The diagonal of the rhombus is equal to its side, its length is 10cm. Find the second diagonal and the corners of the rhombus.

Suppose AB = AC = 10 cm.

The diagonals of the rhombus intersect at right angles and the intersection point is halved:

AO = OC = AC / 2;

AO = OC = 10/2 = 5 cm.

Consider a right-angled triangle ∆ABO.

Using the Pythagorean theorem, we find the side of the ВD, which is half of the second diagonal of the ВD:

AB ^ 2 = BO ^ 2 + AO ^ 2;

BO ^ 2 = AB ^ 2 – AO ^ 2;

ВO ^ 2 = 10 ^ 2 – 5 ^ 2 = 100 – 25 = 75;

BO = √75 = 8.66 cm.

BD = BO · 2;

BD = 8.66 2 = 17.32 cm.

Using the cosine theorem, we can find the angle ∠BAO:

cos BAO = AO / AB;

cos BAO = 5/10 = 0.5, which is equal to an angle of 60 °.

∠A = ∠BAO · 2;

∠А = 60 2 = 120 °.

Since the sum of the degree measures of all the angles of the rhombus is 360 °, and the opposite angles are equal:

∠В = ∠D = (360 – ∠А – ∠С) / 2;

∠В = ∠D = (360 – 120 – 120) / 2 = 120/2 = 60 °.

Answer: the second diagonal is 17.32 cm, and the angles are 120 degrees and 60 degrees.