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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

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

CE

^{2}= CD

^{2}+ DE

^{2}

$\Rightarrow $( 2 +

*x*)

^{2}= 2

^{2}+ ( 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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