# 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.

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