Let's analyze both figures (a) and (b) to find their perimeter and area.

Part (a): Irregular Quadrilateral

• Side lengths:
  • $15 \, \text{m}$,
  • $11 \, \text{m}$,
  • $9 \, \text{m}$,
  • $9.85 \, \text{m}$

Perimeter of Figure (a):

The perimeter $P$ is the sum of all the side lengths: $P = 15 + 11 + 9 + 9.85 = 44.85 \, \text{m}$

Area of Figure (a):

This figure can be split into a rectangle and a right triangle:

1. Rectangle:
   • Length = $9 \, \text{m}$
   • Width = $11 \, \text{m}$ $\text{Area of rectangle} = 9 \times 11 = 99 \, \text{m}^2$
2. Right triangle:
   • Base = $9 \, \text{m}$
   • Height = $15 - 11 = 4 \, \text{m}$ $\text{Area of triangle} = \frac{1}{2} \times 9 \times 4 = 18 \, \text{m}^2$

Total area is the sum of the rectangle and triangle: $\text{Total area} = 99 + 18 = 117 \, \text{m}^2$

Part (b): Triangle

• Known sides:
  • $9 \, \text{m}$ (base),
  • $14 \, \text{m}$ (hypotenuse),
  • $7 \, \text{m}$ (height)

Perimeter of Figure (b):

We need to find the unknown side (adjacent side) using the Pythagorean theorem: $\text{Adjacent side} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} \approx 12.12 \, \text{m}$ Now, the perimeter is: $P = 9 + 14 + 12.12 \approx 35.12 \, \text{m}$

Area of Figure (b):

The area of the triangle is: $\text{Area} = \frac{1}{2} \times 9 \times 7 = 31.5 \, \text{m}^2$

Summary:

• Perimeter of (a): $44.85 \, \text{m}$
• Area of (a): $117 \, \text{m}^2$
• Perimeter of (b): $35.12 \, \text{m}$
• Area of (b): $31.5 \, \text{m}^2$

Would you like more details on any part of this solution?

Here are some related questions for further exploration:

1. How can you break down more complex polygons into simpler shapes for area calculations?
2. What other methods could you use to find the area of an irregular shape?
3. How does the Pythagorean theorem help in calculating unknown sides in triangles?
4. What are alternative ways to calculate the area of a triangle besides base and height?
5. Can you explore the differences in perimeter calculations between regular and irregular shapes?

Tip: Always look for ways to simplify complex shapes into basic geometric forms (like rectangles or triangles) when solving for area and perimeter.