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Geometric Constructions
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Midpoints & Perpendicular Bisectors
Constructing Midpoints & Perpendicular Bisectors © T Madas
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Constructing the Midpoint & Perpendicular bisector of a line segment
Construction Lines C A B Construction Lines D © T Madas
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Constructing Angle Bisectors
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Constructing an angle bisector
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Equilateral Triangles
Constructing Equilateral Triangles © T Madas
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Constructing an equilateral triangle
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Constructing a perpendicular to a given point on a line © T Madas
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Constructing a perpendicular to a given point on a line segment
Construct the perpendicular bisector of this segment © T Madas
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Constructing a perpendicular to a given point on a line segment
Alternative Construction Why does it work? © T Madas
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Constructing a perpendicular to a given point outside a line © T Madas
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Constructing a perpendicular from a given point outside a line segment
Construct the perpendicular bisector of this segment © T Madas
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an identical triangle to one given
Constructing an identical triangle to one given © T Madas
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Constructing an identical triangle to one given
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an identical angle to one given
Constructing an identical angle to one given © T Madas
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Constructing an identical angle and sides to one given
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Constructing an identical angle to one given
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a parallel line to a given line,
Constructing a parallel line to a given line, through a given point © T Madas
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Constructing a parallel line to a given line, through a given point
Why does it work? © T Madas
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Constructing a parallel line to a given line, through a given point
Alternative Construction Why does it work? © T Madas
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Circumscribing a triangle
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Circumscribing a triangle
The 3 perpendicular bisectors of the sides of a triangle are concurrent. The point they meet is called the Circumcentre. The Circumcentre has the property of being equidistant from all three vertices of the triangle. For this construction: We need the intersection of two perpendicular bisectors (since all three all concurrent). © T Madas
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Inscribing a triangle in a circle
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Inscribing a triangle in a circle
The 3 angle bisectors of a triangle are concurrent. The point they meet is called the Incentre. The Incentre has the property of being equidistant from all three sides of the triangle. For this construction: We need the intersection of two angle bisectors (since all three all concurrent). © T Madas
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Dividing a line into a given number of equal segments © T Madas
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Dividing a line into a given number of equal segments
Suppose we want to divide AB into 3 equal segments A B © T Madas
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The midpoint of a line segment
Construct: The midpoint of a line segment The perpendicular bisector of a line segment The angle bisector of an acute angle The angle bisector of an obtuse angle An equilateral triangle A triangle with sides: 3 cm, 4 cm and 6 cm A 45° angle A 60° angle A 120° angle The perpendicular to a line segment through a given point on the line segment The perpendicular to a line segment through a given point outside the line segment A parallel line to a line segment which passes through a point outside the line segment © T Madas
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Constructing Regular Hexagons
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Constructing a Regular Hexagon
Why does it work? © T Madas
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Constructing a Regular Hexagon
Why does it work? © T Madas
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Constructing a Regular Hexagon
Why does it work? © T Madas
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Constructing Regular Pentagons
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Constructing a Regular Pentagon
Start with the circumscribing circle Draw a diameter Draw the perpendicular bisector of that diameter Mark its intersection with the circle Bisect the radius as shown Draw arc as shown and mark its intersection with the diameter This is the required side length for a regular pentagon, which is circumscribed by the circle © T Madas
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Constructing Regular Octagons
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Constructing a Regular Octagon
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Constructing Regular Decagons
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Constructing a Regular Decagon
To construct a decagon you need a regular pentagon first Bisect one of the sides of the pentagon Use the chord of the new arc produced to construct a regular decagon You can use this idea to constuct a dodecagon from a hexagon A 16 – sided regular polygon using an octagon An eicosagon from a decagon and so on © T Madas
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© T Madas
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