#$&* course Mth 158 1/30/13 around 5:45PM 003. `* 3
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Given Solution: * * ** The Pythagorean Theorem tells us that c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse. Substituting 14 and 48 for a and b we get c^2 = 14^2 + 48^2, so that c^2 = 196 + 2304 or c^2 = 2500. This tells us that c = + sqrt(2500) or -sqrt(2500). • Since the length of a side can't be negative we conclude that c = +sqrt(2500) = 50. ** ********************************************* Question: * R.3.22 \ 18 (was R.3.12). Is a triangle with legs of 10, 24 and 26 a right triangle, and how did you arrive at your answer? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10^2 = 100, 24^2 = 576, 26^2 = 676 100 + 576 = 676 10^2 + 24^2 = 26^2 I used the Converse of the pythagorean theorem. confidence rating #$&*: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** To find the volume and surface are a sphere we use the given formulas: Volume = 4/3 * pi * r^3 V = 4/3 * pi * (3 m)^3 V = 4/3 * pi * 27 m^3 V = 36pi m^3 Surface Area = 4 * pi * r^2 S = 4 * pi * (3 m)^2 S = 4 * pi * 9 m^2 S = 36pi m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.3.50 \ 42 (was R.3.36). A pool of diameter 20 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The diameter of whole enclosing deck is: 20ft + 3ft + 3ft = 26ft r = 26ft / 2 = 13ft Area of enclosing deck: A = 13ft ^ 2 * pi A = 169 ft^2 * pi or A = 530.66 ft^2 Area of pool: r = d / 2 r = 20ft / 2 = 10ft A = 10ft^2 * pi A = 100ft^2 * pi or A = 314ft^2 Area of deck = Area of whole deck - Area of pool: 530.66ft^2 - 314ft^2 = 216.66ft^2 confidence rating #$&*: 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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!