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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 .

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12 7 surface area of spheres

12.7 Surface Area of Spheres

Angela Isac

Abby Kern

1st hour

objectives
Objectives
  • Recognize and define basic properties of spheres.
  • Find surface areas of spheres.
what is a sphere
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
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
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
Hemispheres
  • Each great circle separates a sphere into two congruent halves, each called a hemisphere.
  • A hemisphere has a circular base.
example 1
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
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.

Answer: AB is approximately 9.5cm.

surface area of spheres
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
Example #2

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

example 2 cont
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.

Answer: 804.4 in2

area of a hemisphere
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
Example #3

Find the area of the hemisphere.

example 3 cont
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.

Answer: 166.3cm2

assignment
Assignment

Pre-AP Geometry: Page 674 #10 - 29