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The length of a rectangle is 1 meter less than its width. The area of the rectangle is 42 square meters. Find the dimensions of the rectangle.

44. The length of a rectangle is 11 meter less than its width. The area of the rectangle is 4242 square meters. Find the dimensions of the rectangle.

Full solution

Q. 44. The length of a rectangle is 11 meter less than its width. The area of the rectangle is 4242 square meters. Find the dimensions of the rectangle.
  1. Define Variables: Let's call the width of the rectangle ww and the length ll. We know that the length is 11 meter less than the width, so we can write l=w1l = w - 1.
  2. Calculate Area: The area of the rectangle is given by the formula Area=length×width\text{Area} = \text{length} \times \text{width}, which is 42=l×w42 = l \times w. Substituting ll with w1w - 1, we get 42=(w1)×w42 = (w - 1) \times w.
  3. Solve Quadratic Equation: Now we need to solve the quadratic equation 42=w2w42 = w^2 - w. Moving all terms to one side, we get w2w42=0w^2 - w - 42 = 0.
  4. Factorize Equation: To solve the quadratic equation, we can factor it. We're looking for two numbers that multiply to 42-42 and add up to 1-1. Those numbers are 7-7 and 66, so we can write (w7)(w+6)=0(w - 7)(w + 6) = 0.
  5. Find Width: Setting each factor equal to zero gives us w7=0w - 7 = 0 or w+6=0w + 6 = 0. Solving for ww gives us w=7w = 7 or w=6w = -6. Since a width can't be negative, we discard w=6w = -6.
  6. Find Length: Now that we have the width w=7w = 7, we can find the length by substituting ww into l=w1l = w - 1. So l=71=6l = 7 - 1 = 6.

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