A quadrilateral ABCD such that ∠A = 90°. A circle with center O touches the sides AB, BC, CD, DA at E, F, G and H respectively. If CD = 30cm, AD = 15cm and CF = 20cm, find the radius of the circle
Answers
Answer:
Since tangents to a circle is perpendicular to the radius through the point.
∴∠ORD=∠OSD=90
o
It is given that ∠D=90
o
. Also, OR=OS. Therefore, ORDS is a square.
Since tangents from an exterior point to a circle are equal in length.
∴BP=BQ CQ=CR and, DR=DS.
Now,
BP=BQ
⇒ BQ=27 [∵BP=27 cm (Given)]
⇒BC−CQ=27
⇒38−CQ=27
∵ BC=38cm ⇒CQ=11 cm
⇒CR=11 [∵CR=CQ]
⇒CD−DR=11
⇒25−DR=11 [∵CD=25cm]
⇒DR=14 cm
But, ORDS is a square. ⇒ OR=DR=14 cm.
⇒ r=14 cm
Answer:
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ABCD is a quadrilateral such that ∠D=90
o
. A circle C(O,r) touch the sides AB, BC, CD and DA at P,Q,R and S respectively. If BC=38cm, CD=25cm and BP=27cm, find r.
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Solution
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Since tangents to a circle is perpendicular to the radius through the point.
∴∠ORD=∠OSD=90
o
It is given that ∠D=90
o
. Also, OR=OS. Therefore, ORDS is a square.
Since tangents from an exterior point to a circle are equal in length.
∴BP=BQ CQ=CR and, DR=DS.
Now,
BP=BQ
⇒ BQ=27 [∵BP=27 cm (Given)]
⇒BC−CQ=27
⇒38−CQ=27
∵ BC=38cm ⇒CQ=11 cm
⇒CR=11 [∵CR=CQ]
⇒CD−DR=11
⇒25−DR=11 [∵CD=25cm]
⇒DR=14 cm
But, ORDS is a square. ⇒ OR=DR=14 cm.
⇒ r=14 cm.
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