Rectangle ABCD: Finding Diagonal Length m with Side Ratio √(x/2)

Start by writing the ratio as $\frac{x}{BC} = \frac{\sqrt{x}}{\sqrt{2}}$. Then cross-multiply to get $\sqrt{2} \cdot x = \sqrt{x} \cdot BC$, and solve for BC.

We divide both sides by $\sqrt{x}$ to isolate BC. Remember: $\frac{\sqrt{2} \cdot x}{\sqrt{x}} = \sqrt{2} \cdot \sqrt{x}$ because x can be written as $\sqrt{x} \cdot \sqrt{x}$.

Since ABCD is a rectangle, triangle ABC is a right triangle. The diagonal AC is the hypotenuse, so we use: $AB^2 + BC^2 = AC^2$ or $x^2 + (\sqrt{2x})^2 = m^2$.

When you square a square root, they cancel out! So $(\sqrt{2x})^2 = 2x$. This is why our final equation becomes $x^2 + 2x = m^2$.

Yes! Try x = 4: then $BC = \sqrt{2 \cdot 4} = \sqrt{8} = 2\sqrt{2}$. Check the ratio: $\frac{4}{2\sqrt{2}} = \sqrt{2} = \sqrt{\frac{4}{2}}$ ✓