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• Hi Let the width = x, then Length = x + 1, from the Pythagoras theorem it follows that (Diagonal of rectangle)^2 = (its length)^2 + (its width)^2 4^2 = (x +1)^2 + x^2 .............|Diagonal. = 4cms x^2 + 1 + 2x + x^2 = 16 2x^2 + 2x - 1…

  in math rectangle problem? Comment by sudhakarbabu May 2021
• Hi Let the width = x, then Length = x + 1, from the Pythagoras theorem it follows that (Diagonal of rectangle)^2 = (its length)^2 + (its width)^2 4^2 = (x +1)^2 + x^2 .............|Diagonal. = 4cms x^2 + 1 + 2x + x^2 = 16 2x^2 + 2x - 1…

  in math rectangle problem? Comment by sudhakarbabu May 2021
• Hi Let the width = x, then Length = x + 1, from the Pythagoras theorem it follows that (Diagonal of rectangle)^2 = (its length)^2 + (its width)^2 4^2 = (x +1)^2 + x^2 .............|Diagonal. = 4cms x^2 + 1 + 2x + x^2 = 16 2x^2 + 2x - 1…

  in math rectangle problem? Comment by sudhakarbabu May 2021