Question

in the given quadrilateral find ABCD is circumscribe touching P,Q,R,S such that DAB=90 if CS is 27 cm and CB is 38cm and radius of the circle is 10cm then AB =

Answer 1

Answer:

option b) 21

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Answer 2

Answer:

Option (b) 21cm is the required length of AB

Step-by-step explanation:

Explanation :

Given , ABCD is a quadrilateral in which

circumscribe touching P,Q,R,s

∠DAB = 90°

CS = 27cm , CB = 38cm and radius of circle is 10 cm .

Since the length of the tangents drawn from an external point to the

circle are equal .

Step1:

Then, CR =CS = 27cm  (where CS = 27 cm given )

Now , BR = (BC -CR)

put the value of BC = 38cm and CR = 27cm .

        = 38 -27 = 11cm

Step2:

Join OQ then PAQO is a rectangle .

∠A = 90° (given)

∠ P =∠Q = ∠90° (radius is perpendicular at point contact)

⇒∠O = ∠90°

Step 3:

In Δ OPQ       (pythagoras theorem)

⇒$PQ^{2} =OP^{2} +OQ^{2}$

          = $(10)^{2} +(10)^{2} =200$

$PQ = \sqrt{200} =10\sqrt{2}$cm

Step4:

In ΔAPQ where ∠ A = 90 °

Let AP = AQ = x

⇒$PQ^{2} =AP^{2} +AQ^{2}$  (Pythagoras theorem)

 ⇒        $(10\sqrt{2}) ^{2} = 2x^{2}$

⇒   $200 = 2x^{2}$  ⇒  $100 = x^{2}$

⇒       $x =10cm$.

∴AQ = PQ = 10 cm (where radius of circle is 10cm given in the question).

Hence ,AB = (AQ +BQ ) = (10 +11)cm = 21 cm.

Final answer :

Hence , the required length of AB is 21cm .