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,

Please find below the solution to the asked query:

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

$\Rightarrow$( 2 + x ) ²  = 2² + ( 2 - x ) ² 

$\Rightarrow$4 + x ² + 4 x  = 4 + 4 + x ² - 4 x 

$\Rightarrow$8 x  = 4

$\Rightarrow$x  = $\frac{1}{2}$ 

Therefore,

CE  =  2 + $\frac{1}{2}$  = $\frac{4 + 1}{2} = \frac{\mathbf{5}}{\mathbf{2}}$ cm

So,

Option ( B )                                                                     ( Ans )


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