In the previous section we saw the method of construction of a triangle when any two angles and their included side are given. In this section we will learn another method.
Construction of a Right angled Triangle when Hypotenuse and one side are given
Mr. A now puts forward one more challenge. He has a drawing of a triangle ABC with him. As before, 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 wants us to draw an exact replica of the triangle. He will give us one information: The triangle is a right angled one. He also gives the lengths of the hypotenuse and one side. With this information, can we draw an exact replica? Let us try:
• The length of the hypotenuse is 5 cm. One side is 3 cm. The triangle is right angled
• We must draw a rough sketch with this given data. This is shown in the fig.11.14. Such a rough sketch will give us an idea about how to proceed. It is important to mark the hypotenuse as the side which is opposite to the 90o angle.
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| Fig.11.14 |
• Let us begin the construction. The steps are shown in the fig.11.15
• First we draw a horizontal line 3.0 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’.
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| Fig.11.15 |
• From the rough fig., it is clear that, C lies some where on a line which is at right angle to AB.
• So we draw a line AC' (of any convenient length) at an angle of 90o to AB
• From the rough fig., it is also clear that, C is at a distance of 5.0 cm from B
• So with B as center, we draw an arc of radius 5.0 cm
• This arc cuts the line AC' at C
• ΔABC is the required triangle
It may be noted that, this method of constructing a triangle is related to the RHS criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The ΔABC that we have constructed, and the ΔABC which Mr. A is holding, are both right angled. Also, they have the hypotenuse and one side in common. If two right angled triangles have the hypotenuse and one side 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 a hypotenuse and one side, we can easily do the construction.
Solved example 11.7
Construct ΔPQR in which ∠P = 90o, RQ = 8.1 cm, and one side = 3.9 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig. 11.16(a)
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| Fig.11.16 |
• Based on the rough sketch, we can proceed to do the construction:
• First draw a horizontal line PQ of length 3.9 cm
• Draw a line PR' (of any convenient length) at an angle of 90o to PQ
• With Q as center, draw an arc of radius 8.1 cm
• This arc cuts the line PR' at R
• ΔPQR is the required triangle
Solved example 11.8
ΔABC is an isosceles right angled triangle. Construct ΔABC, if AC = 7.5 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig.11.17(a)
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| Fig.11.17 |
• We are given one side: AB = 7.5 cm.
• In an isosceles triangle, two sides are equal. None of them can be the hypotenuse. This is because, the hypotenuse is the longest side, and it is unique. There cannot be two sides with the length of the hypotenuse
• So the rough fig. will be as shown in Fig.11.17(a)
• Now we can proceed to do the construction The steps are shown in fig.(b)
• Draw a horizontal line CA' of any convenient length
• At C draw a perpendicular CB' to CA'
• With C as center, draw two arcs, each of radius 7.5 cm
• These arcs cut CA' at A, and CB' at B
• Join A and B. ΔABC is the required triangle
So we have learned the method to construct a right angled triangle when it's hypotenuse and one side are given. In the next chapter, we will learn the construction of Quadrilaterals.
In the previous section we saw the method of construction of a triangle when any two sides and their included angle are given. In this section we will learn another method.
Construction of a Triangle when two angles and the included side are given
Mr. A now puts forward another challenge. He has a drawing of a triangle PQR with him. As before, 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 wants us to draw an exact replica of the triangle. He will give us one information: 'Two angles', and the 'length of the side included in between those two angles'. With this information, can we draw an exact replica? Let us try:
• The given angles are P = 55o, Q = 40o and PQ = 8.2 cm
• We must draw a rough fig. with this given data. This is shown in the fig.11.10. Such a rough fig. will give us an idea about how to proceed
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| Fig.11.10 |
• Let us begin the construction. The steps are shown in the fig.11.11
• First we draw a horizontal line 8.2 cm in length, and name it as PQ. This is one side of the required triangle. It also fixes two corners P and Q.
• Now, if we can locate the correct position of ‘R’, the problem is solved. So our next aim is to locate ‘R’.
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| Fig.11.11 |
• From the rough fig., it is clear that, R lies some where on a line which is inclined at an angle of 55o to PQ.
• So we draw a line PP' (of any convenient length) at an angle of 55o to PQ
• From the rough fig., it is also clear that, R lies some where on a line which is inclined at an angle of 40o to PQ.
• So we draw a line QQ' (of any convenient length) at an angle of 40o to PQ
• The two lines PP' and QQ' intersect at the point R. So ΔPQR is the required triangle
It may be noted that, this method of constructing a triangle is related to the ASA criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The ΔPQR that we have constructed, and the ΔPQR which Mr. A is holding, have two angles and their included side in common. If two triangles have two angles and their included side 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 any two angles, and their included side, we can easily do the construction.
Solved example 11.5
Construct ΔABC in which ∠A = 37o, ∠B = 78o and AB = 6.4 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig. 11.12(a)
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| Fig.11.12 |
• Based on the rough sketch, we can proceed to do the construction:
• First draw a horizontal line AB of length 6.4 cm
• Draw a line AA' (of any convenient length) at an angle of 37o to AB
• Draw a line BB' (of any convenient length) at an angle of 78o to AB
• These two lines will intersect at C. ΔABC is our required triangle
Solved example 11.6
In ΔMNO, MN = 4.4 cm, ∠N = 101o and ∠O = 48o. Construct the triangle.
Solution:
• First of all we have to draw a rough fig., using the given data. It is shown in the fig.11.13(a)
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| Fig.11.13 |
• We are given one side: MN = 4.4 cm.
• ∠N = 101o. ∠M is not given. We can use the 'ASA congruence' only if we get ∠M
• But ∠O is given as 48o. So M = 180 - (101 + 48) [∵ sum of the three interior angles of a triangle = 180o]
• So we get ∠M = 180 - 149 = 31o
• Now we can use the ASA property:
• First draw a horizontal line MN of length 4.4 cm
• Draw a line MM' (of any convenient length) at an angle of 31o to MN
• Draw a line NN' (of any convenient length) at an angle of 101o to MN
• These two lines will intersect at O. ΔMNO is our required triangle
So we have learned the method to construct a triangle when any two of it's angles and their included side are given. In the next section, we will learn one more method.
In the previous section we saw the method of construction of a triangle when it's three sides are given. In this section we will learn another method.
Construction of a Triangle when lengths of 2 sides and included angle are given
Mr. A has now put forward a new challenge. He has a drawing of a triangle XYZ with him. As before, 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 two sides, and the included angle between those two sides. With this information, can we draw an exact replica? Let us try:
• The given lengths are XY = 8.0 cm, XZ = 6.6 cm and ∠X = 65o
• We must draw a rough fig. with this given data. This is shown in the fig.11.6. Such a rough fig. will give us an idea about how to proceed
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| Fig.11.6 |
• Let us begin the construction. The steps are shown in the fig.11.7
• First we draw a horizontal line 8.0 cm in length, and name it as XY. This is one side of the required triangle. It also fixes two corners X and Y.
• Now, if we can locate the correct position of ‘Z’, the problem is solved. So our next aim is to locate ‘Z’. The construction steps are shown in the fig.11.7.
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| Fig.11.7 |
• From the rough fig., it is clear that, Z lies some where on a line which is inclined at an angle of 65o to XY.
• So we draw a line XZ' (of any convenient length) at an angle of 65o to XY
• With X as center, draw an arc with radius = 6.6 cm
• This arc (shown in green colour) will intersect XZ' at a point. This point of intersection will be at a distance of 6.6 cm from X. So this is our required point Z.
• Join Z to X. This gives us the required ΔXYZ
It may be noted that, this method of constructing a triangle is related to the SAS 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 two sides and their included angle in common. If two triangles have two sides and their included angle 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 any two sides, and their included angle, we can easily do the construction.
Solved example 11.3
Construct ΔABC in which AB = 4.8 cm, AC = 6.2 cm, and ∠A = 120o
Solution:
• First of all we have to draw a rough fig. using the given data. It is shown in the fig. 11.8(a)
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| Fig.11.8 |
• Based on the rough fig., we can proceed to do the construction:
• First draw a horizontal line AB of length 4.8 cm
• Draw a line AC' (of any convenient length) at an angle of 120o to AB
• With A as center, draw an arc of radius 6.2 cm
• This arc (shown in green colour) will intersect AC' at a point. This point of intersection will be at a distance of 6.2 cm from A. So this is our required point C.
• Join B to C. This completes the required ΔABC
Solved example 11.4
ΔABC is an isosceles triangle. The equal sides are AC and BC with lengths of 7.0 cm each. Construct the triangle if ∠B is 42o
Solution:
• First of all we have to draw a rough fig., using the given data. It is shown in the fig. 11.9(a)
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| Fig.11.9 |
• We are given two sides: AC and BC. But their included angle is not given. With out the included angle, we cannot use the two sides to construct the triangle.
• ΔABC is an isosceles triangle. ∠B is one 'base angle'. So the other base angle ∠A will also be same as ∠B. (Details here) So we get ∠A = ∠B = 42o
• ∠So C = 180 - (42 + 42) = 180 -84 = 96o (∵ sum of the three interior angles of a triangle is equal to 180o). Thus the included angle between the two sides = 96o
• Based on this, we can do the construction. Fig.b shows the steps
• First draw a horizontal line AB' of any convenient length
• At A, draw the line AC' at an angle of 42o with AB'
• With A as center, draw an arc with a radius of 7.0 cm
• This arc (shown in yellow colour) will cut AC' at a point. This point of intersection is at a distance of 7.0 cm from A. So this point of inter section is C
• With C as center, draw an arc of radius 7.0 cm
• This arc (shown in green colour) will cut the line AB' at B. This point of intersection is at a distance of 7.0 cm from C. So this point is B.
• Join B to C. This completes the required ABC
• An alternate method is to draw a line at an angle of 96o with AC. This will cut the line AB' at B
So we have learned the method to construct a triangle when any two of it's sides and their included angle are given. In the next section, we will learn another method.
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
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| 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’
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| 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'
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| 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)
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| 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)
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| 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 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.
