Part 3 Module 8 Real-world problems involving area

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. rArea W L A h b rea = LW bh = (1/2)bh rr 2

Exercise #1 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. B. C. D. 7 1014 169 2028 Solution #1 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 wont be comparable. Solution #1, page 2 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.

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

32.498 square yards 782 feet 374 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 + rr2 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) + r(62.33)2 = 44,702.91 square yards Our answer doesnt 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 260.67 yards 124.67 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. B. C. D.

6 hours 12 hours 4 hours 9 hours Solution #4 It takes 36 hours to decontaminate a circular puddle of toxic waste with a radius of 6 feet. How long will it take to decontaminate a similar spill with a radius of 2 feet? The answer IS NOT 12 hours , because the larger

circle is not three times as big as the small circle (as the scale drawing below suggests). 2 6 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) = (62 square feet)/(22 square feet) = (36 square feet)/(4 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. 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 95 mm

190 47.5 m 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. 47.5 95 190 142.5 mm mm

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) 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 95 mm 190 47.5 m 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 dont 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. 95 mm 47.5 x190 m

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). 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. 190mm 47.5 95 142.5 x mm

2+ 2 xx22= 95 142 . 5 = 9025 + 20306 . 5 xx2==29331 . 2 5 29331

.2171 5.26 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.

Recently Viewed Presentations

  • Media Bias - Dearborn Public Schools

    Media Bias - Dearborn Public Schools

    Let's Find Out Author's Purpose Activity 6-8ish Student YouTube Videos Author's Purpose PIE 5:32 What is the Author's Purpose in the Following Commercials? Author's Purpose Commercials 13 min . Title: Media Bias ... Bias in the Media and Author's Purpose...
  • KALKULUS I - WordPress.com

    KALKULUS I - WordPress.com

    12 cm, 30 feet, 120 km. cm, feet, and km are examples of units for length quantity. 12 gram, 30 pounds, 2 ton. Gram, pound, and ton are examples of units for mass quantity. 5 second, second is an example...
  • Transition to Socialism 1952- 1963

    Transition to Socialism 1952- 1963

    Mao believed that modernisation of the economy was essential to PRC's survival as a nation. Wanted to build on the model of Stalin's 5 year plans. 1952- 1st 5yr Plan introduced- aimed to develop state- directed growth of heavy industry-...
  • Basic Chemistry, Solutions, Osmosis

    Basic Chemistry, Solutions, Osmosis

    Arial Default Design BASIC CHEMISTRY, SOLUTIONS, OSMOSIS BASIC CHEMISTRYof SOLUTIONS Solutions are made to a certain concentration Dilutions and Concentrations DISSOLVING IS A PHYSICAL CHANGE NOT ALL CHEMICALS DISSOLVE IN WATER Slide 7 Hydrogen Bonds make water cohesive "sticky" Surface...
  • Experiences with DEP teaching Sridhar Iyer Vikram Gadre

    Experiences with DEP teaching Sridhar Iyer Vikram Gadre

    Times New Roman Arial Wingdings Default Design Experiences with DEP teaching Outline Activities in teaching a course Some major differences between an IIT class and a DEP class Planning the course Assumptions v/s reality Preparing for a lecture Assumptions v/s...
  • The 5 or 6 Kingdoms

    The 5 or 6 Kingdoms

    divided living things into one of two "kingdoms" - plant and animal kingdoms . divided each of the kingdoms into smaller groups called "genera" (plural of "genus") divided each genera into smaller groups called "species" designed a system of naming...
  • Grade 5: The Western Hemisphere - Lancaster High School

    Grade 5: The Western Hemisphere - Lancaster High School

    Key Ideas and Concepts . Between 1100 B.C.E. and 1500 C.E., complex societies and civilizations developed in the Western Hemisphere. Although these complex societies and civilizations have certain defining characteristics in common, each is also known for unique cultural achievements...
  • Prokaryote Taxonomy and Diversity - Penn State York

    Prokaryote Taxonomy and Diversity - Penn State York

    Arial Times New Roman Default Design Microsoft Photo Editor 3.0 Photo Prokaryote Taxonomy & Diversity Species Concept for Prokaryotes Phenetic Characters: Phenetic Identification Use of dichotomous keys for bacteria Phenetic Identification Use of multi-test kits and their databases.