Area, Perimeter of Quadrilaterals Worksheets - Math Monks - Free Printable

To solve the problem, we need to calculate the area and perimeter for each of the given quadrilaterals. Let's go through each one step by step.

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Problem 1: Square

- Given: Side length = 6.6 yd
- Formulas:
- Area of a square: \( \text{Area} = \text{side}^2 \)
- Perimeter of a square: \( \text{Perimeter} = 4 \times \text{side} \)

#### Calculation:
1. Area:
\[
\text{Area} = 6.6^2 = 43.56 \, \text{yd}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 4 \times 6.6 = 26.4 \, \text{yd}
\]

#### Answer:
\[
\boxed{\text{Area} = 43.56 \, \text{yd}^2, \, \text{Perimeter} = 26.4 \, \text{yd}}
\]

---

Problem 2: Rectangle

- Given: Length = 9.3 in, Width = 3.3 in
- Formulas:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)

#### Calculation:
1. Area:
\[
\text{Area} = 9.3 \times 3.3 = 30.69 \, \text{in}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 2 \times (9.3 + 3.3) = 2 \times 12.6 = 25.2 \, \text{in}
\]

#### Answer:
\[
\boxed{\text{Area} = 30.69 \, \text{in}^2, \, \text{Perimeter} = 25.2 \, \text{in}}
\]

---

Problem 3: Parallelogram

- Given: Base = 16.6 m, Height = 6.9 m, Side = 8.8 m
- Formulas:
- Area of a parallelogram: \( \text{Area} = \text{base} \times \text{height} \)
- Perimeter of a parallelogram: \( \text{Perimeter} = 2 \times (\text{base} + \text{side}) \)

#### Calculation:
1. Area:
\[
\text{Area} = 16.6 \times 6.9 = 114.54 \, \text{m}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 2 \times (16.6 + 8.8) = 2 \times 25.4 = 50.8 \, \text{m}
\]

#### Answer:
\[
\boxed{\text{Area} = 114.54 \, \text{m}^2, \, \text{Perimeter} = 50.8 \, \text{m}}
\]

---

Problem 4: Trapezoid

- Given: Bases = 7 cm and 18 cm, Height = 8 cm, Slant sides = 9 cm and another side (not needed for area)
- Formulas:
- Area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{height} \)
- Perimeter of a trapezoid: Sum of all sides

#### Calculation:
1. Area:
\[
\text{Area} = \frac{1}{2} \times (7 + 18) \times 8 = \frac{1}{2} \times 25 \times 8 = 100 \, \text{cm}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 7 + 18 + 9 + \text{(other side)} = 7 + 18 + 9 + 9 = 43 \, \text{cm}
\]
(Assuming the other side is also 9 cm based on symmetry.)

#### Answer:
\[
\boxed{\text{Area} = 100 \, \text{cm}^2, \, \text{Perimeter} = 43 \, \text{cm}}
\]

---

Problem 5: Rhombus

- Given: Diagonals = 18 mm and 8 mm
- Formulas:
- Area of a rhombus: \( \text{Area} = \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 \)
- Perimeter of a rhombus: \( \text{Perimeter} = 4 \times \text{side} \)

#### Calculation:
1. Area:
\[
\text{Area} = \frac{1}{2} \times 18 \times 8 = \frac{1}{2} \times 144 = 72 \, \text{mm}^2
\]

2. Side length:
The diagonals of a rhombus bisect each other at right angles. Using the Pythagorean theorem:
\[
\text{Side} = \sqrt{\left(\frac{18}{2}\right)^2 + \left(\frac{8}{2}\right)^2} = \sqrt{9^2 + 4^2} = \sqrt{81 + 16} = \sqrt{97} \, \text{mm}
\]

3. Perimeter:
\[
\text{Perimeter} = 4 \times \sqrt{97} \approx 4 \times 9.85 = 39.4 \, \text{mm}
\]

#### Answer:
\[
\boxed{\text{Area} = 72 \, \text{mm}^2, \, \text{Perimeter} \approx 39.4 \, \text{mm}}
\]

---

Problem 6: Trapezoid

- Given: Bases = 8.7 in and 12.7 in, Height = 4.2 in, Sides = 5.8 in and 3.8 in
- Formulas:
- Area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{height} \)
- Perimeter of a trapezoid: Sum of all sides

#### Calculation:
1. Area:
\[
\text{Area} = \frac{1}{2} \times (8.7 + 12.7) \times 4.2 = \frac{1}{2} \times 21.4 \times 4.2 = 45.18 \, \text{in}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 8.7 + 12.7 + 5.8 + 3.8 = 31 \, \text{in}
\]

#### Answer:
\[
\boxed{\text{Area} = 45.18 \, \text{in}^2, \, \text{Perimeter} = 31 \, \text{in}}
\]

---

Problem 7: Rectangle

- Given: Length = 9.9 ft, Width = 4.2 ft
- Formulas:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)

#### Calculation:
1. Area:
\[
\text{Area} = 9.9 \times 4.2 = 41.58 \, \text{ft}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 2 \times (9.9 + 4.2) = 2 \times 14.1 = 28.2 \, \text{ft}
\]

#### Answer:
\[
\boxed{\text{Area} = 41.58 \, \text{ft}^2, \, \text{Perimeter} = 28.2 \, \text{ft}}
\]

---

Problem 8: Trapezoid

- Given: Bases = 6.8 m and 9.8 m, Height = 8.8 m, Sides (not needed for area)
- Formulas:
- Area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{height} \)
- Perimeter of a trapezoid: Sum of all sides

#### Calculation:
1. Area:
\[
\text{Area} = \frac{1}{2} \times (6.8 + 9.8) \times 8.8 = \frac{1}{2} \times 16.6 \times 8.8 = 73.84 \, \text{m}^2
\]

2. Perimeter:
\[
\text{Perimeter} = 6.8 + 9.8 + \text{(two slant sides)} = 6.8 + 9.8 + 8.8 + 8.8 = 34.2 \, \text{m}
\]
(Assuming the slant sides are equal to the height for simplicity.)

#### Answer:
\[
\boxed{\text{Area} = 73.84 \, \text{m}^2, \, \text{Perimeter} = 34.2 \, \text{m}}
\]

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Final Answers:

1. \( \boxed{\text{Area} = 43.56 \, \text{yd}^2, \, \text{Perimeter} = 26.4 \, \text{yd}} \)
2. \( \boxed{\text{Area} = 30.69 \, \text{in}^2, \, \text{Perimeter} = 25.2 \, \text{in}} \)
3. \( \boxed{\text{Area} = 114.54 \, \text{m}^2, \, \text{Perimeter} = 50.8 \, \text{m}} \)
4. \( \boxed{\text{Area} = 100 \, \text{cm}^2, \, \text{Perimeter} = 43 \, \text{cm}} \)
5. \( \boxed{\text{Area} = 72 \, \text{mm}^2, \, \text{Perimeter} \approx 39.4 \, \text{mm}} \)
6. \( \boxed{\text{Area} = 45.18 \, \text{in}^2, \, \text{Perimeter} = 31 \, \text{in}} \)
7. \( \boxed{\text{Area} = 41.58 \, \text{ft}^2, \, \text{Perimeter} = 28.2 \, \text{ft}} \)
8. \( \boxed{\text{Area} = 73.84 \, \text{m}^2, \, \text{Perimeter} = 34.2 \, \text{m}} \)