Chapter 11 - Construction of Triangles

In the previous section we have completed the discussion on the Congruence of Triangles. In this section we will learn the various methods of construction of Triangles.

Construction of a Triangle when lengths of 3 sides are given

Mr. A has a drawing of a triangle ABC with him. All the details (lengths of all 3 sides, and angles at all 3 corners) of the triangle is given in the drawing. He does not want to show us that drawing. But he want us to draw an exact replica of the triangle. He will give us one information: The lengths of all the 3 sides. With this information, can we draw an exact replica? Let us try:

• The lengths are AB = 8.5 cm, BC = 7.2 cm and CA = 4.6.
• We must draw a rough fig. with this given data. This is shown in the fig.11.1. Such a rough fig. will give us an idea about how to proceed

Fig.11.1

• Let us begin the construction. The steps are shown in the animation in fig.11.2
• First we draw a horizontal line 8.5 cm in length, and name it as AB. This is one side of the required triangle. It also fixes two corners A and B. 
• Now, if we can locate the correct position of ‘C’, the problem is solved. So our next aim is to locate ‘C’

Fig.11.2

• From the rough fig., it is clear that, C is at a distance of 4.6 cm from A. - - - condition 1 
• So it can be any where on a circle with A as center and radius 4.6 cm. This is shown by the green circle in the animation in fig.11.2
• From the rough fig., it is also clear that, C is at a distance of 7.2 cm from B. - - - condition 2
• So it can be any where on a circle with B as center and radius 7.2 cm. This is shown by the yellow circle in the animation.
• So our point C is moving around. We need a fix on it.
• Look at the point of intersection of the two circles. At the point of intersection, both the conditions 1 and 2 will be satisfied. Because, 'the point of intersection' is at a distance of 4.6 cm from A, and at the same time, it is at a distance of 7.2 cm from B
• So C is at the point of intersection of the two circles. In fact there is another point of intersection below the segment AB. But to draw the triangle, we need take only one. We will take the point of intersection which lies above the line AB
• Also, we need not draw full circles. We need to draw only convenient ‘parts of these circles’, near the ‘probable point of intersection’.
• ‘Parts of circles’ are Arcs. We need to draw an arc with center A and radius 4.6 cm. We need to draw another arc with center B and radius 7.2 cm. The point of intersection of these two arcs is the position of ‘C’
• The fig.11.3(a) shows the completed ΔABC. Fig.(b) shows ΔABC when the point of intersection below AB is chosen as 'C'

Fig.11.3

• Both the triangles in figs.(a) and (b) are exact replicas of each other. More importantly, each of them are exact replicas of the triangle which Mr. A is holding. So our job is accomplished.

It may be noted that, this method of constructing a triangle is related to the SSS criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The Δ ABC that we have constructed, and the ΔABC which Mr. A is holding, have all the 3 sides the same. If two triangles have all the 3 sides the same, they are congruent to one another. In other words, one is the exact replica of the other. So next time some one gives us the 3 sides of a triangle, we can easily do the construction.

Solved example 11.1
Construct ΔPQR in which PQ = 5.5 cm, QR = 3.8 cm, and PR = 7.5 cm
Solution:
• First of all we have to draw a rough fig. using the given data. It is shown in the fig. 11.4(a)

Fig.11.4

• Based on the rough fig., we can proceed to do the construction:
• First draw a horizontal line PQ of length 5.5 cm
• With P as center, draw an arc of radius 7.5 cm (shown in green colour in fig.(b))
• With Q as center, draw an arc of radius 3.8 cm (shown in yellow colour in fig.(b))
• This arc cuts the first arc at 'R'
• Join R with P. Also join R with Q. We get the required ΔPQR
• The steps in construction, and the final ΔPQR are shown in fig.11.4(b)

Solved example 11.2
Construct ΔABC in which AB = 5.8 cm, BC = 3.6 cm, CA = 4.6 cm. Measure ∠C
Solution:
• First of all we have to draw a rough fig. using the given data. It is shown in the fig. 11.5(a)

Fig.11.5

• Based on the rough fig., we can proceed to do the construction:
• First draw a horizontal line AB of length 5.8 cm
• With A as center, draw an arc of radius 4.6 cm (shown in green colour in fig.(b))
• With B as center, draw an arc of radius 3.6 cm (shown in yellow colour in fig.(b))
• This arc cuts the first arc at 'C'
• Join C with A. Also join C with B. We get the required ΔABC
• The steps in construction, and the final ΔABC are shown in fig.11.5(b)
• When we measure ∠C, we get 90^o
• So ΔABC is a right angled triangle, right angled at C

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In this method, we are using the lengths of the three sides of a triangle. It is worthwhile to recall an important property that we learned about 'the sum of the lengths of two sides': The sum of any two sides of a triangle should be greater than the third side. (Details here) If this condition is not satisfied, the triangle will not even exist. Because, the green circle and the yellow circle (shown in the animation in fig.11.2 above) will never meet. So when we are given three sides, it is better to check whether they satisfy the condition, before starting the construction.

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So we have learned the method to construct a triangle when it's three sides are given. In the next section, we will learn another method.

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Chapter 10.2 - SSS Criterion for the Congruence of Triangles

In the previous section we have learned that congruent triangles completely and exactly cover each other. We have also learned that the correspondence is also important for the congruence to work. In this section we will discuss about the 'Criteria for the Congruence of Triangles'.

Fig.10.8(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, PQR, UVW, MNP and NOP.

Fig.10.8

Our problem is this: Is there any triangle in fig.(b), which is congruent to ABC? If yes, which one?
Solution: Looking at fig.(a) & (b), we find that the measurements of all the 3 sides of all the triangles are given. So here we use a rule known as the SSS criterion. The ‘S’ stands for ‘Side’. So this is the abbreviation for 'Side Side Side criterion'. The 3 sides of a triangle. 
■ This rule is based on the fact that, when two triangles are congruent, their 3 sides will be equal. Let us apply this rule to our problem:
• In fig.(a), We have ΔABC with the 3 sides: AB = 3.0 cm, BC = 2.6 cm and CA = 4.1 cm
• In fig.(b) we have one particular triangle, which is ΔUVW with the 3 sides: UV = 3.0 cm, VW = 4.1 cm and WU = 2.6 cm
• The two triangles are having the same sides. So they are congruent.

But the order of the sides is not correct. It is our job to specify the correct order. We do this by writing the correspondence. How do we write the correspondence with out making a ‘trace copy of UVW’, and placing it over ABC? It can be done as follows:
• Consider ABC. Take any two sides say 3.0 cm and 2.6 cm. They meet at B
• Corresponding to this, in UVW, 3.0 cm and 2.6 cm meet at U
• So we see that B corresponds to U That is., B↔U
• Now, starting from B, segment BA is 3.0. And starting from U, segment UV is 3.0
• So we see that A corresponds to V That is., A↔V
• The only ones remaining are C and W, and obviously, they will correspond to each other because they are the only remaining ones. So we get C↔W
• So we write: A↔V, B↔U and C↔W 
• This is same as writing: ABC↔ VUW. 
Note that, the name of the second triangle is UVW, but in correspondence, it is VUW. We can write: ΔABC ≅ ΔUVW under the correspondence: ABC↔VUW.


The animation given below in fig.10.9, shows the superposition and the correspondence:

Fig.10.9

Based on the above discussion, we can write down the criterion:
■ SSS Congruence criterion:
If under a given correspondence, the three sides of one triangle are equal to the three corresponding sides of another triangle, then the triangles are congruent.

Solved example 10.4
In the fig.10.10 given below, prove that, ΔPRS and ΔPRQ are congruent. Also state the relation between the ∠SPR and ∠QPR

Fig.10.10

Solution:
Part 1:
• Take ΔPRS. Consider the side PS. It is shown to be equal to the side PQ of the other ΔPRQ
• Consider the side SR. It is shown to be equal to RQ of the other ΔPRQ
• PR is common to both the triangles. So the third side PR of ΔPRS is equal to the third side PR of ΔPRQ
• So all the 3 sides are same. The triangles are congruent
• But we have to write the correspondence:
• The equal sides PS and PQ start from P. So P↔P and S↔Q. 
• The only remaining vertex in both the triangles is R. So R↔R
• So we can write: P↔P, S↔Q, and R↔R. This is same as PSR↔PQR
• Thus we can write: ΔPRS ≅ ΔPQR under the correspondence: PSR↔PQR  
Part 2:
• From part 1, we have seen that the two triangles are congruent. That is., they are exact copies. So the angles at the corresponding corners must also be the same. 
• ∠SPR is the angle at P in the ΔPRS
• ∠QPR is the angle at P in the ΔPQR
• We have seen that P↔P. So the angle at P will be same in both the triangles
• Thus the relation is that: ∠SPR and ∠QPR are equal
Solved example 10.5
In the fig.10.11 below, if PM = ON and PN = OM, which of the following is true?
(i) ΔPMN≅ΔMNO  (ii) ΔPNM≅ΔOMN  (iii) ΔPMN≅ΔOMN  (iv) ΔPMN≅ΔONM

Fig.10.11

Solution:
• We have two triangles: ΔMNP and ΔMNO
• Given PM = ON and PN = OM. The third side MN is common to both the triangles. So all 3 sides are the same. 
• Thus ΔMNP and ΔMNO are congruent to one another. But we have to write the correspondence.
• In ΔMNP, take the two sides: PM and PN. They meet at P
• In ΔMNO, the corresponding sides are ON and OM respectively. These two sides meet at O. So we get P↔O
• Consider PM in ΔPMN. It starts from P and moves towards M. The corresponding side ON in ΔMNO starts from O and moves towards N. Thus we get M↔N
• All statements which satisfy P↔O and M↔N are true
(i) ΔPMN≅ΔMNO is not true because, in this statement, P↔M, which is not true
(ii) ΔPNM≅ΔOMN is true because, in this statement, P↔O, N↔M, M↔N, are all true
(iii) ΔPMN≅ΔOMN is not true because, M↔M and N↔N is not true
(iv) ΔPMN≅ΔONM is true because, P↔O, M↔N, N↔M, are all true

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So we have completed the discussion on SSS criterion. In the next section we will discuss about another criterion for the Congruence of Triangles.

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