Let ABCD be a quadrilateral with area 18,with | vedclass

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(B) Let the radius of the circle be $r$. Since the circle touches all four sides,the height of the trapezoid $ABCD$ is the diameter of the circle,so $

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$2$

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(B) Let the radius of the circle be $r$. Since the circle touches all four sides,the height of the trapezoid $ABCD$ is the diameter of the circle,so $AD = 2r$.Given $AB = 2CD$,let $CD = x$,then $AB = 2x$.The area of the trapezoid is given by $	ext{Area} = \frac{1}{2} 	imes (AB + CD) 	imes AD = 18$.Substituting the values,$\frac{1}{2} 	imes (2x + x) 	imes 2r = 18$,which simplifies to $3xr = 18$,or $xr = 6$.Let the center of the circle be $(r, r)$. The side $CD$ lies on the line $y = 2r$ and $AB$ lies on the line $y = 0$. The side $AD$ lies on the $y$-axis $(x = 0)$.The circle is tangent to $AD$ at $(0, r)$,to $AB$ at $(r, 0)$,and to $CD$ at $(r, 2r)$.Let the side $BC$ be tangent to the circle at some point. The distance from the center $(r, r)$ to the line $BC$ must be $r$.Using the property of tangents from an external point,the distance from $B(2x, 0)$ to the point of tangency on $AB$ is $2x - r$,and the distance from $C(x, 2r)$ to the point of tangency on $CD$ is $x - r$.Since the tangents from $B$ and $C$ to the circle are equal,we have $	an 	heta = \frac{x-r}{r}$ and $	an(90^\circ - 	heta) = \frac{2x-r}{r}$.Thus,$\frac{x-r}{r} = \frac{r}{2x-r}$,which implies $(x-r)(2x-r) = r^2$.$2x^2 - 3xr + r^2 = r^2 \Rightarrow 2x^2 - 3xr = 0$.Since $x 
eq 0$,we have $2x = 3r$,or $x = \frac{3r}{2}$.Substituting $x = \frac{3r}{2}$ into $xr = 6$,we get $(\frac{3r}{2})r = 6$ $\Rightarrow r^2 = 4$ $\Rightarrow r = 2$.

Let $ABCD$ be a quadrilateral with area $18$,with side $AB$ parallel to the side $CD$ and $AB = 2CD$. Let $AD$ be perpendicular to $AB$ and $CD$. If a circle is drawn inside the quadrilateral $ABCD$ touching all the sides,then its radius is

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