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.
• 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.
• 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’.
• 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)
• 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)
• 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.
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.
 |
| Fig.11.14 |
• 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’.
 |
| Fig.11.15 |
• 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)
 |
| Fig.11.16 |
• 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)
 |
| Fig.11.17 |
• 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.
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