#$&* course Mth 152 Dec 6 - 1:40am 022. ``q Query 22
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Given Solution: By 'trimming' a triangle from one 'end' of the parallelogram, rotating it 180 deg and pasting it onto the other 'end', we get a rectangle whose base is equal to the base of the parallelogram, and whose width is equal to its altitude (the altitude is the distance between the base and its parallel side; the distance between two parallel lines is measured at a right angle to the lines). The formula for the area of a parallelogram is therefore A = bh where A stands for the area, b for the base and h for the altitude. We take the side of length 4 inches as the base. The segment of length 2.5 inches constitutes an altitude, since it makes a right angle between the base and its parallel side. Thus we have A = 4 in* 2.5 in = 10 in^2. The area is 10 in^2. It is important that you see how this parallelogram can be rearranged into a rectangle with dimensions 4 in by 2.5 in, that this rectangle is easily subdivided into 8 one-inch squares and 4 rectangles each 1 inch by 1/2 inch, that the 4 rectangles can then be rearranged into two 1-inch squares, so that the area of the rectangle is equivalent to the area of ten 1-inch squares (i.e., the area is 10 square inches, expressed as 10 in^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.3.18 A trapezoid has parallel sides of lengths 5 cm and 4 cm. A line segment of length 3 cm runs from one of these sides to the other, making an angle of 90 degrees with the 5 cm side. What is the area of the given trapezoid and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A trapezoid with parallel sides of 5cm and 4cm A line segment of 3cm runs across making a right angle with the 5cm side. The trapezoid in question is set up to become a square with a little adjusting. So to find the area we need to do some readjusting and turning so we can use the area formula easily for a square. The figures we have are 3cm, 5cm, and 4cm Area formula: 1/2 h (b + B) 1/2 (3 cm) (4 cm + 5 cm) 1/2 (3 cm) (9 cm) 1/2 (27 cm^ 2) A = 13.5 cm^2 If we re-adjust the length of the adjusted rectangle the length of the side becomes 4.5cm. We can use this figure with 13.5cm^2 to find the area. 4.5 cm * 5 cm = 13.5 cm^2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A trapezoid is easily rearranged into a rectangle as follows: Cut each of the nonparallel sides along a line, which we will call L, through its midpoint and perpendicular to one of the parallel sides. You will have cut off a triangle from each side. Rotate each triangle 90 degrees and paste it to the same end from which it was cut, with the cut edge running along line L. You end up with a rectangle whose base has a length halfway between the lengths of the original parallel sides. If the parallel sides had lengths b and B, the newly formed rectangle therefore has base 1/2 ( b + B). The width of the rectangle is an altitude of the original trapezoid. The rectangle therefore has base 1/2 (b + B) and altitude h, so its area is the product 1/2 (b + B) * h, which in standard form is 1/2 h ( b + B). The formula for finding the area of a trapezoid is therefore A = 1/2h ( b + B ). For the given trapezoid: h = 3 cm, b = 4 cm, B = 5 cm A = 1/2 (3 cm) (4 cm + 5 cm) A = 1/2 (3 cm) (9 cm) A = 1/2 (27 cm^2) A = 13.5 cm^2 The area is 13.5 cm^2. This is easily reconciled with the given construction: When the sides are trimmed the resulting rectangle has length halfway between the 3 cm and 4 cm lengths of the bases of the trapezoid. So the rectangle has length 4.5 cm. The width of the rectangle is equal to the 5 cm altitude of the trapezoid. So the area of the rectangle is 4.5 cm * 5 cm = 13.5 cm^2. You should be able to show how this rectangle can be divided into 1 cm squares and 1 cm * .5 cm rectangles to geometrically demonstrate how the area must be 13.5 cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.3.24 The length of a rectangle is 20 more than its width, and its perimeter is 176. What are its dimensions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Length of rectangle is 20 more than its width = W+20 Perimeter = 176 If we know what the perimeter is we can use the formula to separate the perimeter to find the sides. The perimeter = 2(1ength) +2(width). Like I wrote before L= W+20 2 (20 + w) +2w = 176 40 + 2w + 2w = 176 40 + 4w = 176 4w = 136 w = 34 The width is 34 Now we have to find the length. We know the perimeter is 2 times the length and 2 times the width. So half the perimeter is 34*2. We already know that the length is the width + 20 so the w is 54. 2(54) + 2(34) = 176 Which is the correct perimeter. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe perimeter of a rectangle is the distance around it, the sum of the lengths of its sides. The perimeter is therefore double the length plus double the width. In symbols we can write this as P = 2l + 2w, where l and w are length and width. We are told that the length is 20 more than the width, so Length = 20 + w Perimeter = 176 This gives us the equation 176 = 2 (20 + w) + 2w which we proceed to solve for w: 176 = 40 + 2w + 2w by the Distributive Law. Simplifying we get 176 = 40 + 4w Subtract 40 from both sides to get 136 = 4w Divide both sides by 4 to get w = 34 l = w + 20 so l = 34 + 20 = 54 We therefore have Length = 54 and Width = 34 Checking, we have perimeter = 2 * length + 2 * width so we should have 176 = 2(54) + 2(34). The right-hand side does give us 176 so the solution checks out. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.3.48 A trapezoid has bases x and x+4. Its altitude is 3 and its area is 30. What is the value of x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A trapezoid has bases x and x+4 Altitude = 3 Area = 30 What is x? First we look for clues to help us. We know the altitude or height is 3. We know what the area of the trapezoid is but not its bases or sides. To find the area we use the formula: A = 1/2h (b + b). ½(3) (x + (x + 4) = 30 1/2 *3 (2x + 4) = 30 3 ( 2x+4) = 60 6 x + 12 = 60 x = 8 The bases then are 8 and 8+4=12 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The formula for finding the area of a trapezoid is A = 1/2h ( b + B). We have A = 30 h = 3 B = x + 4 b = x Substituting into the formula A = 1/2 h ( b + B) we have 30 = 1/2 * 3 ( x + x+4). Since x + x + 4 = 2x + 4 we have 30 = 1/2 * 3 (2x + 4). Multiplying both sides by 2 in order to 'clear' denominators we have 60 = 3 ( 2x+4). By the distributive law we get 60 = 6 x + 12. Subtract 12 from both sides to get 48 = 6x Divide both sides by 6 to obtain x = 8. The answer is 8. To check the answer: The bases of the trapezoid are x and x + 4; if x = 8 the bases are therefore 8 and 12. The altitude is 3. A trapezoid with bases 8 and 12 has 'average' base 10 (i.e., this trapezoid can be rearranged into a rectangle having length 10). Its altitude is 3 (which would correspond to the width of the equivalent rectangle). Its area is therefore 10 * 3 = 30, which agrees with the given area. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.3.54 It costs $60 for enough paint to cover a ceiling of 9 ft x 15 ft. How much does it cost to paint a ceiling with dimensions 18 ft x 30 ft? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A ceiling 9ft by 15ft costs $60 to paint. We want to paint a 18ft x 30ft ceiling. This is easy, its just double. 9*2=18 and 15*2=15 so what cost $60 to paint would logically cost $120 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we double the sides of a rectangle, then it requires four of the original rectangle to cover the new one. Using this fact we reason as follows: The sides are doubled, with width changing from 9 ft. to 18 ft. and length from 15 ft. to 30 ft. When the sides are doubled the area increases by a factor of 4. So the cost is $60 * 4 = $240 The cost for the second ceiling is $240 We could alternatively have figured out the cost per square foot for the first ceiling, and multiplied this by the number of square feet of the second: The first ceiling has area 9 ft x 15 ft = 135 ft^2, so its cost per square foot is $60 / (135 ft^2). The second ceiling has area 18 ft x 30 ft = 540 ft^2, so its cost is cost / sq ft * number of sq ft = ($60 / (135 ft^2) ) * 540 ft^2 = $ 60 * 540 ft^2 / (135 ft^2) = $60 * 4 = $240. It would have been possible to divide $60 by 135 ft^2, obtaining a cost per square foot of $.4444... / ft^2. Multiplying this by 540 ft^2 would yield the total cost. However roundoff error might be a problem. For example, if we round of to $.44 / ft^2, we would get $.44 / ft^2 * 540 ft^2 = $237.60, which is $2.40 less than the accurate estimate. This is about a 1% error, which might or might not be significant. A business which consistently makes 1% errors, for example, might be at a competitive disadvantage with a business that makes accurate estimates. If we used $.4444 / ft^2, we would still end up with a low estimate, but in this case the difference would only be a couple of cents. In either case, the accurate estimate of $240 is easier to find, and questions of roundoff error doesn't enter into this estimate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Wow, was way off on that. I went back and reread it and I missed the dimensions part. Why is it a factor of four? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: Here's a solution by a student from a previous semester: I found the area of each figure and then added those three results together. Area of Triangle = 1/2bh A = 1/2 (10) (9) A = 1/2 (90) A = 45 Area of Rectangle = lw A = (10) (4) A = 40 Area of Parallelogram = bh A = (10) (3) A = 30 Total area is 45 + 40 + 30 = 115 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.3.67 A circle with diameter 26 m is inscribed in a square. To the nearest square meter, what is the area inside the square and outside circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A diameter crosses all the way across a circle = 26m and a radius of 13m. We want to find the area of the square and the outside circle. The squares area is easy. Its just 26*26= 676m^2 The circle as long as we know the radius is easy as well, we just use the formula of circumference = pi r^2 (13 m)^2 = 169 (3.14) m^2 = 531 m^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The circle has diameter 26 m so its radius is 13 m and its area is A_circle = pi r^2 = pi * (13 m)^2 = 169 pi m^2 = 531 m^2. The square has four sides, each with length equal to the diameter of the circle. Thus the square has sides of length 26 m, so its area is the square of its side A_square = (26 m)^2 = 676 m^2. The area of the requested region is therefore the difference of these two areas: A = A_square - A_circle = 676 m^2 - 531 m^2 = 145 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Last part noted. ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `q Query 9.3.72 Pizzas of diameters 10, 12, 14 inches are sold for respective prices of $11.99, $13.99, $14.99 Which pizza is the best buy and how did you obtain your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 = $11.99 12 = $13.99 14 = $14.99 The thickness is the same but the sizes are not. One is 10 inches, twelve inches and fourteen inches. Since pizzas are circles we can find their area of slice-age. A= 3.14 (5)^2 = 78.5 $11.99 / 78.5 = 15.3 A =3.14 (6)^2 = 113.04 $13.99 / 113.04 = 12.4 A = 3.14 (7)^2 = 153.86 $14.99 / 153.86 = 9.7 The best buy is $14.99 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Student Solution: The thickness is about the same for all three pizzas so the amount of pizza can be measured by its area. I therefore found the area of each. A = 3.14 * r^2 and radius is 1/2 the circumference. So we get areas A = 3.14 (5)^2 = 78.5 A = 3.14 (6)^2 = 113.04 A = 3.14 (7)^2 = 153.86 I then divide the prices and these answers to get the price per square inch. $11.99 / 78.5 = 15.3 $13.99 / 113.04 = 12.4 $14.99 / 153.86 = 9.7 Since 9.7 is the least, and since this is the result for the 14 inch pizza, the 14 in. pizza for $14.99 is the best buy. INSTRUCTOR COMMENT: This is a good solution. The following reasoning is also instructive: The area of a circle is proportional to the square of the diameter. The value is the the cost divided by the area. So the value is proportional to cost divided by the square of the diameter. Comparing cost divided by square of diameter: 11.99 / (10^2) = .1199 13.99 / (12^2) = .1097 14.99 / (14^2) = .0764 Therefore the third pizza is the better value. It wasn't necessary to find the actual areas; reasoning by proportionality was sufficient. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: