1. **State the problem:** We have two similar right triangles $\triangle NML \sim \triangle KMO$. Given sides are $NL = 15$ m, $ML = 12$ m, $OM = 16$ m, and we need to find $s = KO$. 2. **Recall the property of similar triangles:** Corresponding sides of similar triangles are proportional. This means: $$\frac{NL}{KO} = \frac{ML}{OM}$$ or equivalently: $$\frac{15}{s} = \frac{12}{16}$$ 3. **Set up the proportion and solve for $s$:** $$\frac{15}{s} = \frac{12}{16}$$ Cross-multiply: $$15 \times 16 = 12 \times s$$ $$240 = 12s$$ 4. **Isolate $s$:** $$s = \frac{240}{12}$$ Show canceling common factors: $$s = \frac{\cancel{240}^{20} \times 12}{\cancel{12}} = 20$$ 5. **Final answer:** $$s = 20$$ meters This means the side $KO$ in triangle $KMO$ is 20 meters long.