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Geometry and Measurement - Geometric Constructions

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Introduction to Geometric Construction: Construction is the process of drawing geometric figures accurately using only two tools: a compass and a straightedge (ruler without markings). In these drawings, we do not use a protractor to measure angles, but rather use geometric properties to create them. A straightedge is used to draw line segments, while a compass is used to draw circles and arcs of a specific radius.

Perpendicular Bisector: This construction produces a line that divides a segment ABAB into two equal halves at a 9090^{\circ} angle. To visualize the process: place the compass point at AA with a radius greater than half the length of ABAB and draw arcs above and below the segment. Repeat this from point BB with the same radius. Connecting the two points where the arcs intersect creates the perpendicular bisector.

Angle Bisector: An angle bisector is a ray that divides an angle into two equal parts. Visually, you place the compass on the vertex VV and draw an arc cutting both arms of the angle at points XX and YY. Then, from XX and YY, draw two arcs with the same radius that intersect at a point ZZ inside the angle. The line VZVZ is the bisector, meaning XVZ=YVZ\angle XVZ = \angle YVZ.

Constructing 6060^{\circ} and 120120^{\circ} Angles: These are fundamental angles built using equilateral triangle properties. Draw a ray with endpoint OO and an arc crossing it at PP. Keep the compass width the same, place the point at PP, and draw another arc crossing the first at QQ. The angle POQ\angle POQ is 6060^{\circ}. If you move the compass to QQ and draw a third arc crossing the original large arc at RR, the angle POR\angle POR is 120120^{\circ}.

Constructing 9090^{\circ} and 4545^{\circ} Angles: A 9090^{\circ} angle is typically constructed by creating a perpendicular bisector on a straight line or by bisecting the 180180^{\circ} angle of a straight line. To create a 4545^{\circ} angle, you first construct a 9090^{\circ} angle and then apply the angle bisector method to split it exactly in half.

Constructing Parallel Lines: To draw a line through point PP parallel to line mm, we use the property of corresponding angles. Draw a transversal line through PP that intersects mm at point AA. Construct an arc at point AA and a matching arc at point PP. Measure the width of the angle at AA with the compass and transfer that width to the arc at PP. Drawing a line through PP and this new intersection point results in parallel lines.

Perpendicular from a Point to a Line: To drop a perpendicular from point PP (not on line LL) to line LL, draw an arc from PP that cuts line LL at two points, XX and YY. Then, find the perpendicular bisector of the segment XYXY. The line passing through PP and the midpoint of XYXY will be perpendicular to line LL at a 9090^{\circ} angle.

📐Formulae

Angle Bisector Equation: θ1=θ2=12θtotal\text{Angle Bisector Equation: } \theta_{1} = \theta_{2} = \frac{1}{2}\theta_{total}

Perpendicularity Condition: Angle between lines=90\text{Perpendicularity Condition: } \text{Angle between lines} = 90^{\circ}

Midpoint of segment AB:M=A+B2\text{Midpoint of segment } AB: M = \frac{A + B}{2}

Sum of angles on a straight line: θ=180\text{Sum of angles on a straight line: } \sum \theta = 180^{\circ}

Interior angles of an equilateral triangle: 60,60,60\text{Interior angles of an equilateral triangle: } 60^{\circ}, 60^{\circ}, 60^{\circ}

💡Examples

Problem 1:

Describe how to construct a 3030^{\circ} angle starting from a line segment OAOA.

Solution:

  1. Place the compass point at OO and draw a large arc that intersects segment OAOA at point PP. \ 2. Keeping the same compass width, place the compass point at PP and draw an arc that intersects the first arc at point QQ. \ 3. Draw a ray from OO through QQ. The angle QOA\angle QOA is 6060^{\circ}. \ 4. Place the compass point at PP and draw an arc in the interior of QOA\angle QOA. \ 5. Place the compass point at QQ and, with the same radius, draw an arc that intersects the arc from step 4 at point RR. \ 6. Draw ray OROR.

Explanation:

A 3030^{\circ} angle is half of a 6060^{\circ} angle. We first construct the 6060^{\circ} angle using the equilateral triangle method and then bisect that angle to reach 3030^{\circ}.

Problem 2:

Given a line segment XYXY of length 1010 cm, construct its perpendicular bisector and identify the length of the resulting segments.

Solution:

  1. Draw line segment XY=10XY = 10 cm using a ruler. \ 2. Open the compass to a width greater than 55 cm (e.g., 77 cm). \ 3. Place the compass at XX and draw arcs above and below the line. \ 4. Place the compass at YY with the same width and draw arcs intersecting the first ones at points AA and BB. \ 5. Draw a line through AA and BB, intersecting XYXY at point MM. \ 6. Calculate the segments: XM=MY=10 cm2=5 cmXM = MY = \frac{10\text{ cm}}{2} = 5\text{ cm}.

Explanation:

The perpendicular bisector divides a segment into two equal parts. Since the original length was 1010 cm, the construction creates a midpoint MM such that the segments on either side are 55 cm each, meeting at a 9090^{\circ} angle.