In the previous section we saw some solved examples on proportionality between two quantities. In this section we will see a few more solved examples.
Solved example 24.5
In the angle shown in fig.24.10 below, an object moves along the slanting line. As the distance of the object from the vertex changes, it's height from the horizontal line also changes.
(i) Prove that height is proportional to the distance
(ii) Calculate the constant of proportionality for 30o, 35o and 60o angles
Solution:
1. The given angle in fig.24.10 is reproduced in fig.24.11(a) below. Some modifications are also made:
• The blue object which moves along the slanting line is marked as Q
• A perpendicular is dropped from Q to the horizontal leg of the given angle. The foot of the perpendicular is marked as P
• So we get a right triangle APQ
2. There is also another right triangle ABC. This is our base triangle. That is., we are going to do the calculations based on ⊿ABC:
• We assume that BC is fixed at it's position.
• Also we assume that all sides of ⊿ABC are known.
3. But QP is not fixed. Because, Q can be at any point along the slanting line.
4. Now we find the relation between ⊿ABC and ⊿APQ:
• Angle at A is denoted as x. It is same for both the triangles
• Angle at P and B are both 90
• Angle at C and Q will both be equal to [180 – (90+x)] = (90-x)
• So the two triangles are have the same angles. They are similar triangles. We can apply theorem 19.2:
5. Side opposite angle x in ABC⁄Side opposite angle x in APQ =
Side opposite angle (90-x) in ABC⁄Side opposite angle (90-x) in APQ =
Side opposite angle 90 in ABC⁄Side opposite angle 90 in APQ
6. This is same as:
BC⁄PQ = AB⁄AP = AC⁄AQ
7. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ
• AC⁄BC is a constant because ⊿ABC is fixed. We can calculate AC⁄BC. So we can write:
• AQ = A constant × PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
Part (ii)
1. In this part we explore the cases when the angle x at vertex A is 30o, 60o and 45o
2. First we will take 30o. Consider the fig.24.11(b). A base triangle ABC is drawn in it. How is it drawn?
• First mark a point C such that AC = 2 cm. Next drop perpendicular BC onto the horizontal leg. Then:
♦ BC will be 1 cm
♦ AB will be √3 cm
• The above two will naturally occur because: √[(√3)2+12] = √[3+1] = √[4] = 2
3. So we established a base triangle. As we saw in fig.(a), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
BC⁄PQ = AB⁄AP = AC⁄AQ
5. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ ⇒ AQ = 2⁄1 × PQ ⇒ AQ = 2 PQ
9. In the above result, '2' is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is '2'
1. Next we will take 60o. Consider the fig.24.11(c) below.
A base triangle ABC is drawn in it. How is it drawn?
• First mark a point C such that AC = 2 cm. Next drop perpendicular BC onto the horizontal leg. Then:
♦ BC will be √3 cm
♦ AB will be 1 cm
• The above two will naturally occur because: √[(√3)2+12] = √[3+1] = √[4] = 2
3. So we established a base triangle. As we saw in figs.24.11(a) and (b), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
BC⁄PQ = AB⁄AP = AC⁄AQ
5. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ ⇒ AQ = 2⁄√3 × PQ
9. In the above result, 2⁄√3 is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is 2⁄√3
1. Finally we will take 45o. Consider the fig.24.11(d) above.
A base triangle ABC is drawn in it. How is it drawn?
• First mark a point B on the horizontal line such that AB = 1 cm. Next erect perpendicular BC upto the slanting leg. Then:
♦ BC will be 1 cm
♦ AC will be √2 cm
• The above two will naturally occur because: √[12+12] = √[1+1] = √2
3. So we established a base triangle. As we saw in figs.24.11(a),(b) and (c), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
BC⁄PQ = AB⁄AP = AC⁄AQ
5. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ ⇒ AQ = √2⁄1 × PQ = √2PQ
9. In the above result, √2 is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is √2
Based on the above problem, we can have a discussion on a similar problem that is commonly encountered in science and engineering.
1. Consider fig.24.13 below. BC is a vertical line. 'A' is another point to the left of BC. This point is joined to B and C. Thus we get a triangle ABC. This is our base triangle.
2. PQ is a line parallel to BC. This line can move towards the left or towards the right
3. • If it moves towards the left, it's distance from vertex A decreases.
• If it moves towards the right, it's distance from vertex A increases.
4. • Also, if it moves towards the left, it's own length PQ decreases
• If it moves towards the right, it's own length PQ increases
5. We have to check whether there is any proportionality between the two quantities:
(i) Distance of the line PQ from A
(ii) Length of the line PQ
6. For that, we add some details to the given fig.24.13(a). The modified fig. is 24.13(b).
• A line AR is drawn through A, perpendicular to BC
• AR will be perpendicular to PQ also
• AR meets BC at D
7. Based on the calculations that we did in the solved example 24.5 above, we can make the following conclusions:
(i) ⊿ADC and ⊿ARQ are similar.
(ii) ⊿ADB and ⊿ARP are similar
8. From 7(i) we get:
CD⁄QR = AD⁄AR = AC⁄AQ
• From the above, we will take the first and second. Because they connect the distance from A and length of QR
CD⁄QR = AD⁄AR ⇒ QR = AR⁄AD × CD
9. From 7(ii) we get:
BD⁄PR = AD⁄AR = AB⁄AP
• From the above, we will take the first and second. Because they connect the 'distance from A' and 'length of PR'
BD⁄PR = AD⁄AR ⇒ PR = AR⁄AD × BD
10. Now we add the results in (8) and (9):
QR + PR = (AR⁄AD × CD) + (AR⁄AD × BD)
⇒ QR + PR = AR⁄AD × (CD +BD)
⇒ PQ = AR⁄AD × BC [∵ (QR + PR) = PQ AND (CD +BD) = BC]
⇒ PQ = BC⁄AD × AR
11. In the above result, BC⁄AD is a constant because ΔABC is the base triangle, whose dimensions are known.
12. So PQ is proportional to AR
• When AR, which is the distance from vertex A increases, the length PQ increases
• When AR decreases, the length PQ decreases
Solution:
1. The ratio is 10:3:12. So, if we divide a sample of calcium carbonate into (10+3+12 =) 25 equal parts, then:
• 10 such equal parts will be calcium
• 3 such equal parts will be carbon
• 12 such equal parts will be oxygen
2. 150 grams of an unknown compound was analysed. Let us divide it into 25 equal parts. Then each part will be 150/25 = 6 grams
• 10 such equal parts = 10 × 6 = 60 grams. Analysis result also shows 60 grams
• 3 such equal parts = 3 × 6 = 18 grams. Analysis result shows 20 grams. So the unknown compound is not calcium carbonate.
• 12 such equal parts = 12 × 6 = 72 grams. Analysis result shows 70 grams. So the unknown compound is not calcium carbonate.
Solved example 24.7
A person invests Rs. 10000 and Rs. 15000 in two different schemes. After one year, he got Rs. 900 as interest for the first amount and Rs 1500 as interest for the second amount.
(i) Are the interests proportional to the investments?
(ii) What is the ratio of the interest to the amount invested in the first scheme? What about the second?
(iii) What is the annual rate of interest for the two schemes?
Solution:
1. We know that the interest (i) obtained is proportional to the principal amount (p)
• When p increases, i increases
• When p decreases, i decreases
• The constant of proportionality is the rate of interest (r)
2. So we get: i = pr. From this we get: r = i/p
• Thus, r for first scheme = 900/10000 = 9/100
♦ This is usually expressed as a percent. So r = 9/100 x 100 = 9%
• r for the second scheme = 1200/15000 = 12/150 = 4/50
♦ In percentage, it is 4/50 x 100 = 8%
3. So the annual rate of interest are:
First scheme - 9%
Second scheme - 8%
This is the solution for part (iii)
4. Since the rates are different, they are not proportional. This is the solution for part (i)
5. Part (ii):
• We need this ratio:
Interest : Amount invested
• Scheme 1:
900 : 10000 = 9 : 100
• Scheme 2:
1200 : 15000 = 12 :150 = 4 : 50 = 8 : 100
In the next section we will see a different type of proportion.
Solved example 24.5
In the angle shown in fig.24.10 below, an object moves along the slanting line. As the distance of the object from the vertex changes, it's height from the horizontal line also changes.
 |
| Fig.24.10 |
(ii) Calculate the constant of proportionality for 30o, 35o and 60o angles
Solution:
1. The given angle in fig.24.10 is reproduced in fig.24.11(a) below. Some modifications are also made:
• The blue object which moves along the slanting line is marked as Q
• A perpendicular is dropped from Q to the horizontal leg of the given angle. The foot of the perpendicular is marked as P
 |
| Fig.24.11 |
2. There is also another right triangle ABC. This is our base triangle. That is., we are going to do the calculations based on ⊿ABC:
• We assume that BC is fixed at it's position.
• Also we assume that all sides of ⊿ABC are known.
3. But QP is not fixed. Because, Q can be at any point along the slanting line.
4. Now we find the relation between ⊿ABC and ⊿APQ:
• Angle at A is denoted as x. It is same for both the triangles
• Angle at P and B are both 90
• Angle at C and Q will both be equal to [180 – (90+x)] = (90-x)
• So the two triangles are have the same angles. They are similar triangles. We can apply theorem 19.2:
5. Side opposite angle x in ABC⁄Side opposite angle x in APQ =
Side opposite angle (90-x) in ABC⁄Side opposite angle (90-x) in APQ =
Side opposite angle 90 in ABC⁄Side opposite angle 90 in APQ
6. This is same as:
BC⁄PQ = AB⁄AP = AC⁄AQ
7. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ
• AC⁄BC is a constant because ⊿ABC is fixed. We can calculate AC⁄BC. So we can write:
• AQ = A constant × PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
Part (ii)
1. In this part we explore the cases when the angle x at vertex A is 30o, 60o and 45o
2. First we will take 30o. Consider the fig.24.11(b). A base triangle ABC is drawn in it. How is it drawn?
• First mark a point C such that AC = 2 cm. Next drop perpendicular BC onto the horizontal leg. Then:
♦ BC will be 1 cm
♦ AB will be √3 cm
• The above two will naturally occur because: √[(√3)2+12] = √[3+1] = √[4] = 2
3. So we established a base triangle. As we saw in fig.(a), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
BC⁄PQ = AB⁄AP = AC⁄AQ
5. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ ⇒ AQ = 2⁄1 × PQ ⇒ AQ = 2 PQ
9. In the above result, '2' is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is '2'
1. Next we will take 60o. Consider the fig.24.11(c) below.
 |
| Fig.24.12 |
• First mark a point C such that AC = 2 cm. Next drop perpendicular BC onto the horizontal leg. Then:
♦ BC will be √3 cm
♦ AB will be 1 cm
• The above two will naturally occur because: √[(√3)2+12] = √[3+1] = √[4] = 2
3. So we established a base triangle. As we saw in figs.24.11(a) and (b), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
BC⁄PQ = AB⁄AP = AC⁄AQ
5. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ ⇒ AQ = 2⁄√3 × PQ
9. In the above result, 2⁄√3 is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is 2⁄√3
1. Finally we will take 45o. Consider the fig.24.11(d) above.
A base triangle ABC is drawn in it. How is it drawn?
• First mark a point B on the horizontal line such that AB = 1 cm. Next erect perpendicular BC upto the slanting leg. Then:
♦ BC will be 1 cm
♦ AC will be √2 cm
• The above two will naturally occur because: √[12+12] = √[1+1] = √2
3. So we established a base triangle. As we saw in figs.24.11(a),(b) and (c), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
BC⁄PQ = AB⁄AP = AC⁄AQ
5. From the above, we will take the first and last. Because they connect the distance and height
BC⁄PQ = AC⁄AQ
8. From this we get:
• AQ = AC⁄BC × PQ ⇒ AQ = √2⁄1 × PQ = √2PQ
9. In the above result, √2 is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is √2
1. Consider fig.24.13 below. BC is a vertical line. 'A' is another point to the left of BC. This point is joined to B and C. Thus we get a triangle ABC. This is our base triangle.
 |
| Fig.24.13 |
3. • If it moves towards the left, it's distance from vertex A decreases.
• If it moves towards the right, it's distance from vertex A increases.
4. • Also, if it moves towards the left, it's own length PQ decreases
• If it moves towards the right, it's own length PQ increases
5. We have to check whether there is any proportionality between the two quantities:
(i) Distance of the line PQ from A
(ii) Length of the line PQ
6. For that, we add some details to the given fig.24.13(a). The modified fig. is 24.13(b).
• A line AR is drawn through A, perpendicular to BC
• AR will be perpendicular to PQ also
• AR meets BC at D
7. Based on the calculations that we did in the solved example 24.5 above, we can make the following conclusions:
(i) ⊿ADC and ⊿ARQ are similar.
(ii) ⊿ADB and ⊿ARP are similar
8. From 7(i) we get:
CD⁄QR = AD⁄AR = AC⁄AQ
• From the above, we will take the first and second. Because they connect the distance from A and length of QR
CD⁄QR = AD⁄AR ⇒ QR = AR⁄AD × CD
9. From 7(ii) we get:
BD⁄PR = AD⁄AR = AB⁄AP
• From the above, we will take the first and second. Because they connect the 'distance from A' and 'length of PR'
BD⁄PR = AD⁄AR ⇒ PR = AR⁄AD × BD
10. Now we add the results in (8) and (9):
QR + PR = (AR⁄AD × CD) + (AR⁄AD × BD)
⇒ QR + PR = AR⁄AD × (CD +BD)
⇒ PQ = AR⁄AD × BC [∵ (QR + PR) = PQ AND (CD +BD) = BC]
⇒ PQ = BC⁄AD × AR
11. In the above result, BC⁄AD is a constant because ΔABC is the base triangle, whose dimensions are known.
12. So PQ is proportional to AR
• When AR, which is the distance from vertex A increases, the length PQ increases
• When AR decreases, the length PQ decreases
• The constant of proportionality is BC⁄AD
Now we will consider a similar problem in circles:
1. Consider fig.24.14(a) below. ABC is a semi circle with centre at "O'. AOC is the diameter. OB is drawn perpendicular to the diameter. OB meets the semicircle at 'B'.
2. Several chords are drawn on the left side of AOC.
Chords nearer to B are shorter.
Chords away from B are longer
3. We have to check whether there is any proportionality between the two quantities:
(i) Distance of the chord from B
(ii) Length of the chord
4. For that, we modify the given fig.24.14(a). The modified fig. is 24.13(b).
• A single chord EF is considered
• It meets OB at D
• EF is parallel to AC. So EF is perpendicular to OB
• OA is the radius 'r'. OE is also equal to 'r'
5. Now we can begin the calculations: Apply Pythagoras theorem to the right triangle OED.
We get: ED = √[OE2 - OD2] = √[r2 - OD2] = √[r2 - (OB-BD)2] = √[r2 - (r-BD)2]
= √[r2 - (r2 - 2rBD + BD2)] = √[r2 - r2 + 2rBD - BD2] = √[2rBD - BD2] = √[BD(2r - BD)].
6. The perpendicular OB from the centre will bisect the chord EF. Details here
So EF = 2ED = 2√[BD(2r - BD)]
7. So we get a relation between two quantities:
(i) The distance BD of the chord from B
(ii) The length of the chord EF
The relation is: EF = 2√[BD(2r - BD)]
8. But on the right side, we are getting square root of BD. So they are not proportional.
In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10:3:12. When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is that compound calcium carbonate?1. Consider fig.24.14(a) below. ABC is a semi circle with centre at "O'. AOC is the diameter. OB is drawn perpendicular to the diameter. OB meets the semicircle at 'B'.
 |
| Fig.24.14 |
Chords nearer to B are shorter.
Chords away from B are longer
3. We have to check whether there is any proportionality between the two quantities:
(i) Distance of the chord from B
(ii) Length of the chord
4. For that, we modify the given fig.24.14(a). The modified fig. is 24.13(b).
• A single chord EF is considered
• It meets OB at D
• EF is parallel to AC. So EF is perpendicular to OB
• OA is the radius 'r'. OE is also equal to 'r'
5. Now we can begin the calculations: Apply Pythagoras theorem to the right triangle OED.
We get: ED = √[OE2 - OD2] = √[r2 - OD2] = √[r2 - (OB-BD)2] = √[r2 - (r-BD)2]
= √[r2 - (r2 - 2rBD + BD2)] = √[r2 - r2 + 2rBD - BD2] = √[2rBD - BD2] = √[BD(2r - BD)].
6. The perpendicular OB from the centre will bisect the chord EF. Details here
So EF = 2ED = 2√[BD(2r - BD)]
7. So we get a relation between two quantities:
(i) The distance BD of the chord from B
(ii) The length of the chord EF
The relation is: EF = 2√[BD(2r - BD)]
8. But on the right side, we are getting square root of BD. So they are not proportional.
Solved example 24.6
Solution:
1. The ratio is 10:3:12. So, if we divide a sample of calcium carbonate into (10+3+12 =) 25 equal parts, then:
• 10 such equal parts will be calcium
• 3 such equal parts will be carbon
• 12 such equal parts will be oxygen
2. 150 grams of an unknown compound was analysed. Let us divide it into 25 equal parts. Then each part will be 150/25 = 6 grams
• 10 such equal parts = 10 × 6 = 60 grams. Analysis result also shows 60 grams
• 3 such equal parts = 3 × 6 = 18 grams. Analysis result shows 20 grams. So the unknown compound is not calcium carbonate.
• 12 such equal parts = 12 × 6 = 72 grams. Analysis result shows 70 grams. So the unknown compound is not calcium carbonate.
Solved example 24.7
A person invests Rs. 10000 and Rs. 15000 in two different schemes. After one year, he got Rs. 900 as interest for the first amount and Rs 1500 as interest for the second amount.
(i) Are the interests proportional to the investments?
(ii) What is the ratio of the interest to the amount invested in the first scheme? What about the second?
(iii) What is the annual rate of interest for the two schemes?
Solution:
1. We know that the interest (i) obtained is proportional to the principal amount (p)
• When p increases, i increases
• When p decreases, i decreases
• The constant of proportionality is the rate of interest (r)
2. So we get: i = pr. From this we get: r = i/p
• Thus, r for first scheme = 900/10000 = 9/100
♦ This is usually expressed as a percent. So r = 9/100 x 100 = 9%
• r for the second scheme = 1200/15000 = 12/150 = 4/50
♦ In percentage, it is 4/50 x 100 = 8%
3. So the annual rate of interest are:
First scheme - 9%
Second scheme - 8%
This is the solution for part (iii)
4. Since the rates are different, they are not proportional. This is the solution for part (i)
5. Part (ii):
• We need this ratio:
Interest : Amount invested
• Scheme 1:
900 : 10000 = 9 : 100
• Scheme 2:
1200 : 15000 = 12 :150 = 4 : 50 = 8 : 100
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