In the previous section we have learned the ASA Criterion for the congruence of Triangles. In this section we will discuss about another criterion.
Fig.10.28(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, UVY, PQR, DEF, and MNO.
Our problem is this: Is there any triangle in fig.(b), which is congruent to ΔABC? If yes, which one?
Solution: Looking at figs.(a) & (b), we find that the measurements of the triangles are incomplete.
• If the triangles had the lengths of all the 3 sides, then we could straight away use the SSS criterion to check the congruence
• If two sides and their included angle is given, we could use the SAS criterion
• If two angles and their included side are given, we could use the ASA criterion
But in this situation, these are not possible because the measurements are incomplete. We cannot find an SSS comparison, SAS comparison or ASA comparison. In such situations, we must try to use other criteria.
We are going to use a rule known as the RHS criterion. First we will see the details about this criterion. Based on that, we will be able to understand how this criterion got it's name.
In the fig.10.28(a), we are given a right angled triangle. In fig.(b), all the triangles that are given for comparison are right angled triangles. Why is that so?
The answer is that, a 'right angled triangle' will be congruent only with another 'right angled triangle'. There is no point in comparing other triangles.
Now, any given right angled triangle will have one hypotenuse. Any other triangle which is congruent to the given triangle should have the same hypotenuse. This is compulsory for congruence. If the triangle that we are comparing does not have the same hypotenuse, we can straight away discard it from our list.
If it does have the same hypotenuse, it could be a congruent triangle. But we need one more detail. Let us see how this is obtained:
Once we find that the hypotenuse of both the triangles are same, there remains two sides. Those are the ‘legs’. Any one of those legs in the given triangle should have it’s exact replica on the other triangle.
That means, for two right angled triangles to be congruent:
• The hypotenuse must be the same
• Any one leg should be the same
■ There is one more condition: The 90o angle. But, as we are comparing only 'right angled triangles', the 90o angle condition will be already satisfied. So we do not have to worry about it.
This RHS criterion gets it’s name thus:
• ‘R’ stands for Right angle
• ‘H’ stands for hypotenuse
• ‘S’ stands for side
Let us apply this rule to our problem:
• Consider fig.10.28(a). We know that, hypotenuse is the side opposite to the 90o angle.
• So in ΔABC, AB with a length of 8 cm is the hypotenuse.
• In fig.(b), we find that, ΔMNO is a right angled triangle, and it’s hypotenuse is also 8 cm.
• We need one more detail: One of the sides.
• In ABC we have a side AC = 6.4 cm. This same length of 6.4 cm is possessed by the side MN in ΔMNO also.
• So ΔABC and ΔMNO are congruent.
■ We are able to establish congruence just by using the hypotenuse and one side. And of course, the 90o angle. All other sides and angles may or may not be given in the problem. But we do not need them to establish congruence.
But establishing a congruence does not solve the problem completely. We have a little more work to do. We have to put the sides and corners of the two triangles in order. We do this by writing the correspondence:
• Consider ΔABC and ΔMNO. In ΔABC, the 90o is at C. In ΔMNO, the 90o is at N. So we get C↔N
• We have considered the hypotenuse of 8 cm length, and a side of 6.4 cm length
• In ΔABC, these two sides intersect at A. In ΔMNO, these two sides intersect at M. So we get A↔M
• The only remaining corners are B and O. So we get B↔O
• So the correspondence is C↔N, A↔M, and B↔O. This can be written as ABC ↔ MON
• So ΔABC and ΔMNO are congruent to one another under the correspondence: ABC ↔ MON
• This is same as writing: ΔABC ≅ ΔMON
So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔMON over ΔABC:
Based on the above discussion, we can write down the criterion:
■ RHS Congruence criterion:
If under a correspondence, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the triangles are congruent.
Solved example 10.13
Given below are some pairs of triangles. In each pair, examine if the triangles are congruent to one another. If they are congruent, write the correspondence.
(i) ΔABC: ∠A = 90o, AB = 3 cm, BC = 4.8 cm. │ΔPQR: ∠P = 90o, PR = 3 cm, RQ = 4.8 cm.
(ii) ΔXYZ: ∠X = 90o, XY = 4.7 cm, XZ = 2.8 cm, YZ = 5.5 cm. │ΔMNO: ∠M = 90o, OM = 2.8 cm, ON = 5.0 cm
Solution:
(i) A rough sketch is shown in the fig.10.30(i) below:
• Both are right angled triangles
■ From the rough sketch, it is clear that, both the triangles have the same 4.8 cm hypotenuse.
■ A side of 3 cm length is present in both the triangles
■ So the triangles are congruent to one another, based on the RHS criterion
• Now we have to write the correspondence:
• The 90o angle is at A and P. So A↔P
• In ΔABC, the 4.8 cm side and 3.0 cm side intersect at B
• In ΔPQR, the 4.8 cm side and 3.0 cm side intersect at R. So we get B↔R
• The only remaining corners are C and Q. So we get C↔Q
• So the correspondence is A↔P, B↔R and C↔Q. This is same as ABC↔PRQ
• So we can write: ΔABC and ΔPQR are congruent to one another under the correspondence: ABC↔PRQ
• This is same as ΔABC ≅ ΔPRQ
(ii) A rough sketch is shown in the fig.10.30(ii)
• Both are right angled triangles
• From the rough sketch, it is clear that, ΔXYZ has a hypotenuse of 5.5 cm, while ΔMNO has a hypotenuse of 5.0 cm
• So the two triangles can never be congruent.
• A side of 2.8 cm length is present in both the triangles. But it does not alter the result.
• If the hypotenuse are different, the triangles can not be congruent
Solved example 10.14
Check whether ΔABC and ΔABD shown in fig.10.31 are congruent or not. Given that AD = BC. If congruent, write the correspondence, and all the corresponding parts.
Solution:
• The two triangles are ΔABC and ΔABD. Both are right angled.
• Hypotenuse of ΔABC is AB. Hypotenuse of ΔABD is also AB
■ So both the triangles have the same Hypotenuse
■ In ΔABC, there is a side BC, with a 'certain length'. In ΔABD, there is AD with the same length
■ So ΔABC and ΔABD are congruent based on the RHS criterion
• Now we have to write the correspondence:
• In ΔABC, we considered the hypotenuse AB and a side BC. These two intersect at B
• In ΔABD, we considered the hypotenuse AB and a side AD. These two intersect at A. So we get A↔B
• In ΔABC, the 90o is at C. In ΔABD, the 90o is at D. So we get C↔D
• So the correspondence is: A↔B and C↔D. This is same as ABC↔BAD - - - (1)
• So we can write: ΔABC and ΔABD are congruent under the correspondence: ABC↔BAD
• This is same as: ΔABC ≅ ΔBAD
• Now we have to write the corresponding parts
■ We have already written the corresponding corners in (1). From that, we will get the corresponding sides also:
■ AB = BA, BC = AD and AC = BD
Solved example 10.15
Check whether ΔABC and ΔABD shown in fig.10.32 are congruent or not. Given that BD = BC. If congruent, write the correspondence, and all the corresponding parts.
Solution:
• The two triangles are ΔABC and ΔABD. Both are right angled.
• Hypotenuse of ΔABC is AB. Hypotenuse of ΔABD is also AB
■ So both the triangles have the same Hypotenuse
■ In ΔABC, there is a side BC, with a 'certain length'. In ΔABD, there is BD with the same length
■ So ΔABC and ΔABD are congruent based on the RHS criterion
• Now we have to write the correspondence:
• In ΔABC, we considered the hypotenuse AB and a side BC. These two intersect at B
• In ΔABD, we considered the hypotenuse AB and a side BD. These two intersect at B. So we get B↔B
• In ΔABC, the 90o is at C. In ΔABD, the 90o is at D. So we get C↔D
• The only remaining corner is A in both the triangles. So we get A↔A
• So the correspondence is: A↔A, B↔B and C↔D. This is same as ABC↔ABD - - - (2)
• So we can write: ΔABC and ΔABD are congruent under the correspondence: ABC↔ABD
• This is same as: ΔABC ≅ ΔABD
• Now we have to write the corresponding parts
■ We have already written the corresponding corners in (2). From that, we will get the corresponding sides also:
■ AB = AB, BC = BD and AC = AD
So we have completed the discussion on congruence of Triangles. In the next chapter, we will learn the construction of Triangles.
Fig.10.28(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, UVY, PQR, DEF, and MNO.
 |
| Fig.10.28 |
Solution: Looking at figs.(a) & (b), we find that the measurements of the triangles are incomplete.
• If the triangles had the lengths of all the 3 sides, then we could straight away use the SSS criterion to check the congruence
• If two sides and their included angle is given, we could use the SAS criterion
• If two angles and their included side are given, we could use the ASA criterion
But in this situation, these are not possible because the measurements are incomplete. We cannot find an SSS comparison, SAS comparison or ASA comparison. In such situations, we must try to use other criteria.
We are going to use a rule known as the RHS criterion. First we will see the details about this criterion. Based on that, we will be able to understand how this criterion got it's name.
In the fig.10.28(a), we are given a right angled triangle. In fig.(b), all the triangles that are given for comparison are right angled triangles. Why is that so?
The answer is that, a 'right angled triangle' will be congruent only with another 'right angled triangle'. There is no point in comparing other triangles.
Now, any given right angled triangle will have one hypotenuse. Any other triangle which is congruent to the given triangle should have the same hypotenuse. This is compulsory for congruence. If the triangle that we are comparing does not have the same hypotenuse, we can straight away discard it from our list.
If it does have the same hypotenuse, it could be a congruent triangle. But we need one more detail. Let us see how this is obtained:
Once we find that the hypotenuse of both the triangles are same, there remains two sides. Those are the ‘legs’. Any one of those legs in the given triangle should have it’s exact replica on the other triangle.
That means, for two right angled triangles to be congruent:
• The hypotenuse must be the same
• Any one leg should be the same
■ There is one more condition: The 90o angle. But, as we are comparing only 'right angled triangles', the 90o angle condition will be already satisfied. So we do not have to worry about it.
This RHS criterion gets it’s name thus:
• ‘R’ stands for Right angle
• ‘H’ stands for hypotenuse
• ‘S’ stands for side
Let us apply this rule to our problem:
• Consider fig.10.28(a). We know that, hypotenuse is the side opposite to the 90o angle.
• So in ΔABC, AB with a length of 8 cm is the hypotenuse.
• In fig.(b), we find that, ΔMNO is a right angled triangle, and it’s hypotenuse is also 8 cm.
• We need one more detail: One of the sides.
• In ABC we have a side AC = 6.4 cm. This same length of 6.4 cm is possessed by the side MN in ΔMNO also.
• So ΔABC and ΔMNO are congruent.
■ We are able to establish congruence just by using the hypotenuse and one side. And of course, the 90o angle. All other sides and angles may or may not be given in the problem. But we do not need them to establish congruence.
But establishing a congruence does not solve the problem completely. We have a little more work to do. We have to put the sides and corners of the two triangles in order. We do this by writing the correspondence:
• Consider ΔABC and ΔMNO. In ΔABC, the 90o is at C. In ΔMNO, the 90o is at N. So we get C↔N
• We have considered the hypotenuse of 8 cm length, and a side of 6.4 cm length
• In ΔABC, these two sides intersect at A. In ΔMNO, these two sides intersect at M. So we get A↔M
• The only remaining corners are B and O. So we get B↔O
• So the correspondence is C↔N, A↔M, and B↔O. This can be written as ABC ↔ MON
• So ΔABC and ΔMNO are congruent to one another under the correspondence: ABC ↔ MON
• This is same as writing: ΔABC ≅ ΔMON
So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔMON over ΔABC:
 |
| Fig.10.29 |
■ RHS Congruence criterion:
If under a correspondence, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the triangles are congruent.
Solved example 10.13
Given below are some pairs of triangles. In each pair, examine if the triangles are congruent to one another. If they are congruent, write the correspondence.
(i) ΔABC: ∠A = 90o, AB = 3 cm, BC = 4.8 cm. │ΔPQR: ∠P = 90o, PR = 3 cm, RQ = 4.8 cm.
(ii) ΔXYZ: ∠X = 90o, XY = 4.7 cm, XZ = 2.8 cm, YZ = 5.5 cm. │ΔMNO: ∠M = 90o, OM = 2.8 cm, ON = 5.0 cm
Solution:
(i) A rough sketch is shown in the fig.10.30(i) below:
 |
| Fig.10.30 |
■ From the rough sketch, it is clear that, both the triangles have the same 4.8 cm hypotenuse.
■ A side of 3 cm length is present in both the triangles
■ So the triangles are congruent to one another, based on the RHS criterion
• Now we have to write the correspondence:
• The 90o angle is at A and P. So A↔P
• In ΔABC, the 4.8 cm side and 3.0 cm side intersect at B
• In ΔPQR, the 4.8 cm side and 3.0 cm side intersect at R. So we get B↔R
• The only remaining corners are C and Q. So we get C↔Q
• So the correspondence is A↔P, B↔R and C↔Q. This is same as ABC↔PRQ
• So we can write: ΔABC and ΔPQR are congruent to one another under the correspondence: ABC↔PRQ
• This is same as ΔABC ≅ ΔPRQ
(ii) A rough sketch is shown in the fig.10.30(ii)
• Both are right angled triangles
• From the rough sketch, it is clear that, ΔXYZ has a hypotenuse of 5.5 cm, while ΔMNO has a hypotenuse of 5.0 cm
• So the two triangles can never be congruent.
• A side of 2.8 cm length is present in both the triangles. But it does not alter the result.
• If the hypotenuse are different, the triangles can not be congruent
Solved example 10.14
Check whether ΔABC and ΔABD shown in fig.10.31 are congruent or not. Given that AD = BC. If congruent, write the correspondence, and all the corresponding parts.
 |
| Fig.10.31 |
• The two triangles are ΔABC and ΔABD. Both are right angled.
• Hypotenuse of ΔABC is AB. Hypotenuse of ΔABD is also AB
■ So both the triangles have the same Hypotenuse
■ In ΔABC, there is a side BC, with a 'certain length'. In ΔABD, there is AD with the same length
■ So ΔABC and ΔABD are congruent based on the RHS criterion
• Now we have to write the correspondence:
• In ΔABC, we considered the hypotenuse AB and a side BC. These two intersect at B
• In ΔABD, we considered the hypotenuse AB and a side AD. These two intersect at A. So we get A↔B
• In ΔABC, the 90o is at C. In ΔABD, the 90o is at D. So we get C↔D
• So the correspondence is: A↔B and C↔D. This is same as ABC↔BAD - - - (1)
• So we can write: ΔABC and ΔABD are congruent under the correspondence: ABC↔BAD
• This is same as: ΔABC ≅ ΔBAD
• Now we have to write the corresponding parts
■ We have already written the corresponding corners in (1). From that, we will get the corresponding sides also:
■ AB = BA, BC = AD and AC = BD
Solved example 10.15
Check whether ΔABC and ΔABD shown in fig.10.32 are congruent or not. Given that BD = BC. If congruent, write the correspondence, and all the corresponding parts.
 |
| Fig.10.32 |
• The two triangles are ΔABC and ΔABD. Both are right angled.
• Hypotenuse of ΔABC is AB. Hypotenuse of ΔABD is also AB
■ So both the triangles have the same Hypotenuse
■ In ΔABC, there is a side BC, with a 'certain length'. In ΔABD, there is BD with the same length
■ So ΔABC and ΔABD are congruent based on the RHS criterion
• Now we have to write the correspondence:
• In ΔABC, we considered the hypotenuse AB and a side BC. These two intersect at B
• In ΔABD, we considered the hypotenuse AB and a side BD. These two intersect at B. So we get B↔B
• In ΔABC, the 90o is at C. In ΔABD, the 90o is at D. So we get C↔D
• The only remaining corner is A in both the triangles. So we get A↔A
• So the correspondence is: A↔A, B↔B and C↔D. This is same as ABC↔ABD - - - (2)
• So we can write: ΔABC and ΔABD are congruent under the correspondence: ABC↔ABD
• This is same as: ΔABC ≅ ΔABD
• Now we have to write the corresponding parts
■ We have already written the corresponding corners in (2). From that, we will get the corresponding sides also:
■ AB = AB, BC = BD and AC = AD
So we have completed the discussion on congruence of Triangles. In the next chapter, we will learn the construction of Triangles.
No comments:
Post a Comment