Maths Project on Displacement and Rotation of a Geometrical Figure
Maths Project on Displacement and Rotation of a Geometrical Figure
Objective:
To study the distances between different points of a geometrical figure when it is displaced and/or rotated. Enhance familiarity with co-ordinate geometry.
Description:
- A cut out of a geometrical figure such as a triangle is made and placed on a rectangular sheet of paper marked with X and Y-axis.
- The co-ordinates of the vertices of the triangle and its centroid are noted.
- The triangular cut out is displaced (along x-axis, along y-axis or along any other direction).
- The new co-ordinates of the vertices and the centroid are noted again.
- The procedure is repeated, this time by rotating the triangle as well as displacing it. The new co-ordinates of vertices and centroid are noted again.
- Using the distance formula, distances between the vertices of the triangle are obtained for the triangle in original position and in various displaced and rotated positions.
- Using the new co-ordinates of the vertices and the centroids, students will obtain the ratio in which the centroid divides the medians for various displaced and rotated positions of the triangle.
Result:
Students will verify that under any displacement and rotation of a triangle the distances between vertices remain unchanged, and also that the centroid divides the medians in ratio 2:1 in all cases.
Conclusion:
In this project the students verify (by the method of co-ordinate geometry) what is obvious geometrically, namely that the lengths of a triangle do not change when the triangle is displaced or rotated. This project will develop their familiarity with co-ordinates, distance formula and section formula of co-ordinate geometry.
Position I (Quadrant I)
When the triangle cut out is kept at Position I (I Quadrant) as shown in Fig. 4.1, the following observations are made:
Vertices of triangle are A (3, 6), B (1, 3), C (6, 3).
By midpoint formula, D is midpoint of BC and its co-ordinates are calculated as shown below:
Midpoint = [(x1 + x2) / 2, (y1 + y2) / 2]
= [(1 + 6) / 2, (3 + 3) / 2]
= D (3.5, 3)
The centroid is calculated by formula:
x co-ordinate of centroid = (3 + 1 + 6) / 3 = 10 / 3 = 3.3
y co-ordinate of centroid = (6 + 3 + 3) / 3 = 12 / 3 = 4
Co-ordinates of centroid are G (3.3, 4).
Distance AG = √((3.3 - 3)² + (4 - 6)²) = √(0.09 + 4) = √4.09 ∴ AG = 2.02
Distance GD = √((3.3 - 3.5)² + (4 - 3)²) = √(0.04 + 1) = √1.04 ∴ GD = 1.01
Ratio = AG / GD = 2.02 / 1.01 = 2 / 1
∴ AG : GD = 2 : 1
Graph: Triangle in Position I (Quadrant I) showing vertices A (3, 6), B (1, 3), C (6, 3), midpoint D, and centroid G (plotted from data above).
Position II (Quadrant II)
When the triangle cut out is kept at Position II (II Quadrant) as shown in Fig. 4.2, the following observations are made:
Vertices of triangle are A (-4, 6), B (-6, 3) and C (-1, 3).
By midpoint formula, D is midpoint of BC and its co-ordinates are calculated as shown below:
Midpoint = [(x1 + x2) / 2, (y1 + y2) / 2]
= [(-6 + (-1)) / 2, (3 + 3) / 2]
= [-7 / 2, 3]
∴ D = (-3.5, 3)
The centroid is calculated by formula:
x co-ordinate of centroid = (-4 + (-6) + (-1)) / 3 = -11 / 3 = -3.6
y co-ordinate of centroid = (6 + 3 + 3) / 3 = 12 / 3 = 4
Co-ordinates of centroid are G (-3.6, 4).
Distance AG = √((-3.6 + 4)² + (4 - 6)²) = √((0.4)² + (-2)²) = √(0.16 + 4) = √4.16 ∴ AG = 2.03
Distance DG = √((-3.6 + 3.5)² + (3 - 4)²) = √(0.01 + 1) = √1.01 ∴ DG = 1.00
Ratio = AG / DG = 2.03 / 1.00 = 2 / 1
∴ AG : DG = 2 : 1
Graph: Triangle in Position II (Quadrant II) showing vertices A (-4, 6), B (-6, 3), C (-1, 3), midpoint D, and centroid G (plotted from data above).
Position III (Quadrant III)
When the triangle cut out is kept at Position III (III Quadrant) as shown in Fig. 4.3, the following observations are made:
Vertices of triangle are A (-3, -6), B (-1, -3) and C (-6, -3).
By midpoint formula, D is midpoint of BC and its co-ordinates are calculated as shown below:
Midpoint = [(x1 + x2) / 2, (y1 + y2) / 2]
= [(-1 + (-6)) / 2, (-3 + (-3)) / 2]
= [-3.5, -3]
The centroid is calculated by formula:
x co-ordinate of centroid = (-3 + (-1) + (-6)) / 3 = -10 / 3 = -3.3
y co-ordinate of centroid = (-6 + (-3) + (-3)) / 3 = -12 / 3 = -4
Co-ordinates of centroid are G (-3.3, -4).
Distance AG = √((-3.3 + 3)² + (-4 + 6)²) = √((-0.3)² + (2)²) = √(0.09 + 4) = √4.09 ∴ AG = 2.03
Distance DG = √((-3.3 + 3.5)² + (-4 + 3)²) = √(0.04 + 1) = √1.04 ∴ DG = 1.01
Ratio = AG / DG = 2.03 / 1.01 = 2 / 1
∴ AG : DG = 2 : 1
Graph: Triangle in Position III (Quadrant III) showing vertices A (-3, -6), B (-1, -3), C (-6, -3), midpoint D, and centroid G (plotted from data above).
Position IV (Quadrant IV)
When the triangle cut out is kept at Position IV (IV Quadrant) as shown in Fig. 4.4, the following observations are made:
Vertices of triangle are A (4, -6), B (1, -3) and C (6, -3).
By midpoint formula, D is midpoint of BC and its co-ordinates are calculated as shown below:
Midpoint = [(x1 + x2) / 2, (y1 + y2) / 2]
= [(1 + 6) / 2, (-3 + (-3)) / 2]
= D (3.5, -3)
The centroid is calculated by formula:
x co-ordinate of centroid = (4 + 1 + 6) / 3 = 11 / 3 = 3.6
y co-ordinate of centroid = (-6 + (-3) + (-3)) / 3 = -12 / 3 = -4
Co-ordinates of centroid are G (3.6, -4).
Distance AG = √((3.6 - 4)² + (-4 - (-6))²) = √((-0.4)² + (-2)²) = √(0.16 + 4) = √4.16 ∴ AG = 2.03
Distance DG = √((3.6 - 3.5)² + (-4 + 3)²) = √(0.01 + 1) = √1.01 ∴ DG = 1.00
Ratio = AG / DG = 2.03 / 1.00 = 2 / 1
∴ AG : DG = 2 : 1
Graph: Triangle in Position IV (Quadrant IV) showing vertices A (4, -6), B (1, -3), C (6, -3), midpoint D, and centroid G (plotted from data above).