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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

arun , 10 Years ago
Grade
anser 1 Answers
Ahalya

Last Activity: 10 Years ago

  • 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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