Assignment 3

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course Mth 158

1/30/13 around 5:45PM

003. `*   3 

 

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Question: *   R.3.16 \ 12 (was R.3.6) What is the hypotenuse of a right triangle with legs 14 and 48 and how did you get your result?

 

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Your solution:

c = length of hypotenuse

 a^2 + b^2 = c^2

 14^2 + 48^2 = c^2

 196 + 2304 = c^2

 2500 = c^2

 sqrt(2500) = 50

 c = 50

 

 I used the pythagorean theorem.

 

 

confidence rating #$&*: 3

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Given Solution:

* *  ** The Pythagorean Theorem tells us that

 

c^2 = a^2 + b^2,

 

where a and b are the legs and c the hypotenuse.

Substituting 14 and 48 for a and b we get

 

c^2 = 14^2 + 48^2, so that

c^2 = 196 + 2304 or

c^2 = 2500.

 

This tells us that c = + sqrt(2500) or -sqrt(2500).

 

• Since the length of a side can't be negative we conclude that c = +sqrt(2500) = 50. **

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Question: 

 

 

*   R.3.22 \ 18 (was R.3.12). Is a triangle with legs of 10, 24 and 26 a right triangle, and how did you arrive at your answer?

 

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Your solution:

 10^2 = 100, 24^2 = 576, 26^2 = 676

 100 + 576 = 676

 10^2 + 24^2 = 26^2

 

 I used the Converse of the pythagorean theorem.

 

 

confidence rating #$&*: 3

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Given Solution:

* *  ** Using the Pythagorean Theorem we have

 

c^2 = a^2 + b^2, if and only if the triangle is a right triangle.

 

Substituting we get

 

26^2 = 10^2 + 24^2, or

676 = 100 + 576 so that

676 = 676

 

This confirms that the Pythagorean Theorem applies and we have a right triangle. **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

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Question: 

 

 

*   R.3.34 \ 30 (was R.3.24). What are the volume and surface area of a sphere with radius 3 meters, and how did you obtain your result?

 

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Your solution:

 V = 4/3 * pi * r^3

 V = 4/3 * pi * 3m^3

V = 4/3 * pi * 27m^3

 V = 36 m^3 * pi

 or

V = 113.04 m^3

 

 S = 4 * pi * r^2

 S = 4 * pi * 3m^2

 S = 4 * pi * 9m^2

 S = 36m^2 * pi

 or

S = 113.04 m^2

 

 

confidence rating #$&*: 3

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Given Solution:

* *  ** To find the volume and surface are a sphere we use the given formulas:

 

Volume = 4/3 * pi * r^3

V = 4/3 * pi * (3 m)^3

V = 4/3 * pi * 27 m^3

V = 36pi m^3

 

Surface Area = 4 * pi * r^2

S = 4 * pi * (3 m)^2

S = 4 * pi * 9 m^2

S = 36pi m^2. **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

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Question: 

 

*   R.3.50 \ 42 (was R.3.36). A pool of diameter 20 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result?

 

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Your solution:

 The diameter of whole enclosing deck is:

 20ft + 3ft + 3ft = 26ft

 r = 26ft / 2 = 13ft

Area of enclosing deck:

 A = 13ft ^ 2 * pi

 A = 169 ft^2 * pi

or

 A = 530.66 ft^2

 Area of pool:

 r = d / 2

 r = 20ft / 2 = 10ft

 A = 10ft^2 * pi

 A = 100ft^2 * pi

 or

A = 314ft^2

 Area of deck = Area of whole deck - Area of pool:

 530.66ft^2 - 314ft^2 = 216.66ft^2

 

 

confidence rating #$&*: 3

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Given Solution:

Think of a circle of radius 10 ft and a circle of radius 13 ft, both with the same center. If you 'cut out' the 10 ft circle you are left with a 'ring' which is 3 ft wide. It is this 'ring' that's covered by the deck.  The 10 ft. circle in the middle is the pool.

 

The deck plus the pool gives you a circle of radius 10 ft + 3 ft = 13 ft.

 

The area of the deck plus the pool is therefore

 

• area = pi r^2 = pi * (13 ft)^2 = 169 pi ft^2.

 

So the area of the deck must be

 

• deck area = area of deck and pool - area of pool = 169 pi ft^2 - 100 pi ft^2 = 69 pi ft^2. **

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Self-critique (if necessary):

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Self-critique rating:

 

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Question: 

 

*   R.3.50 \ 42 (was R.3.36). A pool of diameter 20 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result?

 

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Your solution:

 The diameter of whole enclosing deck is:

 20ft + 3ft + 3ft = 26ft

 r = 26ft / 2 = 13ft

Area of enclosing deck:

 A = 13ft ^ 2 * pi

 A = 169 ft^2 * pi

or

 A = 530.66 ft^2

 Area of pool:

 r = d / 2

 r = 20ft / 2 = 10ft

 A = 10ft^2 * pi

 A = 100ft^2 * pi

 or

A = 314ft^2

 Area of deck = Area of whole deck - Area of pool:

 530.66ft^2 - 314ft^2 = 216.66ft^2

 

 

confidence rating #$&*: 3

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Given Solution:

Think of a circle of radius 10 ft and a circle of radius 13 ft, both with the same center. If you 'cut out' the 10 ft circle you are left with a 'ring' which is 3 ft wide. It is this 'ring' that's covered by the deck.  The 10 ft. circle in the middle is the pool.

 

The deck plus the pool gives you a circle of radius 10 ft + 3 ft = 13 ft.

 

The area of the deck plus the pool is therefore

 

• area = pi r^2 = pi * (13 ft)^2 = 169 pi ft^2.

 

So the area of the deck must be

 

• deck area = area of deck and pool - area of pool = 169 pi ft^2 - 100 pi ft^2 = 69 pi ft^2. **

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Self-critique (if necessary):

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Self-critique rating:

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