The diagonal of a regular quadrangular prism is equal to a and forms an angle of 30 with the plane

The diagonal of a regular quadrangular prism is equal to a and forms an angle of 30 with the plane of the lateral face. Find: a) the side of the base of the prism; b) the angle between the diagonal of the prism and the plane of the base c) the area of the lateral surface of the prism; d) the sectional area of the prism by the plane passing through the diagonal of the base of the parallel diagonal of the prism.

Consider a right-angled triangle ABC1, which, by condition, has an angle C1 = 30, then the leg AB, which lies opposite angle 30, is equal to half the length of the hypotenuse BC1.

AB = BC1 / 2 = a / 2 cm.

Let us determine the length of the diagonal of the AВСD square. BC = AB * √2 = a * √2 / 2 cm.

Consider a right-angled triangle BCC1 and determine the value of the angle CBC1.

CosСВС1 = CB / BC1 = (a * √2 / 2) / 2 = √2 / 2.

Angle CBC1 = arcos (a * √2 / 2) = 450.

From the right-angled triangle CBC1 we determine the height of the prism CC1.

Since the angle CBC1 = 45, the triangle CBC1 is isosceles and BC = CC1 = a * √2 / 2.

Let us determine the area of ​​the lateral surface of the prism.

Side = 4 * AB * CC1 = 4 * (a / 2) * (a * √2 / 2) = a2 * √2 cm2.

Consider the section ADK. The height of the KН section is the middle line of the triangle BCC1, since the point H is the intersection of the diagonals and the KН is parallel to BC1 by condition.

Then KH = BC1 / 2 = a / 2.

Then the cross-sectional area is equal to:

Ssection = AD * KН / 2 = (a * √2 / 2) * (a / 2) / 2 = (a2 * √2 / 8) cm2.

Answer:

a) AB = a / 2 cm.

b) Angle CBC1 = 45.

c) S side = a ^ 2 * √2 cm2.

d) Ssection = (a ^ 2 * √2 / 8) cm2.