By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Engineering quantity surveyors must accurately measure and calculate the perimeter and areas of various figures to estimate materials, costs, and quantities for construction projects. Mastery of mensuration techniques enables precise preparation of bills of quantities and cost plans, essential for budgeting and tendering. This chapter focuses on methods to determine perimeters and areas of common geometric shapes encountered in construction sites and architectural designs.
Perimeter and area calculations are fundamental in quantity surveying for determining lengths of boundary lines and surface extents. These measurements influence costing of materials such as fencing, flooring, plastering, and painting. Kenyan quantity surveyors regularly apply these concepts when quantifying site boundaries, slabs, walls, and roofing.
The perimeter is the total distance around a closed figure. For regular polygons, it is computed by multiplying the number of sides by the length of one side. Composite figures combine basic shapes, requiring summation of all external edges.
$$P = n \times s$$
where \(P\) is perimeter, \(n\) is number of sides, and \(s\) is side length.
Example 1: A rectangular site measures 25 m by 15 m. Find the perimeter.
Given: length \(l = 25\, m\), width \(w = 15\, m\)
$$P = 2(l + w)$$
$$P = 2(25 + 15)$$
$$P = 2(40) = 80\, m$$
Answer: 80 m
Example 2: A regular hexagonal water tank has each side 4 m. Calculate the perimeter.
Given: number of sides \(n=6\), side length \(s=4\, m\)
$$P = n \times s$$
$$P = 6 \times 4 = 24\, m$$
Answer: 24 m
Example 3: A compound plot consists of a rectangle 30 m by 20 m and a semicircular driveway attached along the 20 m side. Find the total perimeter.
Given: rectangle length \(l=30\, m\), width \(w=20\, m\), semicircle diameter \(d=20\, m\)
Rectangle perimeter excluding driveway side:
$$P_r = 2(l + w) - w = 2(30 + 20) - 20 = 2(50) - 20 = 100 - 20 = 80\, m$$
Semicircle perimeter (half circumference):
$$P_s = \frac{\pi d}{2} = \frac{3.1416 \times 20}{2} = 31.416\, m$$
Total perimeter:
$$P = P_r + P_s = 80 + 31.416 = 111.416\, m$$
Answer: 111.42 m
Area calculations for rectangles and squares are essential in quantifying floor slabs, walls, and roofing surfaces. The area of a rectangle is the product of its length and width, while a square is a special case with equal sides.
$$A = l \times w$$
where \(A\) is area, \(l\) is length, and \(w\) is width.
Example 1: Calculate the area of a rectangular floor measuring 12 m by 8 m.
Given: \(l=12\, m\), \(w=8\, m\)
$$A = l \times w$$
$$A = 12 \times 8 = 96\, m^2$$
Answer: 96 m²
Example 2: A square column has side length 0.5 m. Find its base area.
Given: \(s=0.5\, m\)
$$A = s^2$$
$$A = 0.5 \times 0.5 = 0.25\, m^2$$
Answer: 0.25 m²
Example 3: A rectangular slab is 20 m long and 15 m wide. A 5 m by 3 m rectangular opening is cut out for a stairwell. Find the net slab area.
Given: slab \(l=20\, m\), \(w=15\, m\); opening \(l_o=5\, m\), \(w_o=3\, m\)
Slab area:
$$A_s = 20 \times 15 = 300\, m^2$$
Opening area:
$$A_o = 5 \times 3 = 15\, m^2$$
Net area:
$$A = A_s - A_o = 300 - 15 = 285\, m^2$$
Answer: 285 m²
Triangles and trapeziums frequently appear in roof trusses, gable ends, and architectural features. The area of a triangle is half the product of its base and height, while the trapezium area is half the sum of parallel sides multiplied by the height.
Triangle area:
$$A = \frac{1}{2} \times b \times h$$
Trapezium area:
$$A = \frac{1}{2} (a + b) \times h$$
where \(a\) and \(b\) are parallel sides, and \(h\) is height.
Example 1: Calculate the area of a triangular gable end with base 6 m and height 4 m.
Given: \(b=6\, m\), \(h=4\, m\)
$$A = \frac{1}{2} \times b \times h$$
$$A = \frac{1}{2} \times 6 \times 4 = 12\, m^2$$
Answer: 12 m²
Example 2: A trapezium-shaped retaining wall has parallel sides 8 m and 5 m, with height 3 m. Find its area.
Given: \(a=8\, m\), \(b=5\, m\), \(h=3\, m\)
$$A = \frac{1}{2} (a + b) \times h$$
$$A = \frac{1}{2} (8 + 5) \times 3 = \frac{1}{2} \times 13 \times 3 = 19.5\, m^2$$
Answer: 19.5 m²
Example 3: A triangular roof truss has sides 5 m base and 3 m height. A rectangular opening 1.5 m by 1 m is cut out. Find the net area.
Given: triangle \(b=5\, m\), \(h=3\, m\); opening \(l=1.5\, m\), \(w=1\, m\)
Triangle area:
$$A_t = \frac{1}{2} \times 5 \times 3 = 7.5\, m^2$$
Opening area:
$$A_o = 1.5 \times 1 = 1.5\, m^2$$
Net area:
$$A = 7.5 - 1.5 = 6\, m^2$$
Answer: 6 m²
Circular shapes are common in water tanks, columns, and landscaping features. The area of a circle is \(\pi\) times the radius squared, and the circumference is \(2 \pi r\). For semicircles, the area is half that of a full circle, and the perimeter includes the curved edge plus the diameter.
Circle area:
$$A = \pi r^2$$
Circle circumference:
$$C = 2 \pi r$$
Semicircle perimeter:
$$P = \pi r + 2r$$
Example 1: Calculate the area of a circular water tank with radius 3 m.
Given: \(r=3\, m\)
$$A = \pi r^2$$
$$A = 3.1416 \times 3^2 = 3.1416 \times 9 = 28.2744\, m^2$$
Answer: 28.27 m²
Example 2: Find the circumference of a circular column with diameter 0.8 m.
Given: diameter \(d = 0.8\, m\), radius \(r = \frac{d}{2} = 0.4\, m\)
$$C = 2 \pi r$$
$$C = 2 \times 3.1416 \times 0.4 = 2.5133\, m$$
Answer: 2.51 m
Example 3: A semicircular driveway has radius 5 m. Calculate its perimeter including the straight edge.
Given: \(r=5\, m\)
$$P = \pi r + 2r$$
$$P = 3.1416 \times 5 + 2 \times 5 = 15.708 + 10 = 25.708\, m$$
Answer: 25.71 m
Example 4: A circular water tank has radius 2.5 m. A rectangular door of 1 m by 2 m is cut out on its side. Find the net surface area of the circular base.
Given: \(r=2.5\, m\), door \(l=2\, m\), \(w=1\, m\)
Circle area:
$$A_c = \pi r^2 = 3.1416 \times 2.5^2 = 3.1416 \times 6.25 = 19.635\, m^2$$
Door area:
$$A_d = 2 \times 1 = 2\, m^2$$
Net area:
$$A = 19.635 - 2 = 17.635\, m^2$$
Answer: 17.64 m²
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Create a free accountThis chapter focused on calculating the perimeter and areas of various plane figures, emphasizing the application of appropriate formulas for common shapes. It then progressed to determining the volume and surface area of regular solids such as cylinders, cones, and spheres, highlighting the importance of accurate measurements in engineering contexts. Techniques for finding the area of irregular figures were introduced, involving methods like grid approximation and decomposition into simpler shapes. The chapter concluded with the use of Pappus theorem to compute areas and volumes generated by rotating plane figures around an axis. Together, these topics provide essential tools for solving practical mensuration problems encountered in civil engineering and construction projects. Mastery of these concepts enables precise quantification of materials and dimensions critical for design and cost estimation.
Calculate the perimeter of a rectangular site measuring 40 m by 25 m. (2 marks)
Find the area of a circular water tank base with a diameter of 10 m. (2 marks)
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