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The perimeter of a rectangle is 46 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 4646 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 4646 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=2L+2WP = 2L + 2W.\newlineGiven that the perimeter is 4646 meters, we can write the equation:\newline46=2L+2W46 = 2L + 2W
  2. Perimeter Formula: Substitute WW with 2(L+1)2(L + 1) in the perimeter equation:\newline46=2L+2[2(L+1)]46 = 2L + 2[2(L + 1)]
  3. Perimeter Equation: Simplify the equation:\newline46=2L+4L+446 = 2L + 4L + 4\newline46=6L+446 = 6L + 4
  4. Simplify Equation: Subtract 44 from both sides to isolate the term with LL: \newline46−4=6L46 - 4 = 6L\newline42=6L42 = 6L
  5. Isolate Term: Divide both sides by 66 to solve for LL: \newline426=6L6\frac{42}{6} = \frac{6L}{6}\newlineL=7L = 7
  6. Solve for L: Now that we have the value of L, we can find W:\newlineW=2(L+1)W = 2(L + 1)\newlineW=2(7+1)W = 2(7 + 1)\newlineW=2(8)W = 2(8)\newlineW=16W = 16
  7. Find Width: Let's verify that these values of LL and WW give the correct perimeter:\newlineP=2L+2WP = 2L + 2W\newlineP=2(7)+2(16)P = 2(7) + 2(16)\newlineP=14+32P = 14 + 32\newlineP=46P = 46

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