Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD, AB =3CD and the area of the quadrilateral is 4 sq.units. If a circle can be drawn touching all the sides of the quadrilateral , then radius is -?

Dear Student,

Please find below the solution to the asked query:

From given information we form our diagram , As :



Here CD  =  x  , So AB  =  3 x   (  As given AB   =  3 CD )

And Points P , Q , R and S are points where circle meets line AB , BC , CD and DA respectively .
So,

Radius of circle =  OP =  OQ  =  OR  =  OS  =  r

Here we construct CM $\perp$ AB and given CD $\perp$ AD  , So AMCD is a rectangle

AM  =  CD =  x    and CM  =  DA =  2 r                                                  ---- ( A )

 

As given CD $\perp$ AD and OP $\perp$ AB , OR $\perp$ CD ( We know radius is perpendicular to tangent at the point of tangency ) , So APOS and DROS are square ,  Then

AP  = AS  =  OP =  OS = DR  =  DS = OR  = OS  = r                              --- ( B )

And

CR =  CD -  DR  =  x  - r            ( From equation B and we assume CD = x )

We know : Lengths of tangents drawn from external point to the circle are equal . So

CR  = CQ  =  x  - r                                              ---- ( 1 )

And

BP = AB -  AP  =  3 x  - r               ( From equation B and we get AB = 3 x )

So ,

BP  = BQ  =  3 x  - r                                              ---- ( 2 )

And

BC  =  BQ +  CQ  =  3 x  - r + x  - r   = 4 x  - 2 r                      ( From equation 1 and 2 )

And

BM = AB  -  AM  =  3 x  -  x =  2 x        ( From equation A and we get AB = 3 x )

Now we apply Pythagoras theorem in triangle CMB and get

BC² = CM² + BM²   ,  Substitute all values we get

( 4 x  - 2 r )² = ( 2 r )² + ( 2 x )²

$\Rightarrow$16 x ² +  4 r ² - 16 xr = 4 r ² + 4 x ²

$\Rightarrow$12 x ² = 16 xr

$\Rightarrow$12 x = 16 r

$\Rightarrow$3 x = 4 r

$\Rightarrow$x = $\frac{4 r}{3}$                                                           ---- ( 1 )

Here in quadrilateral ABCD , AB  | | CD so ABCD is a trapezium .

And we know area of trapezium = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{parallel} \mathrm{sides}}{2}\times \mathrm{Height}$

Given area of ABCD  =  4 square unit

So,

$$\begin{gathered} \frac{\mathrm{AB} +\hspace{0.17em}\mathrm{CD}}{2}\times \mathrm{DA} = 4 \\ \Rightarrow \frac{3 x +\hspace{0.17em}x}{2}\times 2 r = 4 \\ \Rightarrow \frac{4\hspace{0.17em}x}{2}\times 2 r = 4 \\ \Rightarrow 4 x r = 4 \\ \Rightarrow x = \frac{1}{r} , \mathrm{Substitute} \mathrm{value} \mathrm{from} \mathrm{equation} 3 \mathrm{and} \mathrm{get} \\ \frac{4 r }{3} = \frac{1}{r} \\ \Rightarrow 4 r^2 = 3 \\ \Rightarrow r^2 = \frac{3}{4} \\ \Rightarrow r = \sqrt{\frac{3}{4}} \\ \mathbf{\Rightarrow }\mathit{r}\mathbf{ }\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}} \end{gathered}$$
Therefore,

Radius of given circle  = $\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$ unit                                                                  ( Ans )

Hope this information will clear your doubts about Circles.

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