Question

The side of a square is `(1)/(2)(x + 1)` units, and its diagonals is `(3 - x)/sqrt(2)`

Answer 1

By Pythagoras theorem, Inrt ΔABC:
AC² = AB² + BC²
`[((x + 1))/(2)]^2 + [((x + 1))/(2)]^2 = [(3 -x)/sqrt(2)]^2`

⇒ `[((x + 1))/(4)]^2 + [((x + 1))/(4)]^2 = [((3 -x))/(2)]^2`

⇒ `[((x + 1)^2)/(2)] = [((9 + x^2 - 6x))/(2)]`

⇒ (x² + 1 + 2x) - (9 + x² - 6x) = 0
⇒ 8x - 8 = 0
⇒ x = 1units.