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Geometry and Measurement - Geometric Constructions

Grade 8IB

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

🔑Concepts

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The perpendicular bisector of a line segment is a line that passes through the midpoint of the segment at a 90∘90^{\circ} angle. Every point on the perpendicular bisector is equidistant from the segment's endpoints.

A horizontal line segment AB with a vertical perpendicular bisector L passing through midpoint M.
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An angle bisector is a ray that divides an angle into two equal parts. To construct it, arcs are drawn from the vertex and the points where a primary arc intersects the angle's arms.

An angle with a ray splitting it into two equal halves.
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Constructing a 60∘60^{\circ} angle involves drawing an equilateral triangle framework. By using the same compass width for the base and the intersecting arc, you create a vertex that naturally forms 60∘60^{\circ} with the base.

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A perpendicular from a point to a line is constructed by drawing an arc centered at the point that intersects the line twice, then finding the perpendicular bisector of the segment between those two intersections.

📐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 30∘30^{\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 60∘60^{\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 30∘30^{\circ} angle is half of a 60∘60^{\circ} angle. We first construct the 60∘60^{\circ} angle using the equilateral triangle method and then bisect that angle to reach 30∘30^{\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 90∘90^{\circ} angle.

Problem 3:

Construct a 90∘90^{\circ} angle at point PP on a line segment and then bisect it to create a 45∘45^{\circ} angle.

Geometry construction showing a horizontal base line, a vertical line at 90 degrees, and a bisecting ray at 45 degrees.

Solution:

  1. Draw a line and mark point PP.
  2. Use a compass to mark points AA and BB equidistant from PP on the line.
  3. Construct the perpendicular bisector of ABAB to get the 90∘90^{\circ} line.
  4. Place the compass point at the intersection of the 90∘90^{\circ} line and the arc, and at point BB, to draw intersecting arcs.
  5. Draw a line from PP through the intersection to get 45∘45^{\circ}.

Explanation:

A 45∘45^{\circ} angle is exactly half of a 90∘90^{\circ} angle. The construction uses the property that the angle bisector of a right angle creates two equal 45∘45^{\circ} angles.

Problem 4:

Given a circle with center OO and a chord ABAB, construct the perpendicular bisector of the chord and show that it passes through the center OO.

A circle with center O, a horizontal chord AB, and a vertical perpendicular bisector passing through both the midpoint of AB and the center O.

Solution:

  1. Place the compass on point AA and draw arcs above and below the chord ABAB.
  2. With the same radius, place the compass on point BB and draw arcs intersecting the first ones.
  3. Draw a line through the intersection points.
  4. Observe that this line passes through center OO.

Explanation:

In any circle, the perpendicular bisector of a chord always passes through the center of the circle because the center is equidistant from any two points on the circumference.