Objective 2 Examples Use the perimeter, area, or circumference formulas to solve each problem. 2a. Assume that the perimeter of the Mona Lisa by Leonardo da Vinci is 260 cm. If the length of the painting is 77 cm, find its width. Then find the area of the painting. Solution 2a. P = 2l + 2w State the perimeter formula. 260 = 2(77) + 2w Replace P with 260 and l with 77. 260 = 154 + 2w Simplify. 260 - 154 = 154 + 2w - 154 Subtract 154 from each side. 106 = 2w Simplify. 106 2 = 2w 2 Divide each side by 2. 53 = w Simplify. The width of the painting is 53 cm. Since we know both the length and width of the painting, we can calculate the area using the formula A = lw. A = lw State the area formula. A = (77)(53) Replace l with 77 and w with 53. A = 4081 Multiply. So, the area of the painting is 4081 cm2. 2b. What is the height of a triangle whose area is 16 ft2 and whose base is 8 ft? Solution 2b. A = 1 2 bh State the area formula. 16 = 1 2 (8)h Replace A with 16 and b with 8. 16 = 4h Simplify. 16 4h = 4 4 Divide each side by 4. 4 = h Simplify. So, the height of the triangle is 4 ft. h 2c. Suppose the length of a soccer field is 30 yd less than two times its width. If its perimeter is 390 yd, find the dimensions of the soccer field. Solution 2c. What is unknown? The length and width of the soccer field are unknown. What is known? The length is 30 yd less than two times its width. Let w = width of the field. Then 2w - 30 = length of the field. We also know the perimeter is 390 yd. So, we use the perimeter formula of a rectangle to write the equation. P = 2l + 2w State the perimeter formula. 390 = 2(2w - 30) + 2w Replace l with 2w - 30. 390 = 4w - 60 + 2w Apply the distributive property. 390 = 6w - 60 Combine like terms. 390 + 60 = 6w - 60 + 60 Add 60 to each side. b = 8 t l = 2w – 30 w 138 Chapter 2 Linear Equations and Inequalities in One Variable
hendricks_beginning_intermediate_algebra_1e_ch1_3
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