How much fence? How much floor? Two questions, one lesson.
Building a horse paddock: how much fencing do you need? That's perimeter. How much bedding will cover the stable floor? That's area. Architects, farmers, and game designers use both every day - and a few simple formulas unlock every shape.
Ask what is being measured first. Covering a surface means area. Going around an edge means perimeter.
Key formulas: Rectangle: $A = l \times w$ · $P = 2l + 2w$ Triangle: $A = \frac{1}{2} \times b \times h$ · $P = a + b + c$ (sum of all sides) Parallelogram: $A = b \times h$ · $P = 2a + 2b$ Composite shape: split into known shapes → add (or subtract) their areas. Units: Area is always in square units (m², cm²). Perimeter is in linear units (m, cm).
Worked Example - Composite Shape
Worked Example - L-shaped paddock
An L-shaped paddock can be split into two rectangles. Rectangle 1: $10$ m × $6$ m. Rectangle 2: $4$ m × $3$ m.
Perimeter: trace the outside edges (measure each side carefully from the shape diagram) - do not double-count interior edges.
Warm-Up - Simple shapes (formulas given)
Problem 1
A stable floor is a rectangle $12$ m long and $8$ m wide. (a) Find the area. (b) Find the perimeter. (c) Bedding costs €4.50 per m². What does it cost to cover the whole floor?
$A = 12 \times 8$ · $P = 2(12) + 2(8)$
calculate area → perimeter → multiply by price →
Problem 2
A triangular corner of a horse paddock has base $9$ m and height $6$ m. The three sides measure $9$ m, $7.5$ m, and $7.5$ m. (a) Find the area. (b) Find the perimeter.
A parallelogram-shaped riding arena has base $20$ m, height $14$ m, and slant side $16$ m. (a) Find the area. (b) Find the perimeter.
$A = b \times h$ · $P = 2(20) + 2(16)$
calculate each →
Core Problems - composite shapes
Problem 4
An L-shaped paddock: the outer rectangle is $15$ m × $10$ m. A $5$ m × $4$ m rectangular corner has been fenced off for another use. Find the area of the remaining L-shape (subtract the missing corner).
area of full rectangle → subtract corner → answer →
Problem 5
A stable roof truss is a triangle on top of a rectangle. The rectangle is $8$ m wide and $3$ m tall. The triangle sits on top of the rectangle, with base $8$ m and height $2.5$ m. Find the total area of this cross-section.
area rectangle → area triangle → total →
From here - write the equation yourself, then solve
Problem 6
A rectangular paddock has an area of $180$ m² and a width of $9$ m. (a) Calculate its length. (b) Calculate its perimeter. (c) Calculate the fencing cost at €8.50 per metre.
A triangular art canvas has area $60$ cm² and base $15$ cm. Find the height. (Use the triangle area formula.)
write equation A = ½bh → substitute → solve for h →
Problem 8 Challenge
A square paddock and a rectangular paddock both have the same perimeter: $60$ m. The rectangle's length is twice its width. (a) Find the side length of the square. (b) Find the dimensions of the rectangle. (c) Which has the larger area? By how much?
square: 4s=60 → s → rect: 2(2w)+2w=60 → w → compare areas →
Problem 9 Open
Design a horse paddock with area at least $200$ m², using the least possible fencing. (a) State your shape and dimensions. (b) Calculate its area and perimeter. (c) Explain why your design uses little fencing for its area.
choose shape → calculate → justify →
Annual-review repair checkpoint
Checkpoint A · A decreasing sequence
A paddock-training plan uses $54,48,42,36,\ldots$ metres per round. Write the explicit formula $a_n=a_1+(n-1)d$. First calculate $d$ as next term minus previous term, and keep the negative value in parentheses.
next − previous → signed d → explicit formula →
Checkpoint B · The percent gate
A savings account pays $2.5\%$ simple interest per year. Convert $2.5\%$ to decimal rate $r$ by dividing by $100$.
percent ÷ 100 → decimal rate →
Quick review · turn two values into a rate
1. Name axes2. Find Δy3. Find Δx4. Divide + units
Problem 10 · Area changing over time
On day $1$, a temporary grazing area covers $120$ m². On day $5$, it covers $200$ m². (a) Name the $x$-quantity and $y$-quantity, including units. (b) Calculate $\Delta y$. (c) Calculate $\Delta x$. (d) Calculate the rate of change, with units. (e) Complete: The grazing area ______ by ______ square metres per day.
name axes → Δy → Δx → divide → interpret →
Show answers
Problem 1a
$A = 12 \times 8 = 96$ m².
Problem 1b
$P = 2(12)+2(8) = 40$ m.
Problem 1c
Cost $=96 \times 4.50 = €432$.
Problem 2a
$A = \frac{1}{2}(9)(6) = 27$ m².
Problem 2b
$P = 9 + 7.5 + 7.5 = 24$ m.
Problem 3a
$A = 20 \times 14 = 280$ m².
Problem 3b
$P = 2(20)+2(16) = 72$ m.
Problem 4
$A_\text{outer} = 15 \times 10 = 150$ m² · $A_\text{corner} = 5 \times 4 = 20$ m² · $A_\text{L} = 150 - 20 = 130$ m²
Problem 5
$A_\text{rect} = 8 \times 3 = 24$ m² · $A_\text{triangle} = \frac{1}{2}(8)(2.5) = 10$ m² · Total $= 34$ m²
$4s = 60$ → $s = 15$ m, so the square's area is $225$ m².
Problem 8b
$2(2w)+2w = 6w = 60$ → $w = 10$ m and $l = 20$ m, so the rectangle's area is $200$ m².
Problem 8c
The square is larger by $225-200=25$ m².
Problem 9a
Example: a square with side length $15$ m.
Problem 9b
$A=15\times15=225$ m² and $P=4\times15=60$ m.
Problem 9c
A square is the rectangle with the greatest area for a fixed perimeter, so it uses fencing efficiently.
Checkpoint A
$d=48-54=-6$, so $a_n=54+(n-1)(-6)$.
Checkpoint B
$2.5\div100=0.025$, so $r=0.025$.
Problem 10a
$x=$ time in days; $y=$ grazing area in square metres.
Problem 10b
$\Delta y=200-120=80$ m².
Problem 10c
$\Delta x=5-1=4$ days.
Problem 10d
Rate $=\frac{80}{4}=20$ m²/day.
Problem 10e
The grazing area increases by $20$ square metres per day.
Coming up next → Lesson 22: The Horse Business Story
Time to put it all together! Mia is starting a small horse-care business in Bavaria during a family stay. She needs proportionality, area, equations, rate of change - and a break-even calculation. A multi-step story assignment that pulls in everything from Lessons 12-19.