Areas of compound shapes

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Last update: 2025-08-19

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Crash report

Areas of compound shapes

2025-08-19

Crash report

Unit 1: Angles & Constructions
Unit 2: Shapes & Symmetry
Unit 3: Position & Transformation
Unit 4: Area, Volume & Symmetry

🎯 In this topic you will

Estimate and calculate areas of compound 2D shapes.

📐 Working out areas of compound shapes

You can already work out the areas of compound shapes that are made from rectangles and triangles. You can also use other simple shapes, such as circles, in a compound shape. The method to work out the area is the same:

1 Divide the compound shape into simple shapes.
2 Work out the area of each simple shape.
3 Add or subtract the areas of the simple shapes to find the required area.

📘 Worked example

Work out the shaded area of each compound shape. Give your answers correct to 3 s.f.

Answer:

Area A = $l \times w = 3 \times 9 = 27 \,\text{cm}^2$
Area B = $\tfrac{1}{2} \pi r^2 = \tfrac{1}{2} \pi \times 4.5^2 = 31.8 \,\text{cm}^2$
Total area = $27 + 31.8 = 58.8 \,\text{cm}^2$

Area of circle = $\pi r^2 = \pi \times 5^2 = 78.5 \,\text{cm}^2$
Area of triangle = $\tfrac{1}{2} b h = \tfrac{1}{2} \times 4 \times 7 = 14 \,\text{cm}^2$
Shaded area = $78.5 - 14 = 64.5 \,\text{cm}^2$

Divide the compound shape into two simple shapes, A and B.

You know the length and width of rectangle A.

B is a semicircle with radius $9 \div 2 = 4.5 \,\text{cm}$

Work out the area of rectangle A.

Work out the area of semicircle B.

Add together the areas of the rectangle and the semicircle to find the area of the compound shape.

Work out the area of the circle.

Work out the area of the triangle.

The orange-shaded area is the area of the circle minus the area of the triangle.

🧠 PROBLEM-SOLVING Strategy

Areas of Compound 2D Shapes

Break the figure into simple parts, find each area, then combine carefully. Keep units consistent and round only at the end.

Sketch & label. Mark known lengths and decide a plan to split the shape into rectangles, triangles, circles, or semicircles. If a circle is involved, record its radius $r$ or diameter $d$ (remember $d=2r$).
Decompose. Partition the compound shape into non-overlapping simple shapes (A, B, C, …). Note whether the shaded region requires you to add or subtract any parts.
Choose formulas.
- Rectangle: $A = l\times w$
- Triangle: $A = \tfrac12 b h$
- Circle: $A = \pi r^2$; semicircle: $A = \tfrac12 \pi r^2$
- Trapezium: $A=\tfrac12 (a+b)h$ where $a,b$ are parallel sides
Compute each part. Work out areas for A, B, C, … Keep values exact when possible (e.g., leave $\pi$ in expressions) to avoid rounding error.
Combine.
- Shaded region built from pieces: $A_{\text{shaded}} = A_A + A_B + \cdots$
- Hole/removed region: $A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}}$
Units & rounding. Attach correct units (e.g., $\text{cm}^2$, $\text{m}^2$). Round only once at the end to the required accuracy (e.g., estimate to $1$ s.f.; final answer to $1$ d.p. if asked).
Estimate check. For a quick reasonableness test, round dimensions to $1$ s.f. and recompute mentally. Your exact answer should be close to this estimate.
Common pitfalls.
- Using diameter in $\pi r^2$ by mistake (always use $r$).
- For triangles in circles/semicircles, confirm which lengths are bases/heights (a dashed vertical may be the height).
- Subtracting the wrong region when a hole is present—sketch the “kept” vs “removed” parts.
Algebraic generalisation (optional). If all dimensions scale by a factor $k$, area scales by $k^2$. You can often write a general form such as $A(r)=\alpha r^2+\beta r + \gamma$ or $A(r)=c\,r^2$ (e.g., $A=\tfrac{r^2}{2}(\pi-2)$ for a semicircle minus triangle case).

❓ EXERCISES

1. Copy and complete the workings to calculate the areas of these compound shapes.

👀 Show answer

Area A = $l \times w = 5 \times 4 = 20 \,\text{cm}^2$
Area B = $l \times w = 11 \times 2 = 22 \,\text{cm}^2$
Total area = $20 + 22 = 42 \,\text{cm}^2$

👀 Show answer

Area A = $\tfrac{1}{2} b h = \tfrac{1}{2} \times 12 \times 6 = 36 \,\text{cm}^2$
Area B = $l \times w = 12 \times 3 = 36 \,\text{cm}^2$
Total area = $36 + 36 = 72 \,\text{cm}^2$

👀 Show answer

Area A = $l \times w = 12 \times 5 = 60 \,\text{cm}^2$
Area B = $\tfrac{1}{2} \pi r^2 = \tfrac{1}{2} \pi \times 6^2 = 56.5 \,\text{cm}^2$
Total area = $60 + 56.5 = 116.5 \,\text{cm}^2$

👀 Show answer

Area rectangle = $l \times w = 4 \times 1.5 = 6 \,\text{cm}^2$
Area circle = $\pi r^2 = \pi \times 3^2 = 28.3 \,\text{cm}^2$
Shaded area = $28.3 - 6 = 22.3 \,\text{cm}^2$

2. For each of these compound shapes, work out

i. the missing lengths shown by □

👀 Show answer

Missing length = $11 - 8 = 3 \,\text{cm}$

👀 Show answer

Missing length = $11 - (5 + 8) = 3 \,\text{cm}$

ii. the area of the compound shape.

👀 Show answer

Area A = $14 \times 3 = 42 \,\text{cm}^2$
Area B = $3 \times 7 = 21 \,\text{cm}^2$
Total area = $42 + 21 = 63 \,\text{cm}^2$

❓ EXERCISES

Tip

Remember: to work out an estimate, round all the numbers to one significant figure.

3. For each of these compound shapes, work out

i. an estimate of the area of the shape

ii. the area of the shape correct to one decimal place (1 d.p.).

👀 Show answer

i. Round to 1 s.f.: $7.4\!\to\!7$, $5.2\!\to\!5$, $3.9\!\to\!4$. Model as a trapezium: $A \approx \tfrac12(5+4)\times 7 = 31.5\ \text{cm}^2$.

ii. Trapezium area with exact measures: $A=\tfrac12(5.2+3.9)\times 7.4 = 33.67\ \text{cm}^2 \approx \mathbf{33.7\ \text{cm}^2}$.

👀 Show answer

i. Round to 1 s.f.: $2.68\!\to\!3$. $A \approx 3\times 3 + \tfrac12\pi(1.5)^2 = 9 + 3.53 \approx 12.5\ \text{m}^2$.

ii. Exact: rectangle $=2.68\times2.68=7.1824$, semicircle $=\tfrac12\pi(1.34)^2\approx2.8205$. $A\approx 10.0029\ \text{m}^2 \approx \mathbf{10.0\ \text{m}^2}$ (1 d.p.).

👀 Show answer

Interpretation used: semicircle of radius $4.6\ \text{cm}$ plus a right triangle with legs $4.6\ \text{cm}$ and $3.75\ \text{cm}$ (as suggested by the diagram). If this interpretation is not intended, there is not enough information to give a unique value.

i. Round to 1 s.f.: $r\!\approx\!5$, base $\!\approx\!4$. $A \approx \tfrac12\pi(5)^2 + \tfrac12(5)(4) = 39.27 + 10 \approx 49.3\ \text{cm}^2$.

ii. Exact with given numbers: $A=\tfrac12\pi(4.6)^2 + \tfrac12(4.6)(3.75) \approx 33.2381 + 8.625 \approx \mathbf{41.9\ \text{cm}^2}$.

👀 Show answer

Interpretation used: area = (semicircle of diameter $56\ \text{mm}$) $-$ (semicircle of diameter $28\ \text{mm}$).

i. Round to 1 s.f.: diameters $56\!\to\!60$, $28\!\to\!30$ (radii $30$ and $15$). $A \approx \tfrac12\pi(30)^2 - \tfrac12\pi(15)^2 = \tfrac12\pi(900-225) = 337.5\pi \approx 1.06\times10^3\ \text{mm}^2$.

ii. Exact with given numbers (radii $28$ and $14$): $A=\tfrac12\pi(28)^2 - \tfrac12\pi(14)^2 = 392\pi - 98\pi = 294\pi \approx \mathbf{923.6\ \text{mm}^2}$.

🧠 Think like a Mathematician

4. This is part of Kira’s homework.

a. Critique Kira’s method.

b. Can you improve her method? Explain your answer.

c. Discuss your answers to parts $a$ and $b$ with other learners in your class. (Do this as a solo reflection.)

👀 Show Answer

a. Kira’s decomposition is correct: two semicircles make one full circle plus the square. However, she rounds the semicircle areas to $56.5$ too early. Early rounding can accumulate error. Units should be shown consistently, and writing the exact form first is clearer.
b. Keep exact values until the final step: Exact area = $\text{square} + \text{full circle} = 12^2 + \pi\cdot 6^2 = 144 + 36\pi\ \text{cm}^2$. Rounded to 1 d.p.: $144 + 36\pi \approx 257.1\ \text{cm}^2$ (so $257.1\ \text{cm}^2$, or $257\ \text{cm}^2$ to 3 s.f.). This shows method and accuracy clearly.
c. Reflection points: Why is delaying rounding good practice? Could an alternative decomposition change the result (it shouldn’t)? What unit checks ensure correctness?

❓ EXERCISES

5. Work out the area of each shaded compound shape.

👀 Show answer

a. Outer rectangle $= 8\times 5 = 40\ \text{cm}^2$. Hole $= 3\times 2 = 6\ \text{cm}^2$.
Shaded area $= 40 - 6 = \mathbf{34\ \text{cm}^2}$.

b. Triangle $= \tfrac12 \times 12 \times 9 = 54\ \text{cm}^2$. Circle hole radius $r= \tfrac{5}{2}=2.5$ so area $=\pi r^2 = 6.25\pi \approx 19.6\ \text{cm}^2$.
Shaded area $= 54 - 6.25\pi \approx \mathbf{34.4\ \text{cm}^2}$.

c. The labels $3\ \text{mm}$, $8\ \text{mm}$, $9\ \text{mm}$ do not uniquely determine the shaded region’s dimensions from the image alone (e.g., unclear which are radii/diameters/chords and which region is removed). Not enough information provided to compute a single numerical area without additional text.

6. Sofia draws these two shapes.

Sofia makes this conjecture:

Is Sofia correct? Show clearly how you worked out your answer.

👀 Show answer

Square with circle. Square area $= 15.5^2 = 240.25\ \text{cm}^2$. Circle radius $= \tfrac{14}{2}=7$ so area $= \pi\cdot 7^2 = 49\pi \approx 153.94\ \text{cm}^2$.
Shaded area $= 240.25 - 49\pi \approx \mathbf{86.3\ \text{cm}^2}$.

Annulus. Outer radius $R=8$, inner radius $r=6$. Area $= \pi(R^2-r^2)=\pi(64-36)=28\pi \approx \mathbf{88.0\ \text{cm}^2}$.

Conclusion. The two shaded areas differ by about $|88.0-86.3|\approx 1.7\ \text{cm}^2$ (around $2\%$). Sofia’s conjecture is reasonable: the shaded areas are approximately the same size.

🧠 Think like a Mathematician

7. Work on your own to answer this question.

The diagram shows an isosceles triangle inside a semicircle. The base of the triangle is the diameter of the semicircle. The height of the triangle is the radius of the semicircle.

When the radius is $8\ \text{cm}$, you can work out an expression for the shaded area as shown:

Area of semicircle $=\dfrac12\pi r^2=\dfrac12\pi\times 8^2=32\pi$

Area of triangle $=\dfrac12 b h=\dfrac12\times 16\times 8=64$

Shaded area $=32\pi-64=32(\pi-2)\ \text{cm}^2$

a. Use the method shown to work out an expression for the shaded area when the radius is

i.$6\ \text{cm}$ ii.$10\ \text{cm}$ iii.$12\ \text{cm}$ iv.$3\ \text{cm}$

b. What do you notice about your answers to part a?

c. Use the method shown to write a general expression, using algebra, for the shaded area when the radius is $r$.

d. Reflect: explain in a few sentences why your algebra makes sense with the geometry.

👀 Show Answer

a i. Shaded $=\dfrac12\pi r^2 - \dfrac12(2r)r= \dfrac12\pi r^2 - r^2$. For $r=6$: $\dfrac12\pi\cdot 6^2-6^2=18\pi-36=18(\pi-2)\ \text{cm}^2$.
a ii.$r=10$: $\dfrac12\pi\cdot 10^2-10^2=50\pi-100=50(\pi-2)\ \text{cm}^2$.
a iii.$r=12$: $\dfrac12\pi\cdot 12^2-12^2=72\pi-144=72(\pi-2)\ \text{cm}^2$.
a iv.$r=3$: $\dfrac12\pi\cdot 3^2-3^2=\dfrac{9}{2}\pi-9=\dfrac{9}{2}(\pi-2)\ \text{cm}^2$.
b. Each result factors as $k(\pi-2)$ with $k=\dfrac{r^2}{2}$. So all answers share the same structure and scale with $r^2$.
c. General shaded area: $A=\dfrac12\pi r^2 - \dfrac12(2r)r=\dfrac12\pi r^2 - r^2 = r^2\!\left(\dfrac{\pi}{2}-1\right)=\dfrac{r^2}{2}(\pi-2)\ \text{units}^2$.
d. The base of the triangle is the diameter $2r$ and its height is $r$, so triangle area is $\dfrac12(2r)r=r^2$. The semicircle is half a circle of radius $r$, area $\dfrac12\pi r^2$. Subtracting gives the algebra above and explains why area scales with $r^2$.

❓ EXERCISES

8. Arun draws these two shapes.

The side lengths of the squares are the diameters of the circle and semicircles. Arun makes this conjecture: “I think the shaded area in Shape B is greater than the shaded area in Shape A.” What do you think? Explain your answer. Show any working that you do.

👀 Show answer

Let the square side be the diameter $2r$ (here $10\ \text{cm}$, so $r=5$).

Shape A. Shaded area $= (2r)^2 - \pi r^2 = 4r^2 - \pi r^2 = r^2(4-\pi)$.

Shape B. The two semicircles each have radius $r$, so their total area equals one full circle: $\tfrac12\pi r^2+\tfrac12\pi r^2=\pi r^2$. There is no overlap in area (the semicircles just meet at the centre), hence the unshaded part has area $\pi r^2$ again. Therefore the shaded area is also $(2r)^2-\pi r^2=r^2(4-\pi)$.

Conclusion. The shaded areas are equal, so Arun’s conjecture is false. Numerically (for $r=5$): both are $10^2-25\pi \approx \mathbf{21.5\ \text{cm}^2}$.

9. The diagram shows a circle inside a square. The side length of the square is the diameter of the circle.

a. Show that when the radius of the circle is $4\ \text{cm}$, you can write the shaded area as $16(4-\pi)\ \text{cm}^2$.

b. Work out an expression for the shaded area when the radius is i.$5\ \text{cm}$ii.$3\ \text{cm}$iii.$6\ \text{cm}$iv.$10\ \text{cm}$

c. What do you notice about your answers to parts a and b?

d. Write a general expression, using algebra, for the shaded area when the radius is $r$.

👀 Show answer

a. Square side $=2r=8$, so square area $=8^2=64$. Circle area $=\pi r^2=\pi\cdot 4^2=16\pi$. Shaded area $=64-16\pi=16(4-\pi)\ \text{cm}^2$.

b. In general $A=r^2(4-\pi)$, so i. $A=25(4-\pi)$, ii. $A=9(4-\pi)$, iii. $A=36(4-\pi)$, iv. $A=100(4-\pi)$.

c. Every answer has the same factor $(4-\pi)$ and scales with $r^2$.

d. General expression: $A=r^2(4-\pi)\ \text{units}^2$.

📘 What we've learned

Compound shapes can be split into simple shapes such as rectangles, triangles, circles, and semicircles.
The total area is found by adding or subtracting the areas of these simple parts.
Key formulas used include:
- Rectangle: $A = l \times w$
- Triangle: $A = \tfrac{1}{2} b h$
- Circle: $A = \pi r^2$, Semicircle: $A = \tfrac{1}{2} \pi r^2$
- Trapezium: $A = \tfrac{1}{2}(a+b)h$
It is important to keep units consistent and round answers only at the end (e.g., to 1 decimal place or 3 significant figures as required).
Estimation by rounding dimensions to 1 s.f. provides a useful check for reasonableness.