# The value of one of the angles of the parallelogram is 60 degrees, and the smaller diagonal is 2√31 cm

**The value of one of the angles of the parallelogram is 60 degrees, and the smaller diagonal is 2√31 cm. The length of the perpendicular drawn from the point of intersection of the diagonals to the larger side is 0.5√75 cm. Find the lengths of the sides and the larger diagonal of the parallelogram.**

Let’s draw the height of the ВC parallelogram.

The triangle DBK and DOH are similar, since both are rectangular and the angle D is common to them.

Then BD / BK = OD / OH. Since at point O the diagonals are divided in half, then ОD = ВD / 2 = 2 * √31 / 2 = √31 cm.

ВK = BD * OH / OD = 2 * √31 * 0.5 * √75 / √31 = √75 cm.

In a right-angled triangle ABK, the angle ABK = 180 – 90 – 60 = 30, then the leg AK is equal to half the length of AB. AB = 2 * AK.

In the triangle ABK, according to the Pythagorean theorem, AB ^ 2 = AK ^ 2 + BK ^ 2.

(2 * AK) ^ 2 = AK ^ 2 + VK ^ 2.

3 * AK ^ 2 = ВK ^ 2 = 75.

AK ^ 2 = 25.

AK = 5 cm, then AB = 2 * 5 = 10 cm.

In a right-angled triangle BKD, according to the Pythagorean theorem, we find the length of the leg KD.

KD ^ 2 = BD ^ 2 – ВK ^ 2 = (2 * √31) ^ 2 – (√75) ^ 2 = 124 – 75 = 49.

KD = 7 cm.Then AD = AK + KD = 5 + 7 = 12 cm.

From the top of C, draw the perpendicular CP to the base of AD. Triangles ABK and DСР are equal in hypotenuse and leg, then DP = AK = 5 cm, and AR = AD + AK = 12 + 5 = 17 cm.

In a right-angled triangle ACP, according to the Pythagorean theorem, AC ^ 2 = AP ^ 2 + CP ^ 2 = 17 ^ 2 + (√75) ^ 2 = 364.

AC = √364 = 2 * √91 cm.

Answer: The sides of the parallelogram are 10 cm, 12 cm, the large diagonal is 2 * √91 cm.