Chapter 24.5 - Modifications to achieve proportions

In the previous section we saw some solved examples on proportionality between two quantities. In this section we will see a different type of proportion.

1. Consider a regular polygon. We know how to calculate the sum of all it’s interior angles. 
2. The formula is s = 180(n-2). 
• Where s is the sum 
• n is the number of sides of the regular polygon.
3. Let us use this formula for a triangle:
• For a triangle, n = 3
• So s = 180 × (3-2) = 180 × 1 = 180
4. Let us use the formula for a square:
• For a square, n = 4
• So s = 180 × (4-2) = 180 × 2 = 360
5. Let us use the formula for a pentagon:
• For a pentagon, n = 5
• So s = 180 × (5-2) = 180 × 3 = 540
6. Let us tabulate the results:

From the table we can see that s/n is not a constant. So s is not proportional to n.
7. Let us modify the formula a little:
Let s = 180 × m
• Where s is the sum
• m = (n-2) 
• n is the number of sides of the regular polygon.
Now the tabulation will be as shown below:

We can see that s is proportional to m. The constant of proportionality is 180.
8. In ordinary language, we can say this:
The sum of interior angles of a regular polygon is proportional to ‘2 less than the number of sides’.

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There are many examples where proportionality can be achieved by making modifications to one quantity. Let us see another example:
■ We have seen that the area of a square is not proportional to it’s side. Details here. The table that we saw is shown here again.

 • We know that the area of a square is not proportional to it’s side because a/s ratio is not a constant.
• But if we put p = s², a new table can be formed:

• We can see that a is proportional to p. The constant of proportionality is 1.
• In ordinary language, we can say this:
Area of a square is proportional to the square of it’s side.

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We will now see some solved examples:
Solved example 24.8
For circles, is the area proportional to the square of the radius? If so, what is the constant of proportionality?
Solution:
1. We have seen that the area of a circle is not proportional to it’s radius. Details here. 
But the area may be proportional to the 'square of the radius'. Let us try:
2. We know that area a = πr²
• Put r² = q
• Then a = πq
    ♦ π is a constant
    ♦ when q increases a increases
    ♦ When q decreases a decreases
3. So a is proportional to q. That means, a is proportional to the square of the radius.
4. The constant of proportionality is π.

Solved example 24.9
For equilateral triangles, is the area proportional to the square of the side? If so, what is the constant of proportionality?
Solution:
1. We know that area of an equilateral triangle is given by a = (^√3⁄₄)s² Details here.
Where s is the length of side.
• Put s² = q
• Then a = (^√3⁄₄)q
    ♦ ^√3⁄₄ is a constant
    ♦ when q increases a increases
    ♦ When q decreases a decreases
2. So a is proportional to q. That means, a is proportional to the square of the side.
3. The constant of proportionality is ^√3⁄₄.

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So we find that, in some cases, proportionality can be achieved by modifying one quantity. In the next section we will see Inverse proportions.

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Chapter 24.4 - More solved examples on Proportionality

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.

Fig.24.10

(i) Prove that height is proportional to the distance
(ii) Calculate the constant of proportionality for 30^o, 35^o and 60^o 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

• 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 30^o, 60^o and 45^o
2. First we will take 30^o. 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 60^o. Consider the fig.24.11(c) below. 

Fig.24.12

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 45^o. 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

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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. 

Fig.24.13

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

• The constant of proportionality is ^BC⁄_AD

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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'.    

Fig.24.14

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.

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Solved example 24.6

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?
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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In the next section we will see a different type of proportion.

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Chapter 24.3 - Solved examples on Proportionality

In the previous section we saw some cases where there is proportionality between two quantities. In this section we will see some solved examples.

Solved example 24.2
For each pair of quantities given below, check whether the first is proportional to the second. For the proportional quantities, calculate the constant of proportionality.
(a) Perimeter and radius of circles
(b) Area and radius of circles
(c) Distance travelled and the number of rotations of a circular ring along a line
(d) Interest got in an year and the amount deposited in a scheme. Interest is compounded annually
(e) Volume of water poured into a hollow prism and the height of the water level
Solution:
(a) The two quantities are:
(i) Perimeter of the circle
(ii) Radius of the circle
1. Let the perimeter be 'p' and radius be 'r'
2. We know that p = 2πr
3. We can put whatever value we like for r on the right side of the above equation. So r is a variable
4. What ever value we give for r, that value will be multiplied by a constant 2π. So p will always be proportional to r
5. The constant of proportionality is ^p⁄_r = 2π
(b) The two quantities are:
(i) Area of the circle
(ii) Radius of the circle
1. Let the area be 'a' and the radius be 'r'
2. We know that a = πr² = π × r × r
3. We can put whatever value we like for r on the right side of the above equation. So r is a variable.
4. Whatever value we give for r, that value will be multiplied by a 'constant π' and the 'variable r'. That means, r is not multiplied by the same constant every time. 
5. So a and r are not proportional
(c) The two quantities are:
(i) Distance travelled
(ii) Number of rotations
1. Let the distance travelled be 'd' and the number of rotations be 'n'
2. We know this: The distance travelled in one rotation = 2πr. (where r is the radius of the ring) Details here. So distance travelled in n rotations = d = 2nπr
3. We can put whatever value we like for n on the right side of the above equation. So n is a variable. But r is a constant because, we are considering the movement of a single ring.
4. Whatever value we give for n, that value will be multiplied by a constant 2πr. So d will always be proportional to n
5. The constant of proportionality is ^d⁄_n = 2πr. 
Note that, the constant of proportionality is the perimeter of the ring under consideration.
(d) The two quantities are:
(i) Interest got in an year
(ii) Amount deposited
1. Let the interest got in an year be 'i' and the amount deposited be 'p'
2. We know that i = p×r (where r is the rate of interest)
3. We can put whatever value we like for p on the right side of the above equation. So p is a variable. But r is a constant for a scheme
4. Whatever value we give for p, that value will be multiplied by a constant r. So i will always be proportional to p
5. The constant of proportionality is ⁱ⁄_p = r
(e) The two quantities are:
(i) Volume of water poured
(ii) Height of water level
1. Let volume be 'v' and the height of the water level from the base be 'h'
2. Let the base area of the prism be 'a'
3. The poured volume = volume of the water column formed in the prism
4. But volume of water column in the prism = base area × height = ah
5. So we get v = ah ⇒ h = ^v⁄_a
6. We can put whatever value we like for v on the right side of the above equation. So v is a variable. But a is a constant for a prism
7. Whatever value we give for v, that value will be multiplied by a constant ¹⁄_a. So h will always be proportional to p
8. The constant of proportionality is ^h⁄_v = ¹⁄_a

Solved example 24.3
During rainfall, the volume of water falling in each square metre may be considered equal. 
(a) Prove that, the volume of water falling in a region is proportional to the area of that region
(b) Explain why the heights of rain water collected in different sized hollow prisms kept near one another are equal.
Solution:
1. Consider the blue surface in fig.24.7 below. It is a surface on the ground. 
2. On this surface, we can mark a random number of squares, each of side 1 m. So each of such squares will have an area of 1 m². Three such squares are marked. 

Fig.24.7

3. It is stated in the question that, “the volume of water falling in each square metre may be considered equal”. 
4. So each of the three squares will receive equal volume of water. 
5. If this water is not allowed to run off and if there is no seepage into the ground, the water received will become a water column. There will be 3 water columns as shown in the fig.24.7.
6. These water columns are prisms with square base. And they have the same volume (v), because, according to (3), volume of water are the same. 
7. We know that volume = base area × height. But base area of all prisms = 1 m². 
So we can write: v = 1 × h ⇒ v = h
8. But v is same for all the three prisms. 
■ So h will also be same for all the three prisms

Now we will first consider part (ii) of the question:
9. Fig.24.8 below shows two hollow prisms kept near to one another. They are shown in yellow colour. They have different base areas.

Fig.24.8

10. We can think of random number of water columns inside these prisms. Let the base of these columns be squares of area 1 cm². 
11. Then each water column will be a square prism of base area 1 cm². Two such water prisms are shown inside each of the yellow prisms. 
12. We have seen in (8) that the heights of all the water prisms will be equal. Regardless of whether they are in the first yellow prism or the second yellow prism. Let this height be ‘h’. 
13. Let the base area of first yellow prism be a1
Let the base area of second yellow prism be a2
14. Then total number of water prisms (each with base area 1 cm²) in first yellow prism = a1/1 = a1
Total number of water prisms (each with base area 1 cm 2) in second yellow prism = a2/1 = a2
15. All the a1 number of water prisms in the first yellow prism will have the same height ‘h’
All the a2 number of water prisms in the second yellow prism will also have the same height ‘h’
That means, the water level in both the yellow prisms will be ‘h’
16. So, whatever number of yellow hollow prisms (of different base sizes) we place near one another, after the rainfall, the height of water in all will be the same.

Now we take up the part (i):
17. Consider two paddy fields in a locality. Let their areas be a1 and a2. If the water is not allowed to run off and if there is no seepage into the ground, there will be two water prisms. Each will cover the entire area of the respective field. From (8), both water prisms will have the same height ‘h’. 
18. The volume of water in the first field = v1 = a1 h
The volume of water in the second field = v2 = a2 h
19. ‘h’ is a constant. So, if area increases, volume increases
if area decreases, volume decreases
20. That means volume is proportional to the area.

Solved example 24.4
When a weight is suspended by a spring, the extension is proportional to that weight. Explain how this property can be used to make markings on a spring balance.
Solution:
1. Fig.24.9 below shows the spring used inside a spring balance.

Fig.24.9

 • Position 0 shows the situation when no load is applied on the spring. 
• Position 1 shows the situation when a load of w₁ is applied on the spring. 
    ♦ We can see that there is an extension of x₁ cm from the initial position. 
• Position 2 shows the situation when a load of w₂ is applied on the spring. 
    ♦ We can see that there is an extension of x₂ cm from the initial position.
2. It is given in the question that, the extension is proportional to the applied load.
3. So any extension x will be proportional to the weight w that produces that extension. 
We can write: x = a constant 'k' × w ⇒ x = kw ⇒ k = ^x⁄_w
4. The constant 'k' is called the spring constant. Every spring will have a particular value of spring constant. We want to find this constant for our spring.
5. For that, we adopt the following procedure:
(i) Put a known weight w₁. Measure the extension x₁. Then k = ^x₁ ⁄_w1
(ii) Put another known weight w2. Measure the extension x2. Then k = ^x₂ ⁄_w2
(iii) Since k is a constant, we will get the same value for k in both (i) and (ii)
(iv) Repeat the trial with several known weights. In all cases we will get the same k
6. Once k is obtained, we can make the markings on the spring balance. 
(i) The spring is fixed inside an outer casing. Markings are made on this casing.
(ii) When the spring is at zero load position, mark that position on the casing as '0 kg'
(iii) Now we want to mark the 1 kg position.
• Let the extension for a load of 1 kg be 'x_{1 kg}'
• x_{1 kg} = k × 1 kg = k. We have already calculated k. So we will get 'x_{1 kg}'
• Measure this 'x_{1 kg}' from the '0 kg' mark on the casing. And mark it as 1 kg  
(iv) Now we want to mark the 2 kg position.
• Let the extension for a load of 1 kg be 'x_{2 kg}'
• x_{2 kg} = k × 2 kg = 2k. We have already calculated k. So we will get 'x_{2 kg}'
• Measure this 'x_{2 kg}' from the '0 kg' mark on the casing. And mark it as 2 kg
(v) In this way mark 3 kg, 4 kg, 5 kg etc., Fractions between them can also be marked in this way.
We will learn more details about the procedure in higher classes.

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In the next section we will see a few more solved examples.

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Chapter 24.2 - Examples of Proportionality

In the previous section we saw proportionality in the case of A0, A1, A2 series of writing papers. In this section we will see Proportionality in some more cases.

Let us write a summary of what we have discussed so far in this chapter:
■ The ‘3:2 rectangle family’ has all the rectangles whose length is always ³⁄₂ times the width
• That means length is always proportional to the width
   ♦ ³⁄₂ is the constant of proportionality
■ The ‘16:9 rectangle family’ has all the rectangles whose length is always ¹⁶⁄₉ times the width
• That means length is always proportional to the width
    ♦ ¹⁶⁄₉ is the constant of proportionality
■ The ‘√2:1 rectangle family’ has all the rectangles whose length is always √2 times the width
• That means length is always proportional to the width
    ♦ √2 is the constant of proportionality

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Now we will continue our discussion:
1. Consider a square of side 's' = 1 cm. It’s perimeter 'p' will be 4×1 = 4 cm
2. Let us change the side and see how it affects the perimeter:
• let s be 2 cm. Now p becomes 4×2 = 8 cm 
• let s be 2.5 cm. Now p becomes 4×2.5 = 10 cm
• let s be 0.5 cm. Now p becomes 4×0.5 = 2 cm
3. We will write the above results in a tabular form, and calculate the ^p⁄_s ratio in each case:

4. From the table it is clear that, ^p⁄_s is a constant. It’s value is 4
5. So ^p⁄_s = 4 ⇒ p = 4s
• Perimeter is always a constant times the side
• That means perimeter is proportional to the side.
• And 4 is the constant of proportionality
We can use the result in (5) as an equation to find p for any value of s

Another case:
1. Consider a square of side ‘s’= 1 cm. The length of it’s diagonal ‘d’ will be √2. See fig.16.5
[It is a simple application of the Pythagoras theorem. In this way we can find the diagonal of any given square] 
2. Let us change the side and see how it affects the diagonal:
• let s be 2 cm. Now d becomes  = √[2²+2²] = √[2(2²)] = 2√2 cm 
• let s be 2.5 cm. Now p becomes  = √[2.5²+2.5²] = √[2(2.5²)] = 2.5√2 cm 
• let s be 3.25 cm. Now p becomes  = √[3.25²+3.25²] = √[2(3.25²)] = 3.25√2 cm 
• let s be 0.3 cm. Now p becomes  = √[0.3²+0.3²] = √[2(0.3²)] = 0.3√2 cm 
3. We will write the above results in a tabular form, and calculate the d/s ratio in each case:

4. From the table it is clear that, ^d⁄_s is a constant. It’s value is √2
5. So ^d⁄_s = √2 ⇒ d = √2s
• Diagonal is always a constant times the side
• That means diagonal is proportional to the side.
• And √2 is the constant of proportionality
We can use the result in (5) as an equation to find d for any value of s

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In the above discussion, we were considering squares. We considered:

• Perimeter of squares • Diagonal of squares
Now we consider: • Area of squares
1. Consider a square of side ‘s’= 1 cm. It’s area ‘a’ will be 1² = 1 cm²
2. Let us change the side and see how it affects the area:
• let s be 2 cm. Now a becomes 2² = 4 cm²
• let s be 2.25 cm. Now a becomes 2.25² = 5.0625 cm²
• let s be 3 cm. Now a becomes 3² = 9 cm²
• let s be 0.4 cm. Now a becomes 0.4² = 0.16 cm²
3. We will write the above results in a tabular form, and calculate the a/s ratio in each case:

4. From the table, we can see that ^a⁄_s is not a constant.
We will not get a by multiplying s by a fixed number. So a is not proportional to s

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Now we will consider some examples in physics.

1. Consider an object moving at a steady speed of 10 m/s.
2. Since the speed is steady, it will travel a distance ‘d’ of 10 m in a time ‘t’ of one second
• In a time t of 2 seconds it will travel a distance ‘d’ of 2×10 = 20 m
• In a time t of 2.5 seconds it will travel a distance ‘d’ of 2.5×10 = 25 m
• In a time t of 4.25 seconds it will travel a distance ‘d’ of 4.25×10 = 42.5 m
• In a time t of 3.6 seconds it will travel a distance ‘d’ of 3.6×10 = 36 m
3. We will write the above results in a tabular form, and calculate the ^d⁄_t ratio in each case:

4. From the table it is clear that, ^d⁄_t is a constant. It’s value is 10
5. So ^d⁄_t = 10 ⇒ d = 10t
• Distance is always 'a constant × t'
• That means distance is proportional to the time.
• And 10 is the constant of proportionality. Note that, this 'constant of proportionality' is the 'steady speed'
We can use the result in (5) as an equation to find d for any value of t

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In the above example we saw an object which is moving at a steady speed. Now we will consider an object which is moving at a varying speed. 

1. An 'object dropped from a height' gives an example for such a motion. It’s speed will go on increasing. In physics classes we have derived the formula to calculate the distance travelled by such a object. Let us see the details:
• At the instance when the object is dropped, the distance travelled by it is zero
• After a time of t seconds, the distance travelled by it (from the spot where it is dropped) is given by 4.9t² metres
2. So, after 1 second, it will be at a distance of 4.9×1² = 4.9 m from the spot where it is dropped
• After 2 seconds, it will be at a distance of 4.9×2² = 19.6 m from the spot where it is dropped   
• After 2.5 seconds, it will be at a distance of 4.9×2.5² = 30.625 m from the spot where it is dropped
• After 4 seconds, it will be at a distance of 4.9×4² = 78.4 m from the spot where it is dropped
• After 4.2 seconds, it will be at a distance of 4.9×4.2² = 86.436 m from the spot where it is dropped
We will write the above results in a tabular form, and calculate the ^d⁄_t ratio in each case:

3. From the table, we can see that ^d⁄_t is not a constant.
We will not get d by multiplying t by a fixed number. So d is not proportional to t
4. In the earlier case we got ^d⁄_t as a constant. This is because, in that case, the object was moving with a steady speed. But in the present case, the speed is increasing.

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Another example from physics:

1. The density of a material is 12 kg/m3.
2. The mass 'm' of  volume 'v' 1 m³ of that material will be 1×12 = 12 kg
• The mass 'm' of volume 'v' 2 m³ of that material will be 2×12 = 24 kg
• The mass 'm' of  volume 'v' 15 m³ of that material will be 1.5×12 = 18 kg
• The mass 'm' of  volume 'v' 2.25 m³ of that material will be 2.25×12 = 27 kg
3. We will write the above results in a tabular form, and calculate the ^m⁄_v ratio in each case:

4. From the table it is clear that, ^m⁄_v is a constant. It’s value is 12
5. So ^m⁄_v = 12 ⇒ m = 12v
• Mass is always 'a constant × volume'
• That means mass is proportional to the volume.
• And 12 is the constant of proportionality. Note that, this 'constant of proportionality' is the 'density of the material'
We can use the result in (5) as an equation to find m for any value of v

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■ So we have considered a number of cases. In all those cases, there are two quantities.
• In some cases one quantity is proportional to the other. Some examples that we saw in this category are:
    ♦ In the ‘16:9 rectangle family’, the the length is always 16/9 times the width. The constant of proportionality is 16/9
    ♦ In the ‘family of squares’, the diagonal is always 2 times the side. The constant of proportionality is 2. (Note that, all squares belong to the ‘1:1 rectangle family’)
    ♦ The distance travelled by an object moving at a steady speed is always: the 'steady speed' times the time. The constant of proportionality is the ‘steady speed’ 
• In the rest of the cases, neither quantity is proportional to the other. Some examples that we saw in this category are:
    ♦ In the family of squares, the area is not proportional to the side
    ♦ For a freely falling body, the distance travelled is not proportional to the time

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In this chapter we consider those cases where one quantity is proportional to the other. In the next section we will see some solved examples.

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