A sphere with radius 13 is intersected by a plane that is 5 units from the sphere's center. What is the radius of the resulting circular cross section?
A5
B12
C13.93
D8
Explanation
The center of the sphere, the center of the cross-section circle, and any point on the cross section's edge form a right triangle: (cross-section radius)² + (distance from center)² = R². So cross-section radius = √(R² − d²) = √(13² − 5²) = √144 = 12.
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