Instruction

1

**The perimeter**

**of a rhombus**can be calculated, knowing the length of its side. In this case, by definition the perimeter

**of a rhombus**is equal to the sum of the lengths of its sides, and therefore equal to 4a, where a is the side length

**of the rhombus**.

2

If you know the area

Express BD2 through the area:

S = k*BD*BD/2 = k*BD2/2

BD2 = 2*S/k

Then 4*AB2 = (1 + k2)*2S/k. Hence AB is equal to the square root of S(1 + k2)/2k. And the perimeter

**of a rhombus**and the ratio between the diagonals, the problem of finding the perimeter**of a rhombus**is slightly more complicated. Given the area**of a rhombus**S and the ratio of the diagonals AC/BD = k. Area**of a rhombus**can be expressed through the product of the diagonals: S = AC*BD/2. Triangle AOB is a right triangle because the diagonals**of a rhombus**intersect at a 90° angle. Side**of the rhombus**AB by theorem of Pythagoras can be found from the following expression: AB2 = AO2 + OB2. Since the rhombus is a special case of a parallelogram and in a parallelogram the diagonals bisect the point of intersection, then AO = AC/2 and OB = BD/2. Then AB2 = (AC2 + BD2)/4. The condition AC = k*BD, then 4*AB2 = (1 + k2)*BD2.Express BD2 through the area:

S = k*BD*BD/2 = k*BD2/2

BD2 = 2*S/k

Then 4*AB2 = (1 + k2)*2S/k. Hence AB is equal to the square root of S(1 + k2)/2k. And the perimeter

**of a rhombus**is still equal to 4*AB.