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.
• 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
• 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.
• 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)
• 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)
• 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.
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
 |
| Fig.11.6 |
• 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.
 |
| Fig.11.7 |
• 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)
 |
| Fig.11.8 |
• 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)
 |
| Fig.11.9 |
• 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.
No comments:
Post a Comment