The lateral edge of a straight quadrangular pyramid is 4 cm and forms an angle of 45 degrees with the base

The lateral edge of a straight quadrangular pyramid is 4 cm and forms an angle of 45 degrees with the base plane of the pyramid. Find the height of the pyramid and the area of the lateral surface of the pyramid.

The solution of the problem:

Determine the height of the pyramid.
Angle between rib and height: 900 – 450 = 450. Base diagonal d / 2 = h.

b ^ 2 = 2h2 ^;

h ^ 2 = b ^ 2/2;

h ^ 2 = 4 ^ 2/2 = 8;

h = √8 = 2√2 cm.

Determine the side of the base a.
d = 2 * h = 2 * 2√2.

2a ^ 2 = d ^ 2;

a ^ 2 = d ^ 2/2;

a ^ 2 = 16 * 2: 2 = 16;

a = √16 = 4 cm.

Let’s define the apothem with.
c ^ 2 = b ^ 2 – (a / 2) ^ 2;

c ^ 2 = 4 ^ 2 – 2 ^ 2 = 12;

c = √12 = 2√3 cm.

Let us determine the lateral surface area s.
S = (4a * c) / 2 = 2a * c = 2 * 4 * 2√3 = 16√3 cm2;

Answer: the lateral surface area is 16√3 cm2.