Answer: D
In right-angled triangle ABC: \(BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 8^2} = 10\) cm Also, triangles ABC and ABD are similar. ∴ \(\cfrac{AB}{BC} = \cfrac{BD}{AB}\) i.e., \(BC \times BD = AB^2\) ⇒ \(10 \times BD = 6^2\) ⇒ \(BD = \cfrac{36}{10} = 3.6\)
In right-angled triangle ABC: \(BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 8^2} = 10\) cm Also, triangles ABC and ABD are similar. ∴ \(\cfrac{AB}{BC} = \cfrac{BD}{AB}\) i.e., \(BC \times BD = AB^2\) ⇒ \(10 \times BD = 6^2\) ⇒ \(BD = \cfrac{36}{10} = 3.6\)