Inverse Proportion

Inverse proportionality describes a situation in which, when one term is multiplied by a certain number of times, the second term is decreased by the same number of times. This also occurs in reverse, if one term decreases by a certain number of times, the second term increases by the same number of times.

Let's see an example to illustrate this concept.

Given the following table:

We see two values, $X$ and $Y$ . It can be very clearly seen that, when the value of $X$ increases by $2$ , the value of $Y$ also decreases $2$ times. Therefore, it can be said that there is inverse proportionality here.

Detailed explanation

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Test yourself on ratio!

There are two circles.

The length of the diameter of circle 1 is 4 cm.

The length of the diameter of circle 2 is 10 cm.

How many times larger is the area of circle 2 than the area of circle 1?

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Let's look at an example from everyday life

Imagine traveling in some vehicle while the roads are quite empty, without any traffic jams.

As you travel more kilometers and more time passes, the amount of gasoline decreases.

We can say that, as the distance increases the amount of gasoline decreases.

Let's see a graphical representation of inverse proportionality:

The function: $Y=\frac{a}{X}$

represents inverse proportionality.

As the $X$ increases the $Y$ decreases.

How can we check if there is inverse proportionality?

To find out if there is inverse proportionality, we will examine if, when one term is multiplied by a certain amount of times, the second term is decreased by the same amount of times.

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Question 1

There are two circles.

The length of the radius of circle 1 is 6 cm.

The length of the diameter of circle 2 is 12 cm.

How many times greater is the area of circle 2 than the area of circle 1?

Question 2

There are two circles.

One circle has a radius of 4 cm, while the other circle has a radius of 10 cm.

How many times greater is the area of the second circle than the area of the first circle?

Question 3

How many times longer is the radius of the red circle than the radius of the blue circle?

Let's look at an example

Given the following table:

Let's check if every time $X$ increases by a specific amount, $Y$ also decreases in the same proportion.

If this occurs, it means there is inverse proportionality. If not, then there isn't.

Let's ask:

By how much does $X$ increase from $5$ to $10$ ?

The answer is it multiplies by $2$ .

And by how much does $Y$ decrease from $20$ to $10$ ?

The answer is it divides by $2$ .

Let's continue,

By how much does $X$ increase from $5$ to $20$ ? The answer is it multiplies by $4$ .

And by how much does $Y$ decrease from $20$ to $5$ ?

The answer is it divides by $4$ .

We will continue examining and discover that indeed every time $X$ multiplies by a certain number, $Y$ also decreases divided by the same number.

We will see it in the following way:

Examples and exercises with solutions on inverse proportionality

Exercise #1

Given the rectangle ABCD

AB=X the ratio between AB and BC is equal to $\sqrt{\frac{x}{2}}$

We mark the length of the diagonal $A$ with $m$

Check the correct argument:

Video Solution

Step-by-Step Solution

Let's find side BC

Based on what we're given:

$\frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}$

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$\sqrt{2}x=\sqrt{x}BC$

Let's divide by square root x:

$\frac{\sqrt{2}\times x}{\sqrt{x}}=BC$

$\frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC$

Let's reduce the numerator and denominator by square root x:

$\sqrt{2}\sqrt{x}=BC$

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

$AB^2+BC^2=AC^2$

Let's substitute what we're given:

$x^2+(\sqrt{2}\sqrt{x})^2=m^2$

$x^2+2x=m^2$

Answer

$x^2+2x=m^2$

Exercise #2

Given the rectangle ABCD

AB=X

The ratio between AB and BC is $\sqrt{\frac{x}{2}}$

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

Video Solution

Step-by-Step Solution

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$

Answer

$m^2+1=(x+1)^2$

Exercise #3

There are two circles.

The length of the diameter of circle 1 is 4 cm.

The length of the diameter of circle 2 is 10 cm.

How many times larger is the area of circle 2 than the area of circle 1?

Video Solution

Answer

$6\frac{1}{4}$

Exercise #4

There are two circles.

The length of the radius of circle 1 is 6 cm.

The length of the diameter of circle 2 is 12 cm.

How many times greater is the area of circle 2 than the area of circle 1?

Video Solution

Answer

They are equal.

Exercise #5

There are two circles.

One circle has a radius of 4 cm, while the other circle has a radius of 10 cm.

How many times greater is the area of the second circle than the area of the first circle?

Video Solution

Answer

$6\frac{1}{4}$

Do you know what the answer is?

Question 1

How many times longer is the radius of the red circle, which has a diameter of 24, than the radius of the blue circle, which has a diameter of 12?

Question 2

How many times longer is the radius of the red circle than the radius of the blue circle?

Question 3

How many times longer is the radius of the red circle (14 cm) than the radius of the blue circle, which has a diameter of 7?

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