Remedial Area & Perimeter of Composite Figures Worksheet — The ... - Free Printable

Problem: Find the Area and Perimeter of Each Figure

We will solve each figure step by step.

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#### Figure 1:
The shape is a combination of two rectangles.

- Dimensions:
- Top rectangle: \(4 \, \text{cm} \times 3 \, \text{cm}\)
- Bottom rectangle: \(5 \, \text{cm} \times 1 \, \text{cm}\)

##### Step 1: Calculate the Area
The area of the shape is the sum of the areas of the two rectangles.
\[
\text{Area of top rectangle} = 4 \times 3 = 12 \, \text{cm}^2
\]
\[
\text{Area of bottom rectangle} = 5 \times 1 = 5 \, \text{cm}^2
\]
\[
\text{Total Area} = 12 + 5 = 17 \, \text{cm}^2
\]

##### Step 2: Calculate the Perimeter
To find the perimeter, we need to trace the outer boundary of the shape. The shape has the following sides:
- Two vertical sides of length \(3 \, \text{cm}\) (top rectangle)
- One horizontal side of length \(4 \, \text{cm}\) (top rectangle)
- One horizontal side of length \(5 \, \text{cm}\) (bottom rectangle)
- One vertical side of length \(1 \, \text{cm}\) (bottom rectangle)
- One vertical side of length \(4 \, \text{cm}\) (combined height)

Summing these sides:
\[
\text{Perimeter} = 3 + 3 + 4 + 5 + 1 + 4 = 20 \, \text{cm}
\]

##### Final Answer for Figure 1:
\[
\boxed{17 \, \text{cm}^2, 20 \, \text{cm}}
\]

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#### Figure 2:
The shape is a large rectangle with a smaller rectangle removed from one corner.

- Dimensions:
- Large rectangle: \(5 \, \text{cm} \times 4 \, \text{cm}\)
- Small rectangle (removed): \(2 \, \text{cm} \times 2 \, \text{cm}\)

##### Step 1: Calculate the Area
The area of the shape is the area of the large rectangle minus the area of the small rectangle.
\[
\text{Area of large rectangle} = 5 \times 4 = 20 \, \text{cm}^2
\]
\[
\text{Area of small rectangle} = 2 \times 2 = 4 \, \text{cm}^2
\]
\[
\text{Total Area} = 20 - 4 = 16 \, \text{cm}^2
\]

##### Step 2: Calculate the Perimeter
The perimeter of the shape is the same as the perimeter of the large rectangle because removing the small rectangle does not change the outer boundary.
\[
\text{Perimeter} = 2 \times (5 + 4) = 2 \times 9 = 18 \, \text{cm}
\]

##### Final Answer for Figure 2:
\[
\boxed{16 \, \text{cm}^2, 18 \, \text{cm}}
\]

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#### Figure 3:
The shape is a combination of a large rectangle and a smaller rectangle.

- Dimensions:
- Large rectangle: \(10 \, \text{cm} \times 5 \, \text{cm}\)
- Smaller rectangle: \(4 \, \text{cm} \times 2 \, \text{cm}\)

##### Step 1: Calculate the Area
The area of the shape is the sum of the areas of the two rectangles.
\[
\text{Area of large rectangle} = 10 \times 5 = 50 \, \text{cm}^2
\]
\[
\text{Area of smaller rectangle} = 4 \times 2 = 8 \, \text{cm}^2
\]
\[
\text{Total Area} = 50 + 8 = 58 \, \text{cm}^2
\]

##### Step 2: Calculate the Perimeter
To find the perimeter, we need to trace the outer boundary of the shape. The shape has the following sides:
- Two vertical sides of length \(5 \, \text{cm}\) (large rectangle)
- One horizontal side of length \(10 \, \text{cm}\) (large rectangle)
- One horizontal side of length \(4 \, \text{cm}\) (smaller rectangle)
- Two vertical sides of length \(2 \, \text{cm}\) (smaller rectangle)
- One horizontal side of length \(3 \, \text{cm}\) (gap between rectangles)

Summing these sides:
\[
\text{Perimeter} = 5 + 5 + 10 + 4 + 2 + 2 + 3 = 31 \, \text{cm}
\]

##### Final Answer for Figure 3:
\[
\boxed{58 \, \text{cm}^2, 31 \, \text{cm}}
\]

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#### Figure 4:
The shape is a large rectangle with a smaller square removed from the center.

- Dimensions:
- Large rectangle: \(10 \, \text{cm} \times 6 \, \text{cm}\)
- Small square (removed): \(2 \, \text{cm} \times 2 \, \text{cm}\)

##### Step 1: Calculate the Area
The area of the shape is the area of the large rectangle minus the area of the small square.
\[
\text{Area of large rectangle} = 10 \times 6 = 60 \, \text{cm}^2
\]
\[
\text{Area of small square} = 2 \times 2 = 4 \, \text{cm}^2
\]
\[
\text{Total Area} = 60 - 4 = 56 \, \text{cm}^2
\]

##### Step 2: Calculate the Perimeter
The perimeter of the shape is the same as the perimeter of the large rectangle because removing the small square does not change the outer boundary.
\[
\text{Perimeter} = 2 \times (10 + 6) = 2 \times 16 = 32 \, \text{cm}
\]

##### Final Answer for Figure 4:
\[
\boxed{56 \, \text{cm}^2, 32 \, \text{cm}}
\]

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Final Answers:

1. \(\boxed{17 \, \text{cm}^2, 20 \, \text{cm}}\)
2. \(\boxed{16 \, \text{cm}^2, 18 \, \text{cm}}\)
3. \(\boxed{58 \, \text{cm}^2, 31 \, \text{cm}}\)
4. \(\boxed{56 \, \text{cm}^2, 32 \, \text{cm}}\)