surface area of right prisms and cylinders

Surface area of Right Prisms and Cylinders

SA = 2B + ph

MA202 Sections 5 & 401 Chapter 12-4 Slides

Surface area of Right Prisms and Cylinders

SA = 2B + phMA202 Sections 5 & 401 Chapter 12-4 Slides

Example 12.19

We have a strangely shaped gift box. The long sides of the baseare 20cm, the shorter sides of the base are 5cm. The height of thebox is 10cm. Find the total Surface Area.


Example 12.20

Given a small can of frozen orange juice that is 9 cm tall and has adiameter of 5 cm. If the circular ends are metal, and the sides ofthe can are cardboard: how much metal and how much cardboarddo we need to make a juice can?


Surface area of Right Regular Pyramids

SA = B +1

2ps


Example 12.21

A pyramid has a square base that is 10 cm on a side. The edgesthat meet at the apex have length 13 cm. Find the slant height ofthe pyramid and then calculate the total surface area of thepyramid.


Surface area of Right Circular Cones

SA = πr2 + πrs


Example 12.22

An ice cream cone has a diameter of 2 inches and a slant height of6 inches. What is the lateral surface area of the cone?


Volume of Right Prisms and Cylinders

V = Bh


Example 12.23

Find the volume of the gift box and the juice can.


Volume of General Prisms and Cylinders

V = Bh


Volume of General Pyramids and Cones

V =1

3Bh


Example 12.24

The pyramid of Khafre is 145m high, and its square base is 215mon each side. What is the volume of the pyramid?

http://www.egyptarchive.co.uk/html/khafre_01.html


http://www.egyptarchive.co.uk/html/khafre_01.html

Volume of a Sphere

V =4

3πr3


Example 12.25

An ice cream cone is 5 inches high and has an opening 3 inches indiameter. If the cone is filled with ice cream and given ahemispherical top, how much ice cream is there?


Surface Area of a Sphere

SA = 4πr2


Example 12.26

The diameter of Jupiter is about 11 times larger than that ofEarth. How many times greater is the Surface Area of Jupiter? theVolume of Jupiter?

SA of Jupiter

SA of Earth=

4πR2

4πr2=

(R

r

)2

= 112

V of Jupiter

V of Earth=

43πR3

43πr3

=

(R

r

)3

= 113


The Similarity Principle of Measurement

Theorem

Let Figures I and II be similar. Suppose that some lengthdimension of Figure II is k times the corresponding dimension ofFigure I; that is k is the scale factor:

any length measurement (perimeter, diameter, height, slantheight, etc) of Figure II is k times that of the correspondingmeasurement in Figure I.

any area measurement (surface area, area of base, etc) ofFigure II is k2 times that of the corresponding measurementin Figure I.

any volume measurement (volume, capacity, etc) of Figure IIis k3 times that of the corresponding measurement in Figure I.


Example 12.27

1 The size of television sets are measured by the length of thediagonal of their screen. How many times larger is the screenarea of a 40” TV than a 20” TV?

2 A 2” by 4” by 8” rectangular brick of gold weighs 80 lbs.What are the dimensions of a similarly shaped brick thatweighs 10 lbs?

1 k = 40/20 = 2, so area increases by a factor of k2 = 4.

2 Weight corresponds to volume, so the factor isk3 = 10/80 = 1/8. Thus k = 1/2. So a 10 lb brick wouldhave dimensions 1” by 2” by 4”.


Class Discussion

To find SA of a hemisphere, be sure to add the area of thebase.

When responding to Samuel, remember that he is holding theheight constant.


Homework

Homework 7 (due 4/6/10):

Section 12.3 # 2ac, 3, 4ab, 5, 6, 9, 32

Section 12.4 # 2b, 3c, 4d, 5d, 6ab, 19a, 23


surface area of right prisms and cylinders

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