# In a straight parallelepiped, the sides of the base are 3 cm and 5 cm, one of the diagonals of the base is 4 cm

**In a straight parallelepiped, the sides of the base are 3 cm and 5 cm, one of the diagonals of the base is 4 cm, the smaller diagonal of the parallelepiped makes an angle of 60 degrees with the base plane. Find the diagonals of the parallelepiped and the areas of the diagonal sections.**

Since there is a parallelogram at the base, the sum of the squares of its diagonals is equal to the sum of the squares of its sides.

AC ^ 2 + BD ^ 2 = 2 * AB ^ 2 + 2 * AD ^ 2.

AC ^ 2 = 2 * AB ^ 2 + 2 * AD ^ 2 – BD ^ 2 = 18 + 50 – 16 = 52.

AC = √52 cm.

From the right-angled triangle BDD1, we determine the length of the leg DD1 and the hypotenuse BD1.

DD1 = BD * tg60 = 4 * √3 cm.

BD1 = BD / Cos60 = 4 / (1/2) = 8 cm.

From a right-angled triangle ACC1, AC1 ^ 2 = AC ^ 2 * CC1 ^ 2 = 52 + 48 = 100.

AC1 = 10 cm.

Let us determine the areas of the diagonal sections.

Saa1s1s = АА1 * АС = 4 * √3 * √52 = 4 * √156 = 16 * √39 cm2.

Svv1d1d = АА1 * ВD = 4 * √3 * 4 = 16 * √3 cm2.

Answer: The diagonals of the parallelepiped are 8 cm and 10 cm.The areas of the diagonal sections are equal

16 * √39 cm2, 16 * √3 cm2.