Part4 Areas 

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course Mth 151

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: Finding the area of a rectangle we must multiply the length by the width. So, for this particular rectangle we are multiply 4 meters from the width (or height) of the rectangle by 3 meters from the height (or width) of the rectangle, to get the area, which would be 12meters^2. We not only multiplied the length and width, we also multiplied the units of measure, which is why the answer is left with meters squared (meters to the second power).

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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Self-critique (if necessary): I thought 12 meters to the second was right, I’m glad I stuck with my gut on that.

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Self-critique Rating:ok

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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: I googled the Formula… I’m sorry… It’s been years. Anyway. Area equals Base times height divided by two. So, 4.0 meters times 3.0 meters = 12 meters^2, so we are dividing 12m^2 by 2, which gives us 6 m^2.. Simple enough.

confidence rating #$&*: 2

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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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Self-critique (if necessary):ok, I’m sorry I used google, but it did freshen my memory.

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Self-critique Rating:ok

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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: I’m so familiar with what a parallelogram is, because in gymnastics we used mats in the shape of parallelograms. I also know that parallelograms are just rectangles in a special type of way. With that being said, I’m pretty sure to find their area the same processed is being used as it would be to find the area of a rectangle. So, 5 meters times 2 meters, which equals 10 meters^2.

confidence rating #$&*:

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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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Self-critique (if necessary):ok

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Self-critique Rating:ok

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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: area of a triangle, I think that’s base times height divided by two, right?? So, that would be multiplying the base, 5 centimeters, by the height (or altitude), 2 centimeters, which would give us 10 centimeters^2… Now, dividing 10cm^2 by 2, which would give us 5 cm^2…

confidence rating #$&*:2

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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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Self-critique (if necessary): Did I work mine out a different way?? I think I did. But, I think we were doing the same things… We got the same answer.

@&

Your solution was good, but it's also worth understanding the connection with the parallelogram (the area of the triangle being half the area of the parallelogram, which is divided in half by the diagonal).

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Self-critique Rating:ok

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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'?

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Your solution:

The width from left to right would be the distance from the point on (3,0) to the point at (9,0)

Why does it make sense to say that the figure has 'altitudes' of 7 and 11? If altitude is the height of an object or point in relation to sea level or ground level. In this case the x axis would be our “ground level”, so it only makes sense that the altitude, be 7 to the incline of 11 since that is how the object is off the ground (or axis).

Do you agree that the figure appears to be a quadrilateral 'sitting' on the x axis, with 'altitudes' of 7 and 11? In my sketch, and just by logic it would be clear to see that the object is sitting flat on the x axis seeing how both point have the same y-point

What would you say is the 'average graph altitude' of this trapezoid? To get the average graph altitude I would find the average of 7 and 11, which would be 9. I would say this is the ‘average graph altitude’ of the given 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?

Well, the rectangle would have a width of 6 and a length of 9, this would be multiplied to find the area, 9*6= 54…

Do you think this area is more or less than the area of the 'graph trapezoid'?

I believe the ‘graph trapezoid’ would have a larger area. I believe this because we are only using the average of the trapezoids altitude to find the rectangles area, while, the trapezoid still has that extra point (triangle type thing) at its top we haven’t fully calculated when we found the area of the rectangle.

confidence rating #$&*:2

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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.

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Self-critique (if necessary): Yay!! I was right about the extra triangle! I was so afraid I wasn’t making since… But, I got it right.

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Self-critique Rating:ok

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Great.

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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?

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Your solution: first, we must find the graph altitude average; we would add 3cm to 8cm to get 11cm, then divide that product by 2, to get our graph altitude average, which would be 5.5 cm.

Now, we can find the area of the graph trapezoid by multiplying its width by the average altitude we just found. So, 4cm.*5.5cm=22cm… I still don’t believe this calculating that top triangle, is it?????

confidence rating #$&*::2

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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.

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Self-critique (if necessary): I still don’t believe this calculating that top triangle, is it????? Is this area just the average area of the figure???

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Self-critique Rating: ok

@&

You don't have to calculate the top triangle to get the area of the trapezoid.

However it's good to know that you can, if necessary, get the trapezoid area by calculating the area of that triangle and the rectangle that's left beneath it.

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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: Okay, I know this one! Pi times r squared will give us the area of a circle. I’m not sure how to type Pi in here on Microsoft Word… When we do calculations with Pi should we use the Pi button on our calculators, or use 3.14??? I’m going to use 3.14 until you can confirm that, I just prefer the 3.14… Anyway… 3.14*3cm.^2 this equation will give us the area of our circle. First, 3cm squared is 9cm. Next, we multiply 9 cm. by Pi.. 9cm*3.14= 28.26^2cm.

So, the area of our circle is 28.26^2 cm.

confidence rating #$&*:

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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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Self-critique (if necessary): I did not round to 28.3 cm^2, I forgot about the rules of sig.figs… Don’t we all try to forget those after Chemistry? Anyway, I understand the rule, and now that it’s brought back to my attention I will be sure to play by the rule.

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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: Thank you for giving the formula in the previous question, although my knowledge on circles isn’t too terrible. To find the circumference we multiply 2 by Pi, then by the radius, so Circumference =2*Pi*Radius. Now, we just have to plug our given numbers in, simple. C=2*Pi*3cm… 2*3.14=6.28; 6.28*3cm=18.84cm Since our given radius has 3 sigfigs we must ensure our answer doesn’t exceed three as well, so we will round and make our answer 18.8cm.

This time the units of measurement are not raised to a power because we did not multiply units by units, and we did not square any units.

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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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Self-critique (if necessary):All bases were covered! Wohoo!

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Self-critique Rating:ok

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Way to go.

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Question: `q009. What is the area of a circle whose diameter is exactly12 meters?

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Your solution: Pi times r^2 is typically how we achieve finding the area of a circle. In this particular question we are given diameter, though. Diameter is easily converted to give us our radius. Diameter is just two times what the radius is. So, we may divide 12 by 2 to get our radius of 6 meters. Now, all we have to do is solve. Pi*6meters^2...

6meters^2= 36 meters^2

3.14*36meters^2=113.04meters^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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Self-critique (if necessary): I used 3.14 instead of adding more sigfigs to Pi…I know this doesn’t make my answer wrong. As you’ve stated in your solution, using less sig figs makes my answer less accurate…Which, in some businesses those hundredths of a point can be crucial…

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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: Finding the area we would use pi*r^2

But we are only given the circumference. With a little division, we could use the Circumference formula to find our radius since we already know what the circumference is. C=2*Pi*Radius is the formula we will be utilizing to find the Radius.

So, we will Solve for the radius ensuring that the circumference will be used to find the R… This tells us the equation is going to read R= something… And it does. R equals Circumference over 2 times pi. Next, we plug in 14 pi m. to the given C in the equation… R= 14 pi m over 2*pi

From here, we will multiply 2 and pi to get 6.28 so the equations read R=14 pi m./6.28

So, we will now divide… 14 pi m./6.28 which gives us 2.23 pi m.

Since that was the answer that equaled R we now have our radius we can plug into the Area equation pi*r^2

Pi*2.23 pi m.^2=A

2.23 pi M^2= 4.97

4.97 pi m^2*Pi=15.61

So, our area would be 15.61 pi m^2

confidence rating #$&*:0 I don’t even think it’s right…

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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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Self-critique (if necessary): Okay, after I got my equation R=14 pi m/2pi I multiplied 2 by pi instead of dividing 14 by two and Pi by Pi, which is such a better idea… So, I’m now changing my answer since I fully understand what I’m doing now. Learning! YAY! That’s what we’re here for.

Okay, so. After getting R=14 pi m/2pi I’m going to divide 14 by 2 which will give me 7 and dividing pi by pi will give me one (anything divided by itself will give you 1). This leaves the unit of measure, meters, in its place on the top of the fraction, so no denominators are to be a bother. Now, the equation is Radius=7 meters… We found our Radius! The real one! Next to plug in the radius to the Area formula. A=pi* r^2 A=pi*7 meters^2

7^2=14 meters^2Pi

We are leaving Pi as Pi to not over complicate things as you said in the Instructor response. If you wouldn’t have said to leave it, I surely would have multiplied it with the 14 meters^2….

Anyway, my new and correct answer is: 14meters^2 Pi

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Self-critique Rating:3

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Excellent.

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

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Your solution: Just like the previous question we are going to manipulate one formula to help us find the answer. The formula of choice is A=Pi*r^2 and we will need to solve for r. r= the square root of A over pi… (I don’t know how to show a square root on the word processor, sorry). Plugging in the given 78m. to the Area spot in the formula… R=SquareRoot(78m/pi) Dividing 78m by pi, and getting the square root of that number leaves us with 4.98 m. So, we can say our Radius, 5 (I rounded up), is found…. 5 Meters

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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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Self-critique (if necessary): Maybe it wasn’t like the other equation since we didn’t have to manipulate one question to find the missing piece to put in another, but, I still got the right answer, and went through the same steps you did to get there!

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Self-critique Rating:ok

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

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Your solution: Dividing the rectangle into rows and squares each square equalizing one unit. Then, all we have to do is multiply the number of squares in one row by the number of rows… This is how I was taught when I was younger, looking back now, I guess I’m glad for what I learned and didn’t appreciate back then.

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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Self-critique (if necessary):ok

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Self-critique Rating:ok

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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: A right triangle is just half of a rectangle. So, we can find the area of the triangle by multiplying what would be a rectangles length by using the triangles base, then multiply what would be the rectangles base by using the triangles altitude (or height), then, since we just found the area of two right triangles put together, we will divide it by two since a right triangle is half a rectangle.

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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Self-critique (if necessary):ok

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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= base times height finding the area of parallelogram is similar to finding the area of a rectangle since a parallelogram is a rectangle when the tip is cut off and put on the other end of the shape.

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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Self-critique (if necessary): I think I said something to that nature…

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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 of a trapezoid, is found by length times width, which is just the same as finding the area of a rectangle. But, we have to find the average of its altitude to use as the length, and then multiply that number by the width, which in turn, gives us the area.

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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Self-critique (if necessary):ok

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Self-critique Rating:ok

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

Your solution: When given the radius of the circle, the area may be found by multiplying Pi by the radius squared. Otherwise known as Area= 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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Self-critique (if necessary):ok

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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 is the formula to find the circumference of a circle… If we were to get an answer with square units we would know we are using the Area formula, because the circumference formula will not give us an answer with units squared.

onfidence Assessment: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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Self-critique (if necessary):ok

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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): Luckily I have memorized most of the formulas on the formula sheet I used to get in math class “back in the day.” These questions were basically using the formulas to solve the problems at hand.

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&#Good responses. See my notes and let me know if you have questions. &#