt like the area‑and‑perimeter word problems were speaking a different language?
So you stare at a rectangle, a garden, a fence‑line, and the numbers just won’t line up. Turns out you’re not the only one—most students hit the same snag right around the middle of the school year.
Below is the answer key you’ve been hunting, plus the why‑behind each step, common slip‑ups, and a handful of tricks that actually stick. Grab a pencil, follow along, and you’ll be checking off those problems without a second‑guess.
What Is Unit 13 Homework About?
Unit 13 in most middle‑school math curricula is the “real‑world” stretch where you leave pure formulas behind and start measuring actual spaces.
In plain English, the unit asks you to:
- Find the area of irregular shapes that can be broken down into rectangles, triangles, or circles.
- Compute the perimeter (or sometimes the circumference) of those same shapes.
- Translate a word problem—like “A playground needs a fence that’s 2 m away from the edge”—into a set of equations you can solve.
It’s not just about memorizing A = l × w or P = 2(l + w). You’re expected to read a scenario, pick out the relevant dimensions, and decide whether you need area, perimeter, or both.
The Core Skills
| Skill | What You Do |
|---|---|
| Decompose shapes | Split a complex figure into simpler parts you already know how to handle. But |
| Identify missing lengths | Use clues like “the garden is 3 m longer than the shed” to set up algebraic expressions. Still, |
| Choose the right formula | Area for “how much grass” or perimeter for “how much fence. Which means ” |
| Check units | Convert feet to meters, square feet to square meters, etc. , before you plug numbers in. |
If you can juggle those four, the rest of Unit 13 is just practice.
Why It Matters
Why should you care about a handful of word problems? Because the skill set is the math equivalent of a Swiss Army knife.
- Real‑life relevance – Architects, landscapers, and even video‑game designers constantly calculate area and perimeter.
- Test performance – Standardized tests love word problems; they’re a quick way to separate students who can apply concepts from those who just memorize.
- Confidence builder – Solving a seemingly messy scenario gives you a mental shortcut for any future “messy” problem.
When you skip the “why,” the steps feel arbitrary, and you’ll keep making the same errors. Knowing the purpose makes the process click.
How It Works: Step‑by‑Step Answer Key
Below is a full answer key for the most common Unit 13 worksheet set (the one with 12 problems, each mixing area and perimeter). I’ll walk through each problem, show the reasoning, and then give the final answer. Feel free to skip to the answer list if you just want a quick reference, but I recommend reading the explanations—those are the real nuggets.
Problem 1 – Simple Rectangle
Prompt: A rectangular garden is 8 m long and 5 m wide. Find its area and perimeter.
Solution:
Area = length × width = 8 × 5 = 40 m².
Perimeter = 2 × (length + width) = 2 × (8 + 5) = 26 m.
Answer: Area = 40 m², Perimeter = 26 m It's one of those things that adds up..
Problem 2 – Composite Shape (Rectangle + Triangle)
Prompt: A playground consists of a 12 m × 6 m rectangle with a right‑triangle extension on one side. The triangle’s legs are 4 m (along the rectangle) and 3 m (perpendicular). Find total area and the fence length needed around the entire shape.
Solution:
-
Rectangle area = 12 × 6 = 72 m².
-
Triangle area = ½ × 4 × 3 = 6 m².
Total area = 72 + 6 = 78 m².For perimeter, trace the outer edges:
- Bottom of rectangle: 12 m
- Right side of rectangle: 6 m
- Slanted triangle side (hypotenuse) = √(4² + 3²) = 5 m
- Top of triangle (the 4‑m leg) is already part of the rectangle’s top edge, so we don’t count it twice.
- Left side of rectangle: 6 m
Add them up: 12 + 6 + 5 + 6 = 29 m Worth knowing..
Answer: Area = 78 m², Perimeter = 29 m That's the part that actually makes a difference..
Problem 3 – Missing Dimension
Prompt: A square patio has an area of 144 ft². What is its perimeter?
Solution:
Area of a square = side² → side = √144 = 12 ft.
Perimeter = 4 × side = 4 × 12 = 48 ft.
Answer: Perimeter = 48 ft.
Problem 4 – “Fence 2 m Away” Scenario
Prompt: A rectangular field is 30 m long and 20 m wide. A fence must be built 2 m away from the field on all sides. How much fencing is required?
Solution:
The fence forms a larger rectangle: add 2 m to each side twice (once for each side).
New length = 30 + 2 + 2 = 34 m.
New width = 20 + 2 + 2 = 24 m.
Fence length = perimeter of larger rectangle = 2 × (34 + 24) = 116 m.
Answer: 116 m of fence The details matter here..
Problem 5 – Circle Area & Circumference
Prompt: A circular pond has a radius of 7 m. Find its area and the length of a walkway that goes once around it Practical, not theoretical..
Solution:
Area = πr² ≈ 3.14 × 7² ≈ 154 m².
Circumference = 2πr ≈ 2 × 3.14 × 7 ≈ 44 m.
Answer: Area ≈ 154 m², Circumference ≈ 44 m.
Problem 6 – L‑Shaped Figure
Prompt: An L‑shaped garden is formed by two rectangles: a 10 m × 4 m rectangle attached to a 6 m × 3 m rectangle (sharing a 4 m side). Find total area and perimeter Simple, but easy to overlook..
Solution:
Area = (10 × 4) + (6 × 3) = 40 + 18 = 58 m² Easy to understand, harder to ignore..
Perimeter – draw the outline:
- Bottom edge: 10 m
- Right side of large rectangle: 4 m
- Up the shared side: 4 m (but interior, so ignore)
- Top of small rectangle: 6 m
- Left side of small rectangle: 3 m
- Down the remaining side of large rectangle: 10 m (the part not overlapped)
Add: 10 + 4 + 6 + 3 + 10 = 33 m.
Answer: Area = 58 m², Perimeter = 33 m.
Problem 7 – Word Problem with Algebra
Prompt: A rectangular sandbox is 5 m longer than it is wide. Its area is 150 m². Find its dimensions and the amount of edging needed Easy to understand, harder to ignore. And it works..
Solution:
Let width = w. Then length = w + 5.
Area: w × (w + 5) = 150.
Expand: w² + 5w − 150 = 0.
Factor or use quadratic formula: (w + 15)(w − 10) = 0 → w = 10 m (negative root discarded).
Length = 10 + 5 = 15 m And that's really what it comes down to..
Perimeter = 2 × (10 + 15) = 50 m.
Answer: Width = 10 m, Length = 15 m, Edge needed = 50 m.
Problem 8 – Mixed Units
Prompt: A backyard is 20 ft long and 15 ft wide. A new patio will cover half the area. What is the patio’s area in square yards?
Solution:
Backyard area = 20 × 15 = 300 ft².
Half = 150 ft².
Convert to square yards: 1 yd² = 9 ft² → 150 ÷ 9 ≈ 16.7 yd² It's one of those things that adds up..
Answer: Approximately 16.7 yd² Most people skip this — try not to..
Problem 9 – Perimeter of a Track
Prompt: A running track consists of two straight sections 100 m each and two semicircles with a radius of 25 m. Find the total distance around the track.
Solution:
Straight parts: 2 × 100 = 200 m.
Two semicircles make a full circle: circumference = 2πr = 2 × 3.14 × 25 ≈ 157 m.
Total = 200 + 157 = 357 m.
Answer: 357 m Easy to understand, harder to ignore..
Problem 10 – Area of a Trapezoid
Prompt: A trapezoid garden has bases of 8 m and 12 m, and a height of 5 m. Find its area.
Solution:
Area = ½ × (base₁ + base₂) × height = 0.5 × (8 + 12) × 5 = 0.5 × 20 × 5 = 50 m² Simple as that..
Answer: 50 m².
Problem 11 – “Border” Problem
Prompt: A rectangular poster is 24 in wide and 36 in tall. A 2‑in border will be added all around. What will be the new dimensions and total border area?
Solution:
New width = 24 + 2 + 2 = 28 in.
New height = 36 + 2 + 2 = 40 in.
New total area = 28 × 40 = 1,120 in².
Original area = 24 × 36 = 864 in².
Border area = 1,120 − 864 = 256 in².
Answer: New size 28 × 40 in; border adds 256 in².
Problem 12 – Real‑World Scaling
Prompt: On a map, 1 cm represents 5 m. A rectangular park on the map measures 7 cm by 4 cm. Find the real‑world area and perimeter of the park.
Solution:
Scale factor: 1 cm = 5 m → multiply each dimension by 5.
Real length = 7 × 5 = 35 m.
Real width = 4 × 5 = 20 m.
Area = 35 × 20 = 700 m².
Perimeter = 2 × (35 + 20) = 110 m And that's really what it comes down to..
Answer: Area = 700 m², Perimeter = 110 m The details matter here..
Quick Answer Summary
| # | Area | Perimeter / Other |
|---|---|---|
| 1 | 40 m² | 26 m |
| 2 | 78 m² | 29 m |
| 3 | – | 48 ft |
| 4 | – | 116 m |
| 5 | 154 m² | 44 m |
| 6 | 58 m² | 33 m |
| 7 | 150 m² (10 × 15) | 50 m |
| 8 | 16.7 yd² | – |
| 9 | – | 357 m |
| 10 | 50 m² | – |
| 11 | 256 in² border | – |
| 12 | 700 m² | 110 m |
Common Mistakes / What Most People Get Wrong
- Forgetting to subtract interior edges – In composite shapes, the shared side isn’t part of the outer perimeter. I’ve seen students add it twice and end up with a fence length that’s too long.
- Mixing units – Switching from feet to meters mid‑problem without converting skews every answer. Keep a conversion sheet handy.
- Misreading “half the area” – Some treat it as “half the dimensions.” Remember: area halves, not length or width.
- Skipping the diagram – Word problems become a blur without a quick sketch. Even a rough doodle clarifies which sides are shared and which are external.
- Using π≈3.14 for rough work, then switching to 22/7 for final – Consistency matters. Pick one approximation and stick with it throughout a problem.
Practical Tips – What Actually Works
- Draw first, compute second. A 2‑minute sketch saves ten minutes of guesswork.
- Label every segment. Write “a = 8 m,” “b = ?,” directly on the diagram.
- Set up an equation before plugging numbers. For missing dimensions, translate the word “longer” into + something right away.
- Create a “unit check” column. After you finish, glance at each term: does it read “meters” or “square meters”? If not, you’ve likely mixed them up.
- Use a calculator for π only at the end. Keep the exact π symbol in the work; round only for the final answer.
- Practice reverse‑engineering. Take an answer key, hide the steps, and try to reconstruct the problem. It trains you to see the logic in reverse.
FAQ
Q: How do I know when to use perimeter vs. area?
A: Ask yourself what the problem is measuring. If it’s “how much fence” or “how far around,” it’s perimeter. If it’s “how much grass” or “how many tiles,” it’s area.
Q: Can I use the same formula for irregular shapes?
A: Not directly. Break the shape into rectangles, triangles, circles, or trapezoids you know, solve each piece, then add the results Which is the point..
Q: Why do some problems give the area and ask for dimensions?
A: That’s an algebra step. Set side × other side = given area, then solve the resulting equation—often a quadratic.
Q: What if the problem involves a circle and a rectangle together?
A: Treat them separately. Compute the circle’s area (πr²) and the rectangle’s area, then add. For perimeter, trace the outer edge—sometimes the circle’s diameter replaces a straight side.
Q: My teacher says “show all work.” How detailed should I be?
A: Write the formula, plug in numbers, and simplify. Even a quick note like “area = ½ × 4 × 3 = 6” counts as showing work.
So there you have it—a full answer key, the reasoning behind each step, and a toolbox of tips to keep you from stumbling over the next Unit 13 packet.
Next time you open a worksheet, take a breath, sketch, label, and let the numbers fall into place. Good luck, and may your fences be the right length every time!
