Question

In a rectangle $ABCD$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $ABCD$ has equation $2x-y+4=0$. Then the area of the rectangle is.

• A. $16$
• B. $2\sqrt{10}$
• C. $2\sqrt{5}$
• D. $20$

Answer 1

Answer: (A)

$16$

Slope of the given diameter is $2$

Slope of $AB=\frac{6-2}{3-1}=2$

As the slopes of both $AB$ and diameter are equal, so both are parallel

Perpendicular distance of a point from a line is $\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}$

So length $AF=\frac{2-2+4}{\sqrt{4+1}}=\frac{4}{\sqrt{5}}$

As you can see from the figure $OE=AF$

$AB=\sqrt{{(3-1)}^2+{(6-2)}^2}=2\sqrt{5}AE=\frac{AB}{2}=\sqrt{5}$

In $△AOE,A{O}^2={AE}^2+{OE}^2$

$A{O}^2=\frac{16}{5}+5=\frac{41}{5}$ ..... $[\because OE=AF]$

$AO=\sqrt{\frac{41}{5}}$

$AC=2AO=2\sqrt{\frac{41}{5}}$

In $△ABC,A{C}^2={AB}^2+{BC}^2$

$\frac{164}{5}=20+{BC}^2\Rightarrow BC=\frac{8}{\sqrt{5}}$

Area of rectangle $=AB\times BC$

$=\frac{8}{\sqrt{5}}\times 2\sqrt{5}=16$

Hence, option $A$ is correct.