The side of the square is 10cm. The point equidistant from all vertices is at a distance of 11cm

| April 25, 2021 | education

The side of the square is 10cm. The point equidistant from all vertices is at a distance of 11cm from the point of intersection of the diagonals. Find the distance from this point to the top of the square.

Let a square ABCD be given, with sides 10 cm, its diagonals intersect at point O, point S – equidistant from the tops of the square, at a distance of 11 cm from point O, so СO = 11 cm.We have got a pyramid with base ABCD and vertex S , height СO.
Consider a square ABCD, AB = BC = CD = AD = 10 cm, define its diagonals:
AC = BD = a * √2 = 10 * √2 cm.
Consider triangle AOС, <O = 90 °, legs OS and AO, AС – hypotenuse. OС = 11 cm, AO = AC / 2 = d / 2 = 10 * √2 / 2 = 5 * √2 cm.Let’s define the hypotenuse by the Pythagorean theorem:
AС = √ (OС² + AO²) = √ (11² + (5 * √2) ²) = 3 * √19 cm.
Answer: the distance from point S to the top of the square is 3 * √19 cm or 13.07 cm.



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