Question

Сторона квадрата равна 4 см. Точка, равноудаленная от всех вершин квадрата находится на расстоянии 6 см от пересечения его диагоналей. Найти расстояние от этой точки до вершин квадрата.

Answer 1

4√11 см
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Answer 2

This is a question about geometry. You are asking me to find the distance from a point that is equidistant from all the vertices of a square to the vertices of the square, given that the side of the square is 4 cm and the point is 6 cm away from the intersection of the diagonals of the square. I will try to answer this question in detail, using the information from the web search results.

First, let us draw a sketch of the problem. The square ABCD has a side of 4 cm and a diagonal of √32 cm. The point P is equidistant from all the vertices of the square and is 6 cm away from the point O, which is the intersection of the diagonals of the square. We need to find the distance from P to any vertex of the square, say A.

![Sketch of the problem]

To find the distance PA, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, PA is the hypotenuse of the right triangle AOP, and AO and OP are the other two sides. Therefore, we have:

PA^2 = AO^2 + OP^2

We know that AO is half of the diagonal of the square, which is √32/2 cm, and OP is 6 cm. Substituting these values, we get:

PA^2 = (√32/2)^2 + 6^2

Simplifying, we get:

PA^2 = 8 + 36

PA^2 = 44

Taking the square root of both sides, we get:

PA = √44

PA ≈ 6.63 cm

Therefore, the distance from the point P to any vertex of the square is approximately 6.63 cm.

I hope this answer helps you understand the problem better. If you have any more questions, feel free to ask me.