The problem asks us to find the smallest of three positive numbers given their ratio and the sum of their squares.

Representing the Numbers

Let the three positive numbers be represented based on the given ratio 3 : 7 : 8. We can express them as $3x$, $7x$, and $8x$, where $x$ is a positive constant.

Setting Up the Equation

The problem states that the sum of the squares of these numbers is 7808. We can write this as an equation:

$ (3x)^2 + (7x)^2 + (8x)^2 = 7808 $

Solving for the Constant $x$

Expand the squares:

$ 9x^2 + 49x^2 + 64x^2 = 7808 $

Combine the terms involving $x^2$:

$ (9 + 49 + 64)x^2 = 7808 $

$ 122x^2 = 7808 $

Now, solve for $x^2$ by dividing both sides by 122:

$ x^2 = \frac{7808}{122} $

$ x^2 = 64 $

Since the numbers are positive, $x$ must be positive. Take the square root of both sides:

$ x = \sqrt{64} $

$ x = 8 $

Finding the Smallest Number

The three numbers are $3x$, $7x$, and $8x$. The smallest number corresponds to the smallest part of the ratio, which is 3.

Calculate the smallest number using $x=8$:

Smallest Number $= 3x = 3 \times 8 = 24$.

Verification (Optional)

The three numbers are:

• $3x = 3 \times 8 = 24$
• $7x = 7 \times 8 = 56$
• $8x = 8 \times 8 = 64$

Check the sum of their squares:

$ 24^2 + 56^2 + 64^2 = 576 + 3136 + 4096 = 7808 $

The sum matches the given value, confirming our calculation.

Conclusion

The smallest of the three numbers is 24.