Round Conference Tables

Solve The Problem.??

The area of a rectangular conference table is 133 square feet. If its length is 6 feet longer than its width, find the dimensions of the table. Round each dimension to the nearest tenth of a foot, if necessary. a. 8.9 feet by 14.9 feet b. 9.2 feet by 14.4 feet c. 9.2 feet by 15.2 feet d. 8.9 feet by 14.6 feet

Public Comments

1. solve it yourself
2. a) 8.9 feet by 14.9 feet x(x+6)=133 x^2+6x-133=0 solve to x x= 8.9 x+6 = 14.9
3. Area of conference table = L*w = 133 "length is 6 feet longer than its width" --> L=6 + w L*w = (6 + w)(w) = 133 6w + w^2 = 133 w^2 + 6w - 133 = 0 Since this equation doesn't factor, we must use the quadratic formula w = [-b + sqrt(b^2 - 4ac)]/2a and [-b - sqrt(b^2 - 4ac)]/2a w = [-6 + sqrt(6^2 - 4(1)(-133))]/2(1) = [-6 + sqrt(36 + 532)]/2 = [-6 + sqrt(568)]/2 = [-6 + 23.833]/2 = 8.9165 = 8.9 Also, you have to solve for the minus scenario w = [-6 - sqrt(6^2 - 4(1)(-133))]/2(1) = [-6 - sqrt(36 + 532)]/2 = [-6 - sqrt(568)]/2 = [-6 - 23.833]/2 = -14.9165 Distance cannot be negative, so this is not a solution So w = 8.9 and L = 6 + w = 6 + 8.9 = 14.9 Check L*w = 14.9 * 8.9 = 132.61 which is approx. 133 Answer a) 8.9 ft and 14.9 ft
4. the answer is a. the dimensions multiply to the right area and the difference between the dimensions is 6.