Geometric calculations are essential in various professional fields across Kenya, including construction, agriculture, business layout planning, and healthcare facility design. Understanding the properties and measurements of geometrical figures enables professionals to solve practical problems involving area, perimeter, volume, and spatial arrangements. This chapter focuses on performing geometric calculations for common shapes encountered in everyday work environments, equipping students to apply mathematical principles effectively in their respective fields.

2.1 Geometrical figures

Geometrical figures form the foundation of many calculations in professional settings. Mastery of their properties, formulas for perimeter, area, and volume, and the ability to apply these formulas to real-world problems is crucial. This topic introduces the key geometrical figures and how to perform calculations related to them.

2.1.1 Triangles

Triangles are one of the most basic geometrical figures used in various sectors such as land surveying, building design, and agricultural field layout. Calculations involving triangles often include finding the area, perimeter, and sometimes the lengths of sides or heights.

The fundamental formula for the area of a triangle is:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

The perimeter is the sum of all three sides:

$$\text{Perimeter} = a + b + c$$

where \(a\), \(b\), and \(c\) are the lengths of the sides.

Worked Examples

Example 1: A school plans to build a triangular flower bed with a base of 6 m and height of 4 m. Find the area.

Given: base \(b = 6\,m\), height \(h = 4\,m\)

$$\text{Area} = \frac{1}{2} \times b \times h$$

$$= \frac{1}{2} \times 6 \times 4$$

$$= 3 \times 4$$

$$= 12\,m^{2}$$

Answer: 12 square metres

Example 2: A cooperative farm has a triangular plot with sides measuring 5 m, 7 m, and 9 m. Calculate the perimeter.

Given: \(a=5\,m\), \(b=7\,m\), \(c=9\,m\)

$$\text{Perimeter} = a + b + c$$

$$= 5 + 7 + 9$$

$$= 21\,m$$

Answer: 21 metres

Example 3: A county hospital has a triangular signboard with sides 3 m, 4 m, and 5 m. Calculate its area using Heron's formula.

Given: \(a=3\,m\), \(b=4\,m\), \(c=5\,m\)

First, calculate the semi-perimeter \(s\):

$$s = \frac{a + b + c}{2} = \frac{3 + 4 + 5}{2} = 6\,m$$

Then, calculate the area:

$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$

$$= \sqrt{6(6 - 3)(6 - 4)(6 - 5)}$$

$$= \sqrt{6 \times 3 \times 2 \times 1}$$

$$= \sqrt{36}$$

$$= 6\,m^{2}$$

Answer: 6 square metres

Example 4: A retail business wants to fence a triangular garden with sides 8 m, 15 m, and 17 m. Find the perimeter and area.

Given: \(a=8\,m\), \(b=15\,m\), \(c=17\,m\)

Perimeter:

$$= 8 + 15 + 17 = 40\,m$$

Semi-perimeter:

$$s = \frac{40}{2} = 20\,m$$

Area:

$$= \sqrt{20(20 - 8)(20 - 15)(20 - 17)}$$

$$= \sqrt{20 \times 12 \times 5 \times 3}$$

$$= \sqrt{3600} = 60\,m^{2}$$

Answer: Perimeter = 40 metres, Area = 60 square metres

Example 5: A hotel floor has a triangular section with base 10 m and height 7 m. Calculate the area and the length of the hypotenuse if it is a right-angled triangle.

Given: base \(b=10\,m\), height \(h=7\,m\)

Area:

$$= \frac{1}{2} \times 10 \times 7 = 35\,m^{2}$$

Length of hypotenuse \(c\):

$$c = \sqrt{b^{2} + h^{2}} = \sqrt{10^{2} + 7^{2}} = \sqrt{100 + 49} = \sqrt{149} = 12.21\,m$$

Answer: Area = 35 square metres, Hypotenuse = 12.21 metres

2.1.2 Rectangles and Squares

Rectangles and squares are common shapes in buildings, fields, and furniture layout. Calculations for these shapes involve determining area, perimeter, and sometimes diagonal lengths.

The area of a rectangle is:

$$\text{Area} = \text{length} \times \text{width}$$

The perimeter is:

$$\text{Perimeter} = 2(\text{length} + \text{width})$$

For squares, length equals width, so:

$$\text{Area} = \text{side}^2$$

$$\text{Perimeter} = 4 \times \text{side}$$

The diagonal \(d\) of a rectangle or square is:

$$d = \sqrt{\text{length}^2 + \text{width}^2}$$

Worked Examples

Example 1: A retail store has a rectangular floor measuring 12 m by 8 m. Find the area and perimeter.

Given: length \(l=12\,m\), width \(w=8\,m\)

Area:

$$= 12 \times 8 = 96\,m^{2}$$

Perimeter:

$$= 2(12 + 8) = 2 \times 20 = 40\,m$$

Answer: Area = 96 square metres, Perimeter = 40 metres

Example 2: A county office has a square meeting room with sides 9 m. Calculate the area, perimeter, and diagonal length.

Given: side \(s=9\,m\)

Area:

$$= 9^{2} = 81\,m^{2}$$

Perimeter:

$$= 4 \times 9 = 36\,m$$

Diagonal:

$$= \sqrt{9^{2} + 9^{2}} = \sqrt{81 + 81} = \sqrt{162} = 12.73\,m$$

Answer: Area = 81 square metres, Perimeter = 36 metres, Diagonal = 12.73 metres

Example 3: A farm has a rectangular irrigation plot of length 20 m and width 15 m. Calculate the diagonal and perimeter.

Given: length \(l=20\,m\), width \(w=15\,m\)

Diagonal:

$$= \sqrt{20^{2} + 15^{2}} = \sqrt{400 + 225} = \sqrt{625} = 25\,m$$

Perimeter:

$$= 2(20 + 15) = 2 \times 35 = 70\,m$$

Answer: Diagonal = 25 metres, Perimeter = 70 metres

Example 4: A hotel plans to carpet a square hall with a side of 14 m. Find the area and perimeter.

Given: side \(s=14\,m\)

Area:

$$= 14^{2} = 196\,m^{2}$$

Perimeter:

$$= 4 \times 14 = 56\,m$$

Answer: Area = 196 square metres, Perimeter = 56 metres

Example 5: A classroom has a rectangular window 3 m high and 2 m wide. Calculate the area and diagonal length.

Given: height \(h=3\,m\), width \(w=2\,m\)

Area:

$$= 3 \times 2 = 6\,m^{2}$$

Diagonal:

$$= \sqrt{3^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13} = 3.61\,m$$

Answer: Area = 6 square metres, Diagonal = 3.61 metres

2.1.3 Circles

Circles appear in many contexts such as water tanks, circular gardens, and round tables. Calculations involving circles typically include circumference and area.

The circumference \(C\) of a circle is:

$$C = 2 \pi r$$

The area \(A\) is:

$$A = \pi r^{2}$$

where \(r\) is the radius and \(\pi \approx 3.142\).

Worked Examples

Example 1: A SACCO has a circular water tank with radius 4 m. Find the circumference.

Given: radius \(r=4\,m\)

$$C = 2 \pi r = 2 \times 3.142 \times 4 = 25.136\,m$$

Answer: Circumference = 25.14 metres

Example 2: A hotel has a circular dining table with diameter 2 m. Calculate the area of the tabletop.

Given: diameter \(d=2\,m\), radius \(r = \frac{d}{2} = 1\,m\)

$$A = \pi r^{2} = 3.142 \times 1^{2} = 3.142\,m^{2}$$

Answer: Area = 3.14 square metres

Example 3: A county government office has a circular flower bed with radius 5 m. Find the circumference and area.

Given: radius \(r=5\,m\)

Circumference:

$$= 2 \times 3.142 \times 5 = 31.42\,m$$

Area:

$$= 3.142 \times 5^{2} = 3.142 \times 25 = 78.55\,m^{2}$$

Answer: Circumference = 31.42 metres, Area = 78.55 square metres

Example 4: A retail shop installs circular mats with diameter 1.5 m. Calculate the area of one mat.

Given: diameter \(d=1.5\,m\), radius \(r=0.75\,m\)

$$A = 3.142 \times 0.75^{2} = 3.142 \times 0.5625 = 1.767\,m^{2}$$

Answer: Area = 1.77 square metres

Example 5: A university plans a circular lawn with a circumference of 62.8 m. Find the radius and area.

Given: circumference \(C=62.8\,m\)

Radius:

$$r = \frac{C}{2 \pi} = \frac{62.8}{2 \times 3.142} = \frac{62.8}{6.284} = 10\,m$$

Area:

$$= 3.142 \times 10^{2} = 3.142 \times 100 = 314.2\,m^{2}$$

Answer: Radius = 10 metres, Area = 314.2 square metres

2.1.4 Composite Figures

Composite figures combine simple shapes such as rectangles, triangles, and circles to form complex shapes. Calculating area or perimeter involves breaking down the figure into known shapes and summing or subtracting their measures.

The general approach is:

1. Identify simple shapes that make up the composite figure.
2. Calculate area or perimeter of each simple shape.
3. Add or subtract areas as appropriate to find the total.

Worked Examples

Example 1: A farm has a rectangular plot 20 m by 10 m with a semicircular pond of radius 5 m on one side. Find the total area excluding the pond.

Given: rectangle \(20\,m \times 10\,m\), semicircle radius \(r=5\,m\)

Rectangle area:

$$= 20 \times 10 = 200\,m^{2}$$

Semicircle area:

$$= \frac{1}{2} \times \pi r^{2} = \frac{1}{2} \times 3.142 \times 5^{2} = \frac{1}{2} \times 3.142 \times 25 = 39.275\,m^{2}$$

Total area excluding pond:

$$= 200 - 39.275 = 160.725\,m^{2}$$

Answer: 160.73 square metres

Example 2: A hotel garden has a rectangular lawn 15 m by 10 m with a triangular flower bed (base 6 m, height 4 m) inside it. Find the remaining lawn area.

Given: rectangle \(15\,m \times 10\,m\), triangle base \(6\,m\), height \(4\,m\)

Rectangle area:

$$= 15 \times 10 = 150\,m^{2}$$

Triangle area:

$$= \frac{1}{2} \times 6 \times 4 = 12\,m^{2}$$

Remaining lawn area:

$$= 150 - 12 = 138\,m^{2}$$

Answer: 138 square metres

Example 3: A circular flower bed with radius 7 m has a rectangular path 2 m wide around it. Find the area of the path.

Given: inner circle radius \(r=7\,m\), path width \(w=2\,m\)

Outer radius:

$$= 7 + 2 = 9\,m$$

Area of outer circle:

$$= 3.142 \times 9^{2} = 3.142 \times 81 = 254.502\,m^{2}$$

Area of inner circle:

$$= 3.142 \times 7^{2} = 3.142 \times 49 = 153.958\,m^{2}$$

Area of path:

$$= 254.502 - 153.958 = 100.544\,m^{2}$$

Answer: 100.54 square metres

Example 4: A school has a playground shaped as a rectangle 40 m by 20 m with a semicircular stage of radius 10 m attached along the width. Find the total area.

Given: rectangle \(40\,m \times 20\,m\), semicircle radius \(r=10\,m\)

Rectangle area:

$$= 40 \times 20 = 800\,m^{2}$$

Semicircle area:

$$= \frac{1}{2} \times 3.142 \times 10^{2} = \frac{1}{2} \times 3.142 \times 100 = 157.1\,m^{2}$$

Total area:

$$= 800 + 157.1 = 957.1\,m^{2}$$

Answer: 957.1 square metres

Example 5: A county office has a compound shaped as a square 30 m side with a triangular garden (base 12 m, height 9 m) and a circular fountain (radius 3 m) inside it. Find the remaining area excluding the garden and fountain.

Given: square side \(s=30\,m\), triangle base \(b=12\,m\), height \(h=9\,m\), circle radius \(r=3\,m\)

Square area:

$$= 30^{2} = 900\,m^{2}$$

Triangle area:

$$= \frac{1}{2} \times 12 \times 9 = 54\,m^{2}$$

Circle area:

$$= 3.142 \times 3^{2} = 3.142 \times 9 = 28.278\,m^{2}$$

Remaining area:

$$= 900 - 54 - 28.278 = 817.722\,m^{2}$$

Answer: 817.72 square metres

Chapter Summary

This chapter focused on understanding and working with various geometrical figures, including triangles, rectangles, and circles, to develop foundational knowledge for further calculations. It then progressed to calculating the areas of these figures, demonstrating how to determine the size of surfaces using appropriate formulas for each shape. The chapter emphasized practical application by guiding students through the use of Pythagoras' theorem to find unknown side lengths in right-angled triangles. Each section provided clear methods to approach problems step by step, ensuring accuracy in geometric calculations. Mastery of these concepts is essential for solving real-world civil engineering problems involving measurements and spatial reasoning. The chapter combined theoretical understanding with numerical examples to reinforce learning and build confidence in performing geometric calculations.