By the end of this chapter, you will be able to:
These skills will help you solve real-world engineering problems involving measurements and shapes with confidence and precision.
Engineering Mathematics II is essential for Land Survey professionals in Kenya to accurately quantify boundaries, land areas, and construction site layouts. Mastery of mensuration techniques enables surveyors to calculate lengths, perimeters, and areas of various geometric figures encountered in the field. This chapter focuses on calculating perimeters of figures, which is critical for defining property boundaries and planning construction works.
Perimeters define the total boundary length of a closed figure and are fundamental in land surveying for demarcation and fencing purposes. Surveyors routinely measure perimeters to determine the extent of plots, agricultural lands, and construction sites. Understanding how to calculate perimeters of regular and irregular shapes using formulas and field measurements is crucial.
The rectangle and square are common shapes in land parcels and building footprints. The perimeter of these shapes is the sum of all their sides.
The formula for the perimeter \(P\) of a rectangle with length \(l\) and width \(w\) is:
$$P = 2(l + w)$$
For a square with side length \(s\), the perimeter is:
$$P = 4s$$
Example 1: A rectangular plot measures 50 m in length and 30 m in width. Find the perimeter.
Given: \(l = 50\,m\), \(w = 30\,m\)
$$P = 2(l + w)$$
$$P = 2(50 + 30)$$
$$P = 2(80)$$
$$P = 160\,m$$
Answer: 160 m
Example 2: A square garden has each side measuring 25 m. Calculate the perimeter.
Given: \(s = 25\,m\)
$$P = 4s$$
$$P = 4 \times 25$$
$$P = 100\,m$$
Answer: 100 m
Example 3: A rectangular foundation has length 12.5 m and width 7.8 m. Find its perimeter.
Given: \(l = 12.5\,m\), \(w = 7.8\,m\)
$$P = 2(l + w)$$
$$P = 2(12.5 + 7.8)$$
$$P = 2(20.3)$$
$$P = 40.6\,m$$
Answer: 40.6 m
Example 4: A square plot has a perimeter of 240 m. Find the length of one side.
Given: \(P = 240\,m\)
$$P = 4s$$
$$240 = 4s$$
$$s = \frac{240}{4}$$
$$s = 60\,m$$
Answer: 60 m
Example 5: A rectangular field has a perimeter of 300 m. If the length is 95 m, find the width.
Given: \(P = 300\,m\), \(l = 95\,m\)
$$P = 2(l + w)$$
$$300 = 2(95 + w)$$
$$\frac{300}{2} = 95 + w$$
$$150 = 95 + w$$
$$w = 150 - 95$$
$$w = 55\,m$$
Answer: 55 m
Triangles often occur in land subdivision and boundary demarcation. The perimeter is the sum of all three sides.
The perimeter \(P\) of a triangle with sides \(a\), \(b\), and \(c\) is:
$$P = a + b + c$$
Example 1: A triangular plot has sides measuring 40 m, 55 m, and 60 m. Find the perimeter.
Given: \(a=40\,m\), \(b=55\,m\), \(c=60\,m\)
$$P = a + b + c$$
$$P = 40 + 55 + 60$$
$$P = 155\,m$$
Answer: 155 m
Example 2: A triangular survey area has two sides measuring 120 m and 85 m. The perimeter is 300 m. Find the third side.
Given: \(a=120\,m\), \(b=85\,m\), \(P=300\,m\)
$$P = a + b + c$$
$$300 = 120 + 85 + c$$
$$300 = 205 + c$$
$$c = 300 - 205$$
$$c = 95\,m$$
Answer: 95 m
Example 3: A triangle has sides in the ratio 3:4:5 and a perimeter of 360 m. Find the length of each side.
Given: ratio \(3:4:5\), \(P=360\,m\)
Let the common multiplier be \(x\).
$$P = 3x + 4x + 5x = 12x$$
$$360 = 12x$$
$$x = \frac{360}{12} = 30$$
Sides:
$$3x = 3 \times 30 = 90\,m$$
$$4x = 4 \times 30 = 120\,m$$
$$5x = 5 \times 30 = 150\,m$$
Answer: 90 m, 120 m, 150 m
Example 4: A triangular fence has side lengths 75 m and 95 m. The third side is 20 m longer than the shortest side. Find the perimeter.
Given: \(a=75\,m\), \(b=95\,m\), \(c = a + 20 = 75 + 20 = 95\,m\)
$$P = a + b + c$$
$$P = 75 + 95 + 95$$
$$P = 265\,m$$
Answer: 265 m
Example 5: A triangle has a perimeter of 180 m. Two sides measure 65 m and 55 m. Find the third side.
Given: \(P=180\,m\), \(a=65\,m\), \(b=55\,m\)
$$P = a + b + c$$
$$180 = 65 + 55 + c$$
$$180 = 120 + c$$
$$c = 180 - 120 = 60\,m$$
Answer: 60 m
Circular boundaries occur in water tanks, roundabouts, and curved roads. The perimeter of a circle is called the circumference.
The circumference \(C\) of a circle with radius \(r\) is:
$$C = 2 \pi r$$
The length \(L\) of a circular arc with radius \(r\) and central angle \(\theta\) (in degrees) is:
$$L = \frac{\theta}{360} \times 2 \pi r$$
Where \(\pi \approx 3.142\).
Example 1: A circular water tank has a radius of 7 m. Find the circumference.
Given: \(r = 7\,m\)
$$C = 2 \pi r$$
$$C = 2 \times 3.142 \times 7$$
$$C = 43.988\,m$$
Answer: 43.99 m
Example 2: A circular roundabout has diameter 20 m. Find the perimeter.
Given: \(d = 20\,m\), \(r = \frac{d}{2} = 10\,m\)
$$C = 2 \pi r$$
$$C = 2 \times 3.142 \times 10$$
$$C = 62.84\,m$$
Answer: 62.84 m
Example 3: A circular arc has radius 15 m and subtends a central angle of 60°. Find the arc length.
Given: \(r = 15\,m\), \(\theta = 60^\circ\)
$$L = \frac{\theta}{360} \times 2 \pi r$$
$$L = \frac{60}{360} \times 2 \times 3.142 \times 15$$
$$L = \frac{1}{6} \times 94.26$$
$$L = 15.71\,m$$
Answer: 15.71 m
Example 4: A circular garden has circumference 31.42 m. Find its radius.
Given: \(C = 31.42\,m\)
$$C = 2 \pi r$$
$$31.42 = 2 \times 3.142 \times r$$
$$31.42 = 6.284 r$$
$$r = \frac{31.42}{6.284} = 5\,m$$
Answer: 5 m
Example 5: A circular arc has length 25 m and radius 20 m. Find the central angle \(\theta\) in degrees.
Given: \(L = 25\,m\), \(r = 20\,m\)
$$L = \frac{\theta}{360} \times 2 \pi r$$
Rearranged for \(\theta\):
$$\theta = \frac{L \times 360}{2 \pi r}$$
$$\theta = \frac{25 \times 360}{2 \times 3.142 \times 20}$$
$$\theta = \frac{9000}{125.68}$$
$$\theta = 71.6^\circ$$
Answer: 71.6 degrees
Land parcels and construction sites often have irregular polygonal shapes. The perimeter is the sum of the lengths of all sides.
For an irregular polygon with sides \(s_1, s_2, s_3, ..., s_n\):
$$P = s_1 + s_2 + s_3 + \cdots + s_n$$
Example 1: A land parcel has sides measuring 45 m, 60 m, 32 m, 50 m, and 40 m. Find the perimeter.
Given: \(s_1=45\,m\), \(s_2=60\,m\), \(s_3=32\,m\), \(s_4=50\,m\), \(s_5=40\,m\)
$$P = 45 + 60 + 32 + 50 + 40$$
$$P = 227\,m$$
Answer: 227 m
Example 2: A plot has 6 sides with lengths 20 m, 35 m, 30 m, 25 m, 40 m, and 28 m. Calculate the perimeter.
Given: \(s_1=20\,m\), \(s_2=35\,m\), \(s_3=30\,m\), \(s_4=25\,m\), \(s_5=40\,m\), \(s_6=28\,m\)
$$P = 20 + 35 + 30 + 25 + 40 + 28$$
$$P = 178\,m$$
Answer: 178 m
Example 3: A land survey shows an irregular pentagon with side lengths 55 m, 45 m, 50 m, 60 m, and 40 m. Find the perimeter.
Given: \(s_1=55\,m\), \(s_2=45\,m\), \(s_3=50\,m\), \(s_4=60\,m\), \(s_5=40\,m\)
$$P = 55 + 45 + 50 + 60 + 40$$
$$P = 250\,m$$
Answer: 250 m
Example 4: A construction site boundary is marked by 8 sides with lengths 12 m, 15 m, 20 m, 18 m, 22 m, 25 m, 30 m, and 17 m. Calculate the total perimeter.
Given: \(s_1=12\,m\), \(s_2=15\,m\), \(s_3=20\,m\), \(s_4=18\,m\), \(s_5=22\,m\), \(s_6=25\,m\), \(s_7=30\,m\), \(s_8=17\,m\)
$$P = 12 + 15 + 20 + 18 + 22 + 25 + 30 + 17$$
$$P = 159\,m$$
Answer: 159 m
Example 5: A plot has 7 sides with lengths 35 m, 40 m, 45 m, 50 m, 55 m, 60 m, and 65 m. Find the perimeter.
Given: \(s_1=35\,m\), \(s_2=40\,m\), \(s_3=45\,m\), \(s_4=50\,m\), \(s_5=55\,m\), \(s_6=60\,m\), \(s_7=65\,m\)
$$P = 35 + 40 + 45 + 50 + 55 + 60 + 65$$
$$P = 350\,m$$
Answer: 350 m
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Create a free accountThis chapter covered the calculation of perimeters for various plane figures, emphasizing the importance of accurate measurement in engineering applications. It progressed to determining the areas of different shapes, demonstrating how these calculations are essential for site planning and material estimation. The study then extended to three-dimensional solids, where methods for finding volumes and surface areas were explained in detail. Understanding these measurements is crucial for tasks such as concrete volume estimation and surface treatment calculations. The chapter integrated practical examples to illustrate the application of formulas in real engineering contexts. Mastery of these mensuration concepts enables precise quantification of materials and space, which is fundamental in civil engineering project execution.
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