In $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}, \mathrm{AD} \perp \mathrm{BC}$ at $\mathrm{D}$. If $\mathrm{AB}=12 \mathrm{~cm}$ and $\mathrm{AC}=16 \mathrm{~cm}$, then what is the length (in cm) of $\mathrm{BD}$ ?

7.2

\( { AB}^{2 } \) + \( {AC }^{2 } \) = \( {BC }^{2 } \)

= \( { 12}^{2 } \) + \( {16 }^{2 } \) = \( {BC }^{2 } \)

= 144 + 256 = \( {BC }^{2 } \)

= 400 = \( {BC }^{2 } \)

= BC = 20

Now, using similarity of triangles ABC and DBA,

\(\frac{BC}{AB}\) = \(\frac{AB}{BD}\) = \(\frac{AC}{AD}\)

Length of BD = \(\frac{AB \;×\; AB}{BC}\)

= \(\frac{12 \;×\; 12}{20}\) = 7.2 cm

Therefore, length of BD is 7.2 cm.