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Chapter 12:Areas and Volumes of SolidsGaby PaviaandGaby Pages Section 12-1 Bases: congruent polygons lying in parallel planes

Altitude: segment joining the two base planes and perpendicular to both

the length of the altitude is the height of the prism

Lateral Edges: the intersection of adjacent lateral faces. Are parallel segments.

Lateral Faces: the faces of a prism that are not its bases- The Lateral Faces of a prism are parallelograms

Right Prism: a prism whose lateral faces are rectangles.

Oblique Prism: a prism whose lateral face are parallelograms but not rectangles.

Examples of Prisms

Right PrismOblique PrismDifferent Areas of a Prism Lateral Area of a Prism: the sum of the areas of its lateral faces

Total Area of a Prism: the sum of the areas of all its faces

Finding Lateral Area of a Right Prism: Theorem 12-1: The lateral area of a right prism equals the perimeter of a base times the height of the prism.L.A = P x H ( P = perimeter of the base , H = height) Formula applies to any right prismEXAMPLE : Find the Lateral Area of the Prism

SOLUTION:

L.A = P x H

Perimeter : 10 x 2 + 6 x 2 =

32

Height : 5

L.A = 32 x 5 = 160 units2

Finding the Total Area of a Right Prism: Formula : T.A = L.A + 2B B = the area of the base

EXAMPLE: finding total area

SOLUTION: T.A = L.A + 2B

L.A = P x H = Perimeter: 5 + 5 + 6 = 16

Height = 12

L.A = 192

Area of Base = (6 x 4) / 2 =12

T.A = 192 + 12 = 204 units2

Finding the Volume of a Right Prism Theorem 12-2: The volume of a prism equals the area of a base times the height of the prism. V = B x H

Example: find the volume of this right prism

SOLUTION:

V = BH

Area of Base: (8 x 15)/ 2 = 120 / 2 = 60cm 2

Height of Prism : 9cm

V = 60 x 9 =

540cm3

Section 12.2 : PyramidsPoint S is the vertex of the pyramidSegment SO is the altitude of the pyramid The five triangular faces with point S in common are the lateral faces - SBC , SBA , SAE , SED , and SDC

The lateral faces intersect in segments called lateral edges :

- Segments : SA, SE, SD, SC, and SB Segment SF is the slant height of the pyramid. It is the height of a lateral face, and detonated by l

A pentagonal pyramid:Regular PyramidsAll regular pyramids have the same following 4 properties: 1 ) The base is a regular polygon 2 ) All lateral edges are congruent 3) All lateral faces are congruent isosceles triangles 4) The altitude meets the base at its center

A regular square pyramid has base edges 10cm and lateral edges 13cm. Find the slant height

SOLUTION:

Slant Height:

1) Drop an altitude to point F

2) Use Pythagorean Theorem:

l = 132 52 = 144 = 12

Lateral Area of a pyramidTheorem 12-3 : The lateral area of a regular pyramid equals half the perimeter of the base times the slant heightL.A : (1/2) P x l

There are two methods to find the lateral area of a regular pyramid with n lateral faces Method 1: Find the area of one lateral face and multiply by n Method 2: Use the formula ( L.A = (1/2) P x L)

Example: Find the lateral area of the square pyramid

Method 1 Solution:

Area of Lateral Face : (6 x 8) / 2 = 48/2 = 24

24 x 4 = 96 units 2

Method 2 : L.A = (1/2) P x L

Perimeter: 8 x 4 = 32

Slant Height : 6

L.A = (1/2)32 x 6 = 16 x 6 = 96 units 2

Finding the Volume of a PyramidSince the volume of a prism is base x height, the volume of the pyramid must be less than the base x height.

Theorem 12-4: The volume of a pyramid equals one third the area of the base times the height of the pyramid

V= (1/3) x Area of Base x Height

Section 12-3: CylindersCylinder- a prism with bases as circles.In a right cylinder, the segments joining the centers of both circles is an altitude.The length of an altitude is the height of the cylinder. The radius of the base is the radius of the the cylinder. Finding L.A, T.A and volume 18Example 1: Find the Lateral AreaSOLUTION:(Use 3.14 as ) Circumference of the Base: 2r = 2 (5) = 31.4Height : 8L.A = 2 r x height = 3.14 x 8 = 251.2 units2

Example 2: Find the volume SOLUTION:Area of base: r2 = (5) 2 = x 25 = 78.5

Height : 15

V = r2 x height = V= 78.5 x 15 = 1177.5 units3

Example 35 cm4 cmA cylinder has radius 5 cm and height 4 cm. Find (a) the lateral area, (b) total area, and (c) volume of the cylinder.12-3 Cylinders and cones (cont.) Cones- similar to a pyramid except that its base is a circle instead of a polygon

Slant height, lFinding L.A, T.A and volume Example

Find the (a) lateral area, (b) total area, and (c) volume of the cone showed3 6l Section 12.4 : Spheres Recall from Chapter 9 that a sphere is the set of all points that are a given distance from a given point

Finding the Area of a Sphere:Theorem 12-9: The area of a sphere equals 4 times the square of the radiusA = 4r2

EXAMPLE: Find the area of the sphere ( use 3.14 as ) SOLUTION:

A = 4r2

A = 4 x x 22 = 4 x x 4 = 16 = 50.24 cm2

Finding the Volume of a Sphere: Theorem 12-10: The volume of a sphere equals 4/3 times the cube of the radiusV = 4/3 x r3

EXAMPLE: Find the volume of the sphere. (Use 3.14 as )

SOLUTION:

V = 4/3 x r3

V = 4/3 x 53 = 4/3 x 125 = 4.19 x 125 = 523.75 units3

Section 12.5: Similar SolidsSimilar solids are solids that have the same shape but not the same size.

To decide whether two solids are similar:- determine whether the bases are similar* all circles and all squares are similar*

- determine whether corresponding lengths are proportionalDetermining ratiosExample

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