Let `A B C D` be a quadrilateral with area `18` , side `A B` parallel to the side `C D ,a n dA B=2C D` . Let `A D` be perpendicular to `A Ba n dC D` . If a circle is drawn inside the quadrilateral `A B C D` touching all the sides, then its radius is `a = 3` (b) 2 (c) `3/2` (d) 1

Options

A. 3
B. 2
C. `3//2`
D. 1

Answer

B

Explanation

Let `CD=alpha`. Then, AB=2`alpha`.
The equation of BC is
`y=-(2r)/(alpha)(x-2alpha) or, 2rx+alpha y - 4 r alpha = 0`

It touches the circle `(x-1)^(2)+(y-r)^(2)=r^(2)`
`:. |(2x^(2)+ alpha r - 4 alpha r)/(sqrt(4r^(2)+alpha^(2)))|=r`
`rArr (2 r^(2)-3alpha r)^(2)=r^(2)(alpha^(2)+4r^(2))`
`rArr (2r-3 alpha)^(2)=(alpha^(2)+4r^(2))rArr -12r alpha + 9 alpha^(2)=alpha^(2)rArr alpha=(3r)/(2)`

Now, Area of quadrilateral ABCD = 18
`rArr (1)/(2) (AB+CD)x2r=18`
`rArr 3 alpha r = 18`
`rArr alpha r=6 rArr 3r^(2)=12 [:. alpha = (3r)/(2)]`
`rArr r=2`