In a regular quadrangular pyramid, the lateral rib is inclined to the plane of the base at an angle of 45 degrees, the radius of the circumscribed circle near the base is 4 cm. Find the volume.
Since the pyramid is correct, ABCD is a square, and therefore, the radius of the circle described around it is half the diagonal of the square.
Then AC = BD = 2 * R = 2 * 4 = 8 cm.
Since the side ribs are inclined to the base plane at an angle of 45, the SOC triangle is rectangular and isosceles, SO = OC = R = 4 cm.
Determine the area of the base of the pyramid.
Sb = D ^ 2/2 = AC ^ 2/2 = 64/2 = 32 cm2.
Then the volume of the pyramid is: V = Sosn * SO / 3 = 32 * 4/3 = 128/3 = 42 (2/3) cm3.
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