Area and Perimeter of Compound Shapes with Circles — GCSE Maths

What formulae do you need?

Key reminder: The "perimeter" of a semicircle includes the straight diameter edge as well as the curved arc. Students frequently forget the straight edge — always list every boundary segment.

How do you find the area of a shape with a semicircle attached?

Worked example 1: A running track end-cap shape: a rectangle 20 m long and 14 m wide with a semicircle of diameter 14 m attached to each short end.

The total shape is a "stadium" — a rectangle with two semicircles (= one full circle).

1. Area of rectangle: 20 × 14 = 280 m²
2. Area of two semicircles (radius = 7 m): π × 7² = 49π m²
3. Total area: (280 + 49π) m²

As a decimal: 280 + 153.94 ≈ 433.94 m² (3 s.f.)

As exact form: (280 + 49π) m²

How do you find the perimeter of a composite shape?

Worked example 2: A shape consists of a 10 cm × 6 cm rectangle with a semicircle of diameter 6 cm placed on one short end (6 cm side). Find the perimeter.

Identify every boundary segment:

• Two long sides of rectangle: 2 × 10 = 20 cm
• One short side of rectangle (opposite end from semicircle): 6 cm
• Arc of semicircle (radius = 3 cm): π × 3 = 3π cm (do NOT add the diameter — it is internal)

Total perimeter = 20 + 6 + 3π = (26 + 3π) cm

As a decimal: 26 + 9.42 ≈ 35.42 cm (4 s.f.)

Watch out: The 6 cm diameter at the join between the rectangle and semicircle is inside the shape — it is not a boundary edge. Only trace the outer boundary.

How do you find the area of a shape with a circle removed?

Worked example 3: A square of side 12 cm has a circle of radius 4 cm removed from its centre. Find the shaded area (the remaining region).

1. Area of square: 12² = 144 cm²
2. Area of circle: π × 4² = 16π cm²
3. Shaded area: 144 − 16π = (144 − 16π) cm²

As a decimal: 144 − 50.27 ≈ 93.73 cm² (4 s.f.)

How do you find the area of a quarter-circle cutout?

Worked example 4: A 10 cm × 10 cm square has a quarter-circle of radius 10 cm removed from one corner. Find the remaining area.

1. Area of square: 10² = 100 cm²
2. Area of quarter-circle: ¼ × π × 10² = 25π cm²
3. Remaining area: (100 − 25π) cm²

As a decimal: 100 − 78.54 ≈ 21.46 cm² (4 s.f.)

What does "leave your answer in terms of π" mean?

This instruction means do not substitute a decimal approximation for π. Write the answer in the form (a + bπ) or (a − bπ), where a and b are exact numbers.

Example: Instead of writing 26 + 9.424778…, write (26 + 3π).

Leaving answers in exact form avoids rounding errors and is always accepted as fully correct in GCSE mark schemes. When a decimal is needed (e.g. for a practical context), round to 3 significant figures unless told otherwise.

What mistakes should you avoid?

• Using diameter instead of radius in the area formula. πr² uses the radius — if diameter is given, halve it first.
• Forgetting to add the straight edge of a semicircle's perimeter. The perimeter includes the arc (πr) plus the flat diameter (2r).
• Adding areas when you should subtract (or vice versa). Always visualise whether the curved region is added to the shape or removed from it.
• Mixing exact and decimal values. If the question asks for exact form, the π must remain symbolic throughout — substituting 3.14 midway then writing "+ something π" is not exact.

Frequently asked questions

My answer has both a whole number part and a π part — can I simplify further?

Only if there is a common factor. For example, (20 + 4π) = 4(5 + π). However, 5 and π are not "like terms" — you cannot add them to get a single number. The expression (20 + 4π) is already in simplest form unless you need to factorise for a specific purpose.

Do I always need to find both area and perimeter?

Read the question carefully. "Find the area" and "find the perimeter" are different tasks; exam questions will specify which. Many students automatically calculate both and waste time. Some questions do ask for both — but only when explicitly requested.

How do I find the perimeter of a shape with a sector rather than a full semicircle?

For a sector of angle θ° and radius r: the arc length is (θ/360) × 2πr. Add the arc to the other boundary segments, being careful not to include the two radii if they are internal. For a quarter-circle (θ = 90°): arc length = (90/360) × 2πr = πr/2. The perimeter of a quarter-circle sector as an isolated shape includes the arc (πr/2) plus two radii (2r): total = πr/2 + 2r.

Can a compound shape include both a semicircle and a triangle?

Yes — for example, a "church window" shape: a rectangle with a semicircle on top and a triangle below. Split it into all three components, calculate each area, and add. For the perimeter, trace only the outer edge — the joins between components are interior boundaries and are excluded.

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