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Geometry - Practical Geometry: Construction using Ruler and Compasses

Grade 6ICSE

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

🔑Concepts

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A circle is defined by a center and a radius. Using a compass, keeping the needle at center OO and the pencil at a distance rr allows us to draw the set of all points equidistant from OO.

A circle with center O and radius r showing the path of a compass.
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The perpendicular bisector of a line segment ABAB is a line that divides ABAB into two equal parts at a 90∘90^\circ angle. It is constructed by drawing intersecting arcs of equal radius (greater than half of ABAB) from both points AA and BB.

Construction of a perpendicular bisector PQ of segment AB.
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An angle bisector is a ray that divides an angle into two equal parts. To bisect ∠AOB\angle AOB, draw an arc cutting OAOA and OBOB at PP and QQ. Then, draw arcs of the same radius from PP and QQ to intersect at point RR.

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Standard angles like 60∘,120∘,60^\circ, 120^\circ, and 90∘90^\circ can be constructed without a protractor. A 60∘60^\circ angle is formed by drawing an arc and then marking the same radius along that arc from the intersection point on the base line.

📐Formulae

Diameter(d)=2×r\text{Diameter} (d) = 2 \times r

Radius(r)=d2\text{Radius} (r) = \frac{d}{2}

Measure of Bisected Angle=12×Original Angle\text{Measure of Bisected Angle} = \frac{1}{2} \times \text{Original Angle}

Perpendicularity Condition:∠=90∘\text{Perpendicularity Condition}: \angle = 90^\circ

Sum of angles on a straight line=180∘\text{Sum of angles on a straight line} = 180^\circ

💡Examples

Problem 1:

Construct a line segment XY=7.4 cmXY = 7.4 \text{ cm} and find its perpendicular bisector using a ruler and compasses.

Solution:

  1. Draw a line segment XY=7.4 cmXY = 7.4 \text{ cm} using a ruler. 2. Open the compass to a radius that is clearly more than half of 7.4 cm7.4 \text{ cm} (e.g., 4 cm4 \text{ cm}). 3. Place the compass pointer at XX and draw two arcs, one above the segment XYXY and one below. 4. Keeping the same radius, place the pointer at YY and draw arcs that intersect the previous arcs at points PP and QQ. 5. Use a ruler to draw a line passing through PP and QQ.

Explanation:

The line PQPQ is the perpendicular bisector. It intersects XYXY at a point MM, where XM=MY=3.7 cmXM = MY = 3.7 \text{ cm} and ∠PMX=90∘\angle PMX = 90^\circ.

Problem 2:

Construct an angle of 90∘90^\circ at the end-point AA of a ray ABAB.

Solution:

  1. Draw a ray ABAB. 2. With AA as the center and any convenient radius, draw a semi-circular arc that cuts ABAB at point PP. 3. With the same radius and PP as center, draw an arc cutting the first arc at QQ (this represents 60∘60^\circ). 4. With QQ as center and the same radius, draw another arc cutting the first arc at RR (this represents 120∘120^\circ). 5. From QQ and RR, draw two arcs with the same radius that intersect each other at point SS. 6. Join ASAS.

Explanation:

The angle ∠SAB\angle SAB is 90∘90^\circ. This method works because 90∘90^\circ is exactly halfway between 60∘60^\circ and 120∘120^\circ (60+120−602=9060 + \frac{120-60}{2} = 90).

Problem 3:

Construct an angle of 60∘60^\circ using a ruler and compasses on a ray OAOA.

Construction of a 60 degree angle using a compass.

Solution:

  1. Draw a ray OAOA.
  2. With OO as center and any convenient radius, draw an arc cutting OAOA at point MM.
  3. With MM as center and the same radius, draw another arc intersecting the first arc at point NN.
  4. Join ONON and extend it to BB.
  5. ∠AOB=60∘\angle AOB = 60^\circ.

Explanation:

Since the radius used to mark point MM and point NN is the same, △OMN\triangle OMN would form an equilateral triangle if MNMN were joined, making the angle at OO exactly 60∘60^\circ.

Problem 4:

Construct a circle of radius 3.5 cm3.5 \text{ cm} and draw any chord PQPQ. Construct the perpendicular bisector of PQPQ.

Circle with chord PQ and its perpendicular bisector passing through center O.

Solution:

  1. Mark a point OO as the center and draw a circle with radius 3.5 cm3.5 \text{ cm}.
  2. Draw any chord PQPQ inside the circle.
  3. With PP as center and radius >12PQ> \frac{1}{2} PQ, draw arcs above and below PQPQ.
  4. With QQ as center and the same radius, draw arcs to intersect the previous arcs at XX and YY.
  5. Join XYXY. XYXY is the perpendicular bisector of PQPQ.

Explanation:

The perpendicular bisector of any chord of a circle always passes through the center of the circle.