In the previous section we saw two solved examples based on the similarity of triangles. We also saw an easy method to calculate the scale factor 'k'. In this section, we will see more solved examples.
Solved example 19.3
Solved example 19.3
The perpendicular from the square corner of a right triangle, cuts the opposite side into two parts of lengths 2 cm and 3 cm. See fig.19.9(a) below:
(i) Prove that the two small triangles so obtained have the same angles
(ii) If the length of the perpendicular is h, prove that h⁄2 = 3⁄h .
(iii) Calculate the perpendicular sides of the original triangle
(iv) Prove that, if the perpendicular from the square corner of a right triangle cuts the opposite side into two parts of lengths a and b, and if the length of the perpendicular is h, then h2 = ab
Solution:
1. In fig.b, the two triangles obtained are named as ⊿ADC and ⊿BDC. Both are right angled because, CD is perpendicular to AB
2. In ⊿ADC, let ∠A = ao. Then ∠ACD = 180 – (90+a) = (90-a)o
3. The total angle BCA is 90o. So ∠BCD = 90 – (90-a) = 90 -90 +a = ao.
4. In BCD, ∠C = ao. Then ∠DBC = 180 – (90+a) = (90-a)o. So the two triangles have the same angles. This is solution of part (i)
5. We have two triangles. ⊿ADC and ⊿BDC. The angles are same. So we can apply theorem 19.2.
• Let x = a, y = (90-a) and z = 90. Then we can write:
• h⁄3 = 2⁄h = AC⁄BC = k ⇒ h⁄3 = 2⁄h. Taking reciprocals we get: 3⁄h = h⁄2. This is solution of part (ii)
6. Applying pythagoras theorem to ⊿ADC we get:
22 + h2 = AC2 . But from (5), we get h2 = 6
Thus AC2 = 22 + 6 ⇒ AC2 = 4 + 6 ⇒ AC = √10
7. Applying pythagoras theorem to ⊿BDC we get:
32 + h2 = BC2 . But from (5), we get h2 = 6
Thus BC2 = 32 + 6 ⇒ BC2 = 9 + 6 ⇒ BC = √15
• (6) & (7) is the solution for part (iii)
8. For solving part (iv), all we need to do is, to change 2 and 3 in fig.b to a and b. The modification is shown in fig.c. The steps are the same as in (5). We will get:
• h⁄b = a⁄h ⇒ h2 = ab. This is solution of part (iv)
Solved example 19.4
At two ends of a horizontal line, angles of equal size are drawn, and some points on the slanted line are joined. See fig.19.10(a) below:
(i) Prove that the parts of the horizontal line and parts of the slanted line are in the same ratio
(ii) Prove that the two slanted lines at the ends of the horizontal line are also in the same ratio
(iii) Explain how a line of length 6 cm can be divided in the ratio 3:4 using this
Solution:
1. The horizontal line is shown in red colour. At it's ends, two yellow lines are drawn. Both makes the same angle (shown in blue colour) with the red line. So the yellow lines are parallel to each other.
2. The green line is drawn between any two points. That is.,
• take any point on the top yellow line
• take any point on the bottom yellow line
• join the two points by the green line. So the lengths of the two yellow lines need not be the same. When the green line is drawn, we get two triangles.
3. In fig.b, the two triangles are named: ΔACE and ΔBDE
4. Parallel lines AC and BD are cut by a transversal CD. So ∠C = ∠D. (∵ they are alternate interior angles). They are shown in green colour
5. The two white angles at E are equal (∵ they are opposite angles)
6. The blue angles are equal as given in the problem statement. So ∠A = ∠B
7. So the angles in the two triangles are the same. We can apply theorem 19.2:
8. Let • x = ∠A = ∠B, • y = ∠C = ∠D, • z = ∠E = ∠E
So we can write:
• CE⁄DE = AE⁄BE = AC⁄BD = k
9. Take the first two ratios from the above:
• CE⁄DE = AE⁄BE . This is the solution of part (i)
10. Take the three ratios together:
• CE⁄DE = AE⁄BE = AC⁄BD . This is the solution of part (ii)
11. A 6 cm line cannot be divided in the ratio 3:4 using a ruler. We can use the following procedure:
• Draw a horizontal line AB of length 6 cm.
• Draw AC and BD at the ends. The angle must be the same. In the fig.c, the angle used is 35o.
• AC must be 3 cm, and BD must be 4 cm
• Join CD. The line CD cuts AB at E. Then AE:EB = 3:4
Proof::
Taking the second and third ratios from (8), we get: AE⁄BE = AC⁄BD ⇒ AE⁄BE = 3⁄4.
Solved example 19.5
Fig.19.11(a) below shows a blue square inside a right triangle. One corner of the square is exactly above the square corner of the triangle. The other corners of the square are on the three sides of the triangle.
(i) Calculate the side of the square
(ii) What is the length of the side of the square drawn like this in a 3,4,5 triangle?
Solution:
1. When the square is placed inside the triangle as in fig.a, we get two right triangles. One at top, and the other at bottom. In fig b, they are named as ⊿CFE and ⊿EDB
• FE is parallel to AB. So ∠CEF = ∠EBD. They are shown in magenta colour
• ED is parallel to AC. So ∠ECF = ∠BED. They are shown in green colour
• ∠CFE = ∠EDB = 90o . They are shown in white colour
The angles are the same in the two triangles. We can apply theorem 19.2:
2. Let • x = ∠CEF = ∠EBD, • y = ∠ECF = ∠BED • z = ∠CFE = ∠EDB = 90o
3. So we can write:
• CF⁄DE = EF⁄BD = AC⁄BD ⇒ 2⁄DE = EF⁄1 = CE⁄BE
4. Take the first two ratios from above:
• 2⁄DE = EF⁄1 ⇒ EF × DE = 2 × 1.
• But DE = EF. So we can write: EF2 = 2 ⇒ EF = √2. This is the solution for part (i)
5. In fig.c, the side of the required square is taken as 'x' cm. Then CF = (4-x), and BD = (3-x)
6. Take the first two ratios from (3):
• CF⁄DE = EF⁄BD ⇒ (4-x)⁄x = x⁄(3-x)
⇒ (4-x) × (3-x) = x2 ⇒ 12 -3x - 4x + x2 = x2
⇒ 7x = 12 ⇒x = 12⁄7 = 1.7143
Solved example 19.6
Two poles of height 3 m and 2 m are erected upright on the ground. Ropes are stretched from the top of each pole to the foot of the other.
(i) At what height above the ground, do the ropes cross each other?
(ii) Prove that, this distance would be the same, what ever be the distance between the poles
(iii) If the heights of the poles are a and b, and if the height of the point where the ropes cross each other is h, find the relation between a, b and h
Solution:
 |
| Fig.19.9 |
(ii) If the length of the perpendicular is h, prove that h⁄2 = 3⁄h .
(iii) Calculate the perpendicular sides of the original triangle
(iv) Prove that, if the perpendicular from the square corner of a right triangle cuts the opposite side into two parts of lengths a and b, and if the length of the perpendicular is h, then h2 = ab
Solution:
1. In fig.b, the two triangles obtained are named as ⊿ADC and ⊿BDC. Both are right angled because, CD is perpendicular to AB
2. In ⊿ADC, let ∠A = ao. Then ∠ACD = 180 – (90+a) = (90-a)o
3. The total angle BCA is 90o. So ∠BCD = 90 – (90-a) = 90 -90 +a = ao.
4. In BCD, ∠C = ao. Then ∠DBC = 180 – (90+a) = (90-a)o. So the two triangles have the same angles. This is solution of part (i)
5. We have two triangles. ⊿ADC and ⊿BDC. The angles are same. So we can apply theorem 19.2.
• h⁄3 = 2⁄h = AC⁄BC = k ⇒ h⁄3 = 2⁄h. Taking reciprocals we get: 3⁄h = h⁄2. This is solution of part (ii)
6. Applying pythagoras theorem to ⊿ADC we get:
22 + h2 = AC2 . But from (5), we get h2 = 6
Thus AC2 = 22 + 6 ⇒ AC2 = 4 + 6 ⇒ AC = √10
7. Applying pythagoras theorem to ⊿BDC we get:
32 + h2 = BC2 . But from (5), we get h2 = 6
Thus BC2 = 32 + 6 ⇒ BC2 = 9 + 6 ⇒ BC = √15
• (6) & (7) is the solution for part (iii)
8. For solving part (iv), all we need to do is, to change 2 and 3 in fig.b to a and b. The modification is shown in fig.c. The steps are the same as in (5). We will get:
• h⁄b = a⁄h ⇒ h2 = ab. This is solution of part (iv)
Solved example 19.4
At two ends of a horizontal line, angles of equal size are drawn, and some points on the slanted line are joined. See fig.19.10(a) below:
 |
| Fig.19.10 |
(ii) Prove that the two slanted lines at the ends of the horizontal line are also in the same ratio
(iii) Explain how a line of length 6 cm can be divided in the ratio 3:4 using this
Solution:
1. The horizontal line is shown in red colour. At it's ends, two yellow lines are drawn. Both makes the same angle (shown in blue colour) with the red line. So the yellow lines are parallel to each other.
2. The green line is drawn between any two points. That is.,
• take any point on the top yellow line
• take any point on the bottom yellow line
• join the two points by the green line. So the lengths of the two yellow lines need not be the same. When the green line is drawn, we get two triangles.
3. In fig.b, the two triangles are named: ΔACE and ΔBDE
4. Parallel lines AC and BD are cut by a transversal CD. So ∠C = ∠D. (∵ they are alternate interior angles). They are shown in green colour
5. The two white angles at E are equal (∵ they are opposite angles)
6. The blue angles are equal as given in the problem statement. So ∠A = ∠B
7. So the angles in the two triangles are the same. We can apply theorem 19.2:
So we can write:
• CE⁄DE = AE⁄BE = AC⁄BD = k
9. Take the first two ratios from the above:
• CE⁄DE = AE⁄BE . This is the solution of part (i)
10. Take the three ratios together:
• CE⁄DE = AE⁄BE = AC⁄BD . This is the solution of part (ii)
11. A 6 cm line cannot be divided in the ratio 3:4 using a ruler. We can use the following procedure:
• Draw a horizontal line AB of length 6 cm.
• Draw AC and BD at the ends. The angle must be the same. In the fig.c, the angle used is 35o.
• AC must be 3 cm, and BD must be 4 cm
• Join CD. The line CD cuts AB at E. Then AE:EB = 3:4
Proof::
Taking the second and third ratios from (8), we get: AE⁄BE = AC⁄BD ⇒ AE⁄BE = 3⁄4.
Solved example 19.5
Fig.19.11(a) below shows a blue square inside a right triangle. One corner of the square is exactly above the square corner of the triangle. The other corners of the square are on the three sides of the triangle.
 |
| Fig.19.11 |
(ii) What is the length of the side of the square drawn like this in a 3,4,5 triangle?
Solution:
1. When the square is placed inside the triangle as in fig.a, we get two right triangles. One at top, and the other at bottom. In fig b, they are named as ⊿CFE and ⊿EDB
• FE is parallel to AB. So ∠CEF = ∠EBD. They are shown in magenta colour
• ED is parallel to AC. So ∠ECF = ∠BED. They are shown in green colour
• ∠CFE = ∠EDB = 90o . They are shown in white colour
The angles are the same in the two triangles. We can apply theorem 19.2:
3. So we can write:
• CF⁄DE = EF⁄BD = AC⁄BD ⇒ 2⁄DE = EF⁄1 = CE⁄BE
4. Take the first two ratios from above:
• 2⁄DE = EF⁄1 ⇒ EF × DE = 2 × 1.
• But DE = EF. So we can write: EF2 = 2 ⇒ EF = √2. This is the solution for part (i)
5. In fig.c, the side of the required square is taken as 'x' cm. Then CF = (4-x), and BD = (3-x)
6. Take the first two ratios from (3):
• CF⁄DE = EF⁄BD ⇒ (4-x)⁄x = x⁄(3-x)
⇒ (4-x) × (3-x) = x2 ⇒ 12 -3x - 4x + x2 = x2
⇒ 7x = 12 ⇒x = 12⁄7 = 1.7143
Solved example 19.6
Two poles of height 3 m and 2 m are erected upright on the ground. Ropes are stretched from the top of each pole to the foot of the other.
(i) At what height above the ground, do the ropes cross each other?
(ii) Prove that, this distance would be the same, what ever be the distance between the poles
(iii) If the heights of the poles are a and b, and if the height of the point where the ropes cross each other is h, find the relation between a, b and h
Solution:
1. In fig.19.12(a) below, the yellow line
indicates the ground surface. The posts are erected in an ‘upright’
position. That means, they are perpendicular to the ground. The
magenta lines are the ropes.
 |
| Fig.19.12 |
2. In fig.b, the triangles are
given names. The ropes cross each other at point E. The point E is at a height
of ‘h’ m from the ground. F is the foot of the perpendicular
dropped from E.
3. We have to consider the two
triangles ⊿ABC and ⊿ABD separately. Within each of these triangles,
there are two similar triangles. Let us find them:
4. Consider ⊿ABC first.
• ∠C = ∠E (∵ AC is parallel to
EF) • ∠A = ∠F, • ∠B = ∠B
• So the similar triangles are: ⊿ABC and ⊿FBE. We can apply theorem 19.2:
(i) Let • x = ∠C = ∠E, • y = ∠A = ∠F, • z = ∠B
So we can write:
AB⁄FB = CB⁄EB = AC⁄FE ⇒ AB⁄FB = CB⁄EB = 3⁄h .
(ii). Take out the first and third ratios:
AB⁄FB = 3⁄h . This is an important result
5. Now consider ⊿ABD
• ∠D = ∠E (∵ BD is parallel to
EF) • ∠B = ∠F, • ∠A = ∠A
• So the similar triangles are: ⊿ABD and ⊿AFE. We can apply theorem 19.2:
(i) Let • x = ∠D = ∠E, • y = ∠B = ∠F, • z = ∠A
So we can write:
AB⁄AF = AD⁄AE = BD⁄FE ⇒ AB⁄AF = BD⁄FE = 2⁄h .
(ii). Take out the first and third ratios:
AB⁄AF = 2⁄h . This is an important result
6. Take the reciprocals of 4(ii) and 5(ii):
(i) FB⁄AB + AF⁄AB = h⁄3 + h⁄2 This is an important result
⇒(FB+AF)⁄AB = (2h+3h)⁄6 . But (FB + AF) = AB. So we get:
(ii) AB⁄AB = 5h⁄6 ⇒ 5h⁄6 = 1 ⇒ h = 6⁄5 = 1.2 m. This is the solution for part (i)
7. By analysing the similar triangles, in (6) above, we derived the formula for h:
• 5h⁄6 = 1. This formula is independent of the distance between the poles. So, what ever be the distance between the poles, if the heights are 3 m, and 2 m, E will always be 1.2 m above the ground.
This is the solution for part (ii)
8. Fig.d shows the case when the poles can have any heights 'a' and 'b'
• Analysing the similar triangles we will arrive at the result similar to that in 6(i):
• FB⁄AB + AF⁄AB = h⁄a + h⁄b . ⇒(FB+AF)⁄AB = (bh+ah)⁄ab
⇒ 1 = (bh+ah)⁄ab . ⇒ 1 = h(a+b)⁄ab
⇒ h = ab⁄(a+b) This is the solution for part (iii)
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