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Точки СС і DD лежать на сфері із центром ОО, діаметр якої дорівнює 8см8\,\text{см}. Знайдіть відрізок СDСD, якщо трикутник СОDСОD є прямокутним.

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Q. Точки СС і DD лежать на сфері із центром ОО, діаметр якої дорівнює 8см8\,\text{см}. Знайдіть відрізок СDСD, якщо трикутник СОDСОD є прямокутним.
  1. Identify Given Information: Identify the given information and the property that will be used to solve the problem.\newlineWe are given that the diameter of the sphere is 8cm8\,\text{cm}, which means the radius (r)(r) is half of that, so r=4cmr = 4\,\text{cm}. We also know that triangle CODCOD is a right triangle with the right angle at OO. By the Pythagorean theorem, the sum of the squares of the lengths of the two shorter sides of a right triangle is equal to the square of the length of the longest side (hypotenuse).
  2. Apply Pythagorean Theorem: Apply the Pythagorean theorem to triangle CODCOD. Since OO is the center of the sphere and CC and DD lie on the sphere, OCOC and ODOD are radii of the sphere. Therefore, OC=OD=r=4cmOC = OD = r = 4 \, \text{cm}. CDCD is the hypotenuse of the right triangle CODCOD. Using the Pythagorean theorem: OC2+OD2=CD2OC^2 + OD^2 = CD^2.
  3. Substitute and Solve: Substitute the known values into the Pythagorean theorem and solve for CDCD.OC2+OD2=CD2OC^2 + OD^2 = CD^2 becomes 42+42=CD24^2 + 4^2 = CD^2.16+16=CD216 + 16 = CD^2.32=CD232 = CD^2.
  4. Find Length of CD: Find the length of CD by taking the square root of both sides of the equation.\newlineCD=32CD = \sqrt{32}.\newlineCD=16×2CD = \sqrt{16 \times 2}.\newlineCD=16×2CD = \sqrt{16} \times \sqrt{2}.\newlineCD=4×2CD = 4 \times \sqrt{2}.
  5. Check for Errors: Check the result for any mathematical errors. The calculations are correct, and the use of the Pythagorean theorem is appropriate for a right triangle. No math errors are present.

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