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course Mth 173 8 / 28 2 : 00 Copy and paste this document into a text editor, insert your responses and submit using the Submit_Work_Form..............................................
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: OK
********************************************* Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given lengths of the legs : base = 4 meters Height = 3 meters Area of a right angled triangle = (1/2)*base*height = (1/2) * 3 meters * 4 meters = 6 square meters confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: OK
********************************************* 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: Given base of the parallelogram = 5 meters Since it is a parallelogram the opposite parallel line has the same length = 5 meters Height ( given ) = 2 meters Height is the distance between the 2 parallel line considered Area of a parallelogram = (1/2)*(sum of lengths of 2 parallel lines)*(distance between the same parallel lines ) Area = (½)*( 5 meters + 5 meters ) * 2 meters = 5*2 = 10 square meters confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given base of the triangle = 5 cm Given altitude = 2 cm Area of any triangle = ( ½ ) * base * height = ( ½ ) * 2 cm * 5 cm = 5cm^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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): Is It necessary to mention about the joining and forming a parallelogram as far as I know the formula to calculate it without imagining the parallelogram. ------------------------------------------------ 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: The given co-ordinates to the trapezoid are ( 3 , 0 ) , ( 3 , 7 ) , ( 9 , 11 ) and ( 9 , 0 ) The line segment joining ( 3 , 0 ) and ( 3 , 7 ) is a straight line and so is the line joining ( 9 , 11) and ( 9 , 0 ) and it can be seen that this lines are parallel to each other. The base or the width is the distance between the parallel lines of a trapezoid which is 9 - 3 = 6 units The 2 parallel lines are the altitudes and the length of which are 7 units and 11 units. The average graph altitude = ( sum of the altitude ) / 2 = ( 7 + 11 ) / 2 = 9 units A rectangle is created with its width same as the trapezoid and the length as the average altitude Thus the width of the rectangle = 9 - 3 = 6 units The length of the rectangle = 9 units Thus the area of the rectangle = width * length = 6 units * 9 units = 54 square units In order to compare its area with that of the trapezoid we place the rectangle with the same base as the trapezoid that is on ( 3 , 0 ) and ( 9 , 0 ) points. Now the length of the rectangle is the average of the height of the altitudes of the trapezoid thus it the parallel line to the base of the rectangle will intersect the slant line of the trapezoid at the midpoint. Now if we compare the figures we see that the rectangle covers most part of the trapezoid leaving out a triangle outside the rectangle. The trapezoid leaves out a triangle inside the rectangle. Since this rectangle are formed at the midpoint of the line parallel to the base and with the average height we can realize that it will have equal areas and fits exactly over each other. Thus the trapezoid and the rectangle have equal areas. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: OK
********************************************* 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 area of a trapezoid in given by the formula = (1/2) * ( sum of the altitudes ) * ( distance between the altitudes ) Sum of the altitudes = 8 cm + 3 cm = 11 cm Distance between the altitudes / width = 4 cm Area of the trapezoid = ( ½ ) * 11 * 4 = 22 square cm Another way is saying that the trapezoid will have the same area as a rectangle with the same width and length as the average of the altitudes Width of the rectangle = 4 cm Length of the rectangle = (1/2) * ( 8 + 3 ) = 11/2 cm Area of the rectangle = area of the trapezoid = length * breadth = 11/2 cm * 4 cm = 22 square cm confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `aThe area is equal to the product of the 'graph width' and the average 'graph altitude'. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q007. What is the area of a circle whose radius is 3.00 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given radius = 3 cm Area of the circle = 3.14(pi) * radius^2 = 3.14 * 3 cm * 3 cm = 28.26 square cm = 28.3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given radius = 3 cm Circumference = 2 * pi * radius = 2 * 3.14 * 3cm = 6.28 * 3 cm = 18.84 cm confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `aThe circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.8 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q009. What is the area of a circle whose diameter is exactly 12 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given diameter = 12 meters Radius = diameter / 2 = 12 / 2 = 6 meters Area of the circle = pi * r^2 = pi * 6 cm * 6 cm = 36pi square cm confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Circumference of the circle = 14*pi meters Circumference of the circle by formula = 2 * pi * radius Equating the given circumference and the circumference formula : 14 * pi meters = 2 * pi * radius Radius = 7 meters Area of the circle = pi * radius^2 = pi * 7 meters * 7 meters = 49pi square meters confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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 ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q011. What is the radius of circle whose area is 78 square meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given area = 78 square meters Area by formula = pi*radius^2 Equating the given area and the area obtained by formula 78 = pi * radius ^ 2 Radius = sqrt( 78/pi) Radius = 4.98 approximated to 5 The negative solution of the root is neglected because radius of a circle cannot be negative. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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 ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q012. Summary Question 1: How do we visualize the area of a rectangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Imagine a rectangle to be made up of a number of squares of unit length. Every rectangle can be divided into a number of square of unit length , the area of each square would be 1 square unit. The total area of the rectangle would be the sum of the total area of all the squares in the rectangle. Since the area of 1 imagined unit length square is 1 sq cm , the total area of the rectangle is nothing but the total number of rectangles. Since it is of unit length the number of square along the length is equal to the value of the length and the same with the breadth , thus the total number of square = length * breadth which is the total area of the rectangle. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q013. Summary Question 2: How do we visualize the area of a right triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We imagine a right angle triangle to be a cut part of the rectangle along one of its diagnol , that is if we add an identical right angle triangle to the original right angle triangle along the hypotenuse what we obtain is a rectangle. The area of the right angled triangle thus will be half of the area of the rectangle = ½ * ( length of the one leg ) * ( length of another leg ) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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 the parallelogram is nothing but the product of the base and the altitude. The altitude is the direct distance that is perpendicular distance between the two parallel bases. Area = base * altitude confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of the trapezoid is calculated by multiplying the average length of the base with the altitude. The 2 parallel lines in case of the trapezoid has different lengths , we need to find the average length and then multiply it with the altitude that is the direct distance or perpendicular distance between the 2 parallel lines Area = (½)*( sum of lengths of the 2 parallel lines ) * ( altitude ) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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 ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q016. Summary Question 5: How do we calculate the area of a circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of the circle is found out by the formula area = pi * radius^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `aWe use the formula A = pi r^2, where r is the radius of the circle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We calculate the circumference of a circle by the formula 2 * pi * radius The circumference of the circle can be differentiated form the area by observing its unit. The area contains the factor radius^2 thus its unit is square cm or square m or any other square length unit whereas the circumference contains only the radius factor thus its unit contains only the cm or meter or any other length unit confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: OK
********************************************* Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. Solution : The knowledge of the principles by the exercises are noted in my book with all the formula noted that I encountered through the exercise. Every formulae was already know to me but it is no harm in noting them again and revising them. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! `gr41