#$&* course PHY 121 6/9/13 7:55 pm qa areas etc
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Given Solution: `aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2. FREQUENT STUDENT ERRORS The following are the most common erroneous responses to this question: 4 * 3 = 12 4 * 3 = 12 meters INSTRUCTOR EXPLANATION OF ERRORS Both of these solutions do indicate that we multiply 4 by 3, as is appropriate. However consider the following: 4 * 3 = 12. 4 * 3 does not equal 12 meters. 4 * 3 meters would equal 12 meters, as would 4 meters * 3. However the correct result is 4 meters * 3 meters, which is not 12 meters but 12 meters^2, as shown in the given solution. To get the area you multiply the quantities 4 meters and 3 meters, not the numbers 4 and 3. And the result is 12 meters^2, not 12 meters, and not just the number 12. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for area of a triangle is 1/2 base times height. It is also measured in square units. That means that this triangle would be half of 4 times 3, or 1/2 of 12, which is 6 square meters. confidence rating #$&*: 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: 3 ********************************************* 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: The area of a parallelogram is base times height and is measured in square units. Therefore, the area of this parallelogram is 5 times 2, or 10 square meters. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: One half of 5 times 2 or 1/2 of 10, which is 5 square cm. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* 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: The width of this figure at the bottom is 6 units. It comes to a point at the top. On one side the height of the line segment is 7 units. On the other side of the figure, it is 11 units. I would say that the average graph altitude of this trapezoid is 9 units. If I chopped off the piece of the trapezoid above the 9 and made the trapezoid into a triangle, the area of the trapezoid would be 6 times 9, or 54 square units. I think that this is the same as the graph trapezoid because the tiny triangle that you would chop off would come around and nestle into the top of the figure that is left to make a rectangle. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: The average graph altitude of this trapezoid would be (3+8)/2, or 5.5. 5.5 times 4 (the base measurement) is 22 square units. 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: 3 ********************************************* Question: `q007. What is the area of a circle whose radius is 3.00 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for the area of a circle is pi times the radius squared. The radius is 3 cm. The radius squared is 9 square cm. 9 times 3.14 = 28.26 square cm. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just didn't round to the nearest tenth, but I do understand the concept. I just need to be more confident about remembering the formula. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Circumference is 2r times pi. This means 6 times 3.14 or 18.84, which would probably be 18.8 square cm. confidence rating #$&*: 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 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q009. What is the area of a circle whose diameter is exactly 12 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for area of a circle is pi times r squared. In order to find the radius, you have to cut the diameter in half. Therefore you have pi times 6 squared. This is would be 113.04, which I imagine will round to 113 square meters. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I used 3.14 for my multiplier, so I got 113.04. When I multiply by 3.1416, my answer was 113.0976. I am unclear why the 76 just got dropped off and it didn't round to 113.10. ------------------------------------------------ Self-critique Rating: 2
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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): So, because it started with pi as a part of the circumference, you don't need to multiply it out. The answer will be more exact as 49pi. I understand and hope I remember in the future.
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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): I was guessing that the final answer might be about 5, but I am unsure when to round and when to keep things as they are. I understood the basic process, though. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q012. Summary Question 1: How do we visualize the area of a rectangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a rectangle is the number of square units that makes up the rectangle. We can find out how many square units there are by multiplying the length times the width of the rectangle. confidence rating #$&*: 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: 3 ********************************************* Question: `q013. Summary Question 2: How do we visualize the area of a right triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a right triangle is the number of squares and partial squares it takes to fill the triangle. You would make off the number of units by which you are measuring up the side and across the bottom and extend the lines across the triangle. It turns out that it will be exactly half of the base times the height. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not say that you should imagine a second congruent right triangle that can be used to create a rectangle, which you would find the area of and then cut in half. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a parallelogram is base times height. Because the parallelogram is made of 2 sets of parallel line segments, you can actually cut off the end of the parallelogram to make the angles right angles and move the triangle you cut off to the other end of the parallelogram. This fits perfectly into the other end, creating a rectangle. And the base / height of the parallelogram are the same measurements as the length/width of the rectangle. confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the area of a trapezoid, you measure the lengths of the two parallel lines. You add the lengths together and divide by two to get the average height. Then you multiply the average height by the width, and come up with the number of square units. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just didn't remember the term average altitude. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q016. Summary Question 5: How do we calculate the area of a circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a circle is calculated by taking the radius and squaring it then multiplying that number by pi (or 3.1416). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* 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 is pi times the diameter. I do not know a good way to avoid confusing this formula with the formula for the area of a circle. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I like remembering that the r square gives us square units and circumferences aren't measured in square units. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. I have taught 3rd and 4th grades for 23 years now. We have worked on area of rectangles and squares, so those I own. I participated in a math conference where we created, over the course of several months, a lesson on finding the area of a triangle, so I remember that. The area of the parallelogram, for some reason, I remember because it makes sense to me. I don't remember learning about the area of a trapezoid, but it makes sense because I can picture manipulating the piece mentally. I remember pi r squared but didn't remember whether it was for area or circumference. I have had to refer to notes and the back of the Physics book to remember that it was for area and what the formula was for circumference. Hopefully remembering that the r squared means area so 2 times pi times the radius (or pi times the diameter) will mean circumference. This will just take some time. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Repetition is what I need on the circle things. ------------------------------------------------ Self-critique Rating: 2" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!