Overview

Sphere problems often reduce to right triangles with radii and distances from the center.

Key Ideas

• Surface area: $4\pi R^2$.
• Volume: $\frac{4}{3}\pi R^3$.
• Cross-section radius at distance $d$: $\sqrt{R^2 - d^2}$.

Core Skills

Use a Radius Triangle

Draw the radius to the plane of the cross-section. The radius, distance $d$, and cross-section radius form a right triangle.

Scale Volumes Carefully

If linear scale changes by $k$, volumes scale by $k^3$ and areas by $k^2$.

Recognize Great Circles

If the plane passes through the center, the cross-section is a great circle with radius $R$.

Worked Example

A sphere has radius $10$. A plane is $6$ units from the center. Find the radius of the cross-section circle.

$r = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$.

More Examples

Example 1: Surface Area

Find the surface area of a sphere with radius $5$.

$4\pi\cdot 25 = 100\pi$.

Example 2: Great Circle

What is the radius of the cross-section when a plane passes through the center?

It is $R$.

Example 3: Volume Scaling

If the radius doubles, by what factor does the volume change?

By $2^3=8$.

Strategy Checklist

• Draw the right triangle with $R$, $d$, and cross-section radius.
• Use $R^2=d^2+r^2$.
• Apply correct scaling for area/volume.

Common Pitfalls

• Using diameter instead of radius.
• Forgetting the right-triangle relationship in cross-sections.
• Confusing surface area with cross-section area.

Practice Problems