1. **Problem Statement:** We are given a right triangle VWX with a right angle at X. Points Y and Z are on sides VX and VW respectively, with segment YZ perpendicular to VX. Given lengths are ZW = 6.1, YZ = 7.8, and XW = 13. We need to find the length VZ rounded to the nearest tenth. 2. **Understanding the setup:** Since VWX is a right triangle with right angle at X, and YZ is perpendicular to VX, triangle YZW is right-angled at Z. We know segment lengths ZW = 6.1 and YZ = 7.8. 3. **Using the Pythagorean theorem in triangle YZW:** $$YW = \sqrt{YZ^2 + ZW^2} = \sqrt{7.8^2 + 6.1^2} = \sqrt{60.84 + 37.21} = \sqrt{98.05} \approx 9.9$$ 4. **Using the Pythagorean theorem in triangle XWZ:** Since XW = 13 and ZW = 6.1, we find XZ: $$XZ = \sqrt{XW^2 - ZW^2} = \sqrt{13^2 - 6.1^2} = \sqrt{169 - 37.21} = \sqrt{131.79} \approx 11.5$$ 5. **Finding VZ:** Since VZ = VW - WZ and VW = VX + XW, but VX is unknown, we use the right triangle VWX with right angle at X: $$VW = \sqrt{VX^2 + XW^2}$$ 6. **Using the fact that YZ is perpendicular to VX and Y lies on VX, Z lies on VW, and YZ = 7.8, we can find VX using similar triangles or coordinate geometry. However, since YZ is perpendicular to VX and YZ = 7.8, and YW = 9.9 (from step 3), then VW = VZ + ZW, and VZ = VW - ZW. 7. **Using the Pythagorean theorem in triangle VWX:** Let VX = a, then: $$VW = \sqrt{a^2 + 13^2}$$ 8. **Using the right triangle YZW:** Since YZ is perpendicular to VX, and YZ = 7.8, ZW = 6.1, and YW = 9.9, the length VW = VZ + ZW, so: $$VZ = VW - ZW = \sqrt{a^2 + 169} - 6.1$$ 9. **Using the fact that Y lies on VX, and YZ is perpendicular to VX, the length YZ = 7.8 is the height from Y to Z, so the length VX = a = YW = 9.9 (from step 3). So: $$VX = 9.9$$ 10. **Calculate VW:** $$VW = \sqrt{9.9^2 + 13^2} = \sqrt{98.01 + 169} = \sqrt{267.01} \approx 16.3$$ 11. **Calculate VZ:** $$VZ = VW - ZW = 16.3 - 6.1 = 10.2$$ **Final answer:** $$\boxed{10.2}$$ units (rounded to the nearest tenth).