Chapter 19.4 - Third method to obtain scaled versions of triangles.

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.

1. Fig.19.19(a) shows ΔABC. Two of its sides and included angle are given:

AB = 6 cm, AC = 4 cm, ∠CAB = 30^o

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.

Fig.19.20

 • The intermediary ΔMNO is drawn in the middle. It is drawn by the following procedure:

(i) First draw MN = 4.5 cm

(ii) Complete ΔMNO by drawing lines at 30^o at M and x^o (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 30^o 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:

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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’

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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.

Fig.19.21

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

(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.

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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

Fig.19.22

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

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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.

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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

In fig. 19.24 below, ΔABC has it's circumcentre at O. Circumcentre of ΔPQR is also at O.

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)

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In the next section, we will see the relation between perimeters of similar triangles.

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Chapter 19.5 - Perimeters and areas of Scaled triangles

In the previous section we saw the third method for obtaining scaled versions. In this section, we will see the relation between perimeters of scaled versions. Later we will see the relation of areas also. We will also see angle bisectors, medians and circumradii.

• Let the length of sides a given ΔABC be a, b and c cm. So perimeter will be equal to (a+b+c). 

• Let the scale factor be ‘k’. Then the length of sides of the scaled version will be ka, kb and kc cm. So the perimeter of the scaled version will be equal to (ka+kb+kc). This is same as k(a+b+c). 

We can write:

■ The perimeter of the scaled version is equal to ‘k’ times the perimeter of original triangle.

Next we will see the relation between areas. 

1. Fig.19.25(a) below shows a ΔABC. Length of sides are a, b and c cm.

Fig.19.25

2. Fig.b shows the scaled version ΔA’B’C’. The scale factor is ‘k’. So the length of it’s sides are ka, kb and kc. 

3. Drop perpendiculars CD and C’D’ for both triangles. 

4. Now consider the two smaller triangles on the left sides: ΔADC and ΔA’D’C’. Both have x^o and 90^o the same. So the third angle (90-x)^o will also be the same. That means: ΔADC and ΔA’D’C’ are 'similar' or 'scaled versions'.

5. In the smaller triangles ΔADC and ΔA’D’C, the side with length b is transformed to a side with length kb. In 'scaled versions' the scale factor will be the same for all sides. So the height h will be transformed to kh. That means, the altitude of  ΔA’D’C is kh.

6. Now we can calculate the areas of ΔADC and ΔA'D'C. We have: Area = ¹⁄₂ × base × altitude.

• So area of ΔADC = ¹⁄₂ × c × h = ^(ch)⁄₂

• Area of ΔA’D’C = ¹⁄₂ × kc × kh = ^(k2ch)⁄₂ .
7. We can write:
■ Area of the scaled version is equal to ‘ k² ’ times the Area of original triangle.

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Now we will see angle bisectors. In fig.19.26(a), ΔABC is the given triangle. It's scaled version ΔPQR is given in fig.b.

Fig.19.26

1. Consider ∠ACB. It's value is x^o. CD is the angle bisector of this angle. So we get: ∠ACD = ∠BCD = (x/2)^o

2. Since ΔABC and ΔPQR are similar (scaled versions), ∠ACB = ∠PRQ

3. This ∠PRQ is bisected by RS. So we get ∠PRS = ∠QRS = (x/2)^o

4. Now consider the two triangles ΔACD and ΔPRS. Two angles y^o and (x/2)^o are present in both the triangles. So the third angle must be the same. That is., ∠ADC = ∠PSR

5. So all the three angles are same. They are similar triangles.

6. We have information about one pair of sides. They are AC and PR. We know that PR = k × AC. So the other two sides must also be scaled by the same factor.

7. Thus we get: RS = k × CD and PS = k × AD

8. We want the first result in (7). It shows that, when ΔABC is scaled by a factor 'k', the angle bisector at C is also scaled by the same factor. The same result can be obtained at the other vertices A and B also. In general we can write:

■ When a triangle is scaled by a factor 'k', it's angle bisectors are also scaled by the same factor

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Now we will see medians

In fig.19.27(a), ΔABC is the given triangle. It's scaled version ΔPQR is given in fig.b.

Fig.19.27

1. CD is the median from vertex C and RS is the median from vertex R. That means D is the midpoint of AB, and S is the midpoint of PQ

2. Consider the two triangles ACD and PRS:

PR is the scaled version of AC. (since ΔABC and ΔPQR are scaled versions) That is., PR = k × AC

3. What about sides AD and PS?

(i) We know that PQ = k × AB. 

(ii) So (1/2 × PQ) = [1/2 × (k × AB)]  ⇒  (1/2 × PQ) = [k × (1/2 × AB)]

(iii) But 1/2 × PQ = PS and 1/2 × AB = AD

(iv) Substitute these values in (ii).The result is: PS = k × AD

4. Consider the results in 3(iv) and (2)

• We have two pairs:  ♦ AD is scaled to PS   ♦ AC is scaled to PR

• The angle between them remains unchanged. So ΔPRS is a scaled version of ΔACD. The scale factor is 'k'

• So we get RS = k × CD

• The same result can be obtained for medians drawn from other vertices also. Thus we can write:

■ When a triangle is scaled by a factor 'k', it's medians are also scaled by the same factor

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So in this section, we saw perimeter, area, angle bisector and median. 

There is one more: Circumradius. But we already saw it in solved example 19.11 in the previous section. We can write:

■ When a triangle is scaled by a factor 'k', it's circumradius is also scaled by the same factor

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Now we will see some solved examples related to the topics that we saw in this chapter in general.

Solved example 19.12

In the fig.19.28(a) below, ∠ABC = ∠CEF

Fig.19.28

Prove that EC × AC = FC × BC

Solution:

1. The fig.a is separated into two different triangles. The inner ΔCEF (fig.b) and the outer ΔABC (fig.c)  

2. We find that the two new triangles have x^o in common. The angle at vertex C, whose value is shown as y^o, is also common to both the triangles. This is because, vertex C is common.

3. Thus we have two angles same in the two triangles. So the third angle, whose value is shown as z^o, will also be same for both.

4. That means, all the three angles are same. The triangles are scaled versions of each other. We can apply  theorem 19.2:

^{• CF}⁄_AC  =  ^EF ⁄_AB = ^CE⁄_BC 
Take the first and the third ratios. We get: ^FC⁄_AC  = ^EC⁄_BC  ⇒EC × AC = FC × BC

Solved example 19.13

The length of the shadow of a tree of height 10 m is 4 m. At the same time, the length of the shadow of a tower is 14 m.

(i) Draw a rough sketch based on the given details

(ii) Calculate the height of the tower

Solution:

Fig.19.29

1. In fig.a, RQ is the tree. RP is the ray of the sun. It passes through the top end R of the tree, and then falls on the ground at P. So PQ is the shadow of the tree.

2. In fig.b, BC is the tower. CA is the ray of the sun. It passes through the top end C of the tower, and then falls on the ground at A. So AB is the shadow of the tower.

3. The ground on which shadow is formed is horizontal

4. The tree and tower are 'upright'. That is., the tree and tower makes 90^o with the ground

5. The shadows are measured at the same time. That means, the sun’s rays that causes the shadows will be making the same angle with the horizontal. This is marked as x^o

6. Thus we can complete the rough sketch. The height of the tower is denoted as ‘h’

7. x^o and 90^o is present in both the triangles. So the third angle (90-x)^o will also be the same in both the triangles.
8. We have two triangles. ⊿ABC and ⊿PQR. The angles are same. So we can apply theorem 19.2.

• Let x = x, y = (90-x) and z = 90. Then we can write:
^{• h}⁄₁₀  =  ¹⁴⁄₄ = ^AC⁄_PR = k ⇒ ^h⁄₃  =  ²⁄_h. This is solution of part (ii)
Take the first and the second. We get: ^h⁄₁₀  =  ¹⁴⁄₄  ⇒h = ¹⁴⁰⁄₄ = 35 m

Solved example 19.14

In fig.19.30(a), ∠ADB = ∠BCD = x^o. 

Fig.19.30

Prove that AB × AC = AD²

Solution:

1. The fig.a is separated into two different triangles. The inner ΔABD (fig.b) and the outer ΔACD (fig.c)  

2. We find that the two new triangles have x^o in common. The angle at vertex A, whose value is shown as y^o, is also common to both the triangles. This is because, vertex A is common.

3. Thus we have two angles same in the two triangles. So the third angle, whose value is shown as z^o, will also be same for both.

4. That means, all the three angles are same. The triangles are scaled versions of each other. We can apply  theorem 19.2:

^{• AB}⁄_AD  =  ^BD⁄_CD = ^AD⁄_AC 
Take the first and the third ratios. We get: ^AB⁄_AD  = ^AD⁄_AC  ⇒AB × AC = AD²

Solved example 19.15

In ΔABC (fig.19.31), D is the midpoint of AB. E, F G are the midpoints of AD, BD and CD respectively. 

Fig.19.31

Prove that ΔABC and ΔEFG are similar

Solution:
1. Consider ΔADC and it's inner ΔEDG on the left.
2. E is the midpoint of AD. So ED is half of AD. That means:
■ ED is scaled by a factor '2' to get AD
3. Similarly, since G is the midpoint of CD, We can write:
■ GD is scaled by a factor '2' to get DC
4. The angle at D remains the same. So, applying theorem 19.4, ΔADC is a scaled version of ΔEDG. The scale factor is '2'
5. Thus we get the result:
■ EG is scaled by a factor '2' to get AC
6. Considering ΔBDC, and it's inner ΔFDG, we will get the result:
■ FG is scaled by a factor '2' to get BC
7. Now consider the base AB
• We can write: AD = 2ED and BD = 2FD
• But AB = AD + BD = 2ED + 2FD ⇒ AB = 2(ED+FD) ⇒ AB = 2 EF
That means:
■ EF is scaled by a factor '2' to get AB
8. From (5), (6) and (7), we get the result:
Each side of ΔEFG is scaled by the factor '2' to get corresponding sides of ΔABC. So they are similar.

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In the next section, we will see Polynomials.

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Chapter 19.3 - Angles remain same in scaled triangles

So far in this chapter, we have been doing this:

1. We were given two triangles

2. Enough information were also given by which we could determine that 'the angles of both triangles are the same'

3. Based on (2) we could find unknown sides. We saw several solved examples in the previous section .

■ Now we will see another type of problem:

1. We are given two triangles

2. Enough information is also given by which we can determine that 'their sides are scaled by the same factor'

3. No information is given regarding any of the angles

4. In such a case, it becomes our duty to find the information about the angles. 

■ Let us see an example:

In the fig.19.13(a) below, two triangles are shown. 

Fig.19.13

1. Let us take the ratio of sides:

The shortest length in triangle 1 = 2 cm

The shortest length in triangle 2 = 3 cm

• So ratio = ²⁄₃

The medium length in triangle 1 = 4 cm

The medium length in triangle 2 = 6 cm

• So ratio = ⁴⁄₆ = ²⁄₃

The longest length in triangle 1 = 5 cm

The longest length in triangle 2 = 7.5 cm

• So ratio = ⁵⁄_7.5 = ²⁄₃ .

■ Thus we find that each side is scaled by the same factor. They are similar triangles. But we want to know about angles. For that, we can do the following steps:

2. Draw a line MN 7.5 cm in length. Measure angle x from the first triangle. Mark the same angle at M. This is shown in fig.19.14(c) below:

Fig.19.14

3. Measure angle y from the first triangle. Mark the same angle at N

4. So in fig.c, we have a side MN and two angles at it's ends. This much is sufficient to complete a triangle. (ASA construction)

5. Thus the ΔMNO is completed by extending the lines at the ends M and N in fig.c. The completed ΔMNO is shown in fig.d

6. When x and y in ΔMNO are same as those in ΔABC, the z will also be the same. Because x + y + z = 180

7. So, what is the relation between ΔABC in fig.a and ΔMNO in fig.d?

• We find that the angles are the same. So they are similar triangles. Each side of ΔMNO can be obtained by multiplying the corresponding side of ΔABC by a scale factor 'k'.

• Let us find this 'k', and also the remaining sides OM and ON of ΔMNO. This is a problem similar to the solved example 19.1 that we saw in a previous section. Here we will apply theorem 19.2

(i) ⁴⁄_ON  =  ²⁄_OM = ⁵⁄_7.5 = k ⇒ k =  ²⁄₃.

(ii) So we get:  ⁴⁄_ON = ²⁄₃ ⇒ ON = ¹²⁄₂ = 6  

(iii) Also we get: ²⁄_OM = ²⁄₃ ⇒ OM = ⁶⁄₂ = 3 

8. So MNO is a triangle with sides 3 cm, 6 cm and 7.5 cm. Now, PQR is also a triangle with the same sides.

9. When 3 sides are same, the two triangles are congruent. That means, ΔMNO and ΔPQR are congruent. It is a case of SSS congruence. Let us write the correspondance:

• In ΔPQR, the angle opposite the 3 cm side is Q. In ΔMNO, the angle opposite the 3 cm side is N. So we get Q↔N

• In ΔPQR, the angle opposite the 6 cm side is P. In ΔMNO, the angle opposite the 6 cm side is M. So we get P↔M

• In ΔPQR, the angle opposite the 7.5 cm side is R. In ΔMNO, the angle opposite the 7.5 cm side is O. So we get R↔O

10. So ΔMNO and ΔPQR are congruent according to the correspondence: PQR↔MNO.

• Thus the angles in ΔPQR are the same x, y and z in ΔMNO and also ΔABC 

What is the significance of this result in our present discussion?

Ans: • We were given two triangles: ΔABC and ΔPQR. Only sides were given. No angles were given.

• We took the ratios of sides and found that ΔPQR is a scaled version of ΔABC

• After much calculations, we arrived at (10) in which we find that, the angles in ΔPQR are same as those in ΔABC

11. We can write this: If a triangle is the scaled version of another triangle, the  angles in both triangles will be the same. 

■ We derived this result with the help of an intermediary triangle: ΔMNO

■ Now we will see the general case:

1. In the fig.19.15 below, ΔABC and ΔPQR are two given triangles. ΔPQR is a scaled version of ΔABC. The scale factor being 'k'. 

Fig.19.15

2. We want to prove that the angles in both triangles are the same.

3. Draw an intermediary ΔMNO. For that, use the following procedure:

(i) Draw MN = kp

(ii) Measure x from ΔABC and mark it at M. Measure y from ΔABC and mark it at N

(iii) Complete ΔMNO. Angle at O will be z. The same z in ΔABC

4. ΔABC and ΔMNO have the same angles. So ΔMNO is a scaled version of ΔABC. So:

(i) OM will be equal to kr

(ii) ON will be equal to kq

5. The 3 sides in ΔMNO are equal to the 3 sides in ΔPQR. So MNO↔PQR

6. So angles in ΔMNO are same as those in ΔPQR

7. But angles in ΔMNO are same as those in ΔABC

8. Thus ΔABC and ΔPQR have the same angles

We can write it in the form of a theorem:

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Theorem 19.3:

■ Two triangles are given. One is the scaled version of the other

■ Then the angles in the two triangles are the same

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Once we understand this theorem, there is no need to draw an intermediary triangle. We can directely solve problems by quoting the above theorem 19.3, and writing that, 'angles will be the same'. We will now see some solved examples:
Solved example 19.7
Fig.19.16 (a) shows a triangle. 

Fig.19.16

Draw another triangle with the same angles, but sides scaled by a factor 1¹⁄₄.
Solution:

■ We are required to draw a 'scaled version'. The scale factor is 1¹⁄₄ (same as ⁵⁄₄). 
• We know that, when all the sides are scaled by the same factor, the angles remain the same. So we do not need to consider angles in this problem. All we need are the new sides.

• The shortest side 4 cm will become 4 × ⁵⁄₄  = 5 cm

• The medium side 6 cm will become 6 × ⁵⁄₄  = 7.5 cm

• The longest side 8 cm will become 8 × ⁵⁄₄ = 10 cm

• So we have all the 3 new sides. We can begin the construction:

1. Draw a horizontal line 10 cm long. 

2. With it's left end as centre, draw an arc of radius 5 cm (shown in yellow colour in fig.b). 

3. With it's right end as centre, draw an arc of radius 7.5 cm (shown in green colour in fig.b)

4. The point of intersection of the two arcs is the third vertex of the required triangle

Solved example 19.8

Fig. 19.17(a) shows a quadrilateral. 

Fig.19.17

Draw another quadrilateral with the same angles, but sides scaled by 1¹⁄₂.

Solution:

■ We are required to draw a 'scaled version'. The scale factor is 1¹⁄₂ (same as ³⁄₂).
• The given quadrilateral can be considered to be made up of two triangles 
    ♦ The division is along the diagonal of length 5 cm
    ♦ So we get an upper triangle of sides 5, 2 and 4 cm
    ♦ And a lower triangle of sides 6, 3 and 5 cm
• We must construct each triangle separately
• We know that, when all the sides are scaled by the same factor, the angles remain the same. So we do not need to consider angles in this problem. All we need are the new sides. First we take the bottom triangle:

• The shortest side 3 cm will become 3 × ³⁄₂  = 4.5 cm

• The medium side 5 cm will become 5 × ³⁄₂  = 7.5 cm

• The longest side 6 cm will become 6 × ³⁄₂ = 9 cm

• So we have the 3 new bottom sides. We can begin the construction:

1. Draw a horizontal line 9cm long. 

2. With it's left end as centre, draw an arc of radius 7.5 cm (shown in yellow colour in fig.b). 

3. With it's right end as centre, draw an arc of radius 4.5 cm (shown in green colour in fig.b)

4. The point of intersection of the two arcs is the third vertex of the new triangle

5. Now, the 7.5 cm side is the base of the upper triangle

6. With it's left end as centre, draw an arc of radius 6 cm (shown in green colour in fig.b). 

7. With it's right end as centre, draw an arc of radius 3 cm (shown in yellow colour in fig.b)

8. The point of intersection of the two arcs is the third vertex of the new upper triangle

Solved example 19.9

Fig.19.18(a) shows a triangle.

Fig.19.18

Draw another triangle with the same angles, but sides scaled by ³⁄₄.

Solution:

■ We are required to draw a 'scaled version'. The scale factor is ³⁄₄.

• We know that, when all the sides are scaled by the same factor, the angles remain the same.

Let us begin the construction:

The 6 cm length will become 6 × ³⁄₄ = 4.5 cm

1. Draw a horizontal line 4.5 cm long. 

2. At it's left end, draw a line at an angle 40^o. This is shown in fig.b

3. At it's right end, draw a line at an angle 60^o. The two lines will meet at the third vertex of the required triangle

Explanation:

• We have used the same angles and at the ends of the 4.5 cm line. So the third angle in both the triangles will be (180 - 40 - 60) = 80^o.

• That means the two triangles have the same angles.

• It follows that, one triangle is the scaled version of the other.

• Since the 6 cm side is scaled to 4 cm, the other two sides will also be scaled by the same factor

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In the next section, we will see a third method to obtain scaled versions.

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