Constructions

Construction Related to Segment and Triangles

Geometry is a skill of the eyes, hands and that of the mind. Any construction in geometry involves drawing a diagram based on provided data. In this module, we will learn a few skills regarding construction related to the line segment and triangles. We will also revise a few properties of geometry while doing construction.

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Construction Related to Line Segments

You must have seen many times that a point divides a line segment in a particular ratio. In this module, we will learn how to divide a given line segment in the desired ratio x:y without using a ruler.

Dividing a given line segment in a particular ratio

Let’s say we have been given a line segment AB which we want to divide it by a ratio of 3:2. Here, x = 3 and y = 2. To construct such a line segment, follow the steps as given below:

Steps of Construction

  1. Draw any ray AZ making an acute angle with AB.
  2. Locate 5 points (x + y = 3 + 2 = 5), say, A1, A2, A3, A4, and A5 using a compass on AZ such that these points are equally distanced from each other.
    1. Open the compass of a specific width and make an arc on the ray AZ keeping the needle on the point A. The point where the arc intersects with AZ is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the steps 3 more times making AA1 = A1A2 = A2A3 = A3A4 = A4A5.
  3. Join A5 with B making a line segment A5B. (See Figure 1)
  4. Since we need to divide the line segment in 3:2, draw a line through the point A3 which is parallel to A5B by making ∠AA3C = ∠AA5B such that A3C intersects AB at the point C.
    1. To construct a parallel line using Angle Copy Method, adjust the width of the compass to roughly half of the length of A4A5. Keeping the needle at point A5, make an arc cutting the ray AZ and the line segment A5B. This is arc 1.
    2. Make a similar arc from point A3 using the same compass width. This is arc 2.
    3. Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AZ, the pencil should be on the point where arc 1 meets the line segment A5B.
    4. Keeping the same width, cut the arc 2 making an intersecting point G.
    5. Make a line segment from A3 through G meeting the line segment AB at C. The line segment A3C || A5B.
  5. Therefore, AC : CB = 3 : 2. (See Figure 1)

 

Segment

Figure 1: AC : CB = 3 : 2

Rationale and Proof

Since A3C || A5B

∴ \( \frac{AA_3}{A_3A_5} \) = \( \frac{AC}{CB}  \) [Triangle Proportionality TheoremIf a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally]

\( \frac{AA_3}{A_3A_5} \) = \( \frac{3}{2} \)          [By construction]

∴ \( \frac{AC}{CB} \) = \( \frac{3}{2} \)

Alternate Method

Following is the alternative method to divide a given line segment AB into a said ratio (same as above, that is, 3:2).

Steps of Construction

  1. Draw any ray AG making an acute angle with AB.
  2. Draw a ray BH parallel to AG by making ∠ABH equal to ∠BAG using the same method as described above in steps from 4(a) to 4(e).
  3. Locate the pointsA1, A2, and A3 (x = 3) on AG and B1, B2 (y = 2) on BH such that AA1 = A1A2 = A2A3 = BB1 = B1B2 using the same step as described in 2(a) above.
  4. Join A3B2. Let it intersect AB at a point C (See Figure 2).
  5. Then AC : CB = 3 : 2
Segment

Figure 2: AC : CB = 3 : 2 (Alternate method)

Rationale and Proof

Here,

 ∠CAA3 = ∠CBB2                    [By construction]

∠ACA3 = ∠BCB2                     [Vertically opposite angles are always equal]

∠AA3C = ∠BB2C                     [Alternate interior angles within two parallel lines are always equal]

∴ ∆ACA3 ~ ∆BCB2                 [By AAA criteria of triangle similarity]

⇒ \( \frac{AC}{BC} \) = \( \frac{CA_3}{CB_2} \) = \( \frac{AA_3}{BB_2}  \)

Now, using construction \( \frac{AA_3}{BB_2} \) = \( \frac{3}{2} \)

∴ \( \frac{AC}{BC} \) = \( \frac{3}{2} \)

Construction Related to Triangles

We have studied about similar triangles and their properties in previous classes. Have you ever wondered how these similar triangles are designed? Let’s take a look how.

Constructing a triangle similar to a given triangle

There are a few ways by which we can draw a triangle similar to the given one. Here, we shall study how to construct the same using the Scale Factor method without the help of a ruler.

Scale Factor is the ratio of two corresponding sides of the similar triangles. For example, let’s say that ∆ABC is similar to ∆DEF (∆ABC ~ ∆DEF) wherein AB = 3, BC = 4, and AC = 5; whereas DE = 6, EF = 8 and DF = 10. Here, the Scale Factor of two corresponding sides is

\( \frac{AB}{DE} \) = \( \frac{BC}{EF} \) = \( \frac{AC}{DF} \) = \( \frac{1}{2} \)

So, let’s see how we can construct a triangle similar to the given triangle (∆PQR) with a scale factor of \( \frac{3}{4}\), that is, to construct a smaller triangle (∆P′QR′) compared to ∆PQR wherein the length of the sides of ∆P′QR′ will be 3/4th to that of ∆PQR.

Steps of Construction using Scale Factor

  1. Draw any ray QX making an acute angle with QR on the side opposite to the vertex P.
  2. Locate 4 points Q1, Q2, Q3, and Q4 on QX so that QQ1 = Q1Q2= Q2Q3 = Q3Q4 using the same method as described above in step 2(a) (Dividing a line segment in a given ratio). Note: We located 4 points because 4 is the greater value in the Scale Factor ratio of \( \frac{3}{4} \)
  3. Considering the given triangle, ∆PQR, join a line Q4R.
  4. Draw a line through Q3 parallel to Q4R using the same method as described above in steps from 4(a)  to 4(e) (Dividing a line segment in a given ratio) intersecting the line QR at R′. This gives us Q3R′ || Q4R and ∠QQ3R′ = ∠QQ4R (See Figure 3).
  5. Similarly, draw a line through R′ parallel to the line PR intersecting the line PQ at a point, P′. This gives us P′R′ || PR and ∠P′R′Q = ∠PRQ (See Figure 3).
  6. Subsequently, ∆P′QR′ is the required triangle similar to ∆PQR with Scale Factor of \( \frac{3}{4} \).
Segment

Figure 3: ∆ P′QR′ ~ ∆PQR

Rationale and Proof

Since Q3R′ || Q4R and QQ3 is proportionate to QQ4 in the ratio of 3:4 (by construction)

∴ \( \frac{QQ_3}{QQ_4} \) = \( \frac{QR’}{QR} \) = \( \frac{3}{4} \) [By Triangle Proportionality Theorem]

Here,

∠P′QR′ =  ∠PQR         [Common angle]

∠P′R′Q  = ∠PRQ         [By Construction]

∠R′P′Q = ∠RPQ          [If two corresponding angles of two triangles are equal, the third angle will always be equal]

∴ ∆P′QR′ ~ ∆PQR       [By AAA criteria of triangle similarity]

Since ∆P′QR′ ~ ∆PQR and \( \frac{QR’}{QR} \) = \( \frac{3}{4} \)

∴ ∆P′QR′ ~ ∆PQR with Scale Factor of  \( \frac{3}{4} \).

Solved Examples for You

Question 1: Given a line segment AB. Find a point, describing the steps, which divides the line segment in 2:1 without using a ruler. 

Answer : Steps of Construction

  1. Draw any ray AX, making an acute angle with AB.
  2. Locate 3 points (x + y = 2 + 1 = 3), say, A1, A2, and Ausing a compass on AX such that these points are equally distanced from each other.
    1. Open the compass of a specific width and make an arc on the ray AX keeping the needle on the point A. The point where the arc intersects with AX is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the step one more times making AA1 = A1A2 = A2A3.
  3. Join A3 with B making a line segment A3B. (See Figure A)
  4. Since we need to divide the line segment in 2:1, draw a line through the point A2 which is parallel to A3 by making ∠AA2B = ∠AA3B intersecting AB at the point C.
    1. To construct a parallel line, adjust the width of the compass to roughly half of the length of A2A3. Keeping the needle at point A3, make an arc cutting the ray AX and the line segment A3B. This is arc 1.
    2. Make a similar arc from point A2 using the same compass width. This is arc 2.
    3. Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AX, the pencil should be on the point where arc 1 meets the line segment A3B.
    4. Keeping the same width, cut the arc 2 making an intersecting point O.
    5. Make a line segment from A2 through O meeting the line segment AB at C. The line segment A2C || A3B.
  5. This point C divides AB as AC : CB = 2 : 1. (See Figure A)
Segment

Figure A. AC : CB = 2 : 1

Question 2: What do you mean by segment in maths?

Answer: Segment refers to the part of a line that connects two points. It happens to the shortest distance that exists between the two points. Furthermore, segment has a length. The word “segment” is important because a line can go on continuously without end. In contrast, a line segment has definite endpoints.

Question 3: Explain what is meant by segments of circle?

Answer: A chord of a circle divides a circle into two parts. These parts are known as the segments of circle.

Question 4: How can one measure a line segment?

Answer: A line segment is nothing more than a part of a line. In order to measure the length of a line segment, put the segment’s endpoint on the ruler’s zero mark. They must see where it ends. In inches, one must round to the nearest ¼, while in centimeters; one must measure to the nearest centimeter.

Question 5: How can one find out the midpoint of a segment?

Answer: One can find out the midpoint of a segment by dividing the segment’s length by 2 and counting that value from any endpoint.

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