Chapter 12.7 Surface Areas of Spheres. Objectives Recognize and define basic properties of spheres Find surface areas of spheres.

Chapter 12.7

Surface Areas of Spheres

Objectives

Recognize and define basic properties of spheres

Find surface areas of spheres

• Point D is the center of the sphere

• AB is the diameter of sphere D

• DC, DA, and DB and radii

• FG and AB are chords

• JH is a tangent to sphere D at point E

Parts of a Sphere

- The intersection of a plane and a sphere can be or .

- When a plane intersects a sphere so that it contains the center of the sphere, the intersection is called .

(Note: A great circle has the same center as the sphere, and its radii are also radii of the sphere.)

No Intersection

A PointA Circle

A great Circle

Center

Each Great circle

divides a

sphere into two halves, each

called a hemi-

sphere

.

In the figure, C is the center of the sphere, and plane R intersects the sphere in circle X. If XC = 9 centimeters and CY = 30 centimeters, find XY. Triangle CXY is a right triangle. (Angle X = 90°)

R 30 cm

9 cmXY² + XC² = YC² Pythagorean Theorem

XY² + 9² = 30² Plug in numbers

XY² + 81 = 900 Square Numbers

XY² = 900 – 81 Subtract 81 from both sides

XY² = 819 900 – 81 = 819

XY = √819 Find the square root of 819

XY ≈ 28.6 cm Punch it in the calculator…and

you get the approximate answer

Example 1:

If a sphere has a

surface area of A

square units and a

radius of r units,

then A = 4πr².

Area of a Sphere

(A great circle’s area is πr²)

Example 2:

Find the surface area of the sphere given the area of the great circle.

We know that the surface area of a sphere is four times the area of the great circle.

A = 4πr² Surface Area of a sphere

≈ 4(603.3) πr² ≈ 603.3

≈ 2413.2 Multiply

The surface area of this

sphere is ≈ 2413.2 in.²

G ≈ 603.3 in.²

Find the surface area of the hemisphere.

A hemisphere is half of a sphere. To find the surface area, find half of the surface area of the sphere and add the area of the great circle.

8.4 cm

Surface area = ½ (4πr²) + πr²

Surface area of a hemisphere

= ½ [4π(8.4)²] + π(8.4)²

Substitution

≈ 664.7

Use a calculator

The surface area of the hemisphere is approximately 664.7cm²

Example 3:

Find the surface area of a baseball given the circumference of 9 inches to determine how much leather is needed to cover the ball.

First find the radius of the ball.

C = 2πr Circumference of a circle

9 = 2πr Substitution

9/2π = r Division

1.4 ≈ r Use a calculator

Next, find the surface area.

A = 4πr² Surface area of a sphere

≈ 4π(1.4)² Substitution

≈ 25.8 Use a calculator

The surface area of the baseball is approximately 25.8 inches²

Assignment

Pre-AP GeometryPage 674

# 10-29, #34 and #36