1 2 MODULE 6. GEOMETRY AND UNIT CONVERSION 6a Applications The most common units of length in the American system are inch, foot, yard, and mile. Converting from one unit of length to another is a requisite skill in geometry and real world applications. Units of Length American Units of Length. Facts relating common units of length. 1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 feet (ft) 1 mile (mi) = 5280 feet (ft) Change 48 inches to feet. EXAMPLE 1. Change 24 inches to feet. Solution. Multiply by the conversion factor 1 ft/12 in. Answer: 4 feet 24 in = 24 in 1 Multiplicative Identity Property. 1ft = 24 in 12 in Replace 1 with 1 ft/12 in. 1ft =24 in 12 in Cancel common unit. = 24 1 ft 12 Multiply fractions. =2ft Simplify. Hence, 24 inches equals 2 feet. A summary of conversion factors for units of length is given in Table 6.1. Convert Conversion Factor Convert Conversion Factor feet to inches 12 in/1 ft inches to feet 1 ft/12 in yards to feet 3ft/1yd feet to yards 1yd/3ft miles to feet 5280ft/1 mi feet to miles 1 mi/5280 ft Table 6.1: Conversion factors for units of length. Some conversions require more than one application of a conversion factor. Change 8 yards to inches. EXAMPLE 2. Change 4 yards to inches.
2 6A. APPLICATIONS 3 Solution. We multiply by a chain of conversion factors, the first to change yards to feet, the second to change feet to inches. 4yd=4yd 3ft 12 in 1yd 1ft Multiply by conversion factors. =4 yd 3 ft 12 in 1 yd 1 ft Cancel common units. = in 1 1 Multiply fractions. = 144 in Simplify. Hence, 4 yards equals 144 inches. Answer: 288 inches Applications Geometry Solving many real world problems require some geometry. We start our review with the perimeter of a polygon, in particular, a rectangle. Perimeter of a Polygon. In geometry a polygon is a plane figure made up of a closed path of a finite sequence of segments. The segments are called the edges or sides of the polygon and the points where two edges meet are called the vertices of the polygon. The perimeter of any polygon is the sum of the lengths of its sides. EXAMPLE 3. A quadrilateral (four sides) is a rectangle if all four of its A rectangle has length 12 angles are right angles. It can be shown that the opposite sides of a rectangle must be equal. Find the perimeter of the rectangle shown below, where the meters and width 8 meters. Find its perimeter. sides of the rectangle are measured in meters. 3m 5m Solution. To find the perimeter of the rectangle, find the sum of the four sides. Because opposite sides have the same length, we have two sides of length 5 meters and two sides of length 3 meters. Hence, Perimeter = = 16.
3 4 MODULE 6. GEOMETRY AND UNIT CONVERSION Answer: 40 meters Thus, the perimeter of the rectangle is 16 meters. Note that the perimeter of a rectangle can be found by summing twice the length and width of the sides. This is given by the formula P =2L +2W. Application Area Consider the rectangle shown in Figure 6.1. The length of this rectangle is four inches (4 in) and the width is three inches (3 in). 3 in One square inch (1 in 2 ) 4 in Figure 6.1: A rectangle with length 4 inches and width 3 inches. To find the area of the figure, we can count the individual units of area (in 2 ) that make up the area of the rectangle, twelve square inches (12 in 2 ) in all. However, it is much faster to multiply the number of squares in each row by the number of squares in each column: 4 3 = 12 square inches. The argument presented above leads to the rule for finding the area ofa rectangle. Area of a Rectangle. LetL and W represent the length and width of a rectangle, respectively. L W W L To find the area of the rectangle, calculate the product of the length and width. That is, if A represents the area of the rectangle, then the area of the rectangle is given by the formula A = LW.
4 6A. APPLICATIONS 5 ou Try It! EXAMPLE 4. A rectangle has width 5 feet and length 12 feet. Find the A rectangle has width 17 area of the rectangle. inches and length 33 inches. Solution. Substitute L =12ftandW = 5 ft into the area formula. Find the area of the rectangle. A = LW =(12ft)(5ft) =60ft 2 Hence, the area of the rectangle is 60 square feet. Answer: 561 square inches. Volume of a Prism The natural next step is to look at the volume of a solid. The first one we consider is the rectangular solid or prism. EXAMPLE 5. Pictured below is a rectangular prism. The surface area of the prism pictured in this example is given by the following formula: S =2(WH + LH + LW ) H If L =12,W =4,andH =6 feet, respectively, calculate the surface area. L W The volume of the rectangular prism is given by the formula V = LW H, where L is the length, W is the width, and H is the height of the rectangular prism. Find the volume of a rectangular prism having length 12 feet, width 4 feet, and height 6 feet. Solution. Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of of L, W,andH in the formula V = LW H
5 6 MODULE 6. GEOMETRY AND UNIT CONVERSION with open parentheses. ( )( )( ) V = Next, substitute 12 ft for L, 4ftforW,and6ftforH and simplify. V = ( 12 ft )( 4ft )( 6ft ) =288ft 3 Answer: 288 square feet. Hence, the volume of the rectangular prism is 288 cubic feet. We can simplify a number of formulas by combining like terms. A regular hexagon has six equal sides, each with length x. Find its perimeter in terms of x. EXAMPLE 6. Find a formula for the perimeter P of the (a) rectangle and (b) square pictured below. Simplify your answer as much as possible. L s W W s s L Solution. The perimeter of any polygonal figure is the sum of the lengths of its sides. a) To find the perimeter P of the rectangle, sum its four sides. P = L + W + L + W. s Combine like terms. P =2L +2W. b) To find the perimeter P of the square, sum its four sides. P = s + s + s + s. Combine like terms. P =4s. Answer: P =6x Sometimes a variable in a formula is given in terms of a another variable. The following example illustrates replacing a variable with an expression containing another (related) variable.
6 6A. APPLICATIONS 7 ou Try It! EXAMPLE 7. The length of a rectangle is three feet longer than twice its width. Find the perimeter P of the rectangle in terms of only its width. Solution. From the previous problem, the perimeter of the rectangle is given by P =2L +2W, (6.1) where L and W are the length and width of the rectangle, respectively. This equation gives the perimeter in terms of its length and width, but we re asked to get the perimeter in terms of only its width. However, we re also given the fact that the length is three feet longer than twice the width. That is, the length is described in terms of the width. The length L of a rectangle is 5 meters longer than twice its width W. Find the perimeter P of the rectangle in terms of its width W. Length is Three Feet longer than Twice the Width L = 3 + 2W Because L =3+2W, we can replace L with 3 + 2W in the perimeter formula. P =2L +2W P =2(3+2W )+2W Apply the distributive property, then combine like terms. P =6+4W +2W P =6+6W. This last equation gives the perimeter P in terms of only its width W. Answer: P =6W +10 Consecutive Integers In the application of geometry to solve real world problems, the lengths of geometric figures may be expressed by algebraic expressions. Consecutive Integers. Let k represent an integer. The next consecutive integer is the integer k +1. Thus, if k is an integer, then k + 1 is the next integer, k + 2 is the next integer after that, and so on.
7 8 MODULE 6. GEOMETRY AND UNIT CONVERSION The three sides of a triangle are consecutive integers and the perimeter is 57 centimeters. Find the measure of each side of the triangle. EXAMPLE 8. The three sides of a triangle are consecutive integers and the perimeter is 72 inches. Find the measure of each side of the triangle. Solution. We follow the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. In this case, a carefully labeled diagram is the best way to indicate what the unknown variable represents. k +2 k k +1 In our schematic diagram, we ve labeled the three sides of the triangle with expressions representing the consecutive integers k, k +1, and k Set up an Equation. To find the perimeter P of the triangle, sum the three sides. P = k +(k +1)+(k +2) However, we re given the fact that the perimeter is 72 inches. Thus, 72 = k +(k +1)+(k +2) 3. Solve the Equation. On the right, regroup and combine like terms. Now, solve for k. 72 = 3k =3k +3 3 Subtract 3 from both sides. 69 = 3k Simplify = 3k 3 Divide both sides by = k Simplify. 4. Answer the Question. We ve only found one side, but the question asks for the measure of all three sides. However, the remaining two sides can be found by substituting 23 for k into the expressions k +1andk +2. k +1=23+1 and k +2=23+2 =24 =25 Hence, the three sides measure 23 inches, 24 inches, and 25 inches.
8 6A. APPLICATIONS 9 5. Look Back. Does our solution make sense? Well, the three sides are certainly consecutive integers, and their sum is 23 inches + 24 inches + 25 inches = 72 inches, which was the given perimeter. Therefore, our solution is correct. Answer: 18, 19, and 20 cm Area of a Parallelogram Area of a Parallelogram. A parallelogram having base b and height h has area A = bh. That is, to find the area of a parallelogram, take the product of its base and height. EXAMPLE 9. Find the area of the parallelogram pictured below. 5/3 ft The base of a parallelogram measures 14 inches. The height is 8/7 of an inch. What is the area of the parallelogram? 6ft Solution. The area of the parallelogram is equal to the product of its base and height. That is, A = bh =(6ft) ( ) 5 3 ft Area formula for parallelogram. Substitute: 6 ft for b, 5/3 ft for h. = 30 3 ft2. Multiply numerators and denominators. =10ft 2. Divide. Thus, the area of the parallelogram is 10 square feet. Answer: 16 square inches Area of a Triangle Area of a Triangle. A triangle having base b and height h has area A = (1/2)bh. That is, to find the area of a triangle, take one-half the product of the base and height.
9 10 MODULE 6. GEOMETRY AND UNIT CONVERSION The base of a triangle measures 15 meters. The height is 12 meters. What is the area of the triangle? EXAMPLE 10. Find the area of the triangle pictured below. 6cm 13 cm Solution. To find the area of the triangle, take one-half the product of the base and height. A = 1 2 bh Area of a triangle formula. = 1 (13 cm)(6 cm) Substitute: 13 cm for b, 6cmforh cm2 = 2 =39cm 2. Multiply numerators; multiply denominators. Simplify. Answer: 90 square meters Therefore, the area of the triangle is 39 square centimeters. Area of a Trapezoid. Area of a Trapezoid. A trapezoid with bases b 1 and b 2 and height h has area A = 1 2 h (b 1 + b 2 ). That is, to find the area, sum the bases, multiply by the height, and take one-half of the result. A trapezoid has bases measuring 6 and 15 feet, respectively. The height of the trapezoid is 5 feet. Find the area of the trapezoid. EXAMPLE 11. Find the area of the trapezoid pictured below in 3 in in
10 6A. APPLICATIONS 11 Solution. The formula for the area of a trapezoid is A = 1 2 h (b 1 + b 2 ) Substituting the given bases and height, we get A = 1 ( 2 (3) 4 1 ) Simplify the expression inside the parentheses first. Change mixed fractions to improper fractions, make equivalent fractions with a common denominator, then add. A = 1 ( 17 2 (3) ) 2 = 1 ( 17 2 (3) ) 2 2 = 1 ( 17 2 (3) ) 4 = 1 2 ( 3 1 Multiply numerators and denominators. = 81 8 )( 27 4 This improper fraction is a perfectly good answer, but let s change this result to a mixed fraction (81 divided by 8 is 10 with a remainder of 1). Thus, the area of the trapezoid is ) A =10 1 square inches. 8 Answer: square feet Height of a Triangle EXAMPLE 12. The area of a triangle is 20 square inches. If the length of The area of a triangle is 161 the base is 2 1 square feet. If the base of 2 inches, find the height (altitude) of the triangle. the triangle measures Solution. We follow the Requirements for Word Problem Solutions. feet, find the height of the triangle.
11 12 MODULE 6. GEOMETRY AND UNIT CONVERSION 1. Set up a Variable Dictionary. Our variable dictionary will take the form of a well labeled diagram. h in 2. Set up an Equation. The area A of a triangle with base b and height h is A = 1 2 bh. Substitute A =20andb = = 1 2 ( 2 1 ) h Solve the Equation. Change the mixed fraction to an improper fraction, then simplify. 20 = 1 ( 2 1 ) h Original equation = 1 ( ) 5 h Mixed to improper: = 5 2. ( 1 20 = 2 5 ) h Associative property = 5 4 h Multiply: = 5 4. Now, multiply both sides by 4/5 and solve. 4 5 (20) = 4 ( ) h Multiply both sides by 4/ = h Simplify: (20) = 16 5 and =1. 4. Answer the Question. The height of the triangle is 16 inches.
12 6A. APPLICATIONS Look Back. If the height is 16 inches and the base is area is A = 1 ( 2 1 ) (16) 2 2 inches, then the = = (5) ( ) = (2) (2) = =20 This is the correct area (20 square inches), so our solution is correct. Answer: 8 feet Circumference of a Circle EXAMPLE 13. Find the circumference of a circle given its radius is 12 feet. Solution. The circumference of the circle is given by the formula C = πd, or, because d =2r, C =2πr. Substitute 12 for r. C =2πr =2π(12) = 24π Therefore, the circumference of the circle is exactly C = 24π feet. We can approximate the circumference by entering an approximation for π. Let s use π Note: The symbol is read approximately equal to. Find the radius of a circle having radius 14 inches. Use π 3.14 C =24π 24(3.14) feet It is important to understand that the solution C = 24π feet is the exact circumference, while C feet is only an approximation. Answer: inches Area of a Circle EXAMPLE 14. Find the area of a circle having a diameter of 12.5 meters. Use 3.14 for π and round the answer for the area to the nearest tenth of a square meter. Find the area of a circle having radius 12.2 centimeters. Use π 3.14
13 Answer: cm 2 14 MODULE 6. GEOMETRY AND UNIT CONVERSION Solution. The diameter is twice the radius. Substitute 12.5 for d and solve for r. d =2r 12.5 = 2r Substitute 12.5 for d = 2r 2 2 Divide both sides by = r Simiplify. Hence, the radius is 6.25 meters. To find the area, use the formula A = πr 2 and substutite: 3.14 for π and 6.25 for r. A =(3.14)(6.25) 2 Substitute: 3.14 for π, 6.25 for r. =(3.14)( ) Square first: (6.25) 2 = = Multiply: (3.14)( ) = Hence, the approximate area of the circle is A = square meters. To round to the nearest tenth of a square meter, identify the rounding digit and the test digit. Rounding digit Test digit Because the test digit is greater than or equal to 5, add 1 to the rounding digit and truncate. Thus, correct to the nearest tenth of a square meter, the area of the circle is approximately A 122.7m 2. The Pythagorean Theorem We now state one of the most ancient theorems of mathematics, the Pythagorean Theorem. Pythagorean Theorem. Let c represent the length of the hypotenuse of a right triangle, and let a and b represent the lengths of its legs, as pictured in the image that follows.
14 6A. APPLICATIONS 15 c a b The relationship involving the legs and hypotenuse of the right triangle, given by a 2 + b 2 = c 2, is called the Pythagorean Theorem. EXAMPLE 15. Given the following right triangle, find the exact length of the missing side. 7 The hypotenuse and one leg of a right triangle measure 9 and 7 inches, respectively. Find the length of the remaining leg. x 12 Solution. Note that the hypotenuse (across from the right angle) has length 12. This quantity should lie on one side of the Pythagorean equation all by itself. The sum of the squares of the legs go on the other side. Hence, Solve the equation for x. x =12 2 x 2 +49=144 Exponents first: 7 2 =49and12 2 =144. x = x 2 =95 Subtract 49 from both sides. Simplify both sides. x = 95 Take the nonnegative square root of 95. Hence, the exact length of the missing side is 95. Answer: 32 inches
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