The problem asks us to find two numbers based on their mean proportional and third proportional. Let the two unknown numbers be \(a\) and \(b\).

Understanding Mean and Third Proportional

Before solving, let's define the terms:

• Mean Proportional: For two positive numbers \(a\) and \(b\), the mean proportional (\(m\)) is the number such that \(a:m = m:b\). This implies \(\frac{a}{m} = \frac{m}{b}\), which simplifies to \(m^2 = ab\). So, the mean proportional is \(m = \sqrt{ab}\).
• Third Proportional: For two numbers \(a\) and \(b\), the third proportional (\(t\)) is the number such that \(a:b = b:t\). This implies \(\frac{a}{b} = \frac{b}{t}\), which simplifies to \(at = b^2\). So, the third proportional is \(t = \frac{b^2}{a}\).

Setting Up the Equations

We are given that the mean proportional of the two numbers \(a\) and \(b\) is 6. Using the definition:

\(\sqrt{ab} = 6\)

Squaring both sides, we get the first equation:

\(ab = 36 \quad (1)\)

We are also given that the third proportional of the two numbers \(a\) and \(b\) is 20.25. Using the definition \(\frac{b^2}{a} = t\):

\(\frac{b^2}{a} = 20.25 \quad (2)\)

Now we have a system of two equations with two variables \(a\) and \(b\).

Solving the System of Equations

From equation (1), we can express \(a\) in terms of \(b\) (or \(b\) in terms of \(a\)). Let's express \(a\):

\(a = \frac{36}{b}\)

Substitute this expression for \(a\) into equation (2):

\(\frac{b^2}{\frac{36}{b}} = 20.25\)

Simplify the left side:

\(\frac{b^2 \cdot b}{36} = 20.25\)

\(\frac{b^3}{36} = 20.25\)

Now, multiply both sides by 36 to solve for \(b^3\):

\(b^3 = 20.25 \times 36\)

Let's calculate the product \(20.25 \times 36\):

\(20.25 \times 36 = \frac{2025}{100} \times 36 = \frac{81 \times 25}{100} \times 36 = \frac{81 \times 25}{4 \times 25} \times 36 = 81 \times \frac{36}{4} = 81 \times 9 = 729\)

So, \(b^3 = 729\).

To find \(b\), we take the cube root of 729:

\(b = \sqrt[3]{729}\)

We know that \(9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729\). Therefore,

\(b = 9\)

Now substitute the value of \(b\) back into the equation \(a = \frac{36}{b}\) to find \(a\):

\(a = \frac{36}{9}\)

\(a = 4\)

The two numbers are 4 and 9.

Verifying the Solution

Let's check if these numbers satisfy the original conditions:

• Mean Proportional: \(\sqrt{4 \times 9} = \sqrt{36} = 6\). This matches the given mean proportional.
• Third Proportional: The third proportional of 4 and 9 is \(\frac{9^2}{4} = \frac{81}{4} = 20.25\). This matches the given third proportional.

Both conditions are satisfied by the numbers 4 and 9.

Checking Other Options

Let's quickly check the other options to confirm our answer.

As the table shows, only the numbers 4 and 9 satisfy both conditions.

Conclusion

The two numbers such that their mean proportional is 6 and third proportional is 20.25 are 4 and 9.

Revision Table: Mean and Third Proportional

Additional Information on Proportions

Proportion is a statement that two ratios are equal. If \(a:b = c:d\), it is read as "\(a\) is to \(b\) as \(c\) is to \(d\)", and the relationship is \(\frac{a}{b} = \frac{c}{d}\). In this proportion, \(a\) and \(d\) are called the extremes, and \(b\) and \(c\) are called the means.

A continued proportion is a sequence of numbers where the ratio between any two consecutive terms is constant. For example, \(a, b, c\) are in continued proportion if \(a:b = b:c\). In this case, \(b\) is the mean proportional between \(a\) and \(c\), and \(c\) is the third proportional to \(a\) and \(b\).

Understanding these basic definitions is crucial for solving problems involving ratios and proportions in mathematics.