Let's solve the problem step by step:

a) Calculate the area of the figure

The figure can be divided into:

1. A rectangle (bottom left).
2. A triangle (top right).
3. A smaller rectangle (bottom right).

Step 1: Areas of the individual parts

• Left rectangle: Width = 3 cm, Height = 4 cm
  $\text{Area} = \text{width} \times \text{height} = 3 \times 4 = 12 \, \text{cm}^2$

• Triangle: Base = 4 cm, Height = (4 cm - 2.5 cm) = 1.5 cm
  $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 1.5 = 3 \, \text{cm}^2$

• Right rectangle: Width = 2 cm, Height = 2.5 cm
  $\text{Area} = \text{width} \times \text{height} = 2 \times 2.5 = 5 \, \text{cm}^2$

Step 2: Total area $\text{Total area} = 12 + 3 + 5 = 20 \, \text{cm}^2$

b) Use Pythagoras to find the length of $x$

The diagonal $x$ forms a right triangle where:

• Opposite side = 4 cm (base of the triangle)
• Adjacent side = 1.5 cm (height of the triangle)

Using the Pythagorean theorem: $x = \sqrt{\text{opposite}^2 + \text{adjacent}^2} = \sqrt{4^2 + 1.5^2} = \sqrt{16 + 2.25} = \sqrt{18.25}$ $x \approx 4.3 \, \text{cm} \, (\text{to one decimal place})$

c) Calculate the perimeter of the figure

The perimeter is the sum of all sides. Breaking it down:

1. Top horizontal edge = 3 cm + 4 cm + 2 cm = 9 cm
2. Left vertical edge = 4 cm
3. Right vertical edge = 2.5 cm
4. Bottom horizontal edge = 2 cm
5. Diagonal $x = 4.3 \, \text{cm}$

$\text{Perimeter} = 9 + 4 + 2.5 + 2 + 4.3 = 21.8 \, \text{cm}$

Final Answers

a) $A = 20 \, \text{cm}^2$
b) $x \approx 4.3 \, \text{cm}$
c) $O \approx 21.8 \, \text{cm}$

Would you like further clarification on any part? 😊

5 Related Questions

1. How do you derive the Pythagorean theorem formula?
2. Can you explain how to calculate the area of composite shapes with curves?
3. What are some practical applications of finding a perimeter?
4. How do you round values to different decimal places?
5. How does the orientation of a triangle affect its base and height?

Tip: Always break down complex shapes into simpler parts (rectangles, triangles, etc.) to make calculations easier.