Question
Using the following figure, show that $BD =\sqrt{x}$.
Image

Answer

$ \mathrm{AB}=\mathrm{x}$ and $\mathrm{BC}=1$
$ \mathrm{AC}=\mathrm{AB}+\mathrm{BC}$
$ =\mathrm{x}+1$
diameter$=\mathrm{x}+1$
radius $\mathrm{OA}=\mathrm{OD}=\mathrm{OC}=\mathrm{OB}=\mathrm{OC}-\mathrm{BC}$
$ =\frac{x+1}{2}-1=\frac{x+1-2}{2}=\frac{x-1}{2}$
Using pythagoras in $\triangle \mathrm{BOD}$
$\mathrm{P}^2+\mathrm{B}^2=\mathrm{H}^2$
$ \mathrm{P}^2+\left(\frac{x-1}{2}\right)^2=\left(\frac{x+1}{2}\right)^2$
$ \mathrm{P}^2=\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2$
$ =\frac{(x+1)^2-(x-1)^2}{4}$
$ =\frac{\left(x^2+1+2 x\right)-\left(x^2+1-2 x\right)}{4}$
$\mathrm{P}^2=\frac{4 x}{4}$
$P^2=x$
$\mathrm{P}=\sqrt{x}$

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