Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Equal Area of similar Figures
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Equal Area of similar Figures
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Subject: Technical Drawings
Class: Senior Secondary 1
Term: 3rd Term
Week: 11
Theme: Geometrical Constructions
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State the as sumptions or the or ems of equal are as Construct plain figures of equal are as.
Triangle EBC.
4. Therefore, Triangle EBC is equal in area to quadrilateral ABCD. (This transforms a 4-sided figure into a 3-sided figure of equal area.) Diagrammatic Representation (Textual Description): Draw points A, B, C, D to form an irregular quadrilateral. Draw diagonal AC. Through D, draw a line parallel to AC, intersecting the line AB extended at
E. Connect E to
C. The triangle EBC is the required triangle.
Worked Example 2: To construct a square equal in area to a given rectangle. This method uses the geometric mean principle. The side of the square is the geometric mean of the length and width of the rectangle.
Given: A rectangle ABCD with length L and width
W. Required: To construct a square equal in area to rectangle ABC
D. Steps:
1. Draw the given rectangle ABCD. (Let AB = L and BC = W).
2. Extend the side AB to a point E such that BE = BC (the width W). So, AE = L + W.
3. Bisect the line AE to find its midpoint M.
4. With M as centre and MA (or ME) as radius, draw a semi-circle passing through A and E.
5. From point B, draw a perpendicular line to AE, intersecting the semi-circle at point F. (Reasoning: BF is the geometric mean of AB and BE. In a semi-circle, if a perpendicular is drawn from a point on the diameter to the arc, the length of the perpendicular is the geometric mean of the two segments it divides the diameter into. Thus, BF2 = AB BE = L W.)
6. BF is the side length of the required square. Construct a square with side BF (e.g., BFG H). Area of rectangle = L W. Area of square = BF2 = L W.
Therefore, the square BFG H is equal in area to the rectangle ABC
D. Diagrammatic Representation (Textual Description): Draw rectangle ABCD. Extend AB to E such that BE = BC. Find midpoint M of AE. Draw a semi-circle with centre M and radius MA (or ME). Draw a perpendicular from B to AE, intersecting the semi-circle at
F. Construct a square using BF as a side.
Worked Example 3: To construct a triangle equal in area to a given rectangle.
Given: A rectangle ABC
D. Required: To construct a triangle equal in area to rectangle ABC
D. Steps:
1. Draw the given rectangle ABCD.
2. Extend the base AB to a point E such that BE = A
B. So, AE = 2 AB. (Reasoning: A triangle's area is (1/2) base height. A rectangle's area is base height. To make them equal with the same height, the triangle's base must be twice the rectangle's base.)
3. From point C, draw a line to E.
4. Triangle ADE (or BCE depending on how you construct it) or a triangle constructed on a base of 2AB and height BC will have the same area. Consider Triangle AFE where F is a point such that the height from F to AE is equal to BC. If the base of the triangle is AE = 2 AB and its height is BC, then Area(triangle) = (1/2) (2 AB) BC = AB BC = Area(rectangle).
A simpler method: Take AB as the base of the rectangle. The height is BC. Construct a triangle with base 2AB and height BC. For instance, extend AB to E such that BE = AB. Take C as the apex. Then triangle ACE has base AE = 2AB and height (perpendicular distance from C to AE) = BC.
Therefore, Triangle ACE is the required triangle equal in area to the rectangle ABC
D. Diagrammatic Representation (Textual Description): Draw rectangle ABCD. Extend AB to E such that BE = AB. * Connect C to E. Triangle ACE has equal area to rectangle ABCD. The core idea behind "Equal Area of Figures" is the ability to transform a given geometric shape into another, different shape, such that both figures enclose the same amount of surface. This transformation is based on specific geometric theorems and principles, primarily involving parallel lines and common bases.
Key Definitions: Area: The extent of a two-dimensional surface enclosed within a boundary.
Equal Area: Two or more figures are said to have equal areas if they enclose the same amount of surface, irrespective of their shape.
Transformation: In this context, it refers to changing one geometric figure into another of equal area.
Theorems and Assumptions of Equal Areas: The following theorems are fundamental to constructing figures of equal areas:
1. Theorem 1: Figures on the same base and between the same parallel lines are equal in area.
Explanation: This is the most critical theorem for area transformation. If two triangles or a triangle and a parallelogram share the same base, and their vertices opposite to the base lie on a line parallel to the base, then their areas are equal (for triangles) or the triangle's area is half the parallelogram's area. This holds because their heights (perpendicular distance between the parallel lines) are equal.
Example: Consider a triangle ABC and a triangle ABD, where C and D lie on a line parallel to AB. Both triangles have the same base AB and the same height (perpendicular distance between AB and the parallel line passing through C and D).
Therefore, Area(ABC) = Area(ABD).
2. Theorem 2: The area of a triangle is half the area of a parallelogram if they have the same base and are between the same parallel lines.
Explanation: This is a direct consequence of Theorem
1. A parallelogram can be divided into two congruent triangles by its diagonal.
Therefore, if a triangle and a parallelogram share a base and are between the same parallels, the height is the same for both. Area of triangle = (1/2) base height, and Area of parallelogram = base height.
3. Theorem 3: Two triangles having the same base and equal altitudes have equal areas.
Explanation: This re-states the principle of Theorem 1 more directly for triangles. Since Area = (1/2) base height, if both base and height are equal, the areas must be equal. Common Construction Techniques and Worked
Examples: The following examples demonstrate how to apply these theorems to construct figures of equal areas.
Worked Example 1: To construct a triangle equal in area to a given irregular polygon (e.g., an irregular quadrilateral). This is a general method applicable to any polygon. The principle is to reduce the number of sides by one in each step, transforming a vertex and an adjacent side into an equivalent triangle.
Given: An irregular quadrilateral ABC
D. Required: To construct a triangle equal in area to quadrilateral ABC
D. Steps:
1. Draw the given quadrilateral ABCD. (Assume vertices A, B, C, D in counter-clockwise order).
2. Select a diagonal, say AC. Draw a line from vertex D parallel to diagonal A
C. Let this line intersect the extended base AB at point E. (Reasoning: Triangle ADC and Triangle AEC share the same base AC (if we consider AC as a temporary base) and are between the parallel lines AC and DE.
Therefore, Area(ADC) = Area(AEC) based on Theorem 1.)
3. The quadrilateral ABCD can now be considered as Triangle ABC + Triangle AD
C. Since Area(ADC) = Area(AEC), we can substitute: Area(ABCD) = Area(ABC) + Area(AEC). By observation, Area(ABC) + Area(AEC) combine to form Triangle EBC.
4. Therefore, Triangle EBC is equal in area to quadrilateral ABCD. (This transforms a 4-sided figure into a 3-sided figure of equal area.) Diagrammatic Representation (Textual Description): Draw points A, B, C, D to form an irregular quadrilateral. Draw diagonal AC. Through D, draw a line parallel to AC, intersecting the line AB extended at
E. Connect E to
C. The triangle EBC is the required triangle.
Worked Example 2: To construct a square equal in area to a given rectangle.
This method uses the Teacher Activities: Introduction (10 minutes): Review concepts of area for basic shapes (triangle, rectangle, square) and their formulas. Introduce the concept of transforming figures while maintaining equal area, highlighting its practical relevance in surveying and design (e.g., "Imagine needing to divide an irregularly shaped piece of land amongst family members – how do you ensure everyone gets an equal share if shapes are different?"). Present the performance objectives clearly. Explanation and Demonstration (25 minutes): Clearly state and explain the "Theorems of Equal Areas," particularly Theorem 1 using visual aids (drawing on the board, projector). Demonstrate step-by-step the construction of a triangle equal in area to an irregular quadrilateral (Worked Example 1). Use large, clear drawings on the board or with a projector, explaining each step and the underlying theorem. Demonstrate the construction of a square equal in area to a rectangle (Worked Example 2), explaining the geometric mean principle. Demonstrate the construction of a triangle equal in area to a rectangle (Worked Example 3). Emphasise the importance of accurate use of drawing instruments (ruler, compass, set squares, protractor).
Guided Practice (20 minutes): Distribute drawing paper and instruments. Guide students through one of the constructions (e.g., converting a pentagon into a triangle of equal area, applying the same principle as the quadrilateral example). Provide immediate feedback and correct common errors in instrument handling or construction steps. Encourage peer discussion and problem-solving.
Activity-Based Learning (10 minutes): Divide students into small groups. Assign each group a different polygon (e.g., irregular pentagon, hexagon) and instruct them to collectively plan the steps to convert it into a triangle of equal area. Have each group present their planned steps.
Conclusion and Summary (5 minutes): Recap the key theorems and construction methods. Address any lingering questions. Assign independent practice work.
Student Activities: Actively listen to explanations and observe demonstrations. Ask clarifying questions about the theorems and construction steps. Participate in group discussions and problem-solving. Individually and in groups, practice constructing various figures of equal areas using drawing instruments on drawing paper. Sketch and label their constructions accurately. Present their solutions or planned steps to the class.
Instructions: Students are to attempt these constructions using appropriate drawing instruments.
Question 1: Construct a triangle equal in area to a given rectangle ABCD with sides AB = 60mm and BC = 40mm.
Solution: Steps: Draw the rectangle ABCD with AB = 60mm and BC = 40mm. Extend the base AB to a point E such that BE = AB = 60mm. So, AE = 120mm. Join point C to
E. The triangle ACE is the required triangle. (
Commentary: Area of rectangle = AB BC = 60 40 = 2400 sq mm. Area of triangle ACE = (1/2) base height = (1/2) AE BC = (1/2) (2 AB) BC = (1/2) 120 40 = 2400 sq mm. The areas are equal.)
Question 2: Construct a square equal in area to a rectangle with length 80mm and width 45mm.
Solution: Steps: Draw the rectangle ABCD with AB = 80mm and BC = 45mm. Extend the side AB to point E such that BE = BC = 45mm. Find the midpoint M of the line segment AE (AE = AB + BE = 80 + 45 = 125mm. M is at 62.5mm from A). With M as the centre and MA (or ME) as the radius, draw a semi-circle passing through A and E. From point B, draw a perpendicular line to AE, intersecting the semi-circle at point F. BF is the side length of the required square. Construct a square (e.g., BFGH) with side BF. (
Commentary: BF is the geometric mean of AB and BE. BF = sqrt(AB BE) = sqrt(80 45) = sqrt(3600) = 60mm. The area of the square is 60 60 = 3600 sq mm, which equals the area of the rectangle 80 45 = 3600 sq mm.)* Question 3: Transform an irregular pentagon ABCDE into a triangle of equal area.
Solution: Steps: Draw the given irregular pentagon ABCDE. (Label vertices A, B, C, D, E in sequence). First Reduction (Pentagon to Quadrilateral): Draw diagonal AD. From E, draw a line parallel to AD, intersecting the extended base AB at point F. Join F to D. The quadrilateral FBCD has the same area as the pentagon ABCDE. (Area(ADE) = Area(ADF)). Second Reduction (Quadrilateral to Triangle): Now, we need to convert quadrilateral FBCD into a triangle. Draw diagonal FC. From D, draw a line parallel to FC, intersecting the extended base FB at point
G. Join G to
C. The triangle GBC is the required triangle. (Area(FCD) = Area(FCG)). (
Commentary: This process iteratively reduces the number of sides of the polygon by one at each step until a triangle is formed, all while maintaining the original area.)*
Land Partitioning and Inheritance in Nigeria: In many Nigerian communities, land is often inherited or divided among family members. These plots are frequently irregular in shape. The ability to construct figures of equal area allows surveyors or local community leaders to mathematically divide an irregular plot into multiple, simpler plots (e.g., triangles or rectangles) that are equal in area, ensuring fairness even if the shapes differ significantly. For example, ensuring two siblings receive equal land area, even if one plot is a long, narrow strip and the other is a more compact, irregular shape. Architectural Design and Material Estimation: Nigerian architects and builders frequently work with floor plans that include irregular rooms or spaces. To accurately estimate materials like tiles, carpet, paint, or roofing sheets, these complex shapes can be transformed into equivalent simple shapes (e.g., squares or rectangles) of equal area. This simplifies calculations, reduces waste, and helps in precise budgeting for construction projects common across Nigeria. Infrastructure Development and Road Design: When planning for new roads, bridges, or public facilities, land acquisition is critical. Land compensation might be based on the area of land taken. If plots are irregular, converting them into equivalent simpler areas aids in fair assessment and compensation. This is vital in Nigeria's ongoing infrastructure development projects, where land acquisition issues are common.