12.7 Surface Area of Spheres

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# 12.7 Surface Area of Spheres - PowerPoint PPT Presentation

12.7 Surface Area of Spheres. Angela Isac Abby Kern 1st hour. Objectives. Recognize and define basic properties of spheres. Find surface areas of spheres. . What is a Sphere?. A sphere is the locus of all points that are a given distance from a given point called the center .

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### 12.7 Surface Area of Spheres

Angela Isac

Abby Kern

1st hour

Objectives
• Recognize and define basic properties of spheres.
• Find surface areas of spheres.
What is a Sphere?
• A sphere is the locus of all points that are a given distance from a given point called the center.
• To help visualize a sphere, consider infinitely congruent circles in space that all have the same point for their center.
Properties of Spheres
• A radius of a sphere is a segment with endpoints that are the center of the sphere and a point on the sphere. In the figure, DA, DB, and DC are radii.
• A chord of a sphere is a segment with endpoints that are points on the sphere. In the figure, GF and AB are chords.
• A diameter is a chord that contains the center of the sphere. In the figure, AB is the diameter.
• A tangent to a sphere is a line that intersects the sphere in exactly one point. In the figure, JH is a tangent to the sphere at point E.
Great Circles
• When a plane intersects a sphere so that it contains the center of the sphere, the intersection is called a great circle.
• A great circle has the same center as the sphere, and its radii are also radii of the sphere.
Hemispheres
• Each great circle separates a sphere into two congruent halves, each called a hemisphere.
• A hemisphere has a circular base.
Example #1

In the figure, O is the center of the sphere, and plane R intersects the sphere in circle A. If AO = 3cm and OB=10cm, find AB.

Example #1 Cont.

Use the Pythagorean Theorem to solve for AB.

OB2 = AB2 + AO2Pythagorean Theorem

102 = AB2 + 32OB = 10, AO = 3

100 = AB2 + 9 Simplify.

91 = AB2 Subtract 9 from each side.

9.5 = AB Use a calculator.

Surface Area of Spheres

If a sphere has a surface area of T square units and a radius of r units, then

T= 4 r2

This is simply saying that the surface area (T) of the sphere is 4 times the area of the great circle ( r2).

Example #2

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

Example #2 Cont.

Use the formula for surface area to solve.

T = 4 r2Surface are of the sphere.

T = 4(201.1) r2 = 201.1

T = 804.4 Multiply.

Area of a Hemisphere

Since a hemisphere is half a sphere, to find its surface area, find half of the surface area of the sphere and add the area of the great circle.

T = ½(4 r2) + r2

Example #3

Find the area of the hemisphere.

Example #3 Cont.

Use the formula for the surface area of a

hemisphere to solve.

T = ½(4 r2) + r2 Surface are of a hemisphere

T = ½[4 (4.2)2] + (4.2)2 Substitution

T = 166.3 Use a calculator.