course PHY 201 June 7 around 6:00pm 001. Areas*********************************************
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Given Solution: `aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2. FREQUENT STUDENT ERRORS The following are the most common erroneous responses to this question: 4 * 3 = 12 4 * 3 = 12 meters INSTRUCTOR EXPLANATION OF ERRORS Both of these solutions do indicate that we multiply 4 by 3, as is appropriate. However consider the following: 4 * 3 = 12. 4 * 3 does not equal 12 meters. 4 * 3 meters would equal 12 meters, as would 4 meters * 3. However the correct result is 4 meters * 3 meters, which is not 12 meters but 12 meters^2, as shown in the given solution. To get the area you multiply the quantities 4 meters and 3 meters, not the numbers 4 and 3. And the result is 12 meters^2, not 12 meters, and not just the number 12. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK I didn’t put the meters^2 (squared), which I completely forgot about. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If both legs are 4 and 3 meters, then you times 4 by 3 and get 12. But 12 is the area of two triangles together, so you divide 12 into 2, which gives you 6 meters^2 for the area of the triangle. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aA right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h. STUDENT QUESTION Looking at your solution I think I am a bit rusty on finding the area of triangles. Could you give me a little more details on how you got your answer? INSTRUCTOR RESPONSE As explained, a right triangle is half of a rectangle. There are two ways to put two right triangles together, joining them along the hypotenuse. One of these ways gives you a rectangle. The common hypotenuse thus forms a diagonal line across the rectangle. The area of either triangle is half the area of this rectangle. If this isn't clear, take a blade or a pair of scissors and cut a rectangle out of a piece of paper. Make sure the length of the rectangle is clearly greater than its width. Then cut your rectangle along a diagonal, to form two right triangles. Now join the triangles together along the hypotenuse. They will either form a rectangle or they won't. Either way, flip one of your triangles over and again join them along the hypotenuse. You will have joined the triangles along a common hypotenuse, in two different ways. If you got a rectangle the first time, you won't have one now. And if you have a rectangle now, you didn't have one the first time. It should be clear that the two triangles have equal areas (allowing for a little difference because we can't really cut them with complete accuracy). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the area of the parallelogram whose base is 5 meters and whose altitude is 2 meters, you simply times the two together: 5*2= 10 meters^2. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the area of a triangle, you can use the formula A=(bh)/2 or A=1/2bh. So I used the second and most proper, A=.5*5*2= 5 cm^2 for the area of the triangle. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A trapezoid is a shape that has one base smaller than the other, so one width is not the only information I need from the question. If both of the bases are the same then it would be consider a square or rectangle not a trapezoid. But you can still get the basic area. A = 4*5 = 20 km^2 is the area for the trapezoid. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you find the area for the first altitude, which is A = 4*3= 12 cm^2, and to find the area of the second altitude which is A = 4*8 = 32 cm^2. Then you add the two final areas together to get the final area of the trapezoid, which is 44 cm^2. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):2 I am a little rusty on my areas and am very grateful for these exercises to get me brushed up on the material that I forgotten. I see where I messed up and understand the real answer. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `q007. What is the area of a circle whose radius is 3.00 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the area of a circle you use the formula: A = (pi)*r^2. So, in order to find the area of a circle with a radius of 3 cm, you plug it into the formula and solve: A = (pi)*3^2 = 28.27 cm^2 is the area of the circle. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the circumference of a circle you use the formula: C = 2*pi*radius. So, to get the circumference of this circle you plug 3 into the radius and solve: C=2*pi*3= 18.9 cm^2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.8 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q009. What is the area of a circle whose diameter is exactly 12 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the diameter is 12 meters, the radius is half of the diameter: 12/2=6 meters. You plug 6 into the circle area formula which is: A = pi * radius^2 = pi*6^2 = 113.1 cm^2 is the area of the circle. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q010. What is the area of a circle whose circumference is 14 `pi meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, you change the circumference formula around to where you can solve for the diameter: d = C / pi. So, you plug the circumference into the formula and solve for the diameter: d=(14 pi)/ pi = 14 meters. Then divide the 14 be 2 to get the radius, which is 7 meters. Next, use the area for a circle formula to find the area: A = pi*7^2 = 153.9 meters^2 is the area of the circle. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2. STUDENT QUESTION: Is the answer not 153.86 because you have multiply 49 and pi???? INSTRUCTOR RESPONSE 49 pi is exact and easier to connect to radius 7 (i.e., 49 is clearly the square of 7) than the number 153.86 (you can't look at that number and see any connection at all to 7). You can't express the exact result with a decimal. If the radius is considered exact, then only 49 pi is an acceptable solution. If the radius is considered to be approximate to some degree, then it's perfectly valid to express the result in decimal form, to an appropriate number of significant figures. 153.86 is a fairly accurate approximation. However it's not as accurate as it might seem, since you used only 3 significant figures in your approximation of pi (you used 3.14). The first three figures in your answer are therefore significant (though you need to round); the .86 in your answer is pretty much meaningless. If you round the result to 154 then the figures in your answer are significant and meaningful. Note that a more accurate approximation (though still just an approximation) to 49 pi is 153.93804. An approximation to 5 significant figures is 153.94, not 153.86. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: `q011. What is the radius of circle whose area is 78 square meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you switch up the area formula to solve for the radius: r = Square Root (A / pi). Then you plug the 78 meters^2 into the formula and solve for the radius: r = square root (78 / pi) = 5 meters is the radius of the circle. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q012. Summary Question 1: How do we visualize the area of a rectangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Example: if the width and length is 5 meters and 4 meters, you visualize the rectangle for having 5 rows and 4 columns being 1 meter each, which is 5*4= 20 meters^2. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q013. Summary Question 2: How do we visualize the area of a right triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When visualizing the area of the right triangle, we just put another right triangle right over top of the hypotenuse and use the rectangle formula and then divide by two to get the area of the triangle. A = ˝ b*h confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A parallelogram is solved for the area just like a rectangle would be solved: A = b*h. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ok I did not put that the altitude is measured perpendicular to its base though. ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We treat the area of a trapezoid as we do a rectangle. A = l*w confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q016. Summary Question 5: How do we calculate the area of a circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We use the radius^2, which is half of the diameter, and times it by pi: A = r^2 * pi. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe use the formula A = pi r^2, where r is the radius of the circle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To calculate the circumference of a circle, we use the formula: C = 2*pi*r. Finding the circumference does not use squared units, as the area does. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have organized the information by the shapes and specific problems. All of the exercise are different and require the same formulas, but maybe sometimes needs turning around and solving backwards. ------------------------------------------------ Self-critique rating #$&*3 "