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Find the area and perimeter of irregular shapes with this geometry worksheet.

Geometry worksheet featuring eight irregular shapes with labeled dimensions in inches, feet, and yards, designed to find area and perimeter.

Geometry worksheet featuring eight irregular shapes with labeled dimensions in inches, feet, and yards, designed to find area and perimeter.

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Show Answer Key & Explanations Step-by-step solution for: Finding Area and Perimeter of Irregular shapes - Math Worksheets ...
To solve the problem of finding the area and perimeter of irregular shapes, we need to break each shape into simpler geometric figures (like rectangles) and then calculate the area and perimeter step by step. Let's go through each shape one by one.

---

Shape 1:


#### Dimensions:
- Top rectangle: \(10 \, \text{in} \times 4 \, \text{in}\)
- Bottom rectangle: \(25 \, \text{in} \times 15 \, \text{in}\)

#### Area:
The total area is the sum of the areas of the two rectangles.
\[
\text{Area of top rectangle} = 10 \times 4 = 40 \, \text{in}^2
\]
\[
\text{Area of bottom rectangle} = 25 \times 15 = 375 \, \text{in}^2
\]
\[
\text{Total Area} = 40 + 375 = 415 \, \text{in}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(25 \, \text{in}\)
- Bottom side: \(25 \, \text{in}\)
- Left side: \(10 \, \text{in}\)
- Right side: \(15 \, \text{in}\)
- Vertical side between the two rectangles: \(10 - 4 = 6 \, \text{in}\)

\[
\text{Perimeter} = 25 + 25 + 10 + 15 + 6 + 4 = 85 \, \text{in}
\]

#### Final Answer for Shape 1:
\[
\boxed{415 \, \text{in}^2, 85 \, \text{in}}
\]

---

Shape 2:


#### Dimensions:
- Top rectangle: \(33 \, \text{in} \times 13 \, \text{in}\)
- Bottom rectangle: \(23 \, \text{in} \times 12 \, \text{in}\)

#### Area:
The total area is the sum of the areas of the two rectangles.
\[
\text{Area of top rectangle} = 33 \times 13 = 429 \, \text{in}^2
\]
\[
\text{Area of bottom rectangle} = 23 \times 12 = 276 \, \text{in}^2
\]
\[
\text{Total Area} = 429 + 276 = 705 \, \text{in}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(33 \, \text{in}\)
- Bottom side: \(23 \, \text{in}\)
- Left side: \(13 \, \text{in}\)
- Right side: \(12 \, \text{in}\)
- Vertical side between the two rectangles: \(33 - 23 = 10 \, \text{in}\)

\[
\text{Perimeter} = 33 + 23 + 13 + 12 + 10 + 10 = 101 \, \text{in}
\]

#### Final Answer for Shape 2:
\[
\boxed{705 \, \text{in}^2, 101 \, \text{in}}
\]

---

Shape 3:


#### Dimensions:
- Left rectangle: \(5 \, \text{in} \times 6 \, \text{in}\)
- Right rectangle: \(18 \, \text{in} \times 15 \, \text{in}\)

#### Area:
The total area is the sum of the areas of the two rectangles.
\[
\text{Area of left rectangle} = 5 \times 6 = 30 \, \text{in}^2
\]
\[
\text{Area of right rectangle} = 18 \times 15 = 270 \, \text{in}^2
\]
\[
\text{Total Area} = 30 + 270 = 300 \, \text{in}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(18 \, \text{in}\)
- Bottom side: \(18 \, \text{in}\)
- Left side: \(5 \, \text{in}\)
- Right side: \(15 \, \text{in}\)
- Vertical side between the two rectangles: \(15 - 6 = 9 \, \text{in}\)

\[
\text{Perimeter} = 18 + 18 + 5 + 15 + 9 + 6 = 71 \, \text{in}
\]

#### Final Answer for Shape 3:
\[
\boxed{300 \, \text{in}^2, 71 \, \text{in}}
\]

---

Shape 4:


#### Dimensions:
- Main rectangle: \(26 \, \text{ft} \times 38 \, \text{ft}\)
- Cut-out section: \(11 \, \text{ft} \times 19 \, \text{ft}\)

#### Area:
The total area is the area of the main rectangle minus the area of the cut-out section.
\[
\text{Area of main rectangle} = 26 \times 38 = 988 \, \text{ft}^2
\]
\[
\text{Area of cut-out section} = 11 \times 19 = 209 \, \text{ft}^2
\]
\[
\text{Total Area} = 988 - 209 = 779 \, \text{ft}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(38 \, \text{ft}\)
- Bottom side: \(38 \, \text{ft}\)
- Left side: \(26 \, \text{ft}\)
- Right side: \(26 \, \text{ft}\)
- Additional vertical sides from the cut-out: \(11 \, \text{ft} + 19 \, \text{ft}\)

\[
\text{Perimeter} = 38 + 38 + 26 + 26 + 11 + 19 = 158 \, \text{ft}
\]

#### Final Answer for Shape 4:
\[
\boxed{779 \, \text{ft}^2, 158 \, \text{ft}}
\]

---

Shape 5:


#### Dimensions:
- Top rectangle: \(27 \, \text{yd} \times 16 \, \text{yd}\)
- Bottom rectangle: \(10 \, \text{yd} \times 3 \, \text{yd}\)

#### Area:
The total area is the sum of the areas of the two rectangles.
\[
\text{Area of top rectangle} = 27 \times 16 = 432 \, \text{yd}^2
\]
\[
\text{Area of bottom rectangle} = 10 \times 3 = 30 \, \text{yd}^2
\]
\[
\text{Total Area} = 432 + 30 = 462 \, \text{yd}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(27 \, \text{yd}\)
- Bottom side: \(10 \, \text{yd}\)
- Left side: \(16 \, \text{yd}\)
- Right side: \(3 \, \text{yd}\)
- Horizontal side between the two rectangles: \(27 - 10 = 17 \, \text{yd}\)

\[
\text{Perimeter} = 27 + 10 + 16 + 3 + 17 + 17 = 90 \, \text{yd}
\]

#### Final Answer for Shape 5:
\[
\boxed{462 \, \text{yd}^2, 90 \, \text{yd}}
\]

---

Shape 6:


#### Dimensions:
- Rectangle: \(34 \, \text{yd} \times 9 \, \text{yd}\)

#### Area:
\[
\text{Area} = 34 \times 9 = 306 \, \text{yd}^2
\]

#### Perimeter:
\[
\text{Perimeter} = 2 \times (34 + 9) = 2 \times 43 = 86 \, \text{yd}
\]

#### Final Answer for Shape 6:
\[
\boxed{306 \, \text{yd}^2, 86 \, \text{yd}}
\]

---

Shape 7:


#### Dimensions:
- Top rectangle: \(38 \, \text{in} \times 9 \, \text{in}\)
- Bottom rectangle: \(15 \, \text{in} \times 4 \, \text{in}\)

#### Area:
The total area is the sum of the areas of the two rectangles.
\[
\text{Area of top rectangle} = 38 \times 9 = 342 \, \text{in}^2
\]
\[
\text{Area of bottom rectangle} = 15 \times 4 = 60 \, \text{in}^2
\]
\[
\text{Total Area} = 342 + 60 = 402 \, \text{in}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(38 \, \text{in}\)
- Bottom side: \(15 \, \text{in}\)
- Left side: \(9 \, \text{in}\)
- Right side: \(4 \, \text{in}\)
- Horizontal side between the two rectangles: \(38 - 15 = 23 \, \text{in}\)

\[
\text{Perimeter} = 38 + 15 + 9 + 4 + 23 + 9 = 98 \, \text{in}
\]

#### Final Answer for Shape 7:
\[
\boxed{402 \, \text{in}^2, 98 \, \text{in}}
\]

---

Shape 8:


#### Dimensions:
- Main rectangle: \(26 \, \text{yd} \times 48 \, \text{yd}\)
- Cut-out section: \(11 \, \text{yd} \times 24 \, \text{yd}\)

#### Area:
The total area is the area of the main rectangle minus the area of the cut-out section.
\[
\text{Area of main rectangle} = 26 \times 48 = 1248 \, \text{yd}^2
\]
\[
\text{Area of cut-out section} = 11 \times 24 = 264 \, \text{yd}^2
\]
\[
\text{Total Area} = 1248 - 264 = 984 \, \text{yd}^2
\]

#### Perimeter:
The perimeter is the sum of all the outer edges. We can count the sides directly:
- Top side: \(48 \, \text{yd}\)
- Bottom side: \(48 \, \text{yd}\)
- Left side: \(26 \, \text{yd}\)
- Right side: \(26 \, \text{yd}\)
- Additional vertical sides from the cut-out: \(11 \, \text{yd} + 24 \, \text{yd}\)

\[
\text{Perimeter} = 48 + 48 + 26 + 26 + 11 + 24 = 183 \, \text{yd}
\]

#### Final Answer for Shape 8:
\[
\boxed{984 \, \text{yd}^2, 183 \, \text{yd}}
\]

---

Final Answers:


1. \(\boxed{415 \, \text{in}^2, 85 \, \text{in}}\)
2. \(\boxed{705 \, \text{in}^2, 101 \, \text{in}}\)
3. \(\boxed{300 \, \text{in}^2, 71 \, \text{in}}\)
4. \(\boxed{779 \, \text{ft}^2, 158 \, \text{ft}}\)
5. \(\boxed{462 \, \text{yd}^2, 90 \, \text{yd}}\)
6. \(\boxed{306 \, \text{yd}^2, 86 \, \text{yd}}\)
7. \(\boxed{402 \, \text{in}^2, 98 \, \text{in}}\)
8. \(\boxed{984 \, \text{yd}^2, 183 \, \text{yd}}\)
Parent Tip: Review the logic above to help your child master the concept of area and perimeter of irregular shapes worksheet.
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