Areas

course Phy 201

6/2 11:45

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:

Area = length*width = 4*3 = 12 square meters

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.

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Self-critique (if necessary): OK

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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:

Area = ½ * b*h = .5*4*3 = 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.

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Self-critique (if necessary): OK

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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:

Area = base*altitude = 5*2 = 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.

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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 = ½ *b*h = .5*5*2 = 5.0 square meters

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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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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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:

Area = average of bases*height = 5 km*4 km = 20 square kilometers

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.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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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:

Area = ½*(sum of bases)*width = .5*(3cm+8cm)*4cm = .5*11cm*4cm = .5*44 sq cm = 22 square cm

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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Self-critique (if necessary): OK

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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:

Area = pi*r^2 = pi *3^2 = 9pi square cm or 28.3 square cm

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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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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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 or 18.8 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.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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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:

Diameter = 12 m; radius = 6 m

Area = pi (6 m)^2 = 36 pi square meters or 113 sq m

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): OK

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Self-critique rating #$&* OK

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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:

Circumference = 2 pi r = 14 pi solve for r r= 14 pi/2 pi = 7 meters

Area = pi (r^2) = pi*49 = 49 pi square meters

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.

Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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

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Your solution:

Area = pi* r^2 pi r^2 = 78 m^2

r^2 = 78 m^2/pi r = sqrt(78/pi) meters or 4.98 meters

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.

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Self-critique (if necessary): OK

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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:

The area of a rectangle is the number of units in a row on one side (width) of the rectangle multiplied by the number of rows in the whole figure (length). You use multiplication to find the area to avoid counting each individual square unit.

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:

For the right triangle, you use half of the rectangle formula. This is because two of the triangles will make a single rectangle. You find the number of units in the whole rectangle, with the given dimensions and then take half of it to get the triangle.

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): I think I explained that right.

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Self-critique rating #$&* 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:

A parallelogram is basically a squished rectangle. You use the formula for the rectangle, but instead of the length of the slanted side, you use the altitude. This allows you to ‘cut off’ the triangle edge of the figure and put it on the other end to make a normal 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.

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Self-critique (if necessary): OK

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

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Your solution:

The area of the trapezoid is the average of the uneven sides * the altitude. The average makes the sides even and makes the figure into a normal rectangle, so that you can use the rectangle formula on it.

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?

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Your solution:

The area of the circle is pi times the radius of the circle squared. You measure the radius and then multiply it by pi to revolve it all the way around the circle, to see how many units are contained in the circle.

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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Self-critique rating #$&* 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:

The circumference of the circle is 2 pi r, or pi d units. The formula of the circle will give single units when solved; the area formula will give units squared. Remember to put in the units and you won’t mix them up.

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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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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

With the exercises, I drew lots of pictures labeled with the measurements so that I could visualize the area formulas. I also wrote down the formulas with the figures so that I could associate one with the other. Drawing pictures help a lot with these kinds of questions.

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Self-critique (if necessary): I don’t really know what else to put…

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