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Practical Geometry - Construction of a Line Segment

Grade 6CBSE

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

🔑Concepts

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A line segment is a fixed portion of a line with two definite endpoints. The distance between these endpoints is called its length. For example, a segment AB‾\overline{AB} has endpoints AA and BB.

A line segment with endpoints A and B
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Using a ruler and compasses is more accurate than using a ruler alone. To construct a segment of length rr, we fix the compass pointer at 00 on the ruler and open the pencil arm to the rr mark.

Measuring a 5 cm gap using compasses on a ruler
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To copy a line segment PQ‾\overline{PQ} without measuring its numerical length, place the compass pointer on PP and the pencil point on QQ. Then, without changing the setting, draw an arc on a new line ll from a point AA to find point BB.

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The difference between two segments AB‾\overline{AB} and CD‾\overline{CD} (where AB>CDAB > CD) is a segment XZ‾\overline{XZ} such that XZ=AB−CDXZ = AB - CD.

📐Formulae

LengthofsegmentAB‾=Distance between point A and point BLength of segment \overline{AB} = \text{Distance between point } A \text{ and point } B

1 cm=10 mm1\text{ cm} = 10\text{ mm}

1 m=100 cm1\text{ m} = 100\text{ cm}

If point CC lies between AA and BB: AB=AC+CBAB = AC + CB

If PQ‾\overline{PQ} is the difference between AB‾\overline{AB} and CD‾\overline{CD}: PQ=AB−CDPQ = AB - CD

💡Examples

Problem 1:

Construct a line segment PQ‾\overline{PQ} of length 5.6 cm5.6\text{ cm} using a ruler and compasses.

Solution:

Step 1: Draw a line ll and mark a point PP on it. Step 2: Place the metal pointer of the compasses on the 0 mark0\text{ mark} of the ruler. Step 3: Open the compasses so that the pencil point rests on the 5.6 cm5.6\text{ cm} mark (5 cm5\text{ cm} and 66 small millimeter divisions). Step 4: Without changing the opening of the compasses, place the pointer on point PP and swing an arc to cut the line ll at point QQ. Step 5: PQ‾\overline{PQ} is the required line segment of length 5.6 cm5.6\text{ cm}.

Explanation:

This method is preferred over using just a ruler because the compass preserves the exact length while transferring it to the paper, reducing the chances of manual error while marking points.

Problem 2:

Given two line segments AB‾=3.5 cm\overline{AB} = 3.5\text{ cm} and CD‾=2.4 cm\overline{CD} = 2.4\text{ cm}, construct a segment XY‾\overline{XY} such that XY=AB+CDXY = AB + CD.

Solution:

Step 1: Draw a long line mm and mark a point XX on it. Step 2: Open the compasses to measure the length of AB‾\overline{AB} (3.5 cm3.5\text{ cm}). Step 3: Place the pointer at XX and draw an arc to cut line mm at a point, let's call it ZZ. Now XZ=3.5 cmXZ = 3.5\text{ cm}. Step 4: Open the compasses to measure the length of CD‾\overline{CD} (2.4 cm2.4\text{ cm}). Step 5: Place the pointer at ZZ (the end of the first segment) and draw an arc in the same direction to cut line mm at point YY. Step 6: The total length XY=XZ+ZY=3.5 cm+2.4 cm=5.9 cmXY = XZ + ZY = 3.5\text{ cm} + 2.4\text{ cm} = 5.9\text{ cm}.

Explanation:

To add two line segments, we place them end-to-end on a single line. The distance from the starting point of the first segment to the ending point of the second segment represents the sum.

Problem 3:

Construct a line segment RS‾\overline{RS} of length 4.3 cm4.3\text{ cm} and then construct another segment ST‾\overline{ST} of length 2.5 cm2.5\text{ cm} such that R,S,TR, S, T are collinear and SS lies between RR and TT. Find the total length RTRT.

Line segment RT showing points R, S, and T with lengths 4.3 and 2.5

Solution:

  1. Draw a line ll and mark a point RR.
  2. Using a ruler and compasses, mark point SS at a distance of 4.3 cm4.3\text{ cm} from RR.
  3. From point SS, mark point TT at a distance of 2.5 cm2.5\text{ cm} further along the line.
  4. The total length RT=RS+ST=4.3 cm+2.5 cm=6.8 cmRT = RS + ST = 4.3\text{ cm} + 2.5\text{ cm} = 6.8\text{ cm}.

Explanation:

By the segment addition property, if SS is between RR and TT, the total distance is the sum of the individual parts.

Problem 4:

Given a line segment MN‾\overline{MN} of length 7.5 cm7.5\text{ cm}, construct a point PP on it such that MP=3.2 cmMP = 3.2\text{ cm}. Calculate the length of the remaining part PNPN.

Segment MN with point P marked by a compass arc

Solution:

  1. Draw MN‾\overline{MN} of length 7.5 cm7.5\text{ cm}.
  2. Set the compasses to 3.2 cm3.2\text{ cm}.
  3. Place the pointer at MM and draw an arc cutting MNMN at point PP.
  4. PN=MN−MP=7.5 cm−3.2 cm=4.3 cmPN = MN - MP = 7.5\text{ cm} - 3.2\text{ cm} = 4.3\text{ cm}.

Explanation:

When a point PP is placed on a segment MN‾\overline{MN}, it divides the segment into two parts. The length of one part is the total length minus the length of the other part.