A square and a rectangle each have a perimeter of 40 m. The difference between areas of the two figures is 9cm2. What are the possible dimensions of the rectangle....I will mark you as branliest if you follow me
Answers
Step-by-step explanation:
SOLUTION: Consider the square with perimeter 48m. Let x be the length of each side. The perimeter is the distance around, or 4x. 4x = 48 x = 12 The length of each side is 12m. The area of the square is the side length times the side length, or x*x. 12*12 = 144 The area of the square is 144 square meters. We are told that the difference between the area of the square and the rectangle is 4 square meters. Then the area of the rectangle must be 4 more or 4 less. For now, assume that the rectangle has the lesser area, or 140 square meters. The formula for the area of a rectangle is length times width, or L*W. We have L*W = 140. We also know that the perimeter of the rectangle is 48. We have 2L + 2W = 48 We have two equations with two variables. Solve for L and W. Solve the first equation for L. L*W = 140 L = 140/W Substitute 140/W for L in the second equation. 2L + 2W = 48 2(140/W) + 2W = 48 Solve for W. 280/W + 2W = 48 Subtract 2W from both sides. 280/W = -2W + 48 Multiply both sides by W. 280 = -2W^2 + 48W This is a quadratic equation. Solve by factoring. 2W^2 - 48W + 280 = 0 Divide each term by 2. W^2 - 24W + 140 = 0 (W - 14)(W - 10) = 0 W = 14 or W = 10 The width of the rectangle is 10m or 14m. If the width of the rectangle is 10m, then its length is 14m because 2L + 2W = 2(10) + 2(14) = 48 The dimensions of the rectangle are 10m by 14m. You can work out the dimensions of the rectangle if it has the larger area. Then your equations will be L*W = 148 2L + 2W = 48 I'll leave that to you. Hope this helps!