Answer 1

\[\text{ We know that the diagonals of a rhombus bisect each other at right angles }. \]
\[ \therefore BO = \frac{1}{2}BD = (\frac{1}{2} \times 16) cm\]
\[ = 8cm\]
\[AB = 10 \text{ cm and }\angle AOB = 90°\]
\[\text{ From right } ∆ OAB: \]
\[ {AB}^2 = {AO}^2 + {BO}^2 \]
\[ \Rightarrow {AO}^2 = ( {AB}^2 -{BO}^2 )\]
\[ \Rightarrow {AO}^2 = (10 )^2 - (8 )^2 {cm}^2 \]
\[ \Rightarrow {AO}^2 = (100 - 64) {cm}^2 = 36 {cm}^2 \]
\[ \Rightarrow AO = \sqrt{36} cm = 6cm\]
\[ \therefore AC = 2 \times AO = (2 \times 6) cm = 12 cm\]