12. (1 mark) In trapezium $A B C D, B C \perp A B, B C \perp C D$ and $A C \perp B D$. Given $A B=\sqrt{11}$ and $A D=\sqrt{1001}$. Find $B C$.
(1 分) In trapezium $A B C D, B C \perp A B, B C \perp C D$ and $A C \perp B D$. Given $A B=\sqrt{11}$ and $A D=\sqrt{1001}$. Find $B C$.
12. Let $A C$ and $B D$ intersect at $O$. From $\triangle A B C$, get $\frac{O A}{O B}=\frac{O B}{O C}$, from $\triangle B C D$, get $\frac{O B}{O C}=\frac{O C}{O D}$. If $O A=a$, $O B=a r$, then $O C=a r^{2}, O D=a r^{3}$. By Pythagorean theorem, $A B=a \sqrt{1+r^{2}}, A D=a \sqrt{1+r^{6}}$.
Then $91=\frac{1001}{11}=\frac{A D^{2}}{A B^{2}}=\frac{1+r^{6}}{1+r^{2}}=r^{4}-r^{2}+1$.
Hence $r^{4}-r^{2}-90=\left(r^{2}-10\right)\left(r^{2}+9\right)=0$, or $r^{2}=10, r=\sqrt{10}$.
Since $\sqrt{11}=A B=a \sqrt{1+r^{2}}$, get $a=1$, and
$$
B C=\sqrt{O B^{2}+O C^{2}}=\sqrt{a^2 r^2 + a^2 r^4} = \sqrt{10 + 100} = \sqrt{110} = \underline{\underline{110}}
$$

Alternate Solution

Assign the coordinates $A(0, \sqrt{11}), B(0,0), C(1,0)$ and $D(c, d)$.

As $A C \perp B D$, we have $\left(\frac{d}{c}\right)\left(-\frac{\sqrt{11}}{c}\right)=-1$, or $c^{2}=\sqrt{11} d$.
Hence, $A D^{2}=c^{2}+(d-\sqrt{11})^{2}$
$$
=c^{2}+\left(\frac{c^{2}}{\sqrt{11}}-\sqrt{11}\right)^{2}=1001 \text {. }
$$

Simplify to get $c^{4}-11 c^{2}-10890=0$.
Let $c^{2}=s$, and observe that $10890=99 \times 110$,
have $(s-110)(s+99)=0$, or $s=110$,
implying $c=\sqrt{110}$.