In the regular quadrangular pyramid SABCD O, the center of the base, S-vertex, SO-15, AC-40. Find the side edge SD.

univerkov | December 31, 2020 | education

Because the pyramid is SABCD-regular, then the base is ABCD-square, AC = BD = 40 (the diagonals of the square are equal);
Triangle SDO-rectangular, hence SD-hypotenuse, OD and SO-legs, OD = BD / 2 = 40/2 = 20, SO = 15;
SD = √15 ^ 2 + 20 ^ 2 = √225 + 400 = √625 = 25
Answer: 25

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