Math worksheet for finding the perimeter and area of triangles.

To solve the problem of finding the perimeter and area of the given triangles, we will use the following formulas:

1. Perimeter of a Triangle:
The perimeter is the sum of the lengths of all three sides of the triangle.
\[
\text{Perimeter} = a + b + c
\]
where \(a\), \(b\), and \(c\) are the lengths of the sides.

2. Area of a Triangle:
The area can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
where the base is one side of the triangle, and the height is the perpendicular distance from the base to the opposite vertex.

Let's solve each triangle step by step.

---

Triangle 1:

- Sides: 8 cm, 7 cm, 6 cm
- Height: 6 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 8 + 7 + 6 = 21 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 6 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 6 \times 6 = \frac{1}{2} \times 36 = 18 \, \text{cm}^2
\]

Answer for Triangle 1:
\[
\text{Perimeter} = 21 \, \text{cm}, \quad \text{Area} = 18 \, \text{cm}^2
\]

---

Triangle 2:

- Sides: 6 cm, 6 cm, 3 cm
- Height: 5 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 6 + 6 + 3 = 15 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 3 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 3 \times 5 = \frac{1}{2} \times 15 = 7.5 \, \text{cm}^2
\]

Answer for Triangle 2:
\[
\text{Perimeter} = 15 \, \text{cm}, \quad \text{Area} = 7.5 \, \text{cm}^2
\]

---

Triangle 3:

- Sides: 8 cm, 7 cm, 6 cm
- Height: 6 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 8 + 7 + 6 = 21 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 6 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 6 \times 6 = \frac{1}{2} \times 36 = 18 \, \text{cm}^2
\]

Answer for Triangle 3:
\[
\text{Perimeter} = 21 \, \text{cm}, \quad \text{Area} = 18 \, \text{cm}^2
\]

---

Triangle 4:

- Sides: 5 cm, 5 cm, 5 cm (equilateral triangle)
- Height: 4 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 5 + 5 + 5 = 15 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 5 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 5 \times 4 = \frac{1}{2} \times 20 = 10 \, \text{cm}^2
\]

Answer for Triangle 4:
\[
\text{Perimeter} = 15 \, \text{cm}, \quad \text{Area} = 10 \, \text{cm}^2
\]

---

Triangle 5:

- Sides: 5 cm, 5 cm, 4 cm
- Height: 4 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 5 + 5 + 4 = 14 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 4 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 4 \times 4 = \frac{1}{2} \times 16 = 8 \, \text{cm}^2
\]

Answer for Triangle 5:
\[
\text{Perimeter} = 14 \, \text{cm}, \quad \text{Area} = 8 \, \text{cm}^2
\]

---

Triangle 6:

- Sides: 8 cm, 8 cm, 8 cm (equilateral triangle)
- Height: 7 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 8 + 8 + 8 = 24 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 8 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 8 \times 7 = \frac{1}{2} \times 56 = 28 \, \text{cm}^2
\]

Answer for Triangle 6:
\[
\text{Perimeter} = 24 \, \text{cm}, \quad \text{Area} = 28 \, \text{cm}^2
\]

---

Triangle 7:

- Sides: 5 cm, 6 cm, 4 cm
- Height: 5 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 5 + 6 + 4 = 15 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 4 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 4 \times 5 = \frac{1}{2} \times 20 = 10 \, \text{cm}^2
\]

Answer for Triangle 7:
\[
\text{Perimeter} = 15 \, \text{cm}, \quad \text{Area} = 10 \, \text{cm}^2
\]

---

Triangle 8:

- Sides: 8 cm, 6 cm, 6 cm
- Height: 5 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 8 + 6 + 6 = 20 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 6 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 6 \times 5 = \frac{1}{2} \times 30 = 15 \, \text{cm}^2
\]

Answer for Triangle 8:
\[
\text{Perimeter} = 20 \, \text{cm}, \quad \text{Area} = 15 \, \text{cm}^2
\]

---

Triangle 9:

- Sides: 7 cm, 7 cm, 7 cm (equilateral triangle)
- Height: 6 cm (given for area calculation)

#### Perimeter:
\[
\text{Perimeter} = 7 + 7 + 7 = 21 \, \text{cm}
\]

#### Area:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Assuming the base is 7 cm (since the height is given perpendicular to it):
\[
\text{Area} = \frac{1}{2} \times 7 \times 6 = \frac{1}{2} \times 42 = 21 \, \text{cm}^2
\]

Answer for Triangle 9:
\[
\text{Perimeter} = 21 \, \text{cm}, \quad \text{Area} = 21 \, \text{cm}^2
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Triangle 1:} & \text{Perimeter} = 21 \, \text{cm}, \, \text{Area} = 18 \, \text{cm}^2 \\
\text{Triangle 2:} & \text{Perimeter} = 15 \, \text{cm}, \, \text{Area} = 7.5 \, \text{cm}^2 \\
\text{Triangle 3:} & \text{Perimeter} = 21 \, \text{cm}, \, \text{Area} = 18 \, \text{cm}^2 \\
\text{Triangle 4:} & \text{Perimeter} = 15 \, \text{cm}, \, \text{Area} = 10 \, \text{cm}^2 \\
\text{Triangle 5:} & \text{Perimeter} = 14 \, \text{cm}, \, \text{Area} = 8 \, \text{cm}^2 \\
\text{Triangle 6:} & \text{Perimeter} = 24 \, \text{cm}, \, \text{Area} = 28 \, \text{cm}^2 \\
\text{Triangle 7:} & \text{Perimeter} = 15 \, \text{cm}, \, \text{Area} = 10 \, \text{cm}^2 \\
\text{Triangle 8:} & \text{Perimeter} = 20 \, \text{cm}, \, \text{Area} = 15 \, \text{cm}^2 \\
\text{Triangle 9:} & \text{Perimeter} = 21 \, \text{cm}, \, \text{Area} = 21 \, \text{cm}^2 \\
\end{array}
}
\]