a quadrilateral ABCD circumscribing a circle of radius 10 cm with sides AB=x cm;BC=38cm;CD=27cm;angle 'A'=90.find x

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Ahalya · Answer

ABCD is a quad with angle a=90 degee. join BD you will get a right triangle ABD. let the centre of the circle be O and the point at which the radius of the circle touches the AB be P.[POINT OF CONTACT] join OP. now angle OPA=90.[angle between a radii and a tangent.] therfore this forms a right triangle. therfore OA=10 root over 2 therefore AC equals 20root 2. then you can apply the pythagoras theorem.

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a quadrilateral ABCD circumscribing a circle of radius 10 cm with sides AB=x cm;BC=38cm;CD=27cm;angle 'A'=90.find x

a quadrilateral ABCD circumscribing a circle of radius 10 cm with sides AB=x cm;BC=38cm;CD=27cm;angle 'A'=90.find x

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