Printable math worksheet for calculating area and perimeter of L-shaped figures.

To solve the problem of finding the area and perimeter of each L-shaped figure, we need to break down each shape into simpler rectangular components. Let's go through each problem step by step.

---

Problem 1:

Dimensions:
- Outer rectangle: \(36 \, \text{cm} \times 28 \, \text{cm}\)
- Inner rectangle (cutout): \(30 \, \text{cm} \times 22 \, \text{cm}\)

#### Area:
The area of the L-shape is the area of the outer rectangle minus the area of the inner rectangle.
\[
\text{Area of outer rectangle} = 36 \times 28 = 1008 \, \text{cm}^2
\]
\[
\text{Area of inner rectangle} = 30 \times 22 = 660 \, \text{cm}^2
\]
\[
\text{Area of L-shape} = 1008 - 660 = 348 \, \text{cm}^2
\]

#### Perimeter:
The perimeter of the L-shape is the sum of all its outer edges. We can count the edges directly:
- Top edge: \(28 \, \text{cm}\)
- Bottom edge: \(28 \, \text{cm}\)
- Left edge: \(36 \, \text{cm}\)
- Right edge: \(36 \, \text{cm}\)
- Inner vertical edge: \(30 \, \text{cm}\)
- Inner horizontal edge: \(22 \, \text{cm}\)

However, the inner edges are not part of the perimeter. So, we only count the outer edges:
\[
\text{Perimeter} = 2 \times (36 + 28) = 2 \times 64 = 128 \, \text{cm}
\]

Answer for Problem 1:
\[
\boxed{\text{Area: } 348 \, \text{cm}^2, \text{ Perimeter: } 128 \, \text{cm}}
\]

---

Problem 2:

Dimensions:
- Outer rectangle: \(70 \, \text{yd} \times 55 \, \text{yd}\)
- Inner rectangle (cutout): \(60 \, \text{yd} \times 42 \, \text{yd}\)

#### Area:
\[
\text{Area of outer rectangle} = 70 \times 55 = 3850 \, \text{yd}^2
\]
\[
\text{Area of inner rectangle} = 60 \times 42 = 2520 \, \text{yd}^2
\]
\[
\text{Area of L-shape} = 3850 - 2520 = 1330 \, \text{yd}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (70 + 55) = 2 \times 125 = 250 \, \text{yd}
\]

Answer for Problem 2:
\[
\boxed{\text{Area: } 1330 \, \text{yd}^2, \text{ Perimeter: } 250 \, \text{yd}}
\]

---

Problem 3:

Dimensions:
- Outer rectangle: \(60 \, \text{m} \times 50 \, \text{m}\)
- Inner rectangle (cutout): \(48 \, \text{m} \times 15 \, \text{m}\)

#### Area:
\[
\text{Area of outer rectangle} = 60 \times 50 = 3000 \, \text{m}^2
\]
\[
\text{Area of inner rectangle} = 48 \times 15 = 720 \, \text{m}^2
\]
\[
\text{Area of L-shape} = 3000 - 720 = 2280 \, \text{m}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (60 + 50) = 2 \times 110 = 220 \, \text{m}
\]

Answer for Problem 3:
\[
\boxed{\text{Area: } 2280 \, \text{m}^2, \text{ Perimeter: } 220 \, \text{m}}
\]

---

Problem 4:

Dimensions:
- Outer rectangle: \(37 \, \text{ft} \times 20 \, \text{ft}\)
- Inner rectangle (cutout): \(25 \, \text{ft} \times 15 \, \text{ft}\)

#### Area:
\[
\text{Area of outer rectangle} = 37 \times 20 = 740 \, \text{ft}^2
\]
\[
\text{Area of inner rectangle} = 25 \times 15 = 375 \, \text{ft}^2
\]
\[
\text{Area of L-shape} = 740 - 375 = 365 \, \text{ft}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (37 + 20) = 2 \times 57 = 114 \, \text{ft}
\]

Answer for Problem 4:
\[
\boxed{\text{Area: } 365 \, \text{ft}^2, \text{ Perimeter: } 114 \, \text{ft}}
\]

---

Problem 5:

Dimensions:
- Outer rectangle: \(13 \, \text{in} \times 15 \, \text{in}\)
- Inner rectangle (cutout): \(8 \, \text{in} \times 5 \, \text{in}\)

#### Area:
\[
\text{Area of outer rectangle} = 13 \times 15 = 195 \, \text{in}^2
\]
\[
\text{Area of inner rectangle} = 8 \times 5 = 40 \, \text{in}^2
\]
\[
\text{Area of L-shape} = 195 - 40 = 155 \, \text{in}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (13 + 15) = 2 \times 28 = 56 \, \text{in}
\]

Answer for Problem 5:
\[
\boxed{\text{Area: } 155 \, \text{in}^2, \text{ Perimeter: } 56 \, \text{in}}
\]

---

Problem 6:

Dimensions:
- Outer rectangle: \(26 \, \text{cm} \times 22 \, \text{cm}\)
- Inner rectangle (cutout): \(21 \, \text{cm} \times 2 \, \text{cm}\)

#### Area:
\[
\text{Area of outer rectangle} = 26 \times 22 = 572 \, \text{cm}^2
\]
\[
\text{Area of inner rectangle} = 21 \times 2 = 42 \, \text{cm}^2
\]
\[
\text{Area of L-shape} = 572 - 42 = 530 \, \text{cm}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (26 + 22) = 2 \times 48 = 96 \, \text{cm}
\]

Answer for Problem 6:
\[
\boxed{\text{Area: } 530 \, \text{cm}^2, \text{ Perimeter: } 96 \, \text{cm}}
\]

---

Problem 7:

Dimensions:
- Outer rectangle: \(14 \, \text{m} \times 12 \, \text{m}\)
- Inner rectangle (cutout): \(9 \, \text{m} \times 6 \, \text{m}\)

#### Area:
\[
\text{Area of outer rectangle} = 14 \times 12 = 168 \, \text{m}^2
\]
\[
\text{Area of inner rectangle} = 9 \times 6 = 54 \, \text{m}^2
\]
\[
\text{Area of L-shape} = 168 - 54 = 114 \, \text{m}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (14 + 12) = 2 \times 26 = 52 \, \text{m}
\]

Answer for Problem 7:
\[
\boxed{\text{Area: } 114 \, \text{m}^2, \text{ Perimeter: } 52 \, \text{m}}
\]

---

Problem 8:

Dimensions:
- Outer rectangle: \(60 \, \text{yd} \times 30 \, \text{yd}\)
- Inner rectangle (cutout): \(40 \, \text{yd} \times 20 \, \text{yd}\)

#### Area:
\[
\text{Area of outer rectangle} = 60 \times 30 = 1800 \, \text{yd}^2
\]
\[
\text{Area of inner rectangle} = 40 \times 20 = 800 \, \text{yd}^2
\]
\[
\text{Area of L-shape} = 1800 - 800 = 1000 \, \text{yd}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (60 + 30) = 2 \times 90 = 180 \, \text{yd}
\]

Answer for Problem 8:
\[
\boxed{\text{Area: } 1000 \, \text{yd}^2, \text{ Perimeter: } 180 \, \text{yd}}
\]

---

Problem 9:

Dimensions:
- Outer rectangle: \(17 \, \text{ft} \times 9 \, \text{ft}\)
- Inner rectangle (cutout): \(14 \, \text{ft} \times 3 \, \text{ft}\)

#### Area:
\[
\text{Area of outer rectangle} = 17 \times 9 = 153 \, \text{ft}^2
\]
\[
\text{Area of inner rectangle} = 14 \times 3 = 42 \, \text{ft}^2
\]
\[
\text{Area of L-shape} = 153 - 42 = 111 \, \text{ft}^2
\]

#### Perimeter:
The perimeter is the sum of all outer edges:
\[
\text{Perimeter} = 2 \times (17 + 9) = 2 \times 26 = 52 \, \text{ft}
\]

Answer for Problem 9:
\[
\boxed{\text{Area: } 111 \, \text{ft}^2, \text{ Perimeter: } 52 \, \text{ft}}
\]

---

Final Answers:

1. \(\boxed{\text{Area: } 348 \, \text{cm}^2, \text{ Perimeter: } 128 \, \text{cm}}\)
2. \(\boxed{\text{Area: } 1330 \, \text{yd}^2, \text{ Perimeter: } 250 \, \text{yd}}\)
3. \(\boxed{\text{Area: } 2280 \, \text{m}^2, \text{ Perimeter: } 220 \, \text{m}}\)
4. \(\boxed{\text{Area: } 365 \, \text{ft}^2, \text{ Perimeter: } 114 \, \text{ft}}\)
5. \(\boxed{\text{Area: } 155 \, \text{in}^2, \text{ Perimeter: } 56 \, \text{in}}\)
6. \(\boxed{\text{Area: } 530 \, \text{cm}^2, \text{ Perimeter: } 96 \, \text{cm}}\)
7. \(\boxed{\text{Area: } 114 \, \text{m}^2, \text{ Perimeter: } 52 \, \text{m}}\)
8. \(\boxed{\text{Area: } 1000 \, \text{yd}^2, \text{ Perimeter: } 180 \, \text{yd}}\)
9. \(\boxed{\text{Area: } 111 \, \text{ft}^2, \text{ Perimeter: } 52 \, \text{ft}}\)