Grade 6 Area Worksheets | Find the Area of Compound Shapes - Free Printable

To solve the problem of finding the area for each shape, we will break down each figure into simpler geometric shapes (rectangles) and calculate their areas individually. Then, we will sum up the areas to get the total area for each figure.

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Problem 1:

The given shape can be divided into two rectangles:
1. A rectangle with dimensions \(8 \, \text{m} \times 8 \, \text{m}\).
2. A rectangle with dimensions \(3 \, \text{m} \times 4 \, \text{m}\) (since the total length is \(12 \, \text{m}\) and one part is already \(8 \, \text{m}\), the remaining length is \(12 - 8 = 4 \, \text{m}\)).

#### Step 1: Calculate the area of the first rectangle
\[
\text{Area}_1 = 8 \, \text{m} \times 8 \, \text{m} = 64 \, \text{m}^2
\]

#### Step 2: Calculate the area of the second rectangle
\[
\text{Area}_2 = 3 \, \text{m} \times 4 \, \text{m} = 12 \, \text{m}^2
\]

#### Step 3: Sum the areas
\[
\text{Total Area} = \text{Area}_1 + \text{Area}_2 = 64 \, \text{m}^2 + 12 \, \text{m}^2 = 76 \, \text{m}^2
\]

Answer for Problem 1:
\[
\boxed{76}
\]

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Problem 2:

The given shape can be divided into two rectangles:
1. A rectangle with dimensions \(20 \, \text{cm} \times 13 \, \text{cm}\) (the full height is \(20 \, \text{cm}\), and the width is \(18 \, \text{cm} - 5 \, \text{cm} = 13 \, \text{cm}\)).
2. A rectangle with dimensions \(5 \, \text{cm} \times 10 \, \text{cm}\).

#### Step 1: Calculate the area of the first rectangle
\[
\text{Area}_1 = 20 \, \text{cm} \times 13 \, \text{cm} = 260 \, \text{cm}^2
\]

#### Step 2: Calculate the area of the second rectangle
\[
\text{Area}_2 = 5 \, \text{cm} \times 10 \, \text{cm} = 50 \, \text{cm}^2
\]

#### Step 3: Sum the areas
\[
\text{Total Area} = \text{Area}_1 + \text{Area}_2 = 260 \, \text{cm}^2 + 50 \, \text{cm}^2 = 310 \, \text{cm}^2
\]

Answer for Problem 2:
\[
\boxed{310}
\]

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Problem 3:

The given shape can be divided into three rectangles:
1. A large rectangle with dimensions \(10 \, \text{m} \times 9 \, \text{m}\) (the full height is \(10 \, \text{m}\), and the width is \(14 \, \text{m} - 5 \, \text{m} = 9 \, \text{m}\)).
2. A smaller rectangle with dimensions \(5 \, \text{m} \times 5 \, \text{m}\).
3. Subtract the overlapping area of the middle rectangle (which is \(4 \, \text{m} \times 5 \, \text{m}\)) since it is counted twice.

#### Step 1: Calculate the area of the first rectangle
\[
\text{Area}_1 = 10 \, \text{m} \times 9 \, \text{m} = 90 \, \text{m}^2
\]

#### Step 2: Calculate the area of the second rectangle
\[
\text{Area}_2 = 5 \, \text{m} \times 5 \, \text{m} = 25 \, \text{m}^2
\]

#### Step 3: Calculate the overlapping area to subtract
\[
\text{Overlapping Area} = 4 \, \text{m} \times 5 \, \text{m} = 20 \, \text{m}^2
\]

#### Step 4: Sum the areas and subtract the overlapping area
\[
\text{Total Area} = \text{Area}_1 + \text{Area}_2 - \text{Overlapping Area} = 90 \, \text{m}^2 + 25 \, \text{m}^2 - 20 \, \text{m}^2 = 95 \, \text{m}^2
\]

Answer for Problem 3:
\[
\boxed{95}
\]

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Final Answers:

1. \(\boxed{76}\)
2. \(\boxed{310}\)
3. \(\boxed{95}\)