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Part 3 Module 8 Real-world problems involving area

Part 3 Module 8 Real-world problems involving area. Useful facts. The following facts will be provided on a formula sheet when you take quizzes or tests on this material in lab. www.math.fsu.edu/~wooland/Geo56/Geo14.html

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Part 3 Module 8 Real-world problems involving area

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  1. Part 3 Module 8Real-world problems involving area

  2. Useful facts The following facts will be provided on a formula sheet when you take quizzes or tests on this material in lab.

  3. www.math.fsu.edu/~wooland/Geo56/Geo14.html A rectangular section of wall measuring 22 feet by 8 feet will be covered with isosceles right triangular tiles with legs measuring 5 inches. Approximately how many tiles are needed to cover the section of wall? A. 7 B. 1014 C. 169 D. 2028 Exercise #1

  4. A rectangular section of wall measuring 22 feet by 8 feet will be covered with isosceles right triangular tiles with legs measuring 5 inches. Approximately how many tiles are needed to cover the section of wall? To answer a question of the form “How many of these smaller objects are needed to ‘equal’ this larger object, we divide the size of the larger object by the size of the smaller object. For a two-dimensional object, like a rectangle or triangle, “size” means “area.” The area of a rectangle is LW. The area of a triangle is (1/2)bh. For the rectangular floor, L = 22 ft and W = 8 ft. For the isosceles right triangular tiles, b = 5 inches and h = 5 inches. Before we calculate the areas, we must reconcile the discrepancy in units. If we use the units as currently listed, the area of the rectangle will be in square feet, but the area of the triangle will be in square inches, so the two areas won’t be comparable. Solution #1

  5. The area of a rectangle is LW. The area of a triangle is (1/2)bh. For the rectangular floor, L = 22 ft and W = 8 ft. For the isosceles right triangular tiles, b = 5 inches and h = 5 inches. We will convert the measurements for the rectangle from feet to inches before we calculate the area. L = 22 ft = 22 x 12 inches = 264 inches W = 8 ft = 8 x 12 inches = 96 inches Now, calculate the area of the rectangle and the area of the triangle. Area of rectangle = (264 inches)(96 inches) = 25,344 square inches Area of triangle = (1/2)(5 inches)(5 inches) = 12.5 square inches (Area of rectangle)/(Area of triangle) = 25344/12.5 = 2027.5 It will take roughly 2028 tiles to cover the wall, because the wall is 2028 times as big as one tile. Solution #1, page 2

  6. Exercise #2 www.math.fsu.edu/~wooland/Geo56/Geo19.html The figure below shows the plan for the new parking lot a Southwestdale Mall. It is estimated that such construction will cost $9.50 per square yard. Find the total cost. A. $4360.50 B. $164888.33 C. $1483995 D. $494665

  7. Solution #2 To answer this question, we need to find the area of the parallelogram, and then multiply the area by $9.50. We want the area to be in square yards. This means that we will convert the two measurements (b = 615 feet, h = 762 feet) from feet to yards before we use the formula A=bh. Convert measurements from feet to yards: b = 615 feet = 615/3 yards = 205 yards h = 762 feet = 762/3 yards = 254 yards Area = (205 yards)(254 yards) = 52,070 square yards Cost = 52,070 square yards x $9.50 per square yards = $494,665

  8. Find the area contained within the race track shown below. A. 97,489 square yards B. 296,378 square yards C. 135,412 square yards D. 45,137 square yards E. 32.498 square yards Exercise #3

  9. Solution #3 Because the want to calculate the area of this figure in square yards, but the measurements are in feet, we will convert the two measurements from feet to yards, first. 782 feet = 782/3 yards = 260.67 yards 374 feet = 374/3 yards = 124.67 yards We will assume that the figure consists of a rectangle with a half-circle joined to each end. Thus, Area of figure = area of rectangle + areas of two half-circles = area of rectangle + area of one complete circle Area = LW + πr2 From the figure, we see that L = 260.67 yards, W = 124.67 yards, r = 62.33 yards (radius is half of the diameter, and the diameter is 124.67 yards). A = (260.67)(124.67) + π(62.33)2 = 44,702.91 square yards Our answer doesn’t exactly match any of the multiple-choice options, because we rounded some numbers early in the calculation. The correct choice is D. 45,137 square yards

  10. Exercise #4 Suppose that a tank truck hauling a liquid toxic substance springs a leak, leaving in the parking lot a circular puddle with a radius of 6 feet. It takes 36 hours for the specialists to decontaminate this spill. How long would it take to clean up a similar spill with a radius of 2 feet? A. 6 hours B. 12 hours C. 4 hours D. 9 hours E. 18 hours

  11. Solution #4 There are several ways to arrive at the one correct answer. Any correct solution must take into account the area of the large circle, the area of the small circle, and the fact that the large circle requires 36 hours. We will take the same approach that we did with Exercise #1: compare the areas of the two circles, by dividing: (Area of large circle)/(area of small circle) = (π62 square feet)/(π22 square feet) = (36π square feet)/(4π square feet) = 9 (we have simplified the fraction by canceling common factors and units). This means that the large circle is 9 times as big as the small circle, so it takes 9 times as long to decontaminate. The small circle takes 1/9 as much time as the large circle. 1/9 of 36 hours = 4 hours. The correct answer is C.

  12. Exercise #5 www.math.fsu.edu/~wooland/Geo56/Geo23.html The figure below shows the parcel of land on which Gomer the rancher confines his weasels. His rule of thumb dictates that each weasel requires 25 square meters of space. Approximately how many weasels can the parcel accommodate? A. 6516050 B. 733281 C. 1173 D. 28

  13. Solution #5 Each weasel requires 25 square meters of space. Approximately how many weasels can the parcel accommodate? We need to find the area of the figure, in square meters, and then divide by 25. To find the area of the figure, we need to visualize the figure in terms of simpler figures, such as triangles and rectangles. One way to do this (not the only correct way) is to recognize that the figure is formed by cutting a right triangle off of the upper right-hand corner of a square. Area of figure = Area of square – Area of triangle = 190 x 190 – [(1/2) x 95 x 142.5] = 36100 – 6768.75 = 29331.25 square meters Thus, the number of weasels is 29331.25/25 = 1173.25 The best choice is C. (1173)

  14. Exercise #6 www.math.fsu.edu/~wooland/Geo56/Geo22.html The figure below shows the parcel of land on which Gomer the rancher confines his weasels. He is going to enclose the parcel with a fence. Find the total cost, assuming that the fence will cost $3.00 per linear meter. A. $1568 B. $1743 C. $2081 D. $87994 E. $65634

  15. Solution #6 He is going to enclose the parcel with a fence. Find the total cost, assuming that the fence will cost $3.00 per linear meter. In this case, we don’t need to find the area of the figure. This is a problem involving distance. We need to find the perimeter ( P ) of the figure, and then multiply by $3.00. P = 95 + 190 + 190 + 47.5 + x = 522.5 + x To find x, we recognize that x is the hypotenuse of an “invisible” right triangle, whose legs measure 95 meters and 142.5 meters, respectively (see next slide).

  16. Solution #6, page 2 P = 95 + 190 + 190 + 47.5 + x = 522.5 + x where x is the hypotenuse of an “invisible” right triangle, whose legs measure 95 meters and 142.5 meters, respectively, so we find x by using the Pythagorean Theorem. P = 95 + 190 + 190 + 47.5 + x = 522.5 + 171.26 = 693.76 meters Cost of fence = 693.76 meters x $3.00 per meter = $2081.

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