A point outside the plane of the square is 40 cm away from each of its vertices; another point is 25 cm away from this point and from each of the vertices of the square. Find the area of the square
Since KS = KA = KD = KA, and MA = MV = MC = MD, then point K is projected into point O, the point of intersection of the diagonals of the square, and point M lies on the segment OK.
The length of the segment OK = 40 cm, the length of the segment KM = 25 cm, then the length OM = OK – KM = 40 – 25 = 15 cm.
The OCM triangle is rectangular, in which OC ^ 2 = CM ^ 2 – OM ^ 2 = 625 – 225 = 400.
OS = 20 cm.
Then AC = BD = 2 * OC = 40 cm.
Savsd = AC2 / 2 = 1600/2 = 800 cm2.
Answer: The area of the square is 800 cm2.
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