The diagonal of a square is $4\sqrt{2}$ cm. The diagonal of another square whose area is double that of the first square, is

### Concept & Scaling factor When the area of a $2$D figure is multiplied by a factor of $k$, its linear dimensions (like side length, diagonal, and perimeter) are multiplied by the square root of that factor ($\sqrt{k}$). ### Step-by-Step Solution * **Step 1:** Let the diagonal of the first square be $d_1 = 4\sqrt{2} \text{ cm}$. * **Step 2:** Area of the first square $A_1 = \frac{1}{2} d_1^2 = \frac{1}{2}(4\sqrt{2})^2 = \frac{1}{2}(16 \times 2) = 16 \text{ cm}^2$. * **Step 3:** The area of the second square is double the first. So, $A_2 = 2 \times 16 = 32 \text{ cm}^2$. * **Step 4:** Let the diagonal of the second square be $d_2$. We know $A_2 = \frac{1}{2} d_2^2$. * **Step 5:** $32 = \frac{1}{2} d_2^2 \implies d_2^2 = 64$. * **Step 6:** Taking the square root gives $d_2 = 8 \text{ cm}$. ### Exam Strategy & Shortcut Use proportional scaling. If $\text{Area}_2 = 2 \times \text{Area}_1$, then any linear measure of square 2 is $\sqrt{2}$ times the linear measure of square 1. Therefore, $d_2 = d_1 \times \sqrt{2} = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8 \text{ cm}$. This avoids area calculation altogether. ### Common Pitfall A standard error is doubling the diagonal because the area doubled, leading to $8\sqrt{2} \text{ cm}$. Remember that the relationship between area and length is quadratic, not linear. ### Final Answer Therefore, the correct answer is **8 cm**.