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Practical Geometry - Constructing a Copy of a Line Segment

Grade 6CBSE

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

🔑Concepts

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To copy a line segment AB‾\overline{AB} onto a new line ll, we use a compass to 'measure' the distance between AA and BB instead of a ruler. This ensures higher precision by transferring the exact physical span to a new location.

A line segment AB with a compass arc showing the measured length.
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Once the compass is set to the length of AB‾\overline{AB}, fix the pointer at a point PP on a line ll. Draw an arc that intersects the line ll at point QQ. The segment PQ‾\overline{PQ} is then congruent to AB‾\overline{AB}, written as AB‾≅PQ‾\overline{AB} \cong \overline{PQ}.

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The ruler is used only to draw the initial straight line ll and to provide a straight edge; the compass is the primary tool for transferring the specific length without reading numerical values.

Line l with points P and Q marked by a compass arc.
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Key Steps: 1. Draw a line ll. 2. Mark point PP on it. 3. Place compass pointer on AA and pencil on BB. 4. Without changing the compass width, place pointer on PP and swing an arc to cut ll at QQ.

📐Formulae

Length(AB‾)=Length(PQ‾)Length(\overline{AB}) = Length(\overline{PQ})

AB‾≅PQ‾\overline{AB} \cong \overline{PQ}

L=∣x2−x1∣L = |x_2 - x_1|

💡Examples

Problem 1:

Given a line segment XY‾\overline{XY} of length 6.46.4 cm, construct a copy MN‾\overline{MN} using only a ruler and a compass.

Solution:

  1. Draw a line ll and mark a point MM on it.
  2. Place the compass pointer on point XX of the given segment XY‾\overline{XY} and open it until the pencil tip reaches point YY.
  3. Maintaining the same compass width, place the metal pointer on point MM on line ll.
  4. Draw an arc that cuts the line ll at a point, and name it NN.
  5. The segment MN‾\overline{MN} is the required copy of XY‾\overline{XY}, such that Length(MN)=6.4Length(MN) = 6.4 cm.

Explanation:

The compass acts as a physical measurement transfer tool, ensuring that the distance between MM and NN is identical to the distance between XX and YY without needing to read numerical values on a ruler twice.

Problem 2:

If a line segment AB‾\overline{AB} is given, how can you construct a segment PX‾\overline{PX} whose length is twice that of AB‾\overline{AB}?

Solution:

  1. Draw a long line mm and mark a point PP on it.
  2. Measure the length of the given segment AB‾\overline{AB} using a compass (pointer on AA, pencil on BB).
  3. Place the compass pointer on PP and mark an arc on line mm to get point QQ. Now PQ=ABPQ = AB.
  4. Without changing the compass width, move the pointer to point QQ and mark another arc further along the line mm to get point XX.
  5. The segment PX‾\overline{PX} is the required segment where PX=PQ+QX=AB+AB=2×ABPX = PQ + QX = AB + AB = 2 \times AB.

Explanation:

This construction uses the addition of segments property. By placing two copies of the same segment end-to-end on a straight line, the total length becomes 2×Length(AB)2 \times Length(AB).

Problem 3:

Given two line segments AB‾\overline{AB} and CD‾\overline{CD}, construct a line segment EF‾\overline{EF} such that the length of EF‾\overline{EF} is equal to the sum of the lengths of AB‾\overline{AB} and CD‾\overline{CD}.

Construction of a segment EF which is the sum of AB and CD on a straight line.

Solution:

  1. Draw a line mm and mark a point EE on it.
  2. Open the compass to match the length of AB‾\overline{AB}.
  3. Place the compass pointer on EE and mark an arc on line mm to find point XX.
  4. Now, open the compass to match the length of CD‾\overline{CD}.
  5. Place the pointer on XX and mark another arc on line mm (away from EE) to find point FF.
  6. EF‾\overline{EF} is the required segment where EF=AB+CDEF = AB + CD.

Explanation:

This method uses the property of addition of segments by placing them end-to-end (adjacent) on the same line.

Problem 4:

Given a line segment RS‾\overline{RS} of length 88 cm and TV‾\overline{TV} of length 33 cm, construct a segment XY‾\overline{XY} equal to the difference RS−TVRS - TV.

Construction showing the subtraction of segment TV from RS to get segment XY.

Solution:

  1. Draw a line kk and mark point XX on it.
  2. Adjust the compass to the length of RS‾\overline{RS} and mark point ZZ from XX on the line.
  3. Adjust the compass to the length of TV‾\overline{TV}.
  4. Place the pointer on ZZ and mark an arc back towards XX. The point where it cuts the line is YY.
  5. The segment XY‾\overline{XY} represents RS−TVRS - TV.

Explanation:

To find the difference, we first layout the larger segment and then 'cut back' the length of the smaller segment.