Area
Interactive lessons, worked examples, and problem-solving techniques to help Kenyan students master KJSEA, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Understand and calculate the area of 2D plane figures including triangles, parallelograms, and composite figures by decomposing them into fundamental units.
(a) Concrete Scenario: Imagine laying out square grass sods (each \(1\text{ m} \times 1\text{ m} = 1\text{ m}^2\)) on a shamba in Nakuru. The total area is simply the count of unit squares needed to completely cover the ground without gaps or overlaps.
(b) Geometric Insight: Every 2-dimensional area is based on the product of two perpendicular dimensions (base and height). When a shape is slanted—like a parallelogram—we can slice off a triangle from one side and slide it to the other to form an equivalent rectangle of the same base and perpendicular height: \[\text{Area} = b \times h\] Similarly, any triangle is exactly half of a parallelogram with the same base and height: \[\text{Area} = \frac{1}{2} \times b \times h\]
(c) Composite Shapes Principle: Any irregular polygon can be partitioned (split) into non-overlapping fundamental shapes (rectangles, triangles, parallelograms), or calculated by enclosing it in a larger boundary and subtracting unshaded regions.
Interactive Composite Explorer: House Gable & Wall
Adjust the height of the triangular roof and the rectangular base wall to see how composite areas combine dynamically.
Shape Splitter
Drag the height slider to split the composite shape into a rectangle and a triangle. Watch each piece's area update live into the total.
Key Formulas
Rectangle:
\[ A = l \times w \]where \(l\) is length and \(w\) is width.
Triangle:
\[ A = \frac{1}{2} \times b \times h \]where \(b\) is the base and \(h\) is the perpendicular height (height at \(90^\circ\) to the base).
Parallelogram:
\[ A = b \times h \]where \(b\) is the base and \(h\) is the vertical/perpendicular height (never the slant side).
Trapezium (Trapezoid):
\[ A = \frac{1}{2}(a + b)h \]where \(a\) and \(b\) are the lengths of the two parallel sides, and \(h\) is the perpendicular distance between them.
Composite Shape (Additive Method):
\[ A_{\text{total}} = A_1 + A_2 + A_3 + \dots \]Divide the complex shape into non-overlapping standard shapes and sum their areas.
Worked Examples
Example 1 (Easy): Area of a Triangular Plot
Problem: A roadside tea kiosk in Kericho sits on a triangular piece of land with a base of \(14\text{ m}\) and a perpendicular height of \(9\text{ m}\). Calculate its area.
Step-by-Step Solution:
- Identify the given dimensions: Base \(b = 14\text{ m}\), perpendicular height \(h = 9\text{ m}\).
- Apply the triangle area formula: \[ A = \frac{1}{2} \times b \times h \]
- Substitute the values: \[ A = \frac{1}{2} \times 14 \times 9 = 7 \times 9 = 63\text{ m}^2 \]
Final Answer: \( 63\text{ m}^2 \)
Example 2 (Medium): Area of a Parallelogram Garden
Problem: A farmer in Eldoret has a field shaped like a parallelogram. The base of the field is \(25\text{ m}\), its slanted side is \(15\text{ m}\), and the perpendicular distance between the parallel bases is \(12\text{ m}\). Find the area of the field.
Step-by-Step Solution:
- Identify the relevant parameters: Base \(b = 25\text{ m}\), perpendicular height \(h = 12\text{ m}\). (Note: The slant length \(15\text{ m}\) is extra information and is not the height).
- Apply the parallelogram formula: \[ A = b \times h \]
- Calculate: \[ A = 25 \times 12 = 300\text{ m}^2 \]
Final Answer: \( 300\text{ m}^2 \)
Example 3 (Hard): Area of a Composite Storefront Wall
Problem: The side wall of a grain store consists of a rectangular base of width \(10\text{ m}\) and height \(4\text{ m}\), topped by a triangular gable with the same base \(10\text{ m}\) and a perpendicular height of \(3\text{ m}\). A rectangular ventilation window of \(2\text{ m} \times 1.5\text{ m}\) is cut into the wall. Find the total paintable surface area.
Step-by-Step Solution:
- Area of rectangular wall (\(A_1\)): \[ A_1 = l \times w = 10 \times 4 = 40\text{ m}^2 \]
- Area of triangular gable (\(A_2\)): \[ A_2 = \frac{1}{2} \times b \times h = \frac{1}{2} \times 10 \times 3 = 15\text{ m}^2 \]
- Gross wall area: \[ A_{\text{gross}} = A_1 + A_2 = 40 + 15 = 55\text{ m}^2 \]
- Area of window to subtract (\(A_{\text{window}}\)): \[ A_{\text{window}} = 2 \times 1.5 = 3\text{ m}^2 \]
- Net paintable area: \[ A_{\text{net}} = 55 - 3 = 52\text{ m}^2 \]
Final Answer: \( 52\text{ m}^2 \)
Common Mistakes
Real World
Practice