Find the perimeter and area of each shape in this math worksheet.

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

---

1. Rectangle (Top Left)

#### Dimensions:
- Length = 66 cm
- Width = 40 cm

#### Perimeter:
The formula for the perimeter of a rectangle is:
\[
P = 2 \times (\text{Length} + \text{Width})
\]
\[
P = 2 \times (66 + 40) = 2 \times 106 = 212 \, \text{cm}
\]

#### Area:
The formula for the area of a rectangle is:
\[
A = \text{Length} \times \text{Width}
\]
\[
A = 66 \times 40 = 2640 \, \text{cm}^2
\]

Answer:
\[
\boxed{212 \, \text{cm}, 2640 \, \text{cm}^2}
\]

---

2. Triangle (Top Right)

#### Sides:
- \(a = 52 \, \text{cm}\)
- \(b = 28 \, \text{cm}\)
- \(c = 18 \, \text{cm}\)

#### Perimeter:
The perimeter of a triangle is the sum of its sides:
\[
P = a + b + c
\]
\[
P = 52 + 28 + 18 = 98 \, \text{cm}
\]

#### Area:
We use Heron's formula to find the area of the triangle. First, calculate the semi-perimeter \(s\):
\[
s = \frac{a + b + c}{2} = \frac{52 + 28 + 18}{2} = \frac{98}{2} = 49 \, \text{cm}
\]

Next, apply Heron's formula:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
A = \sqrt{49(49-52)(49-28)(49-18)}
\]
\[
A = \sqrt{49(-3)(21)(31)}
\]

Since the side lengths do not form a valid triangle (the sum of any two sides must be greater than the third side), this triangle is impossible. Therefore, the area cannot be calculated.

Answer:
\[
\boxed{98 \, \text{cm}, \text{Area cannot be calculated}}
\]

---

3. Triangle (Middle Left)

#### Sides:
- \(a = 32 \, \text{cm}\)
- \(b = 29 \, \text{cm}\)
- \(c = 28 \, \text{cm}\)

#### Perimeter:
The perimeter of a triangle is the sum of its sides:
\[
P = a + b + c
\]
\[
P = 32 + 29 + 28 = 89 \, \text{cm}
\]

#### Area:
We use Heron's formula to find the area of the triangle. First, calculate the semi-perimeter \(s\):
\[
s = \frac{a + b + c}{2} = \frac{32 + 29 + 28}{2} = \frac{89}{2} = 44.5 \, \text{cm}
\]

Next, apply Heron's formula:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
A = \sqrt{44.5(44.5-32)(44.5-29)(44.5-28)}
\]
\[
A = \sqrt{44.5 \times 12.5 \times 15.5 \times 16.5}
\]

Calculate step-by-step:
\[
44.5 \times 12.5 = 556.25
\]
\[
15.5 \times 16.5 = 255.75
\]
\[
556.25 \times 255.75 = 142140.9375
\]
\[
A = \sqrt{142140.9375} \approx 377.01 \, \text{cm}^2
\]

Answer:
\[
\boxed{89 \, \text{cm}, 377.01 \, \text{cm}^2}
\]

---

4. Rectangle (Middle Right)

#### Dimensions:
- Length = 48 cm
- Width = 28 cm

#### Perimeter:
The formula for the perimeter of a rectangle is:
\[
P = 2 \times (\text{Length} + \text{Width})
\]
\[
P = 2 \times (48 + 28) = 2 \times 76 = 152 \, \text{cm}
\]

#### Area:
The formula for the area of a rectangle is:
\[
A = \text{Length} \times \text{Width}
\]
\[
A = 48 \times 28 = 1344 \, \text{cm}^2
\]

Answer:
\[
\boxed{152 \, \text{cm}, 1344 \, \text{cm}^2}
\]

---

5. Triangle (Bottom Left)

#### Sides:
- \(a = 32 \, \text{cm}\)
- \(b = 22 \, \text{cm}\)
- \(c = 22 \, \text{cm}\)

#### Perimeter:
The perimeter of a triangle is the sum of its sides:
\[
P = a + b + c
\]
\[
P = 32 + 22 + 22 = 76 \, \text{cm}
\]

#### Area:
We use Heron's formula to find the area of the triangle. First, calculate the semi-perimeter \(s\):
\[
s = \frac{a + b + c}{2} = \frac{32 + 22 + 22}{2} = \frac{76}{2} = 38 \, \text{cm}
\]

Next, apply Heron's formula:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
A = \sqrt{38(38-32)(38-22)(38-22)}
\]
\[
A = \sqrt{38 \times 6 \times 16 \times 16}
\]

Calculate step-by-step:
\[
38 \times 6 = 228
\]
\[
16 \times 16 = 256
\]
\[
228 \times 256 = 58368
\]
\[
A = \sqrt{58368} \approx 241.6 \, \text{cm}^2
\]

Answer:
\[
\boxed{76 \, \text{cm}, 241.6 \, \text{cm}^2}
\]

---

6. Triangle (Bottom Right)

#### Sides:
- \(a = 28 \, \text{cm}\)
- \(b = 30 \, \text{cm}\)
- \(c = 34 \, \text{cm}\)

#### Perimeter:
The perimeter of a triangle is the sum of its sides:
\[
P = a + b + c
\]
\[
P = 28 + 30 + 34 = 92 \, \text{cm}
\]

#### Area:
We use Heron's formula to find the area of the triangle. First, calculate the semi-perimeter \(s\):
\[
s = \frac{a + b + c}{2} = \frac{28 + 30 + 34}{2} = \frac{92}{2} = 46 \, \text{cm}
\]

Next, apply Heron's formula:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
A = \sqrt{46(46-28)(46-30)(46-34)}
\]
\[
A = \sqrt{46 \times 18 \times 16 \times 12}
\]

Calculate step-by-step:
\[
46 \times 18 = 828
\]
\[
16 \times 12 = 192
\]
\[
828 \times 192 = 159024
\]
\[
A = \sqrt{159024} \approx 398.78 \, \text{cm}^2
\]

Answer:
\[
\boxed{92 \, \text{cm}, 398.78 \, \text{cm}^2}
\]

---

Final Answers:

1. Rectangle (Top Left): \(\boxed{212 \, \text{cm}, 2640 \, \text{cm}^2}\)
2. Triangle (Top Right): \(\boxed{98 \, \text{cm}, \text{Area cannot be calculated}}\)
3. Triangle (Middle Left): \(\boxed{89 \, \text{cm}, 377.01 \, \text{cm}^2}\)
4. Rectangle (Middle Right): \(\boxed{152 \, \text{cm}, 1344 \, \text{cm}^2}\)
5. Triangle (Bottom Left): \(\boxed{76 \, \text{cm}, 241.6 \, \text{cm}^2}\)
6. Triangle (Bottom Right): \(\boxed{92 \, \text{cm}, 398.78 \, \text{cm}^2}\)