Chapter 19.1 - Similarity of Triangles - Solved examples

In the previous section we saw the relation between corresponding sides of similar triangles. In this section, we will see some solved examples.

Solved example 19.1

In the triangle in fig.19.5(a), AB = 10 cm, AC = 8 cm, BC = 6 cm.

Fig.19.5

In the triangle in fig.b, PQ = 4 cm

Given: ∠P = ∠C, ∠Q = ∠A and ∠R = ∠B. Find the lengths of the other two sides of ΔPQR

Solution:

We are given two triangles. The angles are same. So we can apply theorem 19.1.

We have:

• Shortest side of triangle 1 = BC = 6 cm

• Medium side of triangle 1 = AC = 8 cm

• Longest side of triangle 1 = AB = 10 cm

We cannot directely make such a classification for the second triangle PQR. This is because, all lengths are not given. But there is an indirect method:

■ To find the shortest side in ΔPQR:

• The shortest side in ΔABC is BC. So the ∠A which is opposite BC will be the smallest angle in ΔABC

• Since the angles in both the triangles are the same, this will be the smallest angle in ΔPQR also. That means., the smallest angle in ΔPQR is ∠Q

• The side opposite to this smallest angle will be the shortest side. So PR is the smallest side in ΔPQR

■ To find the medium side in ΔPQR:

• The medium side in ΔABC is AC. So the ∠B which is opposite AC will be the medium angle in ΔABC

• Since the angles in both the triangles are the same, this will be the medium angle in ΔPQR also. That means., the medium angle in ΔPQR is ∠R

• The side opposite to this medium angle will be the medium side. So PQ is the medium side in ΔPQR

■ To find the longest side in ΔPQR:

• The longest side in ΔABC is AB. So the ∠C which is opposite AB will be the largest angle in ΔABC

• Since the angles in both the triangles are the same, this will be the largest angle in ΔPQR also. That means., the largest angle in ΔPQR is ∠P

• The side opposite to this largest angle will be the largest side. So QR is the largest side in ΔPQR

1. Now we can apply theorem 19.1

^BC⁄_PR  =  ^AC⁄_PQ = ^BA⁄_QR = k 

⇒ ^BC⁄_PR = ⁸⁄₄ = ^BA⁄_QR  = k 

⇒ ^BC⁄_PR = 2 = ^BA⁄_QR  = k 

Thus we get k = 2

2. We have ^BC⁄_PR = k = 2 ⇒ ⁶⁄_PR = 2. So we get PR = ⁶⁄₂  = 3

3. We have ^AB⁄_QR = k = 2 ⇒ ¹⁰⁄_QR = 2. So we get QR = ¹⁰⁄₂  = 5

■ We can see that, the new triangle is exactly half of the original

Solved example 19.2

In fig.19.6(a), ABCD is a square of side 8 cm. It is folded in such a way that the bottom corner A coincides with the midpoint of the top side CD. 

Fig.19.6

Find all the sides of the two right triangles formed at the top.

Solution:

• When the bottom corner A coincides with the top midpoint E, two right triangles ⊿CFE and ⊿EGD are formed. This is shown in figs.b & c. 

• CE and DE are already known. They are 4 cm each because E is the midpoint. 

• We have to calculate CF, EF, EG and DG. First we will consider the ⊿CFE on the left

1. Let CF = x cm. Then FA = 8-x. 

Since the fold passes through F, and E coincides with A, length FA = FE = 8-x

2. Applying Pythagorus theorem we get:

4² + x² = (8-x)² ⇒ 4² + x² = 64 -16x + x² ⇒16 = 64 -16x ⇒16x = 48. So x = 3 cm

3. So the sides of ΔCFE are: CE = 4 cm, CF = x = 3 cm, EF = 8-x = 5 cm

4. Now consider ΔEGD. Two of it's sides: EG and DG are unknowns

Let us check whether the two triangles are similar:

5. In ΔCFE, let ∠F = a^o. Then ∠CEF = 180 – (90+a) = (90-a)^o

6. Consider the line CD, and the point E on it. Three angles are collinear at the point E. They are: ∠CEF, ∠FEG and ∠GED

• We already found in (5) that, ∠CEF = (90-a)^o

7. ∠FEG = 90^o. (∵ corner A of the original square is folded to coincide with E)

8. The sum of the three angles = 180 (∵ they are collinear)

9. Thus we get: (90-a) + 90 + ∠GED = 180^o ⇒ ∠GED = 180 – 90 – (90-a) ⇒ ∠GED = 180 – 90 – 90 + a ⇒ ∠GED = a^o

10. So the angles in the two triangles are the same. They are similar triangles. We can apply theorem 19.1

11. We have:

• Shortest side of triangle 1 = CF = 3 cm

• Medium side of triangle 1 = EC = 4 cm

• Longest side of triangle 1 = EF = 5 cm

We cannot directly make such a classification for the second triangle EGD. This is because, all lengths are not available. So we will use the same indirect method that we used in the previous example:

■ To find the shortest side in ΔEGD:

• The shortest side in ΔCFE is CF. So the ∠E = (90-a), which is opposite CF will be the smallest angle in ΔCFE

• Since the angles in both the triangles are the same, this will be the smallest angle in ΔEGD also. That means., the smallest angle in ΔEGD is ∠G

• The side opposite to this smallest angle will be the shortest side. So ED is the smallest side in ΔEGD

■ To find the medium side in ΔEGD:

• The medium side in ΔCFE is CE. So the ∠F = a^o, which is opposite CE will be the medium angle in ΔCFE

• Since the angles in both the triangles are the same, this will be the medium angle in ΔEGD also. That means., the medium angle in ΔEGD is ∠E

• The side opposite to this medium angle will be the medium side. So GD is the medium side in ΔEGD

■ To find the longest side in ΔEGD:

• The longest side in ΔCFE is EF. So the ∠C = 90^o, which is opposite EF will be the largest angle in ΔCFE

• Since the angles in both the triangles are the same, this will be the largest angle in ΔEGD also. That means., the largest angle in ΔEGD is ∠D

• The side opposite to this largest angle will be the longest side. So EG is the longest side in ΔEGD

12. Now we can apply theorem 19.1

^CF⁄_ED  =  ^CE⁄_GD = ^EF⁄_EG = k 

⇒ ³⁄₄ = ⁴⁄_GD = ⁵⁄_EG  = k

• Thus we get k = ³⁄₄

• We have ⁴⁄_GD = k = ³⁄₄ ⇒ ⁴⁄_GD = ³⁄₄. So we get GD = ¹⁶⁄₃  = 5 ¹⁄_{3
• We have ⁵⁄EG = k = ³⁄4 ⇒ ⁵⁄EG = ³⁄4. So we get EG = ²⁰⁄3  = 6 ²⁄3}

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So we got all the required sides. The problem is solved. But we have to learn a few common properties when a square sheet of paper is folded in this way.

We will write the lengths of sides first:

• Shortest side of triangle 1 = CF = 3 cm, Shortest side of triangle 2 = ED = 4 cm

• Medium side of triangle 1 = EC = 4 cm, Medium side of triangle 2 = GD = 5 ¹⁄₃ cm

• Longest side of triangle 1 = EF = 5 cm, Longest side of triangle 2 = EG  = 6 ²⁄₃ cm

■ We have seen that the scale factor k = ³⁄₄. The two triangles are connected by the scale factor.
■ If we want the larger triangle from the smaller triangle, we should multiply all the sides of the smaller by ⁴⁄₃. This is shown below:
♦ 3 × ⁴⁄₃ = 4 cm   ♦ 4 × ⁴⁄₃ = ¹⁶⁄₃ = 5 ¹⁄₃ cm   ♦ 5 × ⁴⁄₃ = ²⁰⁄₃ = 6 ²⁄₃ cm
■ If we want the smaller triangle from the larger triangle, we should multiply all the sides of the larger by ³⁄₄. This is shown below:
♦ 4 × ³⁄₄ = 3 cm   ♦ 5 ¹⁄₃ × ³⁄₄ = ¹⁶⁄₃ × ³⁄₄ = 4 cm   ♦ 6 ²⁄₃ × ³⁄₄ = ²⁰⁄₃ × ³⁄₄ = 5 cm

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Now we will see a very interesting result:

In the above example, we folded a square of side 8 cm. Let us now consider any square of side 's' cm. The folded square and the measurements will be as shown in fig.19.7 below:

Fig.19.7

1. First we will find 'x' by applying Pythagoras theorem:

(^s⁄₂)² + x² = (s-x)² ⇒ (^s2⁄₄) + x² = (s-x)² ⇒ ^s2⁄₄ + x² = s²-2sx + x² 
⇒ 2sx = s²- ^s2⁄₄ ⇒ 2sx = ^3s2⁄₄ ⇒ 2x = ^3s⁄₄ ⇒  x = ^3s⁄₈ cm. 
2. So EF = (^s⁄₂) - x = ^s⁄₂ - ^3s⁄₈= ^5s⁄₈.
3. Thus we can write the detaiils about the first ΔCFE:

• Shortest side of triangle 1 = CF = ^3s⁄₈ cm

• Medium side of triangle 1 = EC = ^s⁄₂ cm

• Longest side of triangle 1 = EF = ^5s⁄₈ cm
4. The angles in the second ΔEGD are the same. We can apply theorem 19.1:
^(3s/8)⁄_ED  =  ^(s/2)⁄_GD = ^(5s/8)⁄_EG = k 
⇒ ^(3s/8)⁄_(s/2)  =  ^(s/2)⁄_GD = ^(5s/8)⁄_EG = k (∵ ED = s/2) 
⇒ ^3s⁄₈ × ²⁄_s =  ^(s/2)⁄_GD = ^(5s/8)⁄_EG = k
⇒ ³⁄₄=  ^(s/2)⁄_GD = ^(5s/8)⁄_EG = k
• Thus we get k = ³⁄₄ in the general case also.
5. We have ^(s/2)⁄_GD = k = ³⁄₄ ⇒  ^s⁄_(2GD) = ³⁄₄  ⇒  ^s⁄_GD = ³⁄₂ ⇒ GD = ^2s⁄₃.
6. We have ^(5s/8)⁄_EG = k = ³⁄₄ ⇒  ^5s⁄_(8EG) = ³⁄₄  ⇒ EG = ^5s⁄₆.
■ From (4) we can write this: When we fold any square sheet of paper in the above manner, the scale factor between the two triangles will be ³⁄₄.
■ From (3), (5) and (6), we can quickly calculate all the 6 sides of the two triangles for any square.

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Another interesting result:

From (5) above, we have GD = ^2s⁄₃. So BG will be equal to (s - ^2s⁄₃) = ^s⁄₃. This is shown in the fig.9.8 below:

Fig.19.8

■ So for any square that we get, we can fold it in the above manner, and obtain one third of a side.

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We have completed the discussion on 'folding of squares'. Next we have to learn a very important property by which we can fix up the shortest, medium and longest sides of any triangle.
In the examples 19.1 and 19.2, that we saw above, we used an indirect method. If we look at the method carefully, we will get an easier method.
■ In example 19.1:
• The shortest side in both the triangles, are the ones opposite the green angles (∠A & ∠Q)
• The medium side in both the triangles, are the ones opposite the white angles (∠B & ∠R)
• The longest side in both the triangles, are the ones opposite the blue angles (∠C & ∠P)
■ In example 19.2:
• The shortest side in both the triangles, are the ones opposite the blue angles (∠E & ∠G)
• The medium side in both the triangles, are the ones opposite the red angles (∠F & ∠E)
• The longest side in both the triangles, are the ones opposite the white angles (∠C & ∠D)
We can write this:
In any triangle,
• The smallest side will always be the one opposite to the smallest angle
• The medium side will always be the one opposite to the medium angle
• The longest side will always be the one opposite to the largest angle
So we find that, to apply theorem 19.1, all we need to do is, pick equal angles from the two triangles, and take the ratio of the sides opposite to them. We can write it in the form of another theorem .

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Theorem 19.2:

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In the next section, we will see some more solved examples.

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