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The perimeter of a rectangle is 142 meters. Find the length and width if the length is an integer and the width is 2 times the next consecutive integer. " Length "=◻" meters "quad" Width "=◻" meters "

The perimeter of a rectangle is 142142 meters. Find the length and width if the length is an integer and the width is 22 times the next consecutive integer.\newline Length =□ meters  Width =□ meters  \text { Length }=\square \text { meters } \quad \text { Width }=\square \text { meters }

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Q. The perimeter of a rectangle is 142142 meters. Find the length and width if the length is an integer and the width is 22 times the next consecutive integer.\newline Length =□ meters  Width =□ meters  \text { Length }=\square \text { meters } \quad \text { Width }=\square \text { meters }
  1. Define Variables: Let's denote the length of the rectangle as LL (an integer) and the width as WW. According to the problem, the width is 22 times the next consecutive integer of the length. This means W=2(L+1)W = 2(L + 1).\newlineThe formula for the perimeter (PP) of a rectangle is P=2(L+W)P = 2(L + W). We are given that the perimeter is 142142 meters.\newlineSo, we have the equation 142=2(L+W)142 = 2(L + W).\newlineSubstitute WW with 2(L+1)2(L + 1) into the equation: WW00.
  2. Perimeter Formula: Now, let's simplify the equation: 142=2(L+2L+2)142 = 2(L + 2L + 2).\newlineThis simplifies to 142=2(3L+2)142 = 2(3L + 2).\newlineFurther simplification gives us 142=6L+4142 = 6L + 4.
  3. Substitute Width: Next, we need to solve for LL. Subtract 44 from both sides of the equation: 142−4=6L+4−4142 - 4 = 6L + 4 - 4.\newlineThis gives us 138=6L138 = 6L.\newlineNow, divide both sides by 66 to find LL: 138/6=6L/6138 / 6 = 6L / 6.
  4. Simplify Equation: After dividing, we get L=23L = 23. Since LL is an integer, this is a valid solution.\newlineNow we need to find WW using the relationship W=2(L+1)W = 2(L + 1).\newlineSubstitute LL with 2323: W=2(23+1)W = 2(23 + 1).
  5. Solve for L: Calculate W: W=2(24)W = 2(24).\newlineThis gives us W=48W = 48.\newlineNow we have the length and width of the rectangle, and we should check if these values satisfy the original perimeter condition.
  6. Find Width: Check the perimeter with the found values: P=2(L+W)P = 2(L + W). Substitute LL with 2323 and WW with 4848: P=2(23+48)P = 2(23 + 48). Calculate the perimeter: P=2(71)P = 2(71). This gives us P=142P = 142. Since this matches the given perimeter, our solution is correct.

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