One of the diagonals of the rhombus is 30cm. Find another diagonal of the rhombus if its perimeter is 68cm.

ABCD is a rhombus, AB = BC = CD = AD, AB + BC + CD + AD = 68 cm, AC = 30 cm and BD are the diagonals intersecting at point O.
1. The diagonals of the rhombus intersect at right angles and the intersection point is halved. Then:
AO = CO = AC / 2 = 30/2 = 15 cm.
2. Find the length of the side of the rhombus:
P = 4a,
where a is the length of the rhombus side.
4a = 68;
a = 68/4;
a = 17.
Thus, AB = BC = CD = AD = a = 17 cm.
3. Consider the triangle AOB: the angle AOB = 90 degrees (since the diagonals of the rhombus are perpendicular to each other), AB = 17 cm is the hypotenuse (since it lies opposite the right angle AOB), AO = 15 cm and VO are the legs.
By the Pythagorean theorem:
BO = √ (AB ^ 2 – AO ^ 2) = √ (17 ^ 2 – 15 ^ 2) = √ (289 – 225) = √64 = 8 (cm).
4. Since point O divides the diagonal in half, then:
BO = DO = BD / 2.
Then:
BD = 2BO;
BD = 2 * 8 = 16 (cm).
Answer: BD = 16 cm.