1. **Problem statement:** Three rods k, l, and m of equal length are fixed at a common point. Rods k and m are perpendicular, and rod l lies between them. Given: - Distance between free ends of rods k and l is $5\sqrt{2}$ cm. - Distance between free ends of rods l and m is 7 cm. Find the distance between the free ends of rods k and m. 2. **Setup and formula:** Let the length of each rod be $r$. Place the rods in coordinate plane with the common point at origin. - Rod k along x-axis: endpoint at $(r,0)$ - Rod m along y-axis: endpoint at $(0,r)$ - Rod l makes an angle $\theta$ with rod k, endpoint at $(r\cos\theta, r\sin\theta)$ Distance between free ends of rods k and l: $$d_{kl} = \sqrt{(r - r\cos\theta)^2 + (0 - r\sin\theta)^2} = r\sqrt{(1 - \cos\theta)^2 + \sin^2\theta}$$ Distance between free ends of rods l and m: $$d_{lm} = \sqrt{(r\cos\theta - 0)^2 + (r\sin\theta - r)^2} = r\sqrt{\cos^2\theta + (\sin\theta - 1)^2}$$ 3. **Simplify distances:** $$d_{kl} = r\sqrt{1 - 2\cos\theta + \cos^2\theta + \sin^2\theta} = r\sqrt{2 - 2\cos\theta} = r\sqrt{2(1 - \cos\theta)}$$ $$d_{lm} = r\sqrt{\cos^2\theta + 1 - 2\sin\theta + \sin^2\theta} = r\sqrt{2 - 2\sin\theta} = r\sqrt{2(1 - \sin\theta)}$$ 4. **Use given distances:** $$d_{kl} = 5\sqrt{2} = r\sqrt{2(1 - \cos\theta)}$$ $$d_{lm} = 7 = r\sqrt{2(1 - \sin\theta)}$$ Divide both equations by $\sqrt{2}$: $$5 = r\sqrt{1 - \cos\theta}$$ $$\frac{7}{\sqrt{2}} = r\sqrt{1 - \sin\theta}$$ 5. **Square both sides:** $$25 = r^2(1 - \cos\theta)$$ $$\frac{49}{2} = r^2(1 - \sin\theta)$$ 6. **Express $\cos\theta$ and $\sin\theta$ in terms of $r^2$:** $$1 - \cos\theta = \frac{25}{r^2} \Rightarrow \cos\theta = 1 - \frac{25}{r^2}$$ $$1 - \sin\theta = \frac{49}{2r^2} \Rightarrow \sin\theta = 1 - \frac{49}{2r^2}$$ 7. **Use identity $\sin^2\theta + \cos^2\theta = 1$:** $$\left(1 - \frac{49}{2r^2}\right)^2 + \left(1 - \frac{25}{r^2}\right)^2 = 1$$ Expand: $$\left(1 - \frac{49}{2r^2}\right)^2 = 1 - 2 \cdot \frac{49}{2r^2} + \left(\frac{49}{2r^2}\right)^2 = 1 - \frac{49}{r^2} + \frac{2401}{4r^4}$$ $$\left(1 - \frac{25}{r^2}\right)^2 = 1 - 2 \cdot \frac{25}{r^2} + \left(\frac{25}{r^2}\right)^2 = 1 - \frac{50}{r^2} + \frac{625}{r^4}$$ Sum: $$1 - \frac{49}{r^2} + \frac{2401}{4r^4} + 1 - \frac{50}{r^2} + \frac{625}{r^4} = 1$$ Simplify: $$2 - \frac{99}{r^2} + \left(\frac{2401}{4r^4} + \frac{625}{r^4}\right) = 1$$ Combine terms: $$2 - \frac{99}{r^2} + \frac{2401 + 2500}{4r^4} = 1$$ $$2 - \frac{99}{r^2} + \frac{4901}{4r^4} = 1$$ Subtract 1: $$1 - \frac{99}{r^2} + \frac{4901}{4r^4} = 0$$ Multiply both sides by $4r^4$: $$4r^4 - 396r^2 + 4901 = 0$$ 8. **Let $x = r^2$, solve quadratic:** $$4x^2 - 396x + 4901 = 0$$ Use quadratic formula: $$x = \frac{396 \pm \sqrt{396^2 - 4 \cdot 4 \cdot 4901}}{2 \cdot 4}$$ Calculate discriminant: $$396^2 = 156816, \quad 4 \cdot 4 \cdot 4901 = 78416$$ $$\sqrt{156816 - 78416} = \sqrt{78400} = 280$$ So: $$x = \frac{396 \pm 280}{8}$$ Two solutions: $$x_1 = \frac{396 + 280}{8} = \frac{676}{8} = 84.5$$ $$x_2 = \frac{396 - 280}{8} = \frac{116}{8} = 14.5$$ 9. **Check for valid $r^2$:** Since $\cos\theta$ and $\sin\theta$ must be between -1 and 1, test both. For $x=84.5$: $$\cos\theta = 1 - \frac{25}{84.5} \approx 1 - 0.296 = 0.704$$ $$\sin\theta = 1 - \frac{49}{2 \cdot 84.5} = 1 - \frac{49}{169} \approx 1 - 0.29 = 0.71$$ Both valid. For $x=14.5$: $$\cos\theta = 1 - \frac{25}{14.5} = 1 - 1.72 = -0.72$$ $$\sin\theta = 1 - \frac{49}{29} = 1 - 1.69 = -0.69$$ Also valid but less likely given geometry. Choose $r^2 = 84.5$ for simplicity. 10. **Find distance between free ends of rods k and m:** Since rods k and m are perpendicular with endpoints $(r,0)$ and $(0,r)$, $$d_{km} = \sqrt{(r - 0)^2 + (0 - r)^2} = \sqrt{r^2 + r^2} = r\sqrt{2}$$ Calculate $r$: $$r = \sqrt{84.5} \approx 9.19$$ Distance: $$d_{km} = 9.19 \times \sqrt{2} \approx 9.19 \times 1.414 = 13.0$$ **Answer:** 13 cm (Option B)