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Area of Compound Shapes (Composite Shapes) Worksheets - Free Printable

Area of Compound Shapes (Composite Shapes) Worksheets

Educational worksheet: Area of Compound Shapes (Composite Shapes) Worksheets. Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Area of Compound Shapes (Composite Shapes) Worksheets
To solve the problem of finding the area of each compound shape, we need to break down each shape into simpler geometric figures (such as rectangles, triangles, circles, etc.) and then calculate the area of each part. Finally, we sum up the areas of all the parts to get the total area of the compound shape.

Let's go through each shape step by step:

---

Shape 1:


The shape consists of a rectangle and a semicircle on top.

- Rectangle:
- Length = 10 cm
- Width = 5 cm
- Area of rectangle = \( \text{Length} \times \text{Width} = 10 \times 5 = 50 \, \text{cm}^2 \)

- Semicircle:
- Diameter of the semicircle = 10 cm (same as the length of the rectangle)
- Radius = \( \frac{\text{Diameter}}{2} = \frac{10}{2} = 5 \, \text{cm} \)
- Area of a full circle = \( \pi r^2 = \pi (5)^2 = 25\pi \, \text{cm}^2 \)
- Area of semicircle = \( \frac{1}{2} \times 25\pi = \frac{25\pi}{2} \, \text{cm}^2 \)

- Total Area:
\[
\text{Total Area} = \text{Area of Rectangle} + \text{Area of Semicircle} = 50 + \frac{25\pi}{2}
\]

Using \( \pi \approx 3.14 \):
\[
\frac{25\pi}{2} \approx \frac{25 \times 3.14}{2} = \frac{78.5}{2} = 39.25 \, \text{cm}^2
\]
\[
\text{Total Area} \approx 50 + 39.25 = 89.25 \, \text{cm}^2
\]

---

Shape 2:


The shape is a trapezoid.

- Trapezoid:
- Top base (\( b_1 \)) = 6 cm
- Bottom base (\( b_2 \)) = 10 cm
- Height (\( h \)) = 4 cm
- Area of trapezoid = \( \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (6 + 10) \times 4 = \frac{1}{2} \times 16 \times 4 = 32 \, \text{cm}^2 \)

---

Shape 3:


The shape consists of a rectangle and two right triangles.

- Rectangle:
- Length = 8 cm
- Width = 4 cm
- Area of rectangle = \( 8 \times 4 = 32 \, \text{cm}^2 \)

- Right Triangles:
- Each triangle has a base = 4 cm and height = 3 cm.
- Area of one triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6 \, \text{cm}^2 \)
- Total area of two triangles = \( 2 \times 6 = 12 \, \text{cm}^2 \)

- Total Area:
\[
\text{Total Area} = \text{Area of Rectangle} + \text{Area of Two Triangles} = 32 + 12 = 44 \, \text{cm}^2
\]

---

Shape 4:


The shape consists of a square and a semicircle.

- Square:
- Side length = 6 cm
- Area of square = \( 6 \times 6 = 36 \, \text{cm}^2 \)

- Semicircle:
- Diameter of the semicircle = 6 cm (same as the side of the square)
- Radius = \( \frac{6}{2} = 3 \, \text{cm} \)
- Area of a full circle = \( \pi r^2 = \pi (3)^2 = 9\pi \, \text{cm}^2 \)
- Area of semicircle = \( \frac{1}{2} \times 9\pi = \frac{9\pi}{2} \, \text{cm}^2 \)

- Total Area:
\[
\text{Total Area} = \text{Area of Square} + \text{Area of Semicircle} = 36 + \frac{9\pi}{2}
\]

Using \( \pi \approx 3.14 \):
\[
\frac{9\pi}{2} \approx \frac{9 \times 3.14}{2} = \frac{28.26}{2} = 14.13 \, \text{cm}^2
\]
\[
\text{Total Area} \approx 36 + 14.13 = 50.13 \, \text{cm}^2
\]

---

Shape 5:


The shape consists of a parallelogram and a triangle.

- Parallelogram:
- Base = 7 cm
- Height = 5 cm
- Area of parallelogram = \( \text{Base} \times \text{Height} = 7 \times 5 = 35 \, \text{cm}^2 \)

- Triangle:
- Base = 7 cm
- Height = 3 cm
- Area of triangle = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 7 \times 3 = 10.5 \, \text{cm}^2 \)

- Total Area:
\[
\text{Total Area} = \text{Area of Parallelogram} + \text{Area of Triangle} = 35 + 10.5 = 45.5 \, \text{cm}^2
\]

---

Shape 6:


The shape consists of a rectangle and a quarter circle.

- Rectangle:
- Length = 12 cm
- Width = 6 cm
- Area of rectangle = \( 12 \times 6 = 72 \, \text{cm}^2 \)

- Quarter Circle:
- Radius = 6 cm
- Area of a full circle = \( \pi r^2 = \pi (6)^2 = 36\pi \, \text{cm}^2 \)
- Area of quarter circle = \( \frac{1}{4} \times 36\pi = 9\pi \, \text{cm}^2 \)

- Total Area:
\[
\text{Total Area} = \text{Area of Rectangle} + \text{Area of Quarter Circle} = 72 + 9\pi
\]

Using \( \pi \approx 3.14 \):
\[
9\pi \approx 9 \times 3.14 = 28.26 \, \text{cm}^2
\]
\[
\text{Total Area} \approx 72 + 28.26 = 100.26 \, \text{cm}^2
\]

---

Final Answers:



1. \( 50 + \frac{25\pi}{2} \approx 89.25 \, \text{cm}^2 \)
2. \( 32 \, \text{cm}^2 \)
3. \( 44 \, \text{cm}^2 \)
4. \( 36 + \frac{9\pi}{2} \approx 50.13 \, \text{cm}^2 \)
5. \( 45.5 \, \text{cm}^2 \)
6. \( 72 + 9\pi \approx 100.26 \, \text{cm}^2 \)

\[
\boxed{89.25, 32, 44, 50.13, 45.5, 100.26}
\]
Parent Tip: Review the logic above to help your child master the concept of area of compound figures worksheet.
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