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:
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
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.
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 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:
Solved example 19.7
Fig.19.16 (a) shows a triangle.
 |
| Fig.19.16 |
Solution:
• 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 40o. This is shown in fig.b
3. At it's right end, draw a line at an angle 60o. 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) = 80o.
• 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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