Solved: In rhombus ABCD, DB=16 and AC=12 . What is the perim[geometry]

P(BADC) = 40
The diagonals of a rhombus are perpendicular.: \angle {AEB} = 90^{\circ}
The diagonals of a rhombus bisect each other.: \overline {AC} = 2 \overline {AE}
Substitute \overline {AC} = 12 into \overline {AC} = 2 \overline {AE}: 12 = 2 \overline {AE}
Calculate 12 = 2 \overline {AE}: \overline {AE} = 6
The diagonals of a rhombus bisect each other.: \overline {BD} = 2 \overline {BE}
Substitute \overline {BD} = 16 into \overline {BD} = 2 \overline {BE}: 16 = 2 \overline {BE}
Calculate 16 = 2 \overline {BE}: \overline {BE} = 8
Law of Cosines: \overline {AB}^{2} = \overline {BE}^{2} + \overline {AE}^{2} – 2 \overline {BE} \times \overline {AE} \times \cos{\left(\angle {AEB} \right)}
Substitute \angle {AEB} = 90^{\circ} , \overline {AE} = 6 , \overline {BE} = 8 into \overline {AB}^{2} = \overline {BE}^{2} + \overline {AE}^{2} – 2 \overline {BE} \times \overline {AE} \times \cos{\left(\angle {AEB} \right)}: \overline {AB}^{2} = 8^{2} + 6^{2} – 2 \times 8 \times 6 \times \cos{\left(90^{\circ} \right)}
Calculate \overline {AB}^{2} = 8^{2} + 6^{2} – 2 \times 8 \times 6 \times \cos{\left(90^{\circ} \right)}: \overline {AB} = 10
Perimeter of a rhombus: P(BADC) = 4 \overline {AB}
Substitute \overline {AB} = 10 into P(BADC) = 4 \overline {AB}: P(BADC) = 4 \times 10
Calculate P(BADC) = 4 \times 10: P(BADC) = 40