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The lengths of the three sides of a triangle (in meters) are consecutive integers. If the perimeter is 48 meters, find the value of the longest of the three side lengths.
Answer: meters

The lengths of the three sides of a triangle (in meters) are consecutive integers. If the perimeter is 4848 meters, find the value of the longest of the three side lengths.\newlineAnswer: meters

Full solution

Q. The lengths of the three sides of a triangle (in meters) are consecutive integers. If the perimeter is 4848 meters, find the value of the longest of the three side lengths.\newlineAnswer: meters
  1. Denote shortest side: Let's denote the shortest side of the triangle as nn meters. Since the sides are consecutive integers, the other two sides will be n+1n+1 meters and n+2n+2 meters. The perimeter of the triangle is the sum of the lengths of its sides, which is given as 4848 meters. We can set up the following equation to represent this relationship:\newlinen+(n+1)+(n+2)=48n + (n + 1) + (n + 2) = 48
  2. Set up equation: Now, we simplify the equation by combining like terms: 3n+3=483n + 3 = 48
  3. Simplify equation: Next, we subtract 33 from both sides of the equation to isolate the term with nn: \newline3n=4833n = 48 - 3\newline3n=453n = 45
  4. Isolate term with n: Now, we divide both sides of the equation by 33 to solve for n:\newlinen=453n = \frac{45}{3}\newlinen=15n = 15
  5. Solve for nn: Since nn represents the shortest side, the longest side will be n+2n + 2. We already found that n=15n = 15, so the longest side is: 15+2=1715 + 2 = 17 meters

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