Question

a square has a side length of 10 units. mckinney dilates the square using \\( \frac { 2 } { 5 } \\) as her scale factor. then, she writes the area of the dilated square as a fraction of the area of the original square. if she writes the fraction in lowest terms, what fraction does she write? (1 point) \\( \frac { 1 } { 25 } \\) \\( \frac { 16 } { 100 } \\) \\( \frac { 16 } { 25 } \\) \\( \frac { 4 } { 25 } \\)

Explanation:

Step1: Calculate the area of the original square

The area formula of a square is \(A = s^2\), where \(s\) is the side - length. For the original square with \(s = 10\) units, \(A_{original}=10^2=100\) square units.

Step2: Calculate the side - length of the dilated square

If the scale factor is \(k=\frac{2}{5}\), and the original side - length \(s = 10\) units, then the side - length of the dilated square \(s_{dilated}=k\times s=\frac{2}{5}\times10 = 4\) units.

Step3: Calculate the area of the dilated square

Using the area formula \(A = s^2\) again, for the dilated square with \(s = 4\) units, \(A_{dilated}=4^2 = 16\) square units.

Step4: Find the fraction of the areas

The fraction \(\frac{A_{dilated}}{A_{original}}=\frac{16}{100}\). Simplify \(\frac{16}{100}\) by dividing both the numerator and the denominator by their greatest common divisor. The GCD of 16 and 100 is 4. \(\frac{16\div4}{100\div4}=\frac{4}{25}\).

Another way:
If the scale factor of the side - lengths of two similar figures (a square is a regular polygon, and two squares are similar) is \(k\), the scale factor of their areas is \(k^{2}\). Here \(k = \frac{2}{5}\), so the scale factor of the areas \(k^{2}=(\frac{2}{5})^{2}=\frac{2^{2}}{5^{2}}=\frac{4}{25}\)

Answer:

\(\frac{4}{25}\) (the fourth option)