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In a rectangle $A B C D$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $A B C D$ has equation $2 x-y+4=0$. Then, the area of the rectangle is
$16$
$2 \sqrt{10}$
$2 \sqrt{5}$
$20$
Solution
$(a)$
Given,
$A B C D$ is a rectangle $A(1,2)$ and $B(3,6)$ equation of one of the diameter of circle circumscribing the rectangle is
$2 x-y+4=0 \text {. }$
Slope of diameter $=2$
and slope of line $A B=\frac{6-2}{3-1}=\frac{4}{2}=2$
$\therefore$ Side $A B$ is parallel to diameter of circle equation of line $A B$
$y-2=2(x-1) \Rightarrow 2 x-y=0$
Distance between diameter of circle and line $A B$
$d=\left|\frac{4}{\sqrt{2^2+1}}\right|=\frac{4}{\sqrt{5}}$
$\therefore \quad B C=2 d=\frac{8}{\sqrt{5}}$
Now, $\quad A B=\sqrt{(3-1)^2+(6-2)^2}$
$=\sqrt{4+16}=2 \sqrt{\overline{5}}$
Area of rectangle $=A B \times B C$
$=2 \sqrt{5} \times \frac{8}{\sqrt{5}}=16$
