# Square ABCD has side length 2 centimetre. A semi circle with AB as diameter is constructed inside a square and the tangent to the semi circle from C intersect side AD at E , then the length of CE is , (in cm) (A) (2+√5)/2 (B) 5/2 (C) √5 (D) √6

Dear Student,

From given information we form our diagram , As :

Here , AB  =  BC  =  CD  =  DA  =  2 cm  ( Sides of square ABCD )

Let the point of tangency be F of tangent CE . We know tangents drawn from a point to a circle are equal in length , So

BC =  FC =  2 cm ( As we know BC  - 2 cm )

Similarly

AE = EF = x

Thus DE = 2 - x  , CE  =  2  + x

Now we use Pythagorean Theorem in triangle CDE and get

CE2 =  CD2 + DE2

$⇒$( 2 + x ) 2  = 22 + ( 2 - x ) 2

$⇒$4 + x 2 + 4 x  = 4 + 4 + x 2 - 4 x

$⇒$8 x  = 4

$⇒$x  = $\frac{1}{2}$

Therefore,

CE  =  2 + $\frac{1}{2}$  = cm

So,

Option ( B )                                                                     ( Ans )