A is top left, B is bottom left, Z is where the diagnols bisect, C is top right, and D bottom right.

The diagonals of the rectangle have the same length (8x+14) and they bisect one another so AZ=BZ=4x+7.

The question doesn’t state whether the answer is to be in terms of x or an absolute quantity.

The sides AB and CD can be calculated in terms of x and, using Pythagoras Theorem, the length of these sides is √((8x+14)²-(20x-10)²) = √((8x+14+20x-10)(8x+14-20x+10)) = √((28x+4)(-12x+24)) = √(48(7x+1)(2-x)).

Because the quantity under the square root must be positive, this imposes constraints on x.

(7x+1)(2-x) > 0, so either 7x+1 and 2-x are both positive or both negative.

7x+1>0 and 2-x>0 implies x<2 and x>-1/7 so -1/7<x<2; or 7x+1<0 and 2-x<0, implying x>2 and x<-1/7, which is not possible. Therefore -1/7<x<2. Also, 45/7<4x+7<15 are the limits for AZ and BZ in absolute terms. Also, 20x-10>0, so x>½. That narrows the limits: ½<x<2 and 9<4x+7<15.

Another way to approach working out a range of values for x is to work out x as BD approaches zero, that is, x approaches 2; or as AB=CD approaches zero, so 8x+14 approaches 20x-10, that is, when 12x approaches 24, s x approaches ½. The interval for x is (½,2).

[When x=1.2, AB=CD=19 approx, BD=14, AZ=BZ=CZ=DZ=11.8.]

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