#$&* course Mth 163 1/30 - 8:45am Copy and paste this document into a text editor, insert your responses and submit using the Submit_Work_Form.
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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. FREQUENT STUDENT ERRORS The following are the most common erroneous responses to this question: 4 * 3 = 12 4 * 3 = 12 meters INSTRUCTOR EXPLANATION OF ERRORS Both of these solutions do indicate that we multiply 4 by 3, as is appropriate. However consider the following: 4 * 3 = 12. 4 * 3 does not equal 12 meters. 4 * 3 meters would equal 12 meters, as would 4 meters * 3. However the correct result is 4 meters * 3 meters, which is not 12 meters but 12 meters^2, as shown in the given solution. To get the area you multiply the quantities 4 meters and 3 meters, not the numbers 4 and 3. And the result is 12 meters^2, not 12 meters, and not just the number 12. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Area of a triangle = 1/2 (base * height) NOTE: I looked this up after initially having the idea of “figure it out for a square, and then take half of that”. 1/2 (4m * 3m) 1/2 (12m^2) 6m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. STUDENT QUESTION Looking at your solution I think I am a bit rusty on finding the area of triangles. Could you give me a little more details on how you got your answer? INSTRUCTOR RESPONSE As explained, a right triangle is half of a rectangle. There are two ways to put two right triangles together, joining them along the hypotenuse. One of these ways gives you a rectangle. The common hypotenuse thus forms a diagonal line across the rectangle. The area of either triangle is half the area of this rectangle. If this isn't clear, take a blade or a pair of scissors and cut a rectangle out of a piece of paper. Make sure the length of the rectangle is clearly greater than its width. Then cut your rectangle along a diagonal, to form two right triangles. Now join the triangles together along the hypotenuse. They will either form a rectangle or they won't. Either way, flip one of your triangles over and again join them along the hypotenuse. You will have joined the triangles along a common hypotenuse, in two different ways. If you got a rectangle the first time, you won't have one now. And if you have a rectangle now, you didn't have one the first time. It should be clear that the two triangles have equal areas (allowing for a little difference because we can't really cut them with complete accuracy). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Again here, Area = Base * Height. 5m * 2m 10m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Area of a triangle = 1/2 (base * height) (reference prior triangle question for logic analysis) 1/2(5cm*2cm) 1/2(10cm^2) 5cm^2 5.0cm^2 (this is the answer, due to the number of significant figures) -changed upon secondary review- confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: ********************************************* Question: `q005. Sketch on a set of x-y axes the four-sided quadrilateral whose corners are at the points (3, 0), (3, 7), (9, 11) and (9, 0) (just plot these points, then connect them in order with straight lines). What would you say is the width of this figure, as measured from left to right? If the width is measured from left to right, why does it make sense to say that the figure has 'altitudes' of 7 and 11? Do you agree that the figure appears to be a quadrilateral 'sitting' on the x axis, with 'altitudes' of 7 and 11? We will call this figure a 'graph trapezoid'. You might recall from geometry that a trapezoid has two parallel sides, and that its altitude is the distance between those sides. The parallel sides are its bases. There is a standard formula for the area of a trapezoid, in terms of its altitude and its two bases. We are not going to apply this formula to our 'graph trapezoid', for reasons you will understand later in the course. The 'graph trapezoid' you have sketched appears to be 'sitting' on the x axis. An object typically sits on its base. So we will think of its base as the side that runs along the x axis, the side it is 'sitting' on. The 'graph trapezoid' appears to be 'higher' on one side than on the other. We often use the word 'altitude' for height. This 'graph trapezoid' therefore will be said to have two 'graph altitudes', 7 and 11. What therefore would you say is the 'average graph altitude' of this trapezoid? If you constructed a rectangle whose width is the same as that of this trapezoid, and whose length is the 'average graph altitude' of the trapezoid, what would be its area? Do you think this area is more or less than the area of the 'graph trapezoid'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Questions have been re copied for the solution portion: What would you say is the width of this figure, as measured from left to right? The width is the distance from x=3 to x=9 9-3 = 6 If the width is measured from left to right, why does it make sense to say that the figure has 'altitudes' of 7 and 11? It makes sense because at each “ending point” of the base, the altitudes are, respectively, 7 and 11. Do you agree that the figure appears to be a quadrilateral 'sitting' on the x axis, with 'altitudes' of 7 and 11? yes, although I want to call this a figure with an increasing altitude We will call this figure a 'graph trapezoid'. You might recall from geometry that a trapezoid has two parallel sides, and that its altitude is the distance between those sides. The parallel sides are its bases. There is a standard formula for the area of a trapezoid, in terms of its altitude and its two bases. We are not going to apply this formula to our 'graph trapezoid', for reasons you will understand later in the course. The 'graph trapezoid' you have sketched appears to be 'sitting' on the x axis. An object typically sits on its base. So we will think of its base as the side that runs along the x axis, the side it is 'sitting' on. The 'graph trapezoid' appears to be 'higher' on one side than on the other. We often use the word 'altitude' for height. This 'graph trapezoid' therefore will be said to have two 'graph altitudes', 7 and 11. What therefore would you say is the 'average graph altitude' of this trapezoid? I would call the average altitude the average of the two easily identifiable ones, in this case, 7 and 11. 7+11 = 18 18/2 = 9 9 is the average altitude of this graph If you constructed a rectangle whose width is the same as that of this trapezoid, and whose length is the 'average graph altitude' of the trapezoid, what would be its area? The rectangle would have a width of 6 and a length of 9 a = l*w 6 ”units” * 9 “units” =54 units^2 Do you think this area is more or less than the area of the 'graph trapezoid'? I don’t know. Whether more or less, it will be very close to that of the graph trapezoid. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2.5
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Given Solution: The line segment from (3, 0) to (3, 7) is 'vertical', i.e., parallel to the y axis. So is the line segment from (6, 11) to (6, 0). These line segment form what we call here the 'graph altitudes' of the trapezoid. These line segments have lengths of 7 and 11, respectively. The 'graph altitudes' are therefore 7 and 11. The 'average graph altitude' is the average of 7 and 11, which you should easily see is 9. (In case you don't see it, this should be obvious in two ways: 9 is halfway between 7 and 11; also (7 + 11) / 2 = 18 / 2 = 9) The 'base' of the 'graph trapezoid' runs along the x axis from (3, 0) to (9, 0). The distance between these points is 6. So the 'graph trapezoid' has a 'graph width' of 6. A rectangle whose base is equal to that of this 'graph trapezoid' and whose length is equal to the 'average graph altitude' of our 'graph trapezoid' has width 6 and length 9, so its area is 6 * 9 = 54. If this rectangle is positioned on an above the x axis, with one of its widths running along the x axis from (3, 0) to (9, 0), i.e., so that its width corresponds with the 'graph width' of the 'graph trapezoid', then the other width cuts the top of the trapezoid in half. Most of the trapezoid will be inside the rectangle, but a small triangle in the top right corner will be left out. Also the trapezoid will fill most of the rectangle, except for a small triangle in the upper left-hand corner of the rectangle. The area of this triangle is equal to that of the 'left-out' triangle. It follows that the trapezoid and the rectangle have identical areas. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. For the “which has more area” question, I went back to draw a line through y=9, with that, it is pretty easy to see that the two triangles created are equal, and that one could break off to fill that space. HOWEVER: If given a more complex graph figure, I can foresee difficulties in “just seeing that it works”. I assume that it would work given any type of “graph trapezoid” as we have worked with here. At this point I would not be certain how to rearrange / calculate to solve. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. What is the area of a 'graph trapezoid' whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Following the same procedure, we first need to determine the average altitude of this GT “graph trapezoid” 3+8 = 11 11/2 = 5.5 cm is the average height. Now we can calculate area by pretending we have a rectangle with base = width and height = altitude. In this case Base = 4 and height = 5.5 A= 4 units * 5.5 units 22 units^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe area is equal to the product of the 'graph width' and the average 'graph 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: ********************************************* Question: `q007. What is the area of a circle whose radius is 3.00 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Area of a circle = pi * r^2 where pi = 3.14 Area = pi * 3cm^2 ( SELF NOTE: (written AFTER the note below) : the best way to write this to avoid confusion is: pi * (3 cm)^2. Remember the exponent applies to the 3, and the cm SEPARATELY) pi * 9cm^2 (SELF NOTE: The exponent still appears here because we are dealing with the area of something. It’s easier to see WHY if we re-write the equation as as Area = pi * (r * r). Otherwise, it might seem as if we have just arbitrarily, or accidentally left the exponent in the equation 28.26cm^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 with basic math. Concepts to remember / practice: -Significant Figures: Referencing the number of digits that matter in accordance to the way they were measured. This has to do with the way they were written (decimal points, zero’s in certain places. Until you have this mastered, reference the following as a good review / reference when needed: http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs4.html -Exact v. Approximate answers: An exact answer is one that is not rounded, and has an ending point (unlike pi, the number will terminate). In some cases, an exact answer isn’t what we would like to call an “answer”, because it is still in the form of an equation. Nonetheless, if we solved that equation as is, we would receive an exact answer. We get approximate answers when we use rounded figures or we “cut short” numbers (such as using 3.14 for pi in this problem). ANY SOLUTION in which we use 3.14 for pi as we have done here, will result in an approximate solution. ????? Which is common practice, leaving the answer in an exact format, or solving it to an approximation? ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Circumference = 2pi r 2 * pi * 3cm 2 * 3.14 * 3cm 6.28 * 3cm 18.84cm Re-Write after review: While my above solution is correct for an approximate solution, the equation progression should be written as follows, so that we can identify an exact solution as well: 2 * pi * 3cm 6pi This is the exact solution. To solve further would be an approximation because we would substitute 3.14 (an inexact value) for pi. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 = 18.8 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. Just remember to leave pi as “pi” until I must approximate it to solve further. This will allow me to identify an exact and an approximate solution for every problem. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. What is the area of a circle whose diameter is exactly 12 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We need the radius to solve, not the diameter: Radius = 1/2 diameter r = 12m / 2 r = 6m Now we can solve for the area a = pi r^2 pi 6m ^2 pi 36m^2 Here is our exact solution. 3.14 * 36m^2 113.04m^2 Here is our approximate solution. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 ????? Is there a reason that I missed as to why you chose 5 significant figures in your response?
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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. STUDENT QUESTION: Is the answer not 153.86 because you have multiply 49 and pi???? INSTRUCTOR RESPONSE 49 pi is exact and easier to connect to radius 7 (i.e., 49 is clearly the square of 7) than the number 153.86 (you can't look at that number and see any connection at all to 7). You can't express the exact result with a decimal. If the radius is considered exact, then only 49 pi is an acceptable solution. If the radius is considered to be approximate to some degree, then it's perfectly valid to express the result in decimal form, to an appropriate number of significant figures. 153.86 is a fairly accurate approximation. However it's not as accurate as it might seem, since you used only 3 significant figures in your approximation of pi (you used 3.14). The first three figures in your answer are therefore significant (though you need to round); the .86 in your answer is pretty much meaningless. If you round the result to 154 then the figures in your answer are significant and meaningful. Note that a more accurate approximation (though still just an approximation) to 49 pi is 153.93804. An approximation to 5 significant figures is 153.94, not 153.86. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK with process / logic, but questions regarding the determination & use of significant figures: ????? Are significant figures determined by given information in the problem (in this case: 49 is given, having 2 significant figures), or do WE HAVE THE OPTION of determining the number of significant figures based on how far we extend the value of pi? In my case I used 3.14, meaning I should have 3 significant figures in my solution (154 instead of 153.86, with proper rounding). If the latter, then I assume that in any situation in which we are given a value such as pi, that whatever we substitute for that will be our necessary number of significant figures. ????? Also, thinking as practically as possible, I imagine with these problems finding a solution in which we can actually use the information. For example a carpenter building something, or computer-machined parts construction. I understand the difference between an exact and approximate solution, but I am not sure of how to USE an exact solution in the real world, given that we technically can’t have a completely “exact” solution (in a measurable form as we would measure for the examples listed above) due to the extending value of pi. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. What is the radius of circle whose area is 78 square meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this problem we need to use the Area equation. However, we are already given the answer, and just need to solve the equation for “r” Area of a circle = Pi r^2 78m^2 = pi r^2 78m^2 / pi = r^2 SQRT (78m^2 / pi) = r This would be an exact solution, as we cannot solve any further without approximating the value of pi. Also, it is a positive square root, not + / -, as a negative radius doesn’t make any sense if we are thinking about a physical object. SQRT (78m^2 / 3.14) = r As per my question above, I am using 3.14 for the approximated value of pi. Therefore my answer should have 3 significant figures. If I am incorrect, then I should only have 2 significant figures because we are given 78. However, I will be solving with 3 significant figures in this case, and will note corrections later. SQRT (24.8m^2) = r 4.98m = r confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2.5
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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. STUDENT QUESTION Why after all the squaring and dividing is the final product just meters and not meters squared???? INSTRUCTOR RESPONSE It's just the algebra of the units. sqrt( 78 m^2 / pi) = sqrt(78) * sqrt(m^2) / sqrt(pi). The sqrt(78) / sqrt(pi) comes out about 5. The sqrt(m^2) comes out m. This is a good thing, since radius is measured in meters and not square meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ????? Given that I used 3 significant figures instead of 2, is my solution considered to be a closer approximation? Is my answer acceptable? ------------------------------------------------ Self-critique Rating:
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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: ********************************************* Question: `q013. Summary Question 2: How do we visualize the area of a right triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Two right triangles would give us a square, and the area of a square = L * W Given that it takes 2 right triangles to create a square, we simply take half of that value: 1/2 (L * W) VISUALLY speaking, we imagine the “second half of the square” and then use only half of the counted unit value. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a parallelogram is base * height. (CORRECTION AFTER REVIEW: Base * Altitude) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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): Keep altitude and height separate. Height typically refers to the physical line of the object (sides of a rectangle), whereas altitude is the distance from base to top of object. Reference “graph trapezoid” for review if necessary. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Due to a trapezoid having an uneven altitude (when it is turned on it’s side, as with our graph trapezoid) OR uneven sides, we have to get an average value to solve. We do so by adding up the values of the bases (or in our graph trapezoid case, the “sides” because we were working with altitude) and dividing by two. We then use that number (which is our average for the Base value) and multiply Base * Altitude NOTE: In our case, we had turned your “typical” trapezoid on its side, therefore OUR explanation would be as follows: We have to determine an average altitude, because it changes, so we add the altitude of one side to that of the other then divide by 2. We use that number, which is our average altitude, and multiply it by our base value Depending on the rotation of the trapezoid you see in front of you, you’ll either solve for an average base value, OR an average altitude value. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. Despite my rather lengthy explanation / observation, I feel as if I’m understanding of the manipulation of trapezoids to determine area. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q016. Summary Question 5: How do we calculate the area of a circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a circle is equal to pi r^2 In the event that we are not directly given the value of r (radius), we must work / rearrange the equation accordingly to determine its value before we can solve. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Circumference of a circle = 2pi * r To not confuse area and circumference equations, I remember that solutions to area problems will always be in the form of units squared, and the area equation has an exponent of 2. The circumference equation does not. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `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): First, I remember that with area, we are solving for the same thing, just for different shapes. Secondly, I categorize shapes with straight lines as separate from circles. “Straight line” shapes always relate to base * altitude in some fashion. Some cases however, simply have to be averaged due to changes in base or altitude values. Once the necessary averages are determined, we follow the normal procedure of base * altitude. In the case of triangles, however, I remember that it takes TWO of them to create a base * altitude situation. Once that is determined, I know to take half of that value, because I doubled the figure I originally had, in order to make the problem more approachable. In the case of circles, we have area and circumference. I relate everything about circles to r (radius), and make determining the radius a priority. Once r is determined, it is straightforward to solve for circumference or area. I remember area vs. circumference in the manner in which I explained in the previous question. I DO NOT know how to explain the incorporation of pi into the circle equations. I’m not sure how that was determined / came to be / why exactly it works / who had the time to figure out such a thing! While I can solve the equations just fine, I don’t feel as if I can explain WHY the equations work for circle are / circumference. ????? If appropriate for the scope of this course, can you explain OR provide resources that explain the proofs for the area / circumference equations? If i don’t need to worry about it, then I’ll just use the equation as is. I tend to feel more confident with things if I am able to break them down and explain them from the ground up, from multiple angles.