course Phy 201 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters? ********************************************* Your solution: A = 1/2 b*h A = ½(4)(3) = 6 square meters Confidence Assessment: OK
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Given Solution: `aA right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters? ********************************************* Your solution: A = b*h A = 5(2) = 10 square meters Confidence Assessment: 3
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Given Solution: `aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm? ********************************************* Your solution: A = 1/2b*h A = ½(5)(2) = 5 square centimeters Confidence Assessment: 3
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Given Solution: `aIt is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km? ********************************************* Your solution: A = b*h A = 4(5) = 20 square km Confidence Assessment: 2
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Given Solution: `aAny trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the right answer, but it was completely on a whim. I could not remember the formula for area of a trapezoid. It makes sense that you could put two trapezoids together and get a rectangle. I had never thought of it that way. I guess the way I wrote the formula would technically be correct because you would have to have two equal altitudes, so it would just be the base multiplied by the height. Self-critique Rating: 3 ********************************************* Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm? ********************************************* Your solution: 3.0 + 8.0 = 11.0/2 = 5.5 = average height of altitude A = b*average altitude A = 4cm*5.5cm =22 square cm Confidence Assessment: 3
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Given Solution: `aThe area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q007. What is the area of a circle whose radius is 3.00 cm? ********************************************* Your solution: A = π * r^2 A = π * (3)^2 =π * 9 = 9π = 28.27 cm squared Confidence Assessment: 3
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Given Solution: `aThe area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm? ********************************************* Your solution: C = 2π * r =2π * 3 =6π = 18.85 cm Confidence Assessment: 3
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Given Solution: `aThe circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q009. What is the area of a circle whose diameter is exactly 12 meters? ********************************************* Your solution: Radius = diameter/2 Radius = 6 A = π * r^2 =π * 36 = 36π = 113.097 meters squared Confidence Assessment: 3
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Given Solution: `aThe area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q010. What is the area of a circle whose circumference is 14 `pi meters? ********************************************* Your solution: C = 2π * r 14π = 2π * r r = 7 A = π * r^2 A = 49π = 153.94 m squared Confidence Assessment: 3
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Given Solution: `aWe know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK Self-critique Rating:OK ********************************************* Question: `q011. What is the radius of circle whose area is 78 square meters? ********************************************* Your solution: A = π*r^2 = 78 r = sq root (78/π) = 4.98 meters Confidence Assessment: 3
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Given Solution: `aKnowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q012. Summary Question 1: How do we visualize the area of a rectangle? ********************************************* Your solution: You can visualize the rectangle being broken down into 1-unit squares. You will notice that if you count all of the squares, you will get the same number as if you multiplied the number of squares that are along two conjoining sides. Confidence Assessment: 3
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Given Solution: `aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: OK ********************************************* Question: `q013. Summary Question 2: How do we visualize the area of a right triangle? ********************************************* Your solution: If two right triangles of with congruent hypotenuses were joined together, they would form a rectangle. Therefore, you can visualize a right triangle as being ½ of a rectangle. You can use this visualization to say that the area would then be ½(base)(height) which would be the same thing as saying ½(length)(width). Confidence Assessment: 3
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Given Solution: `aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram? ********************************************* Your solution: The area of a parallelogram is equal to the length of the base multiplied by the height of the altitude. Confidence Assessment: 3
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Given Solution: `aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid? ********************************************* Your solution: The area of a trapezoid is found by multiplying the width of the trapezoid by the average of the altitudes. Confidence Assessment: 3
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Given Solution: `aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should’ve mentioned which way the trapezoid needed to be oriented. Self-critique Rating: 3 ********************************************* Question: `q016. Summary Question 5: How do we calculate the area of a circle? ********************************************* Your solution: The area of a circle is found by multiplying π times the value of the radius squared. Confidence Assessment: 3
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Given Solution: `aWe use the formula A = pi r^2, where r is the radius of the circle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle? ********************************************* Your solution: The circumference of a circle is found by multiply the value of the radius by 2π. You can keep these formulas straight because area is measured in square units and the formula for area calls for the value of the radius to be squared. Confidence Assessment: 3
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Given Solution: OK `aWe use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK Self-critique Rating: ********************************************* Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel like this assignment has helped me to jog my memory on all of the different formulas for area. I have not used these formulas probably since classes in high school. I somehow managed to remember most of them. I feel that the way the assignment was set up allowed me to work through the problems and to organize my thoughts. The review questions really drove home the concept of area because I had to spell out the formulas instead of writing down a few symbols and letters. Overall, I feel that the assignment was beneficial. Self-critique Rating: OK