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\[ P = \text{perimeter} \qquad A = l \times b \qquad A = \tfrac{1}{2}(b \times h) \]
Welcome to Grade 8 Perimeter & Area. Perimeter is the distance around the outside of a shape; area is the size of the flat surface inside it. In this lesson you will learn to find the perimeter of squares, rectangles and triangles, convert between mm², cm² and m², calculate the area of rectangles, squares and triangles, and find the circumference and area of circles.
\[ P_{square} = 4s \qquad\qquad P_{rectangle} = 2(l+b) \]
The perimeter (P) of a shape is the total distance around it — add up the lengths of every side. For a square with side \(s\): \(P = s+s+s+s = 4s\). For a rectangle with length \(l\) and breadth \(b\): \(P = l+l+b+b = 2(l+b)\).
\[ P_{triangle} = s_1 + s_2 + s_3 \]
A triangle has three sides, so its perimeter is simply the sum of all three sides: \(P = s_1 + s_2 + s_3\). If the triangle is isosceles (two sides equal), you only need to know one of the equal sides plus the third side.
Piggy's house hunt: find the perimeter of this house shape. It is a rectangle with a triangular roof. Only the outside measurements count — the line where the roof meets the rectangle (13 cm) is inside the shape, so it is not part of the perimeter.
\[ \text{overhang} = \dfrac{16-9}{2} = 3{,}5 \text{ cm each side} \]
Sometimes the roof is wider than the rectangle below it (an overhang), like a real house roof. If the roof base is 16 cm and the rectangle below is only 9 cm wide, the roof overhangs by \(\dfrac{16-9}{2} = 3{,}5\) cm on each side. That overhang length must be added into the perimeter walk — trace the outline all the way around and add every outside edge, including the little overhang ledges.
\[ P_{square}=4s \quad P_{rect}=2(l+b) \quad P_{\triangle}=s_1+s_2+s_3 \]
Perimeter is always found the same way: walk around the outside and add every edge. Watch out for internal lines (like a roofline) that must be excluded, and overhangs that must be included.
A 1 cm × 1 cm square can be divided into 100 small squares of 1 mm × 1 mm. That means \(1\text{ cm}^2 = 100\text{ mm}^2\) — NOT 10, because area is two-dimensional (10 mm × 10 mm = 100 mm²).
\[ \text{cm}^2 \!\to\! \text{mm}^2:\;\times100 \qquad \text{m}^2 \!\to\! \text{cm}^2:\;\times10\,000 \]
Because 1 m = 100 cm and 1 cm = 10 mm, squaring these gives the area conversion factors:
• cm² → mm²: × 100 • mm² → cm²: ÷ 100
• m² → cm²: × 10 000 • cm² → m²: ÷ 10 000
\[ 15\text{ m}^2 = 150\,000\text{ cm}^2 \qquad 5\text{ cm}^2 = 500\text{ mm}^2 \]
To convert m² to cm², multiply by 10 000: \(15 \times 10\,000 = 150\,000\) cm². To convert cm² to mm², multiply by 100: \(5 \times 100 = 500\) mm².
\[ 20\text{ cm}^2 = 0{,}002\text{ m}^2 \qquad 20\text{ mm}^2 = 0{,}2\text{ cm}^2 \]
Going from a smaller unit to a larger one, you divide. \(20\) cm² \(\div\, 10\,000 = 0{,}002\) m². \(20\) mm² \(\div\, 100 = 0{,}2\) cm².
\[ \times100 \;\rightleftharpoons\; \text{cm}^2 \leftrightarrow \text{mm}^2 \qquad \times10\,000 \;\rightleftharpoons\; \text{m}^2 \leftrightarrow \text{cm}^2 \]
Always square the linear conversion factor. Going to a smaller unit → multiply. Going to a bigger unit → divide.
The area of a shape is the size of the flat surface inside it, measured in square units. Counting squares: this rectangle is 6 squares long and 4 squares wide, giving \(6 \times 4 = 24\) squares in total.
\[ A = 12 \times 9 = 108 \text{ cm}^2 \]
Find the area of a rectangle with sides 12 cm and 9 cm. \(A = l \times b = 12 \times 9 = 108\) cm². Always check the units match before multiplying.
\[ A = 50\text{ mm} \times 3\text{ cm} = 5\text{ cm} \times 3\text{ cm} = 15\text{ cm}^2 \]
Calculate the area of a rectangle with length 50 mm and breadth 3 cm, in cm². Convert first: \(50\) mm \(= 5\) cm. Then \(A = 5 \times 3 = 15\) cm². You could also work entirely in mm: \(50 \times 30 = 1500\) mm² — the same area, just a different unit.
\[ 110\text{ mm} = 11\text{ cm} \qquad A = 11^2 = 121 \text{ cm}^2 \]
Find the area of a square bathroom tile with a side of 110 mm, in cm². Convert: \(110\) mm \(= 11\) cm. Then \(A = l^2 = 11^2 = 121\) cm².
\[ 450 = l \times 15 \;\Rightarrow\; l = 450 \div 15 = 30 \text{ cm} \]
If you know the area and one side, divide to find the other. A rectangle has area 450 cm² and width 150 mm (= 15 cm). \(l = 450 \div 15 = 30\) cm.
\[ A = 100 \times 69 = 6\,900 \text{ m}^2 \qquad \text{Cost} = 6\,900 \times R45 = R310\,500 \]
A rugby field is 100 m by 69 m. Its area is \(100 \times 69 = 6\,900\) m². Planting grass at R45 per m² would cost \(6\,900 \times 45 = R310\,500\). Since 1 hectare \(= 100 \times 100 = 10\,000\) m², the field is smaller than 1 hectare — by \(10\,000 - 6\,900 = 3\,100\) m².
\[ A_{rectangle} = l \times b \qquad A_{square} = l^2 \]
Rectangle and square area always come from multiplying two perpendicular sides. Convert units before you multiply, not after.
Take a rectangle and draw one diagonal — it splits into two equal triangles. Since the rectangle's area is \(l \times b\) (or base \(\times\) height), each triangle is exactly half of that: \(A_{\triangle} = \tfrac{1}{2}(\text{base} \times \text{height})\).
\[ A_{\triangle} = \tfrac{1}{2}(b \times h) \]
In this formula, b means base (not breadth), and h means the perpendicular height — the height must be measured at a right angle to the base, shown as a dashed line in diagrams.
Find the area of this triangle. \(A = \tfrac{1}{2}(18 \times 6) = \tfrac{1}{2}(108) = 54\) cm².
Even in an obtuse (leaning) triangle, the same rule applies: use the base and the perpendicular height, not a slanted side. \(A = \tfrac{1}{2}(16 \times 4) = \tfrac{1}{2}(64) = 32\) cm².
\[ A = \tfrac{1}{2}(400 \times 210) = 42\,000 \text{ mm}^2 \]
A right-angled triangle has a base of 400 mm and a perpendicular height of 210 mm. \(A = \tfrac{1}{2}(400 \times 210) = \tfrac{1}{2}(84\,000) = 42\,000\) mm². Keep the units in mm since both measurements were given in mm.
\[ A = \tfrac{1}{2}(8{,}66 \times 10) = 43{,}3 \text{ cm}^2 \]
A triangle has base 8,66 cm and perpendicular height 10 cm. \(A = \tfrac{1}{2}(8{,}66 \times 10) = \tfrac{1}{2}(86{,}6) = 43{,}3\) cm². Decimal values work exactly the same way — just multiply carefully.
\[ A_{\triangle} = \tfrac{1}{2}(\text{base} \times \text{perpendicular height}) \]
A triangle's area is always half of base times perpendicular height. It works for right-angled, obtuse and any other triangle — the perpendicular height is the key.
The centre is the point in the middle of the circle. The radius (r) runs from the centre to any point on the circle. The diameter (d) runs all the way across, through the centre. The circumference (C) is the distance all the way around — it is the circle's perimeter.
\( r = \tfrac{1}{2}d \qquad d = 2r \)
\[ C = \pi d \qquad \text{or} \qquad C = 2\pi r \]
If you divide any circle's circumference by its diameter, you always get the same number: 3,14159... — this constant is called pi (\(\pi\)), and we usually round it to \(\pi \approx 3{,}14\) (or the fraction \(\tfrac{22}{7}\)). Since \(C \div d = \pi\), rearranging gives \(C = \pi d\). Because \(d = 2r\), this is the same as \(C = 2\pi r\).
\[ C = \pi d = 3{,}14 \times 25\text{ cm} = 78{,}5 \text{ cm} \]
Calculate the circumference of a circle with a diameter of 25 cm. \(C = \pi d = 3{,}14 \times 25 = 78{,}5\) cm. If you were given the radius instead, use \(C = 2\pi r\) — for example, a radius of 10 mm gives \(C = 2 \times 3{,}14 \times 10 = 62{,}8\) mm.
Cut a circle into equal sectors and rearrange them — they form a shape close to a rectangle. Its height matches the radius, and its length matches half the circumference (the top and bottom edges are each half the curved edge). So:
\( A = \tfrac{1}{2}C \times r = \tfrac{1}{2}(2\pi r) \times r = \pi r^2 \)
\[ A = \pi r^2 \]
The area of any circle is \(\pi\) times the radius squared. Remember to square the radius before multiplying by \(\pi\) — and if you're only given the diameter, halve it to get the radius first.
\[ A = \pi r^2 = 3{,}14 \times 9^2 = 254{,}34 \text{ cm}^2 \]
Calculate the area of a circle with radius 9 cm. \(A = \pi r^2 = 3{,}14 \times 81 = 254{,}34\) cm². Working backwards: if a circle has area 76 m², then \(r^2 = 76 \div 3{,}14 = 24{,}2\), so \(r = \sqrt{24{,}2} \approx 4{,}92\) m — use the square root to undo the square.
Find the area of the shaded ring: a circle of radius 10 cm with a circle of radius 6 cm removed from its centre. Calculate each circle's area separately, then subtract.
\( A_{large} = 3{,}14 \times 10^2 = 314 \text{ cm}^2 \)
\( A_{small} = 3{,}14 \times 6^2 = 113{,}04 \text{ cm}^2 \)
\( A_{shaded} = 314 - 113{,}04 = 200{,}96 \text{ cm}^2 \)
\[ C = 2\pi r = \pi d \qquad\qquad A = \pi r^2 \]
A circle's circumference (perimeter) is \(2\pi r\) or \(\pi d\). Its area is \(\pi r^2\). Always use \(\pi \approx 3{,}14\), and double-check whether you were given the radius or the diameter before you start.
\[ \begin{aligned} P_{square}&=4s & P_{rect}&=2(l+b) \\ A_{rect}&=l\times b & A_{\triangle}&=\tfrac{1}{2}(b\times h) \\ C_{circle}&=2\pi r & A_{circle}&=\pi r^2 \end{aligned} \]
You now know how to:
• Find the perimeter of squares, rectangles and triangles
• Convert between mm², cm² and m² (multiply going smaller, divide going bigger)
• Calculate the area of rectangles and squares: \(A = l \times b\)
• Calculate the area of triangles: \(A = \tfrac{1}{2}(b \times h)\)
• Calculate the circumference of a circle: \(C = 2\pi r = \pi d\)
• Calculate the area of a circle: \(A = \pi r^2\)
Ready to test yourself?