In the previous section we saw that, if each of the given two triangles is the scaled version of the other, the angles in them will be the same. So far we have seen two methods for obtaining scaled versions of a given triangle. They are:
1. Given all the three sides of a triangle. We can obtain a scaled version by multiplying all the sides by a scale factor 'k'. This is the basis of theorem 19.1 and theorem 19.2.
2. Given a single side of a triangle, and the angles at it's ends. We can obtain a scaled version by drawing one side with the 'scaled side' and the angle at it's ends the same. Based on this method, we developed theorem 19.3
In this section we will learn a third method to obtain a 'scaled version' of a triangle.
Theorem 19.4
Let us now see the practical applications of this theorem:
We will now see a real case which demonstrates the method:
1. Fig.19.22 shows the original ΔABC. We want a scaled version with a scale factor 11⁄2
2. A convenient point 'O' is marked in the interior of ΔABC. Draw OA, OB and OC
3. Scale them by 1.5, to reach P, Q and R. Draw PQ, QR and PR
4. • Then PQ will be 6 × 1.5 = 9.0
• QR will be 3 × 1.5 = 4.5
• PR will be 5 × 1.5 = 7.5
■ ΔPQR is the required triangle
We have to note a point here. 'O' is marked at any convenient point. So lengths of OA, OB, OC etc., may not be perfect whole numbers. Most likely they will be whole numbers plus decimals. In such cases it may not be easy to scale them accurately by a given scale. So we must use geometric mehods. Examples for scaling lines by any given ratio are given in the form of a video presentation here. In this method, measurements of lines is not required.
1. Given all the three sides of a triangle. We can obtain a scaled version by multiplying all the sides by a scale factor 'k'. This is the basis of theorem 19.1 and theorem 19.2.
2. Given a single side of a triangle, and the angles at it's ends. We can obtain a scaled version by drawing one side with the 'scaled side' and the angle at it's ends the same. Based on this method, we developed theorem 19.3
In this section we will learn a third method to obtain a 'scaled version' of a triangle.
1. Fig.19.19(a) shows ΔABC.
Two of its sides and included angle are given:
AB = 6 cm, AC = 4 cm, ∠CAB = 30o
 |
| Fig.19.19 |
2. We want a scaled version of ΔABC. The scale factor being 3⁄4
3. In fig.b, ΔPQR is
shown. • It’s side PQ = 6 × 3⁄4 = 4.5 cm
• Side PR = 4 × 3⁄4 = 3 cm.
• Included angle is same as in ΔABC.
4. Is ΔPQR the required scaled
version?
If PQR is indeed the scaled
version, RQ must be equal to 3⁄4 of BC. But BC is not given.
5. So the question is:
If two sides are scaled by the
same factor, with the included angle remaining the same, will the
third side also be scaled by the same factor?
6. We can find the answer by
using an intermediary triangle. (We used such a method to prove
theorem 19.3 in the previous section).
7. In fig.19.20 below,
• (a) shows the
same ΔABC in fig.19.19a.
• (c) shows the same ΔPQR in fig.19.19b.
• The intermediary ΔMNO is drawn in the middle. It is drawn by the following
procedure:
 |
| Fig.19.20 |
(i) First draw MN = 4.5 cm
(ii) Complete ΔMNO by drawing lines
at 30o at M and xo (the angle at B in the given ΔABC) at N. It is a
simple ASA construction.
8. So the value of y at vertex O
will be the same y at the vertex C in ABC (∵ sum of the angles in any triangle = 180)
9. Thus, all the angles in ΔABC and ΔMNO are the same. They are scaled versions. Consider their lengths:
• AB is scaled by 3⁄4 to get MN
• AC is scaled by 3⁄4 to get OM
• So ON will be equal to 3⁄4 of BC. So, If BC = a cm, ON = 3a⁄4
10. Now consider ΔMNO and ΔPQR:
• They
have two sides and included angle the same:
4.5 cm side, 3.0 cm side, and
the included angle 30.
• It is an SAS congruence. So the
two triangles are equal. Let us write the congruence:
• The angle 30o is at M and P. So
we get M↔P
• N is at 4.5 cm from M. Q is at
the same 4.5 cm from P. So we get N↔Q
• The remaining vertices are O
and R. So we get O↔R
• From the above three, we get
MN↔PQ, NO↔QR and MO↔PR
• From NO↔QR, we can say NO = QR.
11. But from (9), we have seen NO = 3a⁄4.
Thus QR = 3a⁄4
12. So we can say that, if two
sides are scaled by the same factor, with the included angle
remaining the same, the third side will also be scaled by the same
factor. We can write in the form of a theorem:
Theorem 19.4
• We have a ΔABC. Two of
it’s sides and included angle are given
• We draw another ΔPQR
with the two sides scaled by a factor ‘k’, and the included angle
remaining the same
• Then ΔPQR will be a scaled
version of ΔABC, the scale factor being ‘k’
Let us now see the practical applications of this theorem:
1. In fig.19.21(a) below, ΔABC is a given
triangle.
2. Mark any convenient point ‘O’ in the interior of ΔABC, as shown in fig.b.
3. Join OA and OB. Extend OA up to P, in such a way that OP is 3⁄2 times
OA. That means, we are scaling the original length OA by a factor of 3⁄2
2. Mark any convenient point ‘O’ in the interior of ΔABC, as shown in fig.b.
 |
| Fig.19.21 |
(i) For drawing OP, we need the
additional length AP. How much is AP?
Ans: The additional length AP
= OP – OA
(ii) But OP = 3⁄2 × OA = 3OA⁄2
(iii) So AP = 3OA⁄2 – OA = (3OA-2OA)⁄2 = OA⁄2
(iv) So the additional distance AP
= half of OA
(v) With this information, we can
easily draw OP.
4. Using the same method, extend
OB to Q, in such a way that OQ = 3OB⁄2
5. Join PQ. Now we have a large ΔOPQ, and inside it, a small ΔOAB
(i) Side OA is scaled by a factor 3⁄2 to get OP
(ii) Side OB is scaled by the same factor 3⁄2 to get OQ
(iii) The included angle between OA
and OB is the same included angle between OP and OQ
(iv) So ΔOPQ is the scaled version
of ΔOAB, the scale factor being 3⁄2
6. Based on this information, we
can say: the side AB is scaled by the same 3⁄2 to get PQ.
7. Now draw OR in such a way that
OR = 3OC⁄2
8. Join PR and QR. Based on the
above discussion, we can write: PR = 3AC⁄2 and QR = 3BC⁄2
9. That means, each side of ΔABC is scaled by 3⁄2 to get the sides of ΔPQR
10. So ΔPQR is a scaled version of ΔABC, the scale factor being 3⁄2
■ This is a convenient method to
scale any triangle. In this method, we do not have to measure the
angles.1. Fig.19.22 shows the original ΔABC. We want a scaled version with a scale factor 11⁄2
 |
| Fig.19.22 |
3. Scale them by 1.5, to reach P, Q and R. Draw PQ, QR and PR
4. • Then PQ will be 6 × 1.5 = 9.0
• QR will be 3 × 1.5 = 4.5
• PR will be 5 × 1.5 = 7.5
■ ΔPQR is the required triangle
We will now see some solved examples based on the above discussions
Solved example 19.10
The fig.19.23 below shows two concentric circles with centre 'O'. OP is a radius of the outer circle.
This radius intersects the inner circle at 'A'. |
| Fig.19.23 |
OQ is another radius of the outer circle. This radius intersects the inner circle at 'B'. Prove that ΔOAB and ΔOPQ are similar.
Solution:
1. OA and OB are radii of the same circle. So OA = OB
2. OP and OQ are radii of the same circle. So OP = OQ
3. Let OA be scaled by a scale factor k1 to make OP. Then we can write: OP = k1 × OA
4. Let OB be scaled by a scale factor k2 to make OQ. Then we can write: OQ = k2 × OB
5. In (4), substitute for OQ from (2) and substitute for OB from (1). We get: OP = k2 × OA
6. But from (3) we have OP = k1× OA. These two can be true only if k1 = k2
7. That means OA and OB are scaled by the same factor to obtain OP and OQ respectively
8. Now, the angle between OP and OQ is same as the angle between OA and OB.
9. So we have: Two sides scaled by the same factor, with included angle remaining the same. (see theorem 19.4)
10. So ΔOAB and ΔOPQ are similar.
Solved example 19.11
 |
| Fig.19.24 |
(i) Prove that ΔABC and ΔPQR are similar.
(ii) Prove that the 'scale factor of the sides of the triangle' is same as the 'scale factor of the radii of the two circumcircles'.
Solution:
1. OA, OB and OC are radii of the same circle. So OA = OB = OC
2. OP, OQ and OR are radii of the same circle. So OP = OQ = OR
3. Let OA be scaled by a scale factor k1 to make OP. Then we can write: OP = k1 × OA
4. Let OB be scaled by a scale factor k2 to make OQ. Then we can write: OQ = k2 × OB
5. Let OC be scaled by a scale factor k3 to make OR. Then we can write: OR = k3 × OC
• Take any two from (3), (4) and (5). Let us take (3) and (4):
6. In (4), substitute for OQ from (2) and substitute for OB from (1). We get: OP = k2 × OA
7. But from (3) we have OP = k1× OA. These two can be true only if k1 = k2
8. That means OA and OB are scaled by the same factor to obtain OP and OQ respectively. So we get k1 = k2
9. Similarly, by taking (4) and (5), we will get k2 = k3
10. From (8) and (9), we get: k1 = k2 = k3
• Since they are all equal, we will give a common name 'k'. Thus we can write: k1 = k2 = k3 = k
11. Consider triangle OPQ and it's inner triangle OAB. Sides OA and OB are scaled by the same factor k (since k1 = k2 = k), and their included angle remains the same. So ΔOAB and ΔOPQ are similar.
That means, we can obtain each side of ΔOPQ by multiplying the corresponding side of ΔOAB
12. We have already seen that:
• OP is obtained by multiplying OA by k1
• OQ is obtained by multiplying OB by k2
• We have see that k1 = k2 = k
• So PQ will be obtained by multiplying AB by k
That is., PQ = AB × k
13. • In (11) we considered ΔOPQ and it's inner ΔOAB. We obtained the result in (12): PQ = AB × k
• Similarly, if we consider ΔOPR and it's inner ΔOAC, we will get: PR = AC × k
• Also, if we consider ΔOQR and it's inner ΔOBC, we will get: QR = BC × k
14. That means, each side of ΔPQR is obtained by multiplying the corresponding side of ΔABC by 'k'. So ΔABC and ΔPQR are similar. This is the solution of part(i).
15. From (14), it is clear that, 'k' is the scale factor of ABC and PQR. We have to prove that this 'k' is the same scale factor of the circumradii. That is., we have to prove the below three:
• OP = OA×k, • OQ = OB×k, and • OC = OR×k.
16. In (3), (4) and (5), we wrote: OP = k1 × OA, OQ = k2 × OB and OR = k3 × OC
17. In (10), we proved that k1 = k2 = k3 = k
18. This same 'k' is used for scaling ΔABC to ΔPQR. So (15) is proved. This is the solution for part(ii)
• OQ is obtained by multiplying OB by k2
• We have see that k1 = k2 = k
• So PQ will be obtained by multiplying AB by k
That is., PQ = AB × k
13. • In (11) we considered ΔOPQ and it's inner ΔOAB. We obtained the result in (12): PQ = AB × k
• Similarly, if we consider ΔOPR and it's inner ΔOAC, we will get: PR = AC × k
• Also, if we consider ΔOQR and it's inner ΔOBC, we will get: QR = BC × k
14. That means, each side of ΔPQR is obtained by multiplying the corresponding side of ΔABC by 'k'. So ΔABC and ΔPQR are similar. This is the solution of part(i).
15. From (14), it is clear that, 'k' is the scale factor of ABC and PQR. We have to prove that this 'k' is the same scale factor of the circumradii. That is., we have to prove the below three:
• OP = OA×k, • OQ = OB×k, and • OC = OR×k.
16. In (3), (4) and (5), we wrote: OP = k1 × OA, OQ = k2 × OB and OR = k3 × OC
17. In (10), we proved that k1 = k2 = k3 = k
18. This same 'k' is used for scaling ΔABC to ΔPQR. So (15) is proved. This is the solution for part(ii)
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