Question

ABCD and EFGC are squares and the curve y = k√x passes through the origin D and the points B and F. The ratio $\frac{FG}{BC}$ is -

• A. $\frac{\sqrt{5}+1}{2}$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $\frac{\sqrt{5}+1}{4}$
• D. $\frac{\sqrt{3}+1}{4}$

Answer 1

Answer: (A)

$\frac{\sqrt{5}+1}{2}$

Answer 2

Let the side length of square ABCD be $s_1$ and the side length of square EFGC be $s_2$. Since D is the origin (0,0) and ABCD is a square with C on the x-axis, the coordinates are D(0,0), C($s_1$, 0), B($s_1$, $s_1$), and A(0, $s_1$). Thus, BC = $s_1$. Since EFGC is a square with GC on the x-axis and G to the right of C, the coordinates are G($s_1+s_2$, 0), F($s_1+s_2$, $s_2$), and E($s_1$, $s_2$). Thus, FG = $s_2$.

The curve $y = k\sqrt{x}$ passes through D(0,0), B($s_1$, $s_1$), and F($s_1+s_2$, $s_2$). Since B($s_1$, $s_1$) is on the curve: $s_1 = k\sqrt{s_1}$ Squaring both sides, $s_1^2 = k^2 s_1$. Since $s_1 > 0$, we get $s_1 = k^2$, which means $k = \sqrt{s_1}$ (assuming $k>0$ from the graph).

Since F($s_1+s_2$, $s_2$) is on the curve: $s_2 = k\sqrt{s_1+s_2}$ Substitute $k = \sqrt{s_1}$: $s_2 = \sqrt{s_1}\sqrt{s_1+s_2}$ Squaring both sides: $s_2^2 = s_1(s_1+s_2)$ $s_2^2 = s_1^2 + s_1 s_2$ Rearranging the terms, we get a quadratic equation in $s_2$: $s_2^2 - s_1 s_2 - s_1^2 = 0$

We need to find the ratio $\frac{FG}{BC} = \frac{s_2}{s_1}$. Let $r = \frac{s_2}{s_1}$. Divide the equation by $s_1^2$: $(\frac{s_2}{s_1})^2 - \frac{s_1 s_2}{s_1^2} - \frac{s_1^2}{s_1^2} = 0$ $r^2 - r - 1 = 0$

Using the quadratic formula to solve for $r$: $r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$ $r = \frac{1 \pm \sqrt{1+4}}{2}$ $r = \frac{1 \pm \sqrt{5}}{2}$

Since $s_1$ and $s_2$ are lengths, they are positive, so their ratio $r$ must be positive. Therefore, we take the positive root: $r = \frac{1+\sqrt{5}}{2}$

The ratio $\frac{FG}{BC} = \frac{s_2}{s_1} = \frac{1+\sqrt{5}}{2}$.