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So the measures of the three angles are 3(18) or 54, 6(18) or 108, and 1(18) or 18. 7-1 Ratios and Proportions 13. The ratio of the measures of the three sides of a triangle is 9: 7: 5. Its perimeter is 191.1 inches. Find the measure of each side 19. The ratio of the measures of the three angles is 10:8:6. SOLUTION: SOLUTION: Just as the ratio Just as the ratio or 9:7 is equivalent to or or 10:8 is equivalent to or 10x:8x , the extended ratio can be written as 10x:8x:6x. 9x:7x , the extended ratio can be written as 9x:7x:5x. The perimeter is 191.1 inches so the sum of the lengths of the sides is 191.1. Solve for x. We know that sum of the measures of all interior angles in a triangle is 180. Set the sum of the extended ratio equal to 180 and solve for x. So the measures of the three sides are 9(9.1) or 81.9 in., 7(9.1) or 63.7 in., and 5(9.1) or 45.5 in.. So the measures of the three angles are 10(7.5) or 75, 8(7.5) or 60, and 6(7.5) or 45. Find the measures of the angles of each triangle. 17. The ratio of the measures of the three angles is 3:6:1. SOLUTION: Just as the ratio Solve each proportion. 28. or 3:6 is equivalent to or SOLUTION: 3x:6x , the extended ratio can be written as 3x:6x:1x. We know that sum of the measures of all interior angles in a triangle is 180 degrees. Set the sum of the extended ratios equal to 180 and solve for x. Cross multiply. Solve for x. So the measures of the three angles are 3(18) or 54, 6(18) or 108, and 1(18) or 18. ALGEBRA Solve each proportion. Round to the nearest tenth. 19. The ratio of the measures of the three angles is 10:8:6. 31. SOLUTION: Just as the ratio or 10:8 is equivalent to or SOLUTION: 10x:8x , the extended ratio can be written as 10x:8x:6x. We know that sum of the measures of all interior angles in a triangle is 180. Set the sum of the extended ratio equal to 180 and solve for x. eSolutions Manual - Powered by Cognero Page 1 43. CHANGING DIMENSIONS A rectangle has a Solve for x. 7-1 Ratios and Proportions ALGEBRA Solve each proportion. Round to the nearest tenth. 31. Therefore, the area of the new rectangle quadruples. c. SOLUTION: d. The increase in the length to the width of the rectangle in part b is proportional, i.e. . The increase in the length to the width of the rectangle in 43. CHANGING DIMENSIONS A rectangle has a part c is not, i.e. . length and a width w. a. Find the perimeter and area of the rectangle. b. Suppose the length and width of the rectangle are doubled. What is the ratio of the new perimeter to the original perimeter? new area to the original area? c. Suppose the length and width increase by 2 units. What is the ratio of the new perimeter to the original perimeter? new area to the original area? d. What fact about the operations explains the differences between the results in part b and part c? SOLUTION: a. b. Therefore, the perimeter of the new rectangle doubles. Therefore, the area of the new rectangle quadruples. eSolutions Manual - Powered by Cognero c. Page 2