Mensuration is essential to refrigeration and air conditioning professionals in Kenya for accurate measurement of materials, components, and spaces. Calculating perimeters and areas of various figures ensures precise fabrication, installation, and maintenance of ducts, panels, insulation, and equipment footprints. This chapter develops skills to determine these measurements for common geometric shapes encountered in refrigeration and air conditioning tasks.

1.1 Perimeter of Figures

Perimeter refers to the total length of the boundary of a two-dimensional figure. In refrigeration and air conditioning, knowing the perimeter helps estimate lengths of materials such as insulation tape, ductwork edges, or sealing strips. Different shapes require different perimeter formulas depending on their geometry.

1.1.1 Perimeter of Rectangles and Squares

The perimeter of rectangles and squares is the sum of all their sides. Rectangles have opposite sides equal, while squares have all sides equal.

The perimeter \(P\) of a rectangle is calculated by:$$P = 2(l + w)$$ where \(l\) is length and \(w\) is width.

The perimeter of a square is:$$P = 4s$$ where \(s\) is the side length.

Worked Examples

Example 1: A rectangular duct opening measures 1.2 m by 0.6 m. Find its perimeter.
Given: \(l=1.2\,m\), \(w=0.6\,m\)$$P = 2(l + w)$$$$P = 2(1.2 + 0.6)$$$$P = 2(1.8)$$$$P = 3.6\,m$$ Answer: 3.6 m

Example 2: A square insulation panel has a side length of 0.75 m. Find its perimeter.
Given: \(s=0.75\,m\)$$P = 4s$$$$P = 4 \times 0.75$$$$P = 3.0\,m$$ Answer: 3.0 m

Example 3: A rectangular evaporator coil tray measures 2 m by 0.8 m. Calculate the perimeter.
Given: \(l=2\,m\), \(w=0.8\,m\)$$P = 2(l + w)$$$$P = 2(2 + 0.8)$$$$P = 2(2.8)$$$$P = 5.6\,m$$ Answer: 5.6 m

Example 4: A square access panel has perimeter 5.2 m. Find the length of one side.
Given: \(P=5.2\,m\)$$P = 4s$$$$s = \frac{P}{4} = \frac{5.2}{4}$$$$s = 1.3\,m$$ Answer: 1.3 m

Example 5: A rectangular air filter frame has length 1.5 m and perimeter 5.2 m. Find its width.
Given: \(l=1.5\,m\), \(P=5.2\,m\)$$P = 2(l + w)$$$$5.2 = 2(1.5 + w)$$$$\frac{5.2}{2} = 1.5 + w$$$$2.6 = 1.5 + w$$$$w = 2.6 - 1.5 = 1.1\,m$$ Answer: 1.1 m

1.1.2 Perimeter of Circles and Semi-Circles

The perimeter of a circle is called the circumference. It is important for measuring circular duct edges, pipe diameters, and sealing lengths around round openings.

The circumference \(C\) of a circle is given by:$$C = 2 \pi r$$ where \(r\) is the radius.

For a semi-circle, the perimeter \(P\) includes the curved part plus the diameter:$$P = \pi r + 2r$$

Worked Examples

Example 1: A circular duct has a radius of 0.3 m. Find its circumference.
Given: \(r=0.3\,m\), \(\pi = 3.142\)$$C = 2 \pi r$$$$C = 2 \times 3.142 \times 0.3$$$$C = 6.284 \times 0.3$$$$C = 1.8852\,m$$ Answer: 1.885 m

Example 2: A semi-circular window opening has radius 0.5 m. Find the perimeter of the opening.
Given: \(r=0.5\,m\), \(\pi=3.142\)$$P = \pi r + 2r$$$$P = 3.142 \times 0.5 + 2 \times 0.5$$$$P = 1.571 + 1.0$$$$P = 2.571\,m$$ Answer: 2.571 m

Example 3: A circular flange has diameter 0.4 m. Calculate its circumference.
Given: diameter \(d=0.4\,m\), radius \(r = \frac{d}{2} = 0.2\,m\), \(\pi=3.142\)$$C = 2 \pi r$$$$C = 2 \times 3.142 \times 0.2$$$$C = 6.284 \times 0.2$$$$C = 1.2568\,m$$ Answer: 1.257 m

Example 4: A semi-circular insulation strip fits a pipe of diameter 0.6 m. Find length needed to cover the curved edge.
Given: \(d=0.6\,m\), \(r=0.3\,m\), \(\pi=3.142\)$$P = \pi r + 2r$$$$P = 3.142 \times 0.3 + 2 \times 0.3$$$$P = 0.9426 + 0.6$$$$P = 1.5426\,m$$ Answer: 1.543 m

Example 5: A circular cooling fan guard has radius 0.15 m. Calculate the circumference.
Given: \(r=0.15\,m\), \(\pi=3.142\)$$C = 2 \pi r$$$$C = 2 \times 3.142 \times 0.15$$$$C = 6.284 \times 0.15$$$$C = 0.9426\,m$$ Answer: 0.943 m

1.1.3 Perimeter of Triangles and Polygons

Triangles and polygons appear in duct transitions and panel shapes. The perimeter is the sum of all side lengths.

For triangles:$$P = a + b + c$$ where \(a\), \(b\), and \(c\) are side lengths.

For regular polygons with \(n\) sides each of length \(s\):$$P = n \times s$$

Worked Examples

Example 1: A triangular duct has sides 0.8 m, 1.2 m, and 1.0 m. Find its perimeter.
Given: \(a=0.8\,m\), \(b=1.2\,m\), \(c=1.0\,m\)$$P = a + b + c$$$$P = 0.8 + 1.2 + 1.0$$$$P = 3.0\,m$$ Answer: 3.0 m

Example 2: A regular hexagonal panel has side length 0.5 m. Calculate the perimeter.
Given: \(n=6\), \(s=0.5\,m\)$$P = n \times s$$$$P = 6 \times 0.5$$$$P = 3.0\,m$$ Answer: 3.0 m

Example 3: A pentagonal filter frame has sides measuring 0.4 m, 0.5 m, 0.45 m, 0.55 m, and 0.5 m. Find the perimeter.
Given: sides as above$$P = 0.4 + 0.5 + 0.45 + 0.55 + 0.5$$$$P = 2.4\,m$$ Answer: 2.4 m

Example 4: A triangular duct section has two equal sides measuring 0.9 m and the base 1.2 m. Find perimeter.
Given: \(a=0.9\,m\), \(b=0.9\,m\), \(c=1.2\,m\)$$P = a + b + c$$$$P = 0.9 + 0.9 + 1.2$$$$P = 3.0\,m$$ Answer: 3.0 m

Example 5: A regular octagonal air vent has perimeter 6.4 m. Find the length of each side.
Given: \(P=6.4\,m\), \(n=8\)$$P = n \times s$$$$s = \frac{P}{n} = \frac{6.4}{8}$$$$s = 0.8\,m$$ Answer: 0.8 m

Chapter Summary

This chapter focused on the practical application of mensuration in engineering contexts, starting with the calculation of perimeters and areas of various geometric figures as required by specific tasks. It emphasized methods to accurately determine these measurements for both regular and irregular shapes, ensuring precise results. The chapter then explored techniques for finding volumes and surface areas of three-dimensional objects, critical for material estimation and structural design. Special attention was given to irregular figures, where standard formulas do not apply, highlighting the use of decomposition and approximation methods. Throughout, the chapter reinforced the importance of selecting appropriate formulas based on the shape and measurement needed. The integration of these mensuration skills is essential for solving real-world engineering problems related to construction, manufacturing, and resource management.