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Perimeter, Circumference & Area0%

Measurement & Geometry · Topic 5 of 9

Perimeter, Circumference & Area

Video coming soon3 worked examples

Theory

The Golden Rule: When calculating the perimeter of a shape that includes a fraction of a circle (like a semi-circle), candidates constantly calculate the curved arc length and stop. To find the true perimeter, you must always add the straight, flat edges back onto your curved answer.

1. Basic Area Formulae

You must confidently calculate the area of basic 2D shapes:

Rectangle: Area = Length × Breadth
Triangle: Area = $\frac{1}{2} \times \text{Base} \times \text{Height}$ . Ensure you use the strict perpendicular (vertical) height, not the slanted sides.

2. Circles (Circumference & Area)

You must know the difference between the radius (middle to edge) and diameter (edge to edge, through the middle).

Circumference (Curved Length): $C = \pi d$ .
Area: $A = \pi r^2$ .

Non-Calculator Rule: In Paper 1, you must manually use $\pi = 3.14$ .

3. Fractions of a Circle

At National 5 level, you are frequently asked to find the area or curved length of a fraction of a circle (e.g., a semi-circle or a quarter-circle).

Simply write out the full circle formula, and divide it by the fraction required (e.g., for a semi-circle, calculate $\pi d$ and divide by 2).

4. Composite Shapes (Adding and Subtracting)

Composite shapes are made by joining standard shapes together.

Adding Area: If a shape is made of a triangle sitting on top of a rectangle, calculate both areas separately and add them together.
Subtracting Area: If a patio surrounds a circular flower bed, you calculate the area of the whole patio space, and subtract the area of the circular flower bed from it.
Perimeter: Trace your finger around the outside of the shape. Add the curved arc lengths and any straight outer edges together.

⚠️ Common Examiner Traps

The "Forgotten Straight Edges" Trap: When finding the perimeter of a semi-circle, candidates often calculate the curve ( $\frac{1}{2}\pi d$ ) and declare that the final answer, completely forgetting to add the straight diameter line closing the shape at the bottom.
The "Internal Line" Trap: When finding the perimeter of two joined shapes (like a rectangle and a semi-circle), candidates often add up every single number on the diagram. You must ignore the internal lines where the shapes join; perimeter is strictly the outside border.
The "Radius vs Diameter" Mix-Up: Candidates frequently substitute the diameter into the Area formula ( $A = \pi r^2$ ), or the radius into the Circumference formula ( $C = \pi d$ ). Always halve the diameter to find the radius before calculating Area.

Worked examples

Example 1

Zainab designs a new badge. The design is based on a rectangle and a semi-circle as shown in the diagram.

The rectangle is 10 cm wide and 20 cm long.
The semi-circle is attached to one of the 10 cm sides.

She decides to put gold edging around the entire outside border of the badge. Calculate the length of gold edging she needs. (Take $\pi = 3.14$ ) (3 marks)

Step 1: Calculate the curved edge of the semi-circle. The diameter is 10 cm.

C = \pi d \div 2

3.14 \times 10 \div 2 = 31.4 \div 2 = 15.7 \text{ cm}

Step 2: Identify the straight outside edges. The semi-circle covers one of the 10 cm sides (meaning it is an internal line and must be ignored). The remaining outside straight edges are 20 cm, 10 cm, and 20 cm.

20 + 10 + 20 = 50 \text{ cm}

Step 3: Add the curved and straight edges together to find the full perimeter.

50 + 15.7 = 65.7 \text{ cm}

Final Answer: She needs 65.7 cm of gold edging.

Example 2

A flag is in the shape of an isosceles triangle with a rectangle on the top.

The rectangle is 10 cm wide and 4 cm high.
The total height of the flag (from the bottom tip of the triangle to the top of the rectangle) is 13 cm.

Calculate the area of the flag. (4 marks)

Step 1: Calculate the area of the top rectangle.

A = L \times B \rightarrow 10 \times 4 = 40 \text{ cm}^2

Step 2: Find the dimensions of the triangle. The base is 10 cm (matching the rectangle). The perpendicular height of the triangle is the total height minus the rectangle's height ( $13 - 4 = 9 \text{ cm}$ ).

Step 3: Calculate the area of the triangle.

A = \frac{1}{2} \times b \times h \rightarrow 0.5 \times 10 \times 9 = 45 \text{ cm}^2

Step 4: Add the two areas together.

40 + 45 = 85 \text{ cm}^2

Example 3

The reception area in a hotel features a large mirror. The mirror is in the shape of a square with identical semi-circles on each side.

The square has sides of length 1.2 metres.
The semi-circles have a diameter of 0.7 metres.

Calculate the area of the mirror. (2 marks)

Step 1: Calculate the area of the square.

A = L \times B \rightarrow 1.2 \times 1.2 = 1.44 \text{ m}^2

Step 2: Recognise that 4 identical semi-circles make exactly 2 full circles. Find the radius of these circles (half the diameter).

0.7 \div 2 = 0.35 \text{ m}

Step 3: Calculate the area of the 2 full circles.

A = 2 \times \pi \times r^2 \rightarrow 2 \times \pi \times 0.35^2

A = 0.76969... \text{ m}^2

Step 4: Add the areas together for the total.

1.44 + 0.76969... = 2.20969...

Final Answer: The area of the mirror is 2.21 m² (rounded to 2 d.p.).

Ready to practise?

11 Perimeter Circumference and Area questions with answers and video solutions