A plane intersects a sphere that has a radius of 13 cm. The distance from the center of the sphere to the closest point on the plane is 5 cm. What is the radius of the circle that is the intersection of the sphere and the plane?
A. 8 cm
B. 10 cm
C. 12 cm
D. 13 cm
Draw the figure and convince yourself that the radius of the circle is given by Pythagoras theorem
r^2=sqrt(13^2-5^2)
To find the radius of the circle that is the intersection of the sphere and the plane, we need to understand the relationship between the sphere, the plane, and the circle.
First, let's visualize the problem. Imagine a sphere with a radius of 13 cm, and a plane passing through it. The distance from the center of the sphere to the closest point on the plane is given as 5 cm. This means that the plane is not tangent to the sphere but intersects it.
When a plane intersects a sphere, the intersection is always a circle. The radius of this circle, denoted as "r", can be found by drawing a line connecting the center of the sphere and the closest point on the plane where they intersect.
Now, we can use the Pythagorean theorem to find the value of "r".
The hypotenuse of the right triangle formed by the line connecting the center of the sphere to the closest point on the plane is the radius of the sphere, which is 13 cm. One of the legs of the triangle is the distance from the center of the sphere to the plane, which is 5 cm. The other leg is the radius of the circle, denoted as "r".
Using the Pythagorean theorem, we have:
r^2 + 5^2 = 13^2
Simplifying the equation, we get:
r^2 + 25 = 169
r^2 = 169 - 25
r^2 = 144
r = √144
r = 12
Therefore, the radius of the circle that is the intersection of the sphere and the plane is 12 cm.
So, the correct answer is C. 12 cm.