Areas

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course Mth 152

Question: `q001. There are 11 questions and 7 summary questions in this assignment.

What is the area of a rectangle whose dimensions are 4 m by 3 meters.

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Your solution:

4 * 3 = 12 meters ^2

Because to find the area you have to multiply the length times the width.

confidence rating #$&*: 3

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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.

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Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?

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Your solution:

˝ * 12 m = 6.0 m^2 I believe this is the answer because A = L * W = 4.0 m * 3.0 m = 12 m^2 because two of these right triangles make a rectangle and half of that is 6.0 m^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).

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Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?

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Your solution:

A = b * h so A = 5.0 m * 2.0 m , A = 10 m^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.

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Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?

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Your solution:

A = b * h so A = 5.0 cm * 2.0 cm = 10 m^2 so the area of a triangle is A = ˝ * b * h

= ˝ * 5.0 cm * 2.0 cm = 5.0 cm^2

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.

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Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?

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Your solution:

A = b * h = A = 4.0 km * 5.0 km = 20 km^2. Because a trapezoid is a rectangle with two of the sides on a slope.

confidence rating #$&*: 3

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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.

STUDENT SOLUTION ILLUSTRATING NEED TO USE UNITS IN ALL STEPS

A=Base time average altitude therefore………A=4 *5= 20 km ^2

INSTRUCTOR COMMENT

A = (4 km) * (5 km) = 20 km^2.

Use the units at every step. km * km = km^2, and this is why the answer comes out in km^2.

Try to show the units and how they work out in every step of the solution.

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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?

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Your solution:

Altitude = (3 cm + 8 cm) / 2 = 5.5 cm and from this we can find area which is A = b * h, A = 4 cm * 5.5 cm = 22 cm^2.

confidence rating #$&*:: 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.

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Question: `q007. What is the area of a circle whose radius is 3.00 cm?

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Your solution:

The area of a circle is A = pi*r^2 so A = pi * 3.00cm^2

A = 9 pi cm^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.

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Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm?

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Your solution:

C = 2 pi * r = 2 pi * 3 cm = 6 pi 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 = 18.8 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.

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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?

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Your solution:

A = pi*r^2 the radius is half of the diameter so A = pi *(6 m)^2

A= 36 pi m^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. 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.

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Question: `q010. What is the area of a circle whose circumference is 14 `pi meters?

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Your solution:

C = 2 pi r The best way to go about solving this is to solve for “r” so r = C / 2 pi

r = 14 pi m / 2 pi = 14/2 * pi/pi * m = 7 * 1 meters = 7 meters

A = pi * 7 meters ^ 2 = 49 pi meters ^ 2

confidence rating #$&*:: 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.

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.

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Question: `q011. What is the radius of circle whose area is 78 square meters?

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Your solution:

A = pi r^2 so you solve for r A/pi = r^2 you then sqrt both sides making

Sqrt ( A / pi) = r now you take A = 78 m^2 = r = sqrt( 78 m^2 / pi)

= sqrt ( 78 / pi ) = r = 5.0 m

confidence rating #$&*:: 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.

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.

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Question: `q012. Summary Question 1: How do we visualize the area of a rectangle?

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Your solution:

You imagine the rectangle is covered in small squares covering one unit each. You then multiply the number of squares in the width by the number of squares in the length to arrive at the area.

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.

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Question: `q013. Summary Question 2: How do we visualize the area of a right triangle?

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Your solution:

First you should imagine another right triangle put together at the hypotenuse to form a rectangle. Because the area of a rectangle is b * h the area of a triangle must be ˝b * h.

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.

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Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram?

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Your solution:

Area of a parallelogram is found by A = a * b where “a” is altitude and “b” is base.

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.

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Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid?

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Your solution:

Area = average altitude times the width so A = a * w

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.

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Question: `q016. Summary Question 5: How do we calculate the area of a circle?

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Your solution:

A = 2 pi*r^2

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.

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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?

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Your solution:

C = 2 pi r unlike finding the area where the radius is squared.

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.

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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I have written them down into notes on paper so I am able to study them to have a higher success rate of passing the tests.

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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I have written them down into notes on paper so I am able to study them to have a higher success rate of passing the tests.

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Self-critique Rating: 3

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course Mth 173

1/22/11 1:02am

Note: If you do not have the disks for your course, you should not complete this exercise at this time. Complete this once you have your disks.Task: `q000: Verify that you have your disks.

Your response (insert your response beginning in the next line):

Note: If you do not have the disks for your course, you cannot complete this exercise at this time. Complete this once you have your disks.

Checking disks:

I have my disks

Task: `q001. This is for Mth 158 students only.

Disks for your course should have been packaged with your textbook. You should follow the instructions given by the publishers to run your disks.

Please verify below that your disks work and that you have been able to access the material. If not, briefly describe the problem.

Your response (insert your response beginning in the next line):

#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)

Student other than Mth 158 students do as follows:

Task: `q001a. This applies to all courses except Mth 158, which uses publisher disks rather than disks produced by the instructor.

Class notes, in lecture format with video clips, are distributed on the disks you purchased in the bookstore.

If you do not have your disks yet you will have to skip this instruction for now, and you will need to return to this exercise when you receive them. If that is the case you may close this assignment after first entering in the response area below a statement that you do not yet have the disks.

Otherwise run one of the disks for your course

1. Insert the disk you have selected into your drive. Open Windows Explorer and run the HTML file in the root folder (simply locate the file and double-click on it). That file will have a name like disk_1.htm or disk_2.htm, and will be one of very few files in the root folder, so you should be able to locate it easily enough.

• The information on your disk was originally assembled for CD's, and your disk consists of a compilation of a number of CD's. The HTML file in the root folder will give you a list of the CDs collected on your disk.

2. When you run the HTML file you will either get a menu of Class Notes or a series of direct links to video clips.

• If you get a menu of Class Notes, click on one of the links (#3 would be a good choice but any will do). Otherwise go to the instruction #4.

• You will see a page containing notes. If you scroll down the page you will see links to video clips embedded within the notes.

• Click on several of these links to see how they work.

4. If you got a series of direct links, click on one of them in order to see how they work. Click on several more so you will be familiar with the format of these video clips.

Describe what you did and what you saw.

Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)

When I opened the CD, I saw two files for HTML, so I opened both. Both pages have a lot of links. On the first page, all of the links lead me to information that can be found on the other page. So, I suppose the first page I opened has links to the other page I opened, so it is trivial. Although, the second page I opened was actually trivial because all of the links on it did not work. Fortunately, the first page I opened had the same links as the other, and they worked. When I click on the links, I see pictures of examples and questions of the examples, then when I scroll down the page, the questions of the examples are answered. These are the class notes.

For Mth 158 students:

Task: `q001b. If you are in Mth 158 you should have the publisher's lecture series, which should have either been packaged with your text or purchased separately (according to Textbook Information provided elsewhere).

You should follow the publisher's instructions for using those disks. Note that you might need to have QuickTime running before you can access the disks.

Check out some of the disks and describe what you see below:

Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

#$&* (your response should have gone on the line above this one)

For students in all courses other than Mth 158:

Task: `q002. See root folder file and information for a listing of the contents of each disk. You don't need to understand what you're seeing at this point; that will become clear when you begin your content assignments for the course.

Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

#$&* (your response should have gone on the line above this one)

I see the root folder file and information.

Task: `q003.

Check the rest of your disks.

Each disk should be accessed by browsing to the disk and running the file whose name most closely matches the name of the disk, or the HTML file in the root folder of the DVD. Insert each disk in turn into your drive, browse to the appropriate file, and run it.

If all your disks work, indicate that they do. If you have trouble with any of them, or with these instructions, describe your problems in detail.

Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

#$&* (your response should have gone on the line above this one)

All of my disks work.

"

#(*!

@& I'm glad everything appears to be working.

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