The diameter of the ball is 25 cm. On its surface, point A and a circle are given, all points of which are removed

The diameter of the ball is 25 cm. On its surface, point A and a circle are given, all points of which are removed (in a straight line) from A by 15 cm. Find the radius of this circle.

Consider a triangle ABO, in which OB and OA are the radii of the ball, and are equal to half of the AC, and the side AB is, by condition, 15 cm.

Let us determine the area of the triangle through its sides and semiperimeter.

p = (AB + AO + BO) / 2 = (15 + 12.5 + 12.5) / 2 = 20 cm.

Savo = √ (p * (p – AO) * (p – BO) * (p – AB) = √ (20 * (20 – 12.5) * (20 – 12.5) * (20 – 15) = √5625 = 75 cm2.

Let’s define the area of the same triangle through the height and base of the triangle.

Savo = AO * BO1 / 2 = 6.25 * BO1.

75 = 6.25 * BO1.

BO1 = 75 / 6.25 = 12 cm.

Answer: The radius of the circle is 12 cm.




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