The lateral edge of a regular quadrangular pyramid, equal to 8, forms an angle of 45

The lateral edge of a regular quadrangular pyramid, equal to 8, forms an angle of 45 with the base plane. Find the area of the side surface of the pyramid.

1. Let’s omit the height N.

Triangle: edge c – d / 2 – H – isosceles.

The angle between c and H = 180 ° – (90 ° + 45 °) = 45 °, H = d / 2.

2. Define d.

2 (d / 2) ^ 2 = 8 ^ 2;

d / 2 = √32 = 4√2;

d = 8√2.

3. Determine the side of the base a.

2a ^ 2 = d ^ 2;

2a ^ 2 = (8√2) ^ 2;

a = √64;

a = 8.

4. Define the apothem h.

h ^ 2 = c ^ 2 – (a / 2) ^ 2;

h ^ 2 = 8 ^ 2 – (8: 2) ^ 2;

h2 = 48;

h = √48 = 4√3.

5. Let us determine the area s of the lateral surface.

s = 1/2 * P * h;

s = 1/2 * 4 * 8 * 4√3;

s = 64√3.

Answer: the area of the lateral surface is 64√3 sq. units.