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.
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.
$$A_1 = 10 \times 6 = 60 \text{ m}^2$$
$$A_2 = 4 \times 3 = 12 \text{ m}^2$$
$$A_\text{total} = 60 + 12 = 72 \text{ m}^2$$
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)
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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 →
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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.
$A = \frac{1}{2} \times 9 \times 6$ · $P = 9 + 7.5 + 7.5$
calculate each →
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Problem 3
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
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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 →
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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
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Problem 6
A rectangular paddock has an area of $180$ m² and a width of $9$ m. Find the length. Then find the perimeter and the cost of fencing at €8.50 per metre.
write equation → solve for length → perimeter → cost →
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Problem 7
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 →
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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 →
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Problem 9 Open
Design a horse paddock with area at least $200$ m², using the least possible fencing. You may choose any shape (rectangle, triangle, or a composite).
• State your shape and dimensions.
• Calculate the area and perimeter.
• Explain why your design minimises fencing relative to the area.
choose shape → calculate → justify →
Show answers
Problem 1
$A = 12 \times 8 = 96$ m² · $P = 2(12)+2(8) = 40$ m · Cost: $96 \times 4.50 = €432$
Problem 2
$A = \frac{1}{2}(9)(6) = 27$ m² · $P = 9 + 7.5 + 7.5 = 24$ m
Problem 3
$A = 20 \times 14 = 280$ m² · $P = 2(20)+2(16) = 72$ 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
Problem 6
$180 = l \times 9$ → $l = 20$ m · $P = 2(20)+2(9) = 58$ m · Cost $= 58 \times 8.50 = €493$
Problem 7
$60 = \frac{1}{2}(15)(h)$ → $60 = 7.5h$ → $h = 8$ cm
Problem 8
(a) $4s = 60$ → $s = 15$ m, area $= 225$ m² · (b) $2(2w)+2(w) = 6w = 60$ → $w = 10$ m, $l = 20$ m, area $= 200$ m² · (c) Square is larger by $225 - 200 = 25$ m². Among rectangles with equal perimeter, the square always has the greatest area.
Problem 9
Open — for a rectangle, the most efficient shape is a square. A 15 m × 15 m square gives area $225$ m² with perimeter $60$ m. Any rectangle with the same perimeter has a smaller area (see Problem 8). Marks for correct area ≥ 200 m² and a reasonable explanation.
Coming up next → Lesson 20: 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.