In a regular quadrangular pyramid, the height is 7 centimeters, and the lateral

In a regular quadrangular pyramid, the height is 7 centimeters, and the lateral rib is inclined to the plane of the base at an angle of 45 degrees. Find the area of the lateral surface of the pyramid.

SABCD is a regular quadrangular pyramid. AB = BC = CD = AD (since there is a square at the base of a regular quadrangular pyramid).
1. Consider the SOA triangle: SOA angle = 90 degrees (since SO is the height), OAS angle = 45 degrees (by condition). According to the theorem about the sum of the angles of a triangle: angle SOA + angle ОАS + angle АSO = 180 degrees;
90 + 45 + angle АSO = 180;
angle АSO = 180 – 135;
angle АSO = 45 degrees.
The SOA triangle is a rectangular isosceles triangle, then SO = AO = 7 cm – legs. By the Pythagorean theorem, we find the hypotenuse SA:
SA = √ (SO ^ 2 + AO ^ 2);
SA = √ (7 ^ 2 + 7 ^ 2) = √ (49 + 49) = √98 = 7√2 (cm).
2. OA is the radius of a circle around square ABCD. According to the formula:
R = √2t / 2;
√2t / 2 = 7;
t = 7 * 2 / √2;
t = 14 / √2;
t = 14√2 / 2;
t = 7√2 cm.
3. Since SABCD is a regular quadrangular pyramid, all its side faces are equal triangles.
Consider a triangle ASD: AS = SD = AD = 7√2 cm, then ASD is a regular triangle. Regular triangle area:
S = √3a ^ 2/4;
S = √3 * (7√2) ^ 2/4 = 98√3 / 4 = 49√3 / 2 (cm ^ 2).
4. The area of ​​the lateral surface of the SABCD pyramid is:
Side = 4S;
Side = 4 * 49√3 / 2 = 98√3 (cm ^ 2).
Answer: Side = 98√3 cm ^ 2.