Section 2.6 Formulas and Applications of Geometry 2. Geometry Applications In Section A.2, we present numerous facts and formulas relating to geometry.There are also geometry formulas on the inside back cover of the text for quick reference. Solving a Geometry Application Involving Perimeter Example 3 The length of a rectangular lot is 1 m less than twice the width. If the perimeter is 190 m, find the length and width. Solution: Step 1: Read the problem. Let x represent the width of the rectangle. Step 2: Label the variables. Then represents the length. Step 3: Write the formula for perimeter. Step 4: Write an equation in terms of x. Step 5: Solve for x. 2x 1 P 2l 2w 190 212x 12 21x2 190 4x 2 2x 190 6x 2 192 6x 32 x x 32. The width is The length is Step 6: Interpret the results and write the answer in words. The width of the rectangular lot is 32 m and the length is 63 m. Skill Practice 4. The length of a rectangle is 10 ft less than twice the width. If the perimeter is 178 ft, find the length and width. Recall some facts about angles. • Two angles are complementary if the sum of their measures is 90° . • Two angles are supplementary if the sum of their measures is 180° . • The sum of the measures of the angles within a triangle is 180° . • The measures of vertical angles are equal. 2x 1 21322 1 63. 192 6 6x 6 x 2x 1 149 Answer 4. The length is 56 ft, and the width is 33 ft.
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