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Answers
Answer:
✴Given✴
ABCD is a quadrilateral circumscribing a circle with centre O. Sides AB, BC, CD and DA of the quadrilateral are the tangents to the circle at the point of contact P, Q, R and S respectively.
Radius of the circle=10cm, BC=38cm, PB= 27cm
AD is perpendicular to CD.
✴To find ☛ CD
Construction : Radii OS and OR are joined.
Steps :
We have,
BC=38cm and PB=27cm
Now, we know that tangents from an external point are equal.________(1)
BP and BQ are tangents from an external point B.
THEREFORE, BP=BQ
☛ BP=BQ=27cm
Now,
QC=BC-BQ
=38-27
=11cm
THEREFORE, QC=11cm
Again, QC=RC {From (1)}
THEREFORE, RC=11cm
We know that radius is always perpendicular to the tangent.
THEREFORE, OS is perpendicular to AD and OR is perpendicular to DC.
THAT IS,
angle OSR= angle ORD = 90°
ALSO, angle SDR = 90° (Given)
Then, angle SOR =90°
THEREFORE, SDRO is a square.
We know, OS=OR=10cm
THEREFORE, DR=10cm (all the sides of a square
are equal)
Hence, CD = RC+DR
= (11+10)cm
= 21 cm
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