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.
• 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
• 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’
• 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'
• 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)
• 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)
• 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 90o
• So ΔABC is a right angled triangle, right angled at C
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.
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.
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 |
• 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 |
• 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 |
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 |
• 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 |
• 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 90o
• So ΔABC is a right angled triangle, right angled at C
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.
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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